Orthogonality in Normal Linear Spaces Citation James, Robert Clarke (1945) Orthogonality in Normal Linear Spaces. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/c1da-pk72. https://resolver.caltech.edu/CaltechTHESIS:04212025-175730400 Abstract No abstract. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Michal, Aristotle D. Thesis Committee: Unknown, Unknown Defense Date: 2 January 1945 Record Number: CaltechTHESIS:04212025-175730400 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:04212025-175730400 DOI: 10.7907/c1da-pk72 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 17165 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 21 Apr 2025 19:01 Last Modified: 21 Apr 2025 19:20 Thesis Files PDF - Final Version See Usage Policy. 49MB Repository Staff Only: item control page

ORTHOGONALITY IN NORMED LJNEAR SPACFS Thesis by Robert C. James -000- In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1945 ACKNOWLELG l!MENT I wish to express my gratitude for all that the faculty of the Institute has done for me, and for the companionship and inspiration provided by my fellow students. Especially, I am indebted to Professor A. D. Michal for his patient encouragement and assistance. - SUMMARY - Three possible definitions of orthogonality of elements of normed linear spaces are studied. For abstract Euclidean spaces, each is equivalent to the usual requirement that the inner product be zero. These definitions are never vacuous, since for each definition and any elements x and y of a normed linear space there is at least one number a for which x..J..ax+y. It is shovm that neither Pythagorean nor isosceles orthogonality can be either homogeneous or additive in a normed linear space unless that space is abstract Euclidean, although both of these tY:Pes of orthogonality are symmetric. in the isosceles sense, then If x .1-y lµ:+kyll> ½\\idl for all k. This and other similar inequ~lities give a comparison of isosceles and spherical orthogonality. Spherical orthogonality is homo- geneous in any normed linear space, but is neither additive nor synnnetric in general~ It can be additive and symmetric for a space of three or more dimensions only if the space is abstract Euclidean. For two-dimensional spaces, an inner product can be gotten from additivity and a strengthened form of symmetry, expressed in terms of the differential of the norm of of the limits, f±(x;y), of \jx+hY~- )be\\ as h ➔.:t"O. Because of the lack of sy:mmetry of spherical orthogonality, the uniqueness of the number b for which bx+y ...1.. x uniqueness) does not imply that of the number a ( left- for which x..Lax+y {right-unig_uenes·s ). This number b is unique i:f' and only if the space is strictly normed, and can have any value for which IJbx+ylf is minimum. The number a is unique if and only if spherical orthogonality is additive, or if and only if the norm is differentiable at all non-zero points, and can have any value satisfying f ~ {x;y) ~ -a ffxO;:s f_ (x;y). There are many connections between the theories of spherical orthogonality, of linear functionals, and of maximal linear subsets and supporting and tangent hY,Perplanes. For example, for a linear functional h satisfying ff(x) I f(h) ltf'II-\Jx:11 if o. For any and only if x..Lh element there is a maximal linear subset H with x for all f, x.J...H. Right- and left-uniqueness are expressed in terms of conditions on the elements at which a linear functional takes on its ma~imum in the- unit sphere, and of like conditions on the points of contact of supporting hyperplanes of the unit sphere. If the norm of a Banach space is differentiable at non-zero points and some additional assum.:pt1on such as weak compactness, regularity, or uniform convexity is made, then it is possible to give a general form for all linear functionals in terms of this differential or of spherical orthogonality. linear functional f there is an element x f(y) = -alfxll 2 :for all y, where x i ax+y. Then for a such that - NOTATION U Closure of the set x £.. U The element x Uc V The set is contained in the set U lJV The set of all elements belonging to either U the sum of Un V U - (see pg. 1). is a member of the set U and u. V. or V; and V; U v. The set of all elements belonging to both U v. the intersection of U and x+y Sum of the elements • x and x+U The set of all elements :x:+u, where u eu. U+ V The set of all elements u + v, where u E.. U and ax Product of the number au The set of all elements of the form llx!I Norm of the element IJ:rll Modulus of the linear functional T' The conjugate space of - (x ,y) x ..1. y Inner product of The element :x: x a y and x (see Definitions 1.1-1,6). element x v £- V. ( see Def. 1.4). au, where u ~u. (see Def. 1.6). T and f (see pg. 8). (see pg. 8). y (see Def. 1.9). is orthogonal to y~ This may be in the sense of Definition 2.1, 2.2, or 2.3, or of Theorem 3.1, depending on the section in which it is used. :x: :x:.l.U The element is orthogonal to every element of f (:x:;y) The differential of the norm at x with increment u. y, lim llx+hy}/ - llxll unless spec if ied otherwise ( as in Def .• h ' 7.3 and 10,2). See Theorem 7.4 and Corollary 7,4. .1..-+o l:im ,t.,➔+o f_ (:x:;y) lim. ){➔ -o /tx+hYII- !pell h l)x+hyll - llxll . . h (see Corollary 6.6 and pg. 101). - CONTENTS -• 1. Topological Spaces . . . . . . • • • • • • 2. Orthogonality in Normed Linear Spaces • • 11 3. Orthogonality in Abstract Euclidean Spaces 18 \. 4. Isosceles Orthogonality • • • • • • • • • 5. Pythagorean orthogonality • • • • • • • • 20 36 6. Spherical Orthogonality in General Normed Linear Spaces • • • • • • • • • • • • • 43 7. Spherical Orthogonality, Differentiability of the Norm, and Strictly Normed Spaces 63 s. Spherical Orthogonality and the Existence of Inner Products •. 9 •. Spherical Orthogonality, • • • • • • • • • Hyperplanes, and Functionals • • • • • • • • • • • • • • 10. Evaluations of Linear Functionals BIBLIOGRAPHY 89 • • • • 98 115 • • • • • • • • • • • • • • • 129 ~ 1. l. TOPOLOGICAL SPACES The topological spac~s used in this thesis will be normed linear spaces, Banach spaces, and abstract Euclidean spaces. However, these spaces are special cases of more general topological spaces which will be discussed first. Definition 1.1. elements A Hausdorff topological space1 is a set of T such that to each element associated subsets of x of T there are T, called neighborhoods of x, which satisfy the conditions: (1). Each element belongs to all of its neighborhoods. (2). For each of two distinct elements there exists a neighborhood which does not contain the other. If two neighborhoods ( J). U and ~' then there exists a neighborhood of intersection of U and An element borhood of - x x x contained in the v. be open if every element of U. V each contain an element U .has a neighborhood contained in is a limit point of a set contains an element of U U if every neigh- (other than x). The closure, U, of U is the sum of U and all of its limit points. 2 A closed set is a set which is equal to its closure. Also: 1. Hausdorff (III), pp. 228-229; the space satisfies postulates (A), {B), (C), (5). 2. See Pontrjagin (VII), pp. 26-Jo. = u contains a finite number of elements, then U u. {b). If u and V are any sets, then uuvcuvv. (a)• If ( C) • For any set (d) • The complement - = u. u, u 1 Thus u is a closed set. of an open set is closed, and the complement of a closed set is open • .(e). The sum of a finite number of closed sets is closed, and the intersection of a finite number of open sets is open. (f). The sum of any number of open sets is open, and the intersection of any number of closed sets is closed. A Hausdorff space consists only of a set of elements and their neighborhoods, there being no operation between elements defined. Such an operation can be introduced by requiring that the set be an abstract group. A set G of elements is called a group2 if Definition 1.2. there is an operation in elements G which associates. with each pair of x, y a third element (which will be called x+y} and this operation satisfies the conditions: that (1). x+(y+z) = (x+y)+z for all x, y and (2). There exists a zero element such that (3). For each x z. x+O = x for all x. there is an inverse element -x such x+(-x) = o. Definition 1.3. A set G is called a topological group 3 if it is an abstract group and a Hausdorff topological space and the group operations are continuous. That is: 1. The complement of a set consists af all elements not in the set. 2. Pontrjagin (VII), pg. 3. 3. Pontrjagin (VII), pp. 52-53. (1). If x neighborhood V of V of W of y x . and (2). If and x y -are elements of x+ y G, then for any there exist neighborhoods such that u and U+Vc::. w. is an element of G, then for any neighborhood -x • there exists a neighborhood U of x such that -Uc.V. A topological group is called a topological Abelian &i:oup if its abstract group is Abelian1 • Some of the elementary properties of a topological group are: 2 (a). If F is a closed set, U an open set, P an arbitrary set, and :x: any element of the space, then are closed sets, while (b). U+P, P+U, and -U x+F, F +x, and -F a.:re open sets. A topological group is homogeneous; that is, it is sufficient to verify its local properties for a single element only. For example, if the zero element has a neighborhood con- taining only zero, then every element containing the single element (c). of :x: has a neighborhood x. A topological group is regular; that is, for any neighborhood V x U of an element such that x there exists a neighborhood Ve u. In addition to the group operation between elements of a topological group, it will be useful to have multiplication by real numbers. Because of the group operation, multiplication of elements of a topological group by integers is defined. The multiplicat'ion by real numbers will be introduced by requiring that the space be a linear (vector) space. l. A group is Abelian if the group operation is commutative; i.e. if x +y = y+x for all elements :x: and y. 2. Pontrjagin (VII}, pp. 52-53. Definition 1. 4~ A set L is called a linear (vector) ~ /R space it it is an Abelian group and an operation {called multi- L is defined plication) between real numbers and elements of and sat.isfies the conditions: (1). and a(x+y) = ax +ay for all numbers a and elements x Y, (2). (a+ b)x=ax + bx, (J). a{bx)= (ab )x, (4). l•.x :: x. The identity (or zero) of the _Abelian group of a linear space is O•x, since x+O•x=l•x+ O•x = x by (2) and (4) of Definition 1.4. From the theory of groups the element O•x must be the same for all x, 1 although this can be shown directly. For l•x+l•Y = l•x+l•y + O•x = l•x+l•y + O•y by (2) and the commuta- tivity of addition. By subtracting the inverses of 1.y, it follows that O•x == O•Y• (-l)x, since l•x and x is Also, the inverse of x+ (-l)x := l~x + (-l)x == O•x by .(~) and (4). Furthermore: (a). a ;fo O, then (b). If ax = O, either x = O or a = 0. If follows that then ax == O, and For if ax = ay, then i (ax) == O== x by (3 ) and {4) ~ ax = ay and a :;6 0, then x=y. ax + a(-y)=O, since . ay + a(-y) = O by (1). (c). For if If From (1), it also a(x-y)= O and hence from (a) that ax=bx (a-b)x=O and and x=f.:.0, then a= b. x = O by (a) if a :t=- b. 1. PontrjagiK {VII), pg. 4. x =Y• For if ax = bx, Definition 1.5. A set L is called a linear topological space if it is a linear space and a Hausdorff topological space and the operations in a a•x and x+y are continuous simultaneously x, and in x and Y, respectively. and Since a linear topological space is a topological Abelian group, it possesses all the properties discussed below Definitions 1.1 and 1.3. (a). The following will also be useful: U is an open set in a linear topological space, If a .;c O, then au and element of contains <¼,y) au. is an open set. Then For suppose y·:::::ax, for some· x in u. y is an Hence U ¾<Y)= x, and from the continuity of the product at such that ¼vc u. was arbitrary it follows that au is there exists a neighborhood V Then V e. au, and since y of y open .. If U is any neighborhood of the zero of a linear (b). topological spa.ceT there is an open set aU at · (O,O) f'M·t{._, J__JU 1, () that' 'there is a positive number W of zero such that open if such that aVcVcU It follows from the continuity of the product for . la/ ~ 1. --,. · V a =f::. o. awcu if ax S and a neighborhood /a }"' c5 • But by (a), aw is . 1 Since the sum of open sets is open, the .. . sum V of all sets aw for which la l < [ is open. fa/ ~ 1, and vc U, kUMVU. cJ!,€ ~t> Ct W c U .. Clearly ave V if I By introducing the following generalization of absolute value in a linear (vector) space, one obtains a particular type of linear topological space--a normed linear space. 1. See (f) following Definition 1.1. 6. Definition 1 • .6. A set T is a normed linear space it it is a linear space and to each element /pell, the norm of a real number (l). llxl) > O if (2). llx+y/l :s: l/xll+ JlyJI, ( 3). 1/~xl( = /al· 1/xU Theorem 1.1. x of T there corresponds x, satisfying the conditions: x-:/=-0, for all numbers a. A normed linear space is also a linear topological space. Proof: If x is any element of the normed linear space, then £ > 0 for each number the set consisting of those elements satisfying the inequality borhood of x. Then: neighborhoods, 1 since l/x-yll < £ y will be said to be a neigh- (1). An element belongs to each of its IIo)I = 0 by (3) of Definition 1.6; (2) if /lx-yH :::::.£ , then the neighborhood of not contain y; (3) if x defined by /lu-y/1 "" ~' and defined by //x-y)I ~ £ , where x defined by 1/x-yl/ <.S does is in the neighborhoods of u and v /)v-y ll < £.._ , then the neighborhood of £ x is the smaller of £1-/)u-x ll and £~- ltv-x)I, is contained in each of these neighborhoods and hence in their intersection. topological space. The continuity of from (2), and that of If the number elements x and Thus a normed linear space is a Hausdorff ax l}x-ytJ x+ y follows immediately from (2) and (3), of Definition 1.6. is called the distance between the y, then it is easily seen that a normed linear space is also a metrie space: i.e., the distance relation 1. See Definition 1.1. 11:x:-yll ~ p(x,y) satisfies the conditions: (1). p(x,y) = Q it and only if (2). p(:x:,y) (3). p(x,z).:s:p(x,y}+p (y,z). =p (y,x), The neighborhoods of spheres with x defined by :x: IJx-yll , t:. are called as center and of radius £. from an element x of u £- U, then //x-u // x:;: y, for If the distance to a set U is defined as the lower bound X is a limit point only if ·this distance is zero. 1 of U if and Likewise, a sequence of elements is said to have the limit x if lim. 1/x-:x:nl/ l)l~OO =o. A C~:\,1-0:hf sequence is a sequence l:i,m. /~m--xn II /!11,m.~oo = o. xl, x ,· •• such that 2 Such a sequence in a normed linear space does ' not necessarily have a limit.i~ t t...e. Sf°'-C.~ . Definition 1.7. A normed linear space is said to be complete if all Cauchy sequences have a limit~ A complete normed linear 2 space is called a Banach space. 4 function is said to be a functional if its values are real numbers and its arguments are elements of a topological spaceo The norm of Definition 1.6 is an example of a functional in a normed -\('X+"1)=it')')~·\1 ·1) linear space. ' t (c>(~) -;:;,<>1~(-X) Functionals which are linear in the argument will be ~ound useful in studying orthogonality in normed linear spaceso Definition 1.8. A .functional f(x) with argument in a topological Abelian group is linear if it is continuous 3 and 1. See the discussion following Definition 1.1. 2~ · Banach (I), pg. 53. J. That is, for any element x and number > O there exists a neighborhood U of x such that ft (x}-f (u) I< s if U €- u. e 8. 1' (x+y) = f (x) + f {y) for all elements If' a linear functional f'(x} x and y. has its argument in a normed l1near space, it follows that there exists a number such that It (x}J~ M• 1/xll of' such numbers for all elements Xo M is called the modulus of M 1 The lower bound f and written Utll . It can be shown2 that for any element of' a normed linear space there is a linear functional such that f (x0 ) = 1Jx0 II and lftl\ = 1. f (x} The set of all linear functionals defined on a normed linear space T is also a normed linear space (the conjugate space) if the norm of a linear functional is its modulus. By assuming the existence of a bilinear and symmetric functional o~ two variables, and relating this to the norm, it is possible to get a generalization for normed linear spaces of' the inner (scalar) product of vectors. Definition 1.9. An abstract Euclide~n space is a normed linear , space such that to each ordered pair or elements can be associated a number (1). (tx 2 y) = t(x,y) {2). (x,y) = (y,x), (3). (x 1 y+z) (4). llxU 2 (x 2y) x and y there with the following properties: for all numbers t, = (x,y)+(x z), 2 = (x 2x). The number (x,y) product of the elements is the (bilinear and symmetric) inner x and y. 3 The assumption of such an inner product implies that any finite dimensional subset of the 54, Theorem 1. 2. Banach (I), pg. 55, Theorem J. 3. See Frechet (II), pg. 707. 1. Banach (I), pg. space is equivalent 1 to a finite dimensional Euclidean space--. = i.e. to a space with elements of the form. x (x ,x ,•••) and 1 2 2 1 //xii= '- ~(x. ). For example, it is clear from (l) and (3) of • ' l. Definition 1.$, that for any abstract Euclidean space there exist elements x and = 1 12 /1ax+ byJ Then y such that //xii = IJy// =l • - = and ~~ (x,y) (ax+ by,ax + by) -:- a 2 (x,x) + b 2 (y,y) = Q. ~~.t"~:,t a~2 llxll 22-+b 2 /lyJI 2 • 2 =-- Cl i- b - \\ ( CA b) !I is set up, it l' follows that the operations of addition and multiplication by O"l<l '+,$; f-ct-4 Hence if the correspondence ax+ by~ (a, b) --7' ,·e. ;} _;_., ew. ,.,i,,.1.()'""""~1Vect'tri.A'fi1'!:JI..'-' real numbers are preserved under the correspondence and that 1/ax + by ll = ll(a, b))I for all a and b. Euclidean space generated by the elements to the two dimensional Euclidean space. Hence the abstract x and y is equivalent It is evident that the same argument can be used for any finite dimensional abstract Euclidean space. It will be found useful to express the inner product (x,y) directly in terms of the norm: Theorem 1.2. The inner product (x,y} or an abstract Euclidean space is equal to: (1). •D/x+yll 2- 1/x-y/J 21:J , !2,!'. ¼ ( 2). ±½ Dix ±. yJJ - I/xii -J/yl/ :J . ~: - • ( 3) • Proof: 2 /J X 2 21 + ky}/ 2 = //x JI 2 + k 2J} y l} 2+ 2k (X, Y) • F~om (4) of Definition 1.9, /Jx -+ yl/ 2 = (x ± y,x ± y). The additivity and symmetry of (3) and (2) of Definition 1.8 make it possible to expand this as: 1. In the sense that there exists a 1-to-l bilinear correspondence which preserves the norm. (see Banach (I), pg. 180). l.O. II x + y /I 2 = ( x, x} + (y , y } + 2 ( x, y } , 2 • and /Jx-y/1 = (x,x}+ (y,y} -2(:x:,y}. Using (4) of Definition 1.9, these become 2 2 2 (x,y) ~ ±½Dix+ y/J -Jlx/J -lJyJJ ] . 2 (x,y) = ¼UJx + yJl 2 - l}x-yJl Likewise, 2 • )l x+ky J/ = (x + ky, x + ky) = (x,x) + k 2 (y,y)+2k(x,y) Subtracting them gives = l/x/1 2 + k 2 //y// 2+ 2k(x,y). ]! 11. 2. ORTHOGONALITY IN NOR..Tv!ED LINEAR SPACES A definition of orthogonality of elements of a normed 1 linear space has been suggested by B. D. Roberts. two elements x and 1/x +ky /I = ])x -kyll y It is that are orthogonal if and only if identically in k. Unfortunately, there exist normed linear spaces where two elements can be orthogonal by Roberts• definition only if one is zero. 2 In order to avoid this difficulty the following three definitions have been developed. They are equivalent for abstract Euclidean spaces, 3 but are not equivalent in a general normed linear space. Definition 2.1. Isosceles orthogonality: An element of a normed linear space is orthogonal to an element if and only if /Ix+ y ll = /Ix - y/). This is a generalization of the fact that two vectors x and y are perpendic- ular if and only if x +y and x-y are The vectors x+ y and equal in length. x-y can then form the sides of an isosce- les triangle with 2y, or 2x, as base. Be- cause of the nature of geometrical addition ·of vectors, the latter could be taken as the condition tor perpendicularity~ 1. Roberts (VIII), pg. 56. 2. See Example 2.1. 3. See Theorem J.l. y x (X..LY) 12. Definition 2.2. An element Pythagorean orthogonality: x of a normed linear space is orthogonal to an element y (x..Ly) 2 2 2 if and only if /Jx/J + JJy/J = Jlx-y1J • t--- ~·- ::r-"'J o/, I ·•---, h., _, _______. _____: : ,. "1 If a right triangle in ordinary Euclidean space is defined, by virtue of the Pythagorean theorem, as one such that the square · of the hypotenuse is the sum of the squares of the legs, then this is a generalization of the fact that perpendicular vectors can be placed so as to be the legs of a right triangle. Definition 2.3. Spherical orthogonality: An element a normed linear space is orthogonal to an element and only if /Jx + kyll ~ //x ii for all numbers y (x-1.y) x of if k. This definition could have been stated: "x ..L y if and only if every element x +ky lies on or outside the spherical surface consisting of elements z satisfying II z ll = /lxll." It is a generalization of the fact that in ordinary Euclidean space two vectors are perpendicular if, when in the position illustrated, one is tangent to the circle with radius x--or that the vector of this circle for all t} x + ky is longer than the radius k. t i\ ).,~-:,,.__.,_ 1lt6--vv) It is evident that for any of these concepts of orthogonality the zero element is orthogonal to every element of the space, and , Abo O ...:, ·!),. • ~ conversely. 'l ~ <J?Jkc,'jP ""'f .,J·, - ,, . . -1, ' There are other elementary properties of perpendicular vectors whose generalizations will.-be interesting to investigate: 13. I. Symmetry: If II. Homogeneity: and b. III~ x.1.y, then y..1.x. If X.l.Y, then Addit.ivity: If X.J..Y IV. If x y are any two elements, then there exists a number a such that a and and axJ..by for all numbers x.1..z, then x..L(y+z). x..L.(ax+y). The last property is the most important, since it would . .~, f ~) l4~ c,.~o.--·-<-< &.t keep the concept of orthogonality from being vacuous (in the ~-~(.~ of> "tr·!() <Yli!,1£1.,...~ sense that conditions on the orthogonality ,1.., rile-,) - l' i!,r_,.A A(...,,J~ would have no effect on the space). It is clearly satisfied for ordinary Euclidean if'1 ) ' ~u-J' ~ " 4il -CV:-,.,t.c,,v /lf-,\C•. . space, and' it will be shown to hold for ~f- 3 Tjr-11 °D hie orthogonality . of- Definitions 2,i-2~ ·3 . A slightly weaker statement of this property is: "If x are any two elements of a normed linear space, then x onal to some element in the plane of a two-dimensional linear subspace~ x and and y is orthog- y"--where a plane is 1 Both isosceles and Pythagorean orthogonality are clearly symmetric, although it will be seen that spherical orthogonality is not. When the orthogonality is not symmetric, property IV could be stated either·: "There exists an "There exists an a such that a such that ax+y J..x". x.J..ax+y"', or However, spherical orthogonality will be shown to satisfy either statement. Hence the one form will be used for simpiicity. All three types of orthogonality are equivalent if the space is abstract Euclidean, and have properties I-IV in this case. l. That is, the plane of x and y consists of all elements ax+by. 14. Neither isosceles nor Pythag·o rean orthogonality is homogeneous or additive in a normed linear space, although they are both symmetric. The assumption of these properties will be shown to imply that the space is abstract Euclidean. Spherical orthog- onality is clearly homogeneous, although it is not symmetric nor additive, and the assumptions of these properties will be shown to imply that the space is abstract Euclidean. The following example shows the independence of the above types of orthogonality in normed linear spaces, and that tor Roberts• definition of orthogonality there exist normed linear spaces where at least one of two orthogonal elements must be zero. Example 2.1. Let T be the normed linear space consisting '1' o--½ ~ wi.t-',\ 2 of all continuolfs functions of the form f = ax+bx, where ljax + bx2/l .is the<maximum of lax + bx 2 / in the interval fo,i~. 0 (1). Two elements of T are orthogonal by Roberts' defini- tion if and only if one is zero; f i.e. 1/f+kgll= )/f-kgtl = O or g = o. Consider the function f = k only if If f ..Lg,then /lf + kgl / = JJf-kgl/ its maximum only at x = l, and sufficiently small /f + kg } and for all k. /ti If f ;E o, then by taking )f -kg j for all ax + bx2 • takes on k can be made to take on their maxima as near x =l as desired, because of the continuity of g(l) must be zero if for f and k g. small, and not zero. Thus if and by taking f Hence k is a multiple g. must be a multiple x -x~__if it is - ---------,- takes on its maximum only at x = i, / L!,_, ..- I~ I large it follows similarly tha,t f (½) 0 and of x -2x 2 , But j/f + 4g l/ i /I t - 4g l/ , since g ~ o, /I t + 4g /l = max. l5x-6x2 / 4o, I) / of /I f + kg Jj = /I f - kg ll -~\ i,,t = =~ and l/ f -4g /l = ma~. J3x-2x2/ = ,2. .Gl.f, (o, I) 0 15. Henc~ f...Lg is im;possible if either . lff ( or :x: = 1, unless its maximum only at It I suppose f or g reasoning as before, g f..Lg implies = :x:~ ~ f ~ o, and if a:x:+2b:x:2, if !k l is sufficiently small. lkl small a(l±k)x+b(l±2k)x2, and for therefor·e becomes Then 11'1 Since 2'6, \t:1--kg\. must take on its :x:;;: takes on its maximum only for f±-kg Now f' :::: a+2bx~ g(~'5) ; O is either zero o:i;- a multiple of maximum near is zero, takes on its maximum only at . :x: = 2%, the only other possible point for the ma:x::imum, since that )g \ ) takes on I = But J/f+kg)t-=llf-kg\\ a2(1+k)2 1 -1 a2{1 k)2 4b(1+21c) \ -l'b(1:21c) • 1 But this implies k ~ O or :impossible if either of k2 = 1/3. lfl or Jg! takes on its maximum only in the interior of (O,:(J, unless f ing possibility is for _T_h_er_e__f_o_r_e_f_-1.._..g__i_s or g is zero, The only remain- Jf'I and Jgl to take on their max:ilna. at both an interior point of (0,1) and at then both be multiples of .-. x = 1.. But f and g would 2(1-f2)x-+x2, which is clearly impossible. ( 2). The elements t 1 ~ x - x2 in the isosceles sense, since and g 1 ~ :x: are orthogonal 11:x:- (:x:-:x:2 ) ll ~ llx+(x-:x: 2 )1( d: 1. How- ever, they are not orthogonal in the Pythagorean sense,since lfx-x2I 2 + JJ:x:V 2 :::: 17/16, while Jl:r1 - g 1 12 ~ 1. Also, "f1 ~ 1 11 ~ 2 t} ~:x: -:x:2 II ::: (7/16) , which is less than Jf111 ~ ¼. Thus f 1 and g1 are not spherically orthogonal. ( 3). The elements g 2 ~ -2:x:+ ¼'f65 :x:2 are orthogonal in the Pythagorean sense, since Jft2'12 ~ 1, 2 2 Jtg2\\ = 16/65, )Jt2 -g 2 fJ ; 81/65, and hence ,1r2 R2+1~ tl 2 ; llt2-g 1f 2 • 2 2 They are not orthogonal in the isosceles sense, since /)r 2+ g2J) 2 = ~-½ ~65. Also, /lt 2-½g2 IJ = ~ • /2x !}t 2 1/ = l. which is less than Thus f2 and g ½~65 x2 /= . , m-e not 2 spherically orthogonal. The elements (4). J:r 3 orthogonal, since :r g =2(x-x2 ) 3 = 1, and hence : x and 3 (1) + kg (1) / !} t 3+ kg lj ~ lit II = 1, for all 3 arespherioally k. But 1/f -t- g} = /13x-2x2 /J = ~ 3 3 2 and /] f -g )1 : ll x-2x 1l =l. Therefore t and g are not 3 3 3 3 orthogonal in the isosceles sense. They are not orthogonal in the Pythagorean sense since lit 3'/2+//g3 11 2 = while /J r 3 - g.31/2 = i. 3 f, The properties I-IV above could have been stated differ~ ently. For example, homogeneity and additivity could have been combined, and one could have general~zed the theorem: ? ( 4-vi-k,;.,s, -.,'v.5 "Every plane ~ e-t ·1ng a given line contains at least one - -·-- .,, line perpendicular to the given one". Some relations between this and properties I-IV for normed linear spaces are given by the following theorem. Theorem 2.1. (1). Homogeneity and additivity of orthogon- ality are equivalent to: "If for all numbers b. (2) .. a .and x..1..y and X..Lz, then x...1..(ay + bz) If orthogonality is of one of' the types of Defini- tions 2.1-2.J, then homogeneity, additivity, and property Iv1 are equivalent to homogeneity, additivity,~: z are any three elements for which there is a number 1. See page 1.3. a such that x "If x, y and is not orthogonal to X...Lay + z. y, Proof: (1}. all numbers taking if' b a If x ..L y and b, and x J_ z imply x ..L (ay + bz} for then homogeneity can be gotten by = 0 and additivity by taking a= b =l. Conversely, X.l Y and X j_ z_, homogeneity gives Addi ti vi ty then gives x -L (ay + bz}. x _i_ ay (2). If' and x _j_ bz. x, y and z are any three elements, property IV gives the existence of' numbers b and c is not orthogonal to x ..1. (-ex- .2. b y).. such that x ..L bx + y b ¢. 0 y, then Additivity then gives converse can be gotten by taking onal to itself. x .1.. ex + z. If x and homogeneity gives x .L (-bey+ z). The =x, if x is not orthog- If one of the types of orthogonality of Definitions 2.1-2.3 is used, x unless it is zero--in which case y. y and cannot be orthogonal to itself x .J... (ax + y) for all a and 18. 3. ORTHOGONALITY IN ABSTRACT EUCLIDEAN SPACES The most obvious requirement for. . the orthogonality of two elements of an abstract Euclidean ~pace1 is that their inner product be zero. The relation between this and the above types of orthogonality follows imm.ediately from Theorem 1.2 • .. Theorem 3.1. For abstract Euclidean spaces, isosceles, Pythagorean, and spherical orthogonality are equivalent,~ x two elements y are orthogonal if and only if their (x,y) = o. inner product Proof: and It is evident from (1) and (2) of Theor.em 1.2 that x two elements and y are orthogonal in the isosceles sense, or in the Pythagorean sense, if and only if (x,y) = 0~ 2 If' {x,y) = O, it follows from (3) of Theorem 1.2 that //x+kyl l ~ llxl\ for all k, and hence that x and y are spherically orthogonai. 3 Conversely, it also fdllows from (3) that //x + ky/1 > /Ix II ( for der~ ) implies k 2 //y// 2 + 2k{x,y) ?!-- o. k = o, the inequality can hold for all ive with respect to k is zero at {x,y) = o. I e. 'l2.n A. Since this is zero ./..0 • • • f r - -..uvi vW\,.... ~1.,,1,,'f}t--..., . ·"· !,.MU ~,.(. );, U <J--,>L; x and ;,t - U V:, only if the k = o. ~ j•t;J That is, if .e.~ v.t ~¼-. .---~ /t,,q_ \' e o-,vJ1 ;.) /,)--4f-t ~f )< ~ () ~ 0Wo • ' Since two elements k space are orthogonal if and only if 1. See Definition 1 . 99 2. See Definitions 2.1 and 2.2. 3. See Definition 2.3. y l) of an abstract Euclidean (x , y) = o, it is clear that 190 symmetry, homogeneity, and additivity of the orthogonality are equivalent to symmetry, homogeneity, and additivity of the inner product. Definition 1.9, y Also, because of (1), (3) and (4) of (x,ax+y) = a/fxl/ 2 +(x,y).. Hence if are not orthogonal, then x..Lax +Y, Thus if x :/= O, then the number a where such that a x and =~(~ifJ. x ..1_ax +Y is unique'.9 Theorem 3.2. The orthogonality of abstract Euclidean spaces is symmetric, homogeneous, and additive, and for an; two elements x (=# 0) a x.Lax+y. such-that and y there exists a unique nwµ.ber 20. 4. ISOSCELES ORTHOGONALITY A number of relations can be obtained between orthogonal elements of general normed linear spaces. Isosceles orthogonality · is obviously symmetric, and it can be shown that for any elements x and y there is a number a such that xJ..ax+y. l The assumption of homogeneity and additivity of the orthogonality'2 will be shown to imply that the space -is abstract Euclidean. If two vectors perpendicular, then x and y of ordinary Euclidean space are llx+kyll > llxll for k 1:, o. While this is not true for isosceles orthogonality in normed linear spaces, it is possible to establish weaker inequalities. These give an,f idea of the degree of independence of isosceles and spherical orthogonality, as weil as interesting inequalities resulting from the condition llx+yl( = !Ix -yll. Lemma 4.1. If x and y are orthogonal elements of a normed linear space, then: (1) •. 1/x+kyJI~ lk\/~±Y/1 ( 2) • II x+kyll !: /Jx±yl! Proof: then and llx±yll~ 1/x+kyl/, if fkl ~ l, /kl/lX±Yll <- /lx+kyl), if I k I~ 1. x+ky = ½(k+l)(x+y) -½(k.-1) .(.:JE-y) is The triangular inequality of the norm3 there- The equation an identity in k. fore gives and //x+kylls l/½(k+l)(x+y)ll+ll½(k-l)(x-y)/1. /Ix +y ii = 1/x - y I{. Hence i f If x..1..y, /k I > 1, then llx + ky II ::: lkl llx± yl/ • 1. Theorem 4.5 .• 2. Whenever "orthogonality" is used in this section, it will mean the "isosceles orthogonality" of Definition 2.1. J. Condition (2) of Definition 1.6. 21. If' lkl !:: 1, then /j x -t- ky l/ ~ /Ix + y t) . Likewise, the relation J ½(k + l) (x + y) I/.=:: /lx + ky lJ+/l½(k- 1) (x-y) l/ gives the other two inequalities of the Lemma. The inequalities (1) and (2) can also be obtained from each other by replacing k by 1 and k interchanging x and y, since isosceles orthogonality is Theorem 4.1. If x synunetric. y and normed linear space, then are orthogonal elements of a )Jx + kyJI ~ )lx l) for all k such that /kl~ 1. = (x + y) + (x -y), it follows IJx + y)j = IJx - yJJ, this gives Proof: From the identity 2:i . l that 2 !) x l) ~ )I x + y /l+l}x -y,J. Since llxl) s 1/x + Yll• )k l ~ -1. II x + yl/~ /I x+ kyll But from (1) of Lemma 4.1, l/x/1 ~ J/x+kyU Hence if )kl> l. The result of Theorem 4.1 is not valid if lkj< 1. the following example shows that it is possible to have and //x + kyl/ as near to possible to have if In fact, x ..1.. y ½/Ix// as desired--al though it is not 2 //x+kyl/ =½I/xii. These results, and those following, give a comparison between isosceles and spherical - - - - - - ~ - -•--~~••-r.-.•---• --~--- -,1• • ••••--•-•-,-•-• "" " • • • • - , . · - - • - ~ • · • . - • • •, " orthogonality. Example 4.1. of all pairs Let T be the normed linear space consisting (a,b) of real numbers, with (a,b)+ (o 1 d) =(a+c,b+d), = (ka kb), and 1/(a, b)I/ as the larger of la/ and /b l . Let x = (1,0) and . y = (n-1,n). Then llxll = l, (x-y) = (2-n,-n), and (x + y) = (n,n). If n?::.l, then k(a, b) 1 l. Condition (2) of Definition 1.6. 2. Theorem 4.2. 22. //x -y/l = //x+y l) = n, and x and y are orthogonal. But 1 _ (n+l 1,_) and /Ix -l:.. YII= n+l, x--Y-,-2 , 2n · 2n 2n 2n which approaches½ as n becomes infinite. That is, given the element x = (l,O) and e > o, there can be found an element y orthogonal to x and such that /Ix +kyl/"" ½+ £ for some k. However, for no normed linear space can there exist orthogonal elements x (~ O) and y 1/x+kyj) ~ ½I/xii for which for some k. This is shown by the following theorem~ · Theorem 4o 2. If x (,= 0) k. for all k. (k+l)fx-ky} = (l-k)(x+ky)+2k(x-y) The equation is an identity in are orthogonal el em.ents }/x +ky IJ >½)Ix I/ of a normed linear space, then Proof: y and Hence it follows from the triangular 1 inequality of the norm that lk+ll //x-kyl}'!S ll.-k/ Ux+kylj + 2 lkl }lx-y//. Suppose there exists a number k such that There is no loss of generality in taking y could be replaced by that O~k<l. Also, -y. /lx+kyll = ½/Ix//. k>O, since otherwise It then follows from Theo~em 4.1 /lx-ky//~¾/lxl/, since 2)/xll~llx-kyll+}/x+kyll. Substituting in the above inequality, it follows that l(k+l))/xl/~½(l-k)llx!l+2k/lx-yl/, or 2 But from (2) of Lennna 4.1, Therefore if that of //x + ky JI (2k+l)llxll ~ )Ix// (2k+l)//:x://~2k/lx-yl). k}/x- yl{~ }/x+kyll, or and k k/lx-y/J~ ½llxll~ = o. But this is impossible = ½/Ix II. Hence there can not be a number k such /Ix +ky// =½//x//. Since /Ix + ky // is a continuous function k, and therefore must take on all values between its maximum and minimum, it follows that /Jx+ky// > ii/xii 1. Condition (2) of Definition 1.6. for all k. 23. It has been shown by Example 4.1 that Theorem 4.2 is the best result of its type of some kind. as ( f,;-;~yll obtainable without further assumptions However, in this example /I y /) became infinite ~pp-r~-~ch~-d --·· ½)j~-1/ . . By restricting the value of the inequality // y // of Theorem 4.2 can be strengthened, the following theorem being an example of this. Theorem 4.3. If x and y are orthogonal elements of a normed linear space, and IJy/j ~ /Jxll, then for all /Ix+ kyll -2: (2'{2"° -2 ){/xl/ k. Proof: Suppos;e /lx+ky/1 positive, since otherwise =rl/xll, where r < l. Take k to be y could be replaced by -y. Then 0< k....: 1, since by Theorem 4.1, ]lx+kyl/ ~ I/xii if lkl ~ l. Then, since 2llxll ~:dlx+kyll+/Jx-kyll , 1 we have {2-r)/lx1I-::: lfx -kytl. -But {k+ 1)//x - kyll ~ {l - k) )lx+kyll +2k//x -ytl follows from the identity Hence (k+l){x-ky) = {l-k)(x+ky)+2k(x-y). 1 (k+l)(2-r)/lxll ~ {l-k)r)JxlJ+2kl/x-yJI, or {k +l) /Ix II~ r /Ix II +k //x - y /I • Also, from x+y= {x+ky)+(l-k)y condition /Ix -y/J and the orthogonality =// x + yl l, it follows that {Ix -ylJ ~llx+kyl /+(1-k)/ly//. Using /Jx+ky[) =rl/xlf Hence (k + l) /Jxll~ rllx/l + k{l- k + r') /l x tl , which gives But • ~ .L+K Hence and //yl/:5: /lxll, this gives Jlx-yll~ (1-k+r))lxl{. has a minimum (for r~2'/2-2, and k ~ O) of r~ i~• 2'J2 - 2, when k 1/x+kyll ~ (2'12- 2)1/x// for all k. = '12 - 1. As with Theorem 4.2, it is possible to give an example showing that the inequality of Theorem 4.3 can not be strengthened 1. Condition (2) of Definition 1.6. without making further assumptions. have /Jy/1 ~ /lxll, and x.1.. y, That is, it is possible to /I x + ky ll = {2rr - 2)/lxl/. As seen by the method of proof of Theorem 4.3, 1 it is then necessary that Example 4.2. of all pairs k(a 2 b) Let (a 2 b) T be the normed linear space consisting of real numbers, with = (ka kb), and //(a,b)lj = /a/ + lb) • Let x = (l,s} and 2 y = (s ,-1), where Is l + 1 ~ Also , s will be chosen later. ( x + y) = (1 + s , s - 1 ) But to /lx + ky ll = l + s2 = i~s~ Jlxtl • //x + ('12 -1 )y JI = . (2i2 - 2 )// x if. becomes and x (x+ ky) = (l+ks,s-k), and if orthogonal. s, then Then 1/xl) If and k y s = '12 -1, this That is, if /I x t{ = ({ y l/ limiting case of Theorem 4.J. are is taken equal x = (1, f2 -1) y = { f2 -1,-1), then // x + ky ll = (2'12 - 2) JJxll if It is necessary to have = II y l} = {x - y) = (1 - s , s + l) • and /lx + y ll = /Ix -y ll = }l + s l+ll- sf Hence and (a 2 b} + (c 2 d):; (a+c,b-td), k = Y2- 1~ in any example of the This is evident from the proof of that theorem, since for such an example we would have 2 l +k r = ""I+F' and hence every inequality used in the derivation of the \:1ft2 must be valid if made an equality. Among inequality r ~ these was /) y If s. /I x q. Corollary 4.J. If normed linear space, and for all x and y are orthogonal elements of a lj yl/ ~ //x i/ 1 then }/ x + kyl/ > (2V2 - 2) // x ll k. There is an interesting {although very unusual) case where it can be shown that // x +-ky// ;?:. //x ii 1. Because of the inequality for all 'k, r ~ I ~k: 7 and hence that and the following. 25. x and are spherically orthogonal. 1 y This is shown by the following theorem and examples. Theorem 4.4~ If linear space, and lk) ~ l. Hence Proof: If x and y are two elements of a normed = //x+y/1 = 1/x ll, then /}x+ky[j = /lxJJ if //x-y /} /Jx+k.y/1 ~ //xii for all /k/ ::; 1 //x-y/1 and from (2) of Lemma 4.1 that k. = /}x+y/1 = /lxl/, it follows llx:tkyll ~ llxll~ But from = (x-ky) + (x+ky) and the triangular inequal .ity of the norm, 2 it follows that 2/lxll:!S. Vx-kyl) + 1/x+kyl/. Hence /lx+kyll = /!x ii 2x if )k) =s. 1~ l II x+.ky I > llx Ii ByTheorem4.l, for all Example 4.J. /lx+kyll :.: :. Ux ll if lk/~l. Hence k. Let T be the normed linear space consisting' or all continuous functions in the interval (O,l}, with II t!I =max. /t (x)/ • The elements f = x 2 and g = 1-x satisfy the conditions of Theorem 4.4, since Then (as must be true as a result of Theorem 4.4), )/f+kgJI =l if lltll = /Jgll = )Jf+gi J= 1/t-gll = l. +1 lk} .!S 1. That is, the maximum in- (0,1) of /x2 +k(l-x) I is 1 if !k/~ 1, as is evident from the figure. f is orthogonal to g Thus by either Definition 2.1 or Definition 2.J. Example 4.4. (a,b) oi---:=~--7'-~-____;~ -1 Consider the normed linear space of all pairs of real numbers, with ll(a,b)/l l. See Definition 2.J. 2~ Condition (2) of Definition 1.6~ as the larger of )al and 26. • /bl. 1 Let x = (1,0) llxl/ - //x-y// = J/x+y// -:- 1. if /k/ ~ ,1. II x+-kyll if )bl~ 1. Then //x-t-k.yl/ = //( 1, kb) II = //xii = 1, Hence /lx+kyl/ = /Jxl/ Actually, = /) ky J/ y = (O,b), where and )}ky// ~ 1/xl{, and if // ky // . ::. }/xii • For orthogonality in normed linear spaces to have a useful meaning, it is necessary to know that there exist non-zero orthogonal elements. For in that case, assumptions such as homogeneity and additivity of the orthogonality are not vacuous. That is, such assumptions made on the orthogonality will have a restrictive effect on the space and will not necessarily be trivially satisfied, if it is known that there always exist nonzero orthogonal elements. For isosceles orthogonality, it can be shown that for .any elements and y of a normed linear space, there exists at least one number a such that That is, for any elements X x --- y the element X is orthogonal - ,, X - and Y, a~ ( of the form ax+y) and -- to an element in the plane of ......._ _ ~ - - - ••·•--- 'i. It is also possi liYle to give limits to the value of of )/xii and /IY/1. x..Lax+y. a in terms Before establishing these results it is convenient to investigate certain limits which will arise numerous times in this thesis, particularly in the study of spherical orthogonality~ Lemma 4.5. space, then If x and y are elements of a normed linear ¼i~ L/l(n+a.)x+yll - )tnx+yll] = allx\l lim DJ(n+a)x+yl/ - /Jnx+yll1 = -all:x:l/. m~-oo 1. The same space as used for Example 4.1. and Proof: = ...!!...+ ~ 1 identically, //( n+a)x+y }} can n+a n~a . //nx+-1!L.+ all' if n is positive and n a //+a//x+...Ln T • Since be written as large enough that n+a > o. [/J(n+a)x+ y/l - l/nx+y/1] Thus = Dlnx+ n!a I/ - Unx+y/U + a//x+r&a//• But }llnx+-1!L./l- //nx+y///.$ /...!....{llY//, which approaches zero as n +a n+a becomes infinite. Also, lim a}/x+-Lj/ ,,,~oo n+a = a/Jxl/. Hence ~~[//(n+a)x+y// - ·lJnx+yl/] = a//x/1. is independent of -a l/xi/. Since the value of the limit y, it follows that But this is the same as is therefore equal to n lim [ //(n-a)x - yJI - /{nx-y}/7= '..J 11\~CO lim[//(n+-a)x+y//-l/nx+y/l] , which h\?- "" -a}/x/1. --+L_,_ I The above figure illustrates the validity of Lemma 4.5 for ordinary Euclidean space. However two similar results valid in ordinary Euclidean space can not be generalized to normed linear spaces. (1). In an ordinary Euclidean space, lim f°bnx +Y II + //nx -y/1 - 21/nxlTI= o. 11\+oo L!! This is not true in general normed linear spaces. For example, consider the normed linear space of . Example 4.2, consisting of ordered pairs of numbers (a, b) = (1,1). x = (1,0) and y /Jnx-.yU = n if n~l. Since /} nx+y // +- Vnx-y// - 2 /tnxl/ with Then //nx+yH /Jnxl/ = 2 if [/(a, b)// = n+2 = /a/+- /bl • Let if n ~o, and = n, it follows that n ~ 1. This is a special case of a more general relation which is valid in ordinary Euclidean space, 28. but not in a general normed linear space, (2). Letting Namely: lim [/Jnx+ (y+z) II - /Jnx+y l/ ] = lim [ Unx+z /J - 1/ nx /lJ. m~- ~+~ a("'= //nx +(y+z}/1 - }/nx+yll is clear from the figure that and p'"' = //nx-+zll- //nx/1, it lilli o</h = lim nt--,00 /)\~00 Uzl}cos B, for abstract Euclidean spaces, e ---- ---- ---- --- -~--1- - - 1 - - - 1 - -·- The equality (2) is not valid for general normed linear spaces. If the normed linear space of Example 4,2 is used again, with x = (1,0), y = (1,1), and z = (0,-1), then // nx + (y-t-z )l/ - /Jnx+-yl/ = -1 and // nx+z/1- l/nxl{ = l. In fact, if the equality (2) is assumed to hold in a normed linear space, then a differential of the norm, lim hx+hy/1 - )/xi/= f (x·y) exists · at each non-zero point of the h ' ' 1 space. For lim I[ rq.+y/1 - 1/nxl/ = - lim /fnx-yH - {/nx// is equivalent ~~c t)I~ ni ➔ oo to the existence of this differential, and results from (2) if is replaced by z -y. Conversely, the equation (2) is nothing more than an expression of the additivity of this differential 2 --which results from the assumption of the existence or lim ll x+hy// - l/xJ/ .3 .s.-+o h • Thus the equality (2) holds (and the limits involved exist) if and only if this differential exists at each non-zero point of the space. 1 .. Such limits and differentials will be thoroughly studied in relation to sphe~ical brthogonality. 2. That is, it is equivalent to f(x;y)+f(x;z) 3. This is shown by Mazur (VI), pg. 128. = f(x;y+z). Theorem 4.5. If . x and y are elements of a normed linear space, then there exists a number } )I ( I}x +ax+y a such that = tIx - .(ax+y) Ji , p_ri e., , x..1..ax+y. Proof: Define the real function of a real variable, f(n), by f (n) = 1/x+ (nx+y)II - Ux - (nx+y)/l = 1/(n+l}x + y/l - l/(n-l)x+yn. Then 2//xll lim f (n} = t,\lim + c,o Dt(n+2}x + yJI - J/nx+y/U , which is equal to '" ~ 00 by Lemma 4.5. Likewise, M-+-GO lim f(n) =_,~ _D/(n+2)x + yll - ·Jlnx+y/l] . .~~ Then Lermna 4.5 gives for some values of lim f (n) t'l ~-coo = -2JJxl/. Since f (n) is positive n and negative for others, it now follows from the continuity of f(n} 1 that there exists a number a such that • f (a) = o, or ax+y //x + (ax+y)J/ = //x - (ax+y} ll . Then x and are 0rthogonal in the isosceles sense of Definition 2.1. ~ ?C-tj and 2/ of ordinary Euclidean space are 2 orthogonal, then the inner product (x,ax+y) al/x/J +(x,y) is • If two vectors x = zero. (x,y} = J/x)JJ/y]lcoa e , where e is the angle between Also, the vectors x and y., Hence a =- ';~f cos s , and la/~ i~~{ • I t is also clear from the figure that 1/axl/ = Uy)/ /cos e /, and hence that Ja l =~~~ /cos e/. However, these relations do not carry over to normed l i near spaces. While ~i:'h is not valid in general, it i s if // yl/ ~ }lxll , and it can be s ·hown that Jal ~ 2 1/rtn }Jxn if }/y/1 ~ /pcU . j a I~ Theorem 4.6. linear space and If x (;(:. 0) /Jy//.::: [Ix /) , then and y are two ele~emts of a normed Jal ~ ~~~~/l if x .l. a x+y. 1. The continuity of the norm follows fro:i;i. (2) of Definition 1 . 6. 30. Proof: Define the real function of a real variable, f{n), = //x + {nx+y)// - /Ix - {nx+y)II. Since fix+ (nx+y)Jl /l{n+l)x+ylJ, . I it follows from the triangular inequality of the norm that as /Ix+ (nx+y)// .?!. /n+l/ //x)/ - }Jy/1. ~ ln-11 /lxl/ + 1/y//. or Likewise, f {n) =. Qn+ll /Jxl/ - 1/y U] - Dn-11 /}xi/ +/Jy/Q , Hence f(n) =[/n+ll - ln-lD l/x)/ -21/yl/. f(n) :!! 2{nl/xll-l/yR). then if Hence f(n) ;:,. 2(//x}l-//y//). n >-fxff• If O~ n~l, then f(n) > O if Since Now consider l ~ n > ~~{(. If n>l, l}yJ/ ~ ll xJ/, it follows that f(n) > O f (-n) = ... QJx -t- {nx-y)IJ-/Jx - (nx-y) -JU . Since the above argument is valid if only assumption being = J}(n-l)x:+-yJ/ //x - (nx+y)/l y is replaced by llyJ/ ~ !Ix/I, it follows that -y, the f {n)< 0 if n-c:; ; _ JI.IlL Thus the only region for which f (n) can be zero is I/xii ·• for /n/ ~ .J/lJt~ Hence if f (a) O, or x..1-ax+y, then !al ~ /~~1r· = Theorem 4.7 • . If x (:# 0) Proof: Take f' (n) n ?!::. 1. get /12x+ as /Ix + (nx+y )J/-1/x - (nx+ y )/1. From the n~i + ~~i = 1, it fo+lows that identity, if f (n) y are two elements of a. /ly/1 ~ /)xi/, ~ I a/ ~ 2 •1J~~U lg// if x ..Lax+y. normed linear space and and = ll2x+ n,i )I + Q/(n-1 )x+ ~~! y j/ - II (n-1 )x+ylLJ But from the triangular inequality of the norm, we :J.i II~ l/2xf/-~Y,l! • It also follows that }J(n-l)x+y//s. ,,/Jy// + //{n-l)x+~y/), and hence Ii+! n+~ /}(n-l)x + ~~f yl/ - }l(n-l)x + y// ~ -2~~~. f (n) :::.. 2/lx//-4_/JzlJ.. and f (n) > O if n+l ' assumed that n :'! l, but Z //yJ/ - UXII :::. 1 it Thus 1/XQ - f (-n) = - QJx + (nx-y)ll - llx - (nx-y )fil . in the sign of y. n > Z/JyJJ-llx/J It was IJxU • Uy/I ~ //x i/ Al.so, • This differs from f (n) Since the above argument is valid if l. Condition (2) of Definition 1.6. y onl y is -y, the only assumption being //yll ::?::.//xii, it follows f (n )~ 0 if n-<- 2 1 ~ - I/xi/• Since f (n) can not be zero replaced by that if In}> 2 lli~iillxtL, all values of x1-ax+y, must be in the interval a f(a):::: o, or for which la/ -< 2 UYll-llx¥ //:X:// • It was shown above Theorem 4.6 that for ordinary Euclidean lal = ~ lcos 0/ space between the vectors as near to X it x..Lax+y, and y. where is the angle 0 Thus it is possible to have Jal l¾ as desired, whether /IY/1 ~ 1/lell or not. It is I t, l-;:- \\"!Jvit•tf! therefore clear that the inequality of Theorem 4.6 cannot be strengthened without introducing some concept analogous to the I angle between x and y. Moreover, the difficulty of doing this is illustrated by the following Example 4.5, which shows that for normed linear spaces it is possible to have y being a multiple of la I~ 2 ll;rll -1/xJI - y Example 4.5. (a, b) 1 ~. and a multiple of x = (1,0) Hence /af = as the larger of a But of Theorem 4.5 and is the only value of llxll = llyll is = 1. n, Since -1, it is possible ~if/ without y being a multiple of x. 1. The space of Example 4.1. /a1 x +(nx+y) = (2+n,l) Then n= -1 IJx +(nx+y)/1 = 1/x - (nx+y)Jl. in this case the number to have /l(a, b )/1 y = (1,1). and x-(nx+y) = (-n,-1). for which x. Consider the normed linear space of all pairs of real numbers, w-i th Let /{y/1 s Jfx}/, and that the of Theorem 4.7 can be an equality I/XII without having = gj~ without Example 4.6 shows that Theorem 4.6 :x:. is not valid without the assumption relation /aj 32. Example 4.6. (a,b) and Consider the .normed linear space of all pairs of real numbers, with y //(a, b )/j = /a/+ }b/ • 1 Let x = (-1,1) = (a+l,O). Then /lxJ/ = 2 and fly//= Ja+ll. Also, x + (ax+y) = (O,l+a) and /Ix +(ax+y)// follows that x - (ax+y) = {-2,1-a). If' a ~l, it = j/x - (ax+y)// = a+l, and hence that a > l. This ~i~ = ½(a+l), and hence shows that Theorem 4~6 is not valid for normed linear spaces x ..L.ax+y. But without the assumption /lyll ~ llxl/. Moreover, for this example it is seen that a= 2 //xii if a~l, which is the largest value /Jf,/i;; allowed for lal by Theorem 4.7. It has been sh©wn (Th. 3.2) that orthogonality in abstract Euclidean spaces is symmetric, homogeneous, and additive, and that for any elements :x:..Lax+y. 2 that x and y there exists a number a such For normed linear spaces, isosceles orthogonality is clearly symmetri~ and the existence of such a number given by Theorem 4.5. a· is The effect of assuming homogeneity and additivity will now be investigated, it being shown that isosceles orthogonality is homogeneous and additive only for abstract Euclide1n spaces. Lemma 4.8. If isosceles orthogonality is additive in a normed linear space y of T there is a unique number Proof: that T, then for any two elements a such that x (#: 0) X_Lax+y. By Theorem 4.5 there is at least one number x...L.ax+y. Suppose it is also true that }lx+( bx+y)II = )l:x:- (bx+y)II, it is clear that 1. The space of Example 4.20 2. See page 13. and x.J..bx+y. x-1.-(bx-ty). a such Then since Additivity 33. then gives :x:.L(a-b)x. = /1-(a-b)/. or /l+(a-b)/ a-b = O, or a= b. Lemma 4.8'. J/:x:+(a-b)xll = }Ix - (a-b}x/J, That is, But this can only be true if If isosceles orthogonality is homogeneous and additive in a normed linear space where :x: and Proof: y are any elements of and :x: ...Lax+y T, then b/Jy// and 2 y ...L by+:x:, = a/Ix// 2 • Clearly a non-zero element can not be orthogonal to a non-zero multiple of itself. and if nor is zero and either x Hence if y = O, then a= 0 unless =o, then b = 0 unless y = o. If neither x x = O; y T a or b is zero, then :x:..L y and the other is zero because of the uniqueness given by Lemma 4.8. Hence the Lemma is true if one of a, b, x, or y it will be supposed hereafter that none are zero. shown first that a IIYII = ll2ax+y/l • Thus if :x:.L ax+y;, or a Similarly, = = 2 Uxll and J/(:x:+y) - (:x:-y )11 = 21/yll • = }}y}}, .it follows that (x+y) .1- (x-y}. From homogeneity, But also, )I (x+y) + (x-y) II Hence if )):x:J} (2a+l)(x+y)-L.(2a-ll(x-y). That is, //(2a+l)(x+y)+(2a-l){x-y)}j = //2ax+yl/ = Vx+2ayll. /1(2a+l)(x+y) -(2a-l)(x-y)/j, or //2ax+yll = /Jx+2by}j. :x:+2ay =' [x +(a+b)i)+(a-b)y and hence (a-b)y..LE£+(a+b)ij y.J...~+(a+b)~. But Since it was = /lx+2byl/, it now follows that 1Jx+2ay /I But = /l:x:+1(ax+y)// /fx - 1 (ax+y)JI. a a y..L byt:x:, and }/:x:J/ l):x:+2by//. = 1/yll, we get )/2ax+y II = U:x+2byll. I/xii shown that It will be = b if //xi/ == 1/yl/. If :x:..Lax+y, it follows from homogeneity that Thus is zero, and y-L by+x, and If x+2by = l}c+ (a+b) ~ - (a-b)y, a-b-:/,: o, homogeneity gives and the uniqueness given by 34. = 0, contrary to assumpt.ion. Thus a = b Lemma 4. 8 implies . a if /Jy/1 = ~xi/. But if x-Lax+y and y.L by+x, then because of homogeneity, rx.L. %(rx)+y and chosen so that r is = }lyJ/ , then !!.r = br, or a //xJ/2 = b//y)/2 • 1/rxJ/ Theorem 4.8. Hence if y ..L (br) y+rx~ If isosceles orthogonality is homogeneous and additive in a normed linear space T, then T . is an abstract Euclidean space. Proof: Define the inner product ax+y are orthogonal. and 2 as • -a//xJJ , where -( x,y) x It is only necessary to show that this inner product satisfies the conditions of Definition 1.9: (1). (tx,y) = t(x,y). If x.Lax+y, and t ¢ o, then tx.l..%(tx)+y if the orthogonality is homogeneous. Thus 2 2 (tx,y) _ -!//tx/} -atl/xJ/ • Hence (tx,y) t(x,y). If = = t =O, the proof is trivial. (2). = (x,y)=(y,x}. If 2 -alfxl/ and (y,x) (J). (x,y)+(x,z) (x,y) x..Lbx+z. x..Lax+y, and y...L by+x, then 2 -b//y/( • These are equal by Lemma 4.8'. = = (x, y+z). Suppose x..Lax+y and Then x..LITa+b)x+(y+z] if the orthogonality is additive. 2 2 2 Hence (x , y) -a }Ix// , (x, z ) b }/x// , and ( x, y+z ) = - (a +b ) llxJJ ~ 2 (4). (:x:,x) llxll • Since /Ix+ (-x+x) JI J/x - (-x+x)l j , 2 x..1. (-1 )x+x, and (x,x) = JJxJ/ • = =- = = It can be shown that isosceles orthogonality is homogeneous if it is additive. This makes it possible to simplify the assumptions used in Theorem 4. 8,, assuming only add.it ivity. 35. Homogeneity alone can also be shown to be sufficient by using Ficken•s condition for the existence of an inner product. Theorem 4.~. If isosceles orthogonality is homogeneous or additive in a normed linear space T, then T is an abstract Euclidean · space. Proof: It has been proved by Ficken1 that a normed linear space is an abstract Euclidean space if and only if all numbers and elements a for elements x V(x+y}+(x-y)}f and y x and y of T we have 1/{x-t-y} - (x-y}JI and lfax+ylf = 1/x+ayll for for wh ioh lfxll = II YII • If IJxll = llyJI , then {x+y)L(x-y) • 2 If isosceles orthogonality is homogeneous in T, then /l{a+l) {x+y)+ {a-1) (x-y)II = 1/(a+l) {x+y} - (a-1) (x-y}U, or \) ax+y f = llx-+ay fl. Thus a normed linear space for wh ioh isosceles orthogonality is homogeneous is an abstract Euclidean space. x.i..y, then x..L-y and y.J...x for isosceles orthogonality. tive, then If because of the nature of the condition 2 Hence if the orthogonality is addi- nx..Lmy for all integers m and n. Thus llnx+myt\:::: nnx-myl( and ~x+~(y)II = /lx-E(y)II. Since the norm is contin3 uous, it follows that //x+ty~ ; Nx-tyl for all numbers t. That is, x..Lty for all t. Thus isosceles orthogonality is homogeneous if it is additive, and a normed linear space for which isosceles orthogonality is additive is an abstract Euclidean space. Corollary 4.9. Isosceles orthogonality is additive in a normed linear space T 1. Ficken (XIV) • if and only if it is homogeneous in T. 2~ Since x.Ly if and only if /fx+yl( =\}X - YI• 3. As follows from condition (2) of Definition 1.6. 36. 5. PYTHAGORElili ORTHOGONALITY While Pythagorean orthogenality is perhaps the most obvious means of introducing orthogonality into normed linear spaces, it is more difficult to use than the other types 1 because of the squaring of the norms. There is also a lack of symmetry resulting from the possibility of defining 2 }JxJ} +)Jy1) orthogonal if x and y to be = ~ -yJ} , or it )JxJJ2+JJyJJ2 = //x + y/}2. 2 2 One sign is as good as the other, although the two statements are not equivalent. More simply, x.J..y does not imply LL-y., as is the case with isosceles and spherical orthogonality. Pythagorean orthogonality is clearly symmetric, and it will be shown that the assumption that such orthogonality is homogeneous in a normed linear space abstract Euclidean space. T implies T is an Thus homogeneity of the orthogonality implies its additivity. It can also be shown that there exist non-zero orthogonal 2 elements in any normed linear space. Theorem 5.1. If x and y are elements of a normed linear space, then there exists a number a such that 2 1/xJ) 2+)/ax+y// 2 = l x. - (ax+y)J/ , or x..L (ax+y). Proof: :t' ( n) , as Define the real valued function of a real variable, 2 }}xi/ +//nx+yl/ Using the identity 2 - /1 :x: - ( nx+y) // 2 or /lxlJ2+ /Jnx+y/} 2 - / / ( n-1) x+y/,2. (n~l)~+ 2 ~;1 = 1 , we get: 1. See Definitions 2.1-2.J. 2. Whenever "orthogonality" is used in this section, it will mean the "Pythagorean Orthogonality" of Definition 2.2. 370 = /Jx/} 2+ 2~ 1 }lnx+yJI 2+ lJ!(n-1 )x+(n~l~ y// 2 - //(n-1 )x_;y/1~, =• /Jxll 2+( 2n-l) /Ix+ 1nY ,2~ + f (n) 1 Q/(n-1 )x+f-~1} y// -}/(n-1 )x+ yij] LJl{n-1 )x+(n;1 y)/+//{n-1 )x+ yJ[J • The triangular inequality of the norm.1 gives }//(n-l)x+(n~l)y}J - }l{n-l)x+y//}~ J/¼y//, and 2 2 f(n)~ /JxU +(2n-l)//x+½YH llYIJ [ //(n-l)x+(!!~1}y//+//(n-l)x+yJTI, I;I 2 2 ::/JxJ1 +(2n-l)/lx +½yJJ - //yllD/(!!~1) x+(n;}~ y//+ }J(n;1) x+ If ! yjj] • n~O, it follows by using the triangular inequality of the norm th~t JlxJ}~ Likewise if ½}/yl/+Jlx+; y/1 , and //x+ ½y)/ 2 ?: mxJI- ;JlyJO 2 • /}(n~ 1 )x+(n;})y}J+H~~l)x+; yJ/~2f;1~llxl/-+ 2~:1 JJyJJ. n~l, Hence f (n)~ 2n//x}J2+ 2~;-1 )Jy//2 - 2{2r:i-~ Ux// //y//-//y// [2(n;l~ }/xii+ 2~;_1//yf[J , 2 2) }/xii //y}/ , - 2n JJx}/ - 2 p~- =2 llxJI [nJJ:xll-(3~-2~ /}yJLJ , which can be made greater than zero by taking n sufficiently large. 2 ' Now consider 2 f(-n) = ijxJ/ +//nx-y// ~! f - ~1 = 2 -//(n+l)x-yJJ ] . 1 2 1 Using the identity 1, we get: 2 2 2 f(-n) //x}/ - 2~;:1 //nx-yl/ +[l/(n+l)x-~~~ y/1 -//(n+l)x-y}/2], = = /Jxl} 2- ( 2n+l) )Jx - ! y// 2+ {]j ( n+l )x -~:~ y// - //(n+l )x - :r"Jl]X m(n+l)x-f~1~ y}/ + /J{n+l)x -yI!J, 2 _ :% JJxJ } - (2n+l) }Ix - ! y /J 2+ , ~ ) J1Yij m(n+1) x - ~n:1~ y )/+/Xn+l )x - :rJ , 2 = }Jxl} 2- (2n+l) lJx-½y// +//yJ/ W~-;;~ x-(n:~ Y//+//ft1j x-! y}LJ • 2 2 If n> 0, then /Ix yJI ~ }JxJ} - ¾ ltxl} }JyJI +½ J}yll 2 , and 2 2 1 f (-n) .:S -2n/JxJ12+2 (-2 r:7~ J/x/1 llYll - 2~~1 //yJ/ +UYJI [! (~l) J}x//+ ~~ Uyfl] , -! L = r us -2 //xJI f (-n) [n/Jxj/ - 3n,;"2 J/y/D• is negative if .n is sufficiently large. 1. Condition (2) ,o f Definition 1.6. Since 38. f(n) is continuous and has positive a nd negative values, it f (n) = O for some value follows that of a n. That is, //xJ} 2 + //ax+y/}2 = //x - (ax+y)// 2 for some number a. While restrictions on the value of a are not as simple as for isosceles orthogonality, it is possible to obtain them from inequalities derived in the proof of Theorem 5.1~ Thus n;2 //yfil and f (-n)~ -2//x// [ nHxJJ - 3~+2 NYJLJ . f (n}~ 2//xJ/ [ n//xl/ - 3 When }}x/) = }JyJI, these become: 2 f (n)~ ¾/JxJl E12-3n+~ and f (-n)s: -¾JJxJJ For this case it then follows that and a 2 (E2 -3n-~ . must be between -½(3+i17) 2, and that there is a possible value of -½ (3+117 ) and a between The lack of symmetry of these limits is due 1. to the condition for Pythagorean orthogonality not being equivalent to /JxJJ2+ //y/1 2 = // x+y/l 2• That is, x.i y does not imply x ..l.-y. However, Theorem 5.1 is valid if this change is made~ Corollary 5.1. If x y are elements of a normed and linear space, then there exists a number a such that )Jxll2+!Jax+y/1 2 = )/ x+(ax+y}// 2 • By Theorem 5.1, there exists a number Proof: eleIJ}.ents Let x and a= -b. Then -y 2 /}x// + Jfox-yJJ such that 2 2 2 b for the = )Ix - {bx-y)// 2 • 2 }/xi/ + /Jax+y// = /Ix + (ax+y) )/ • However, it is not always possible to find a number for arbitrary x and y such that )]xJ/2+//ax+yl} 2 = }Jx - (ax+y) I} 2 = }} x + (ax+y) )J2. This is Shown by the following example: a .39. Exe.mple 5.1. where //(a, b)}/ and y = (0, 1). Consider the space of all number pairs (a,b), 1 is the larger of /a / and /b/ . Let x - (1,0) Then Consider the equation l +/}(a,1)// 2 ax+y = (a, 1) and x ± ( ax+y) = (1± a ,±1} • }JxJJ 2+JJax+yJJ2 = /Jx - (ax+y)// 2 , or = //(l-a,-1)// 2 • This equation must have at least one solution because of Theorem 5.1. If If it becomes 2 = 1, which has no solution. 1 ~ a ~ 2, it becomes l+a 2 1, which has no solution in 0 ~ a :~ { 1, = the interval (1,2). If for a ~ 2, it becomes l +a 2 = l-2a+a2 , which has no solution a::::?:2. ( 2 - l-2a+a 2 , or a= l-"12 . l+a 2 = l-2a+a 2 , which has no solution If · -l.:$; a ::::: O, it becomes If for a .:S -1, it becomes a ~ -l. Therefore a = l-'-/2 is the only value of a such that 2 2 IJxJJ2+ )Jax+y/} -:- j/x - (ax+y) // • Similarly, a = '12 -1 is the only 2 2 value of a such that //x// +}}ax+y/J2 = //x+(ax+y))/ • Hence there is no value of a such t hat /JxU2+JJax.+yJJ2 )Ix- (ax+y)}/ 2 = 2 IJx+(ax+y)// • = Pythagorean ortho gonality is clearly s ymmetric, and for any x a nd y there exists a number a such that xi.ax+y. H6wever, such orthogonality is neither homogeneous nor additive, and can be homogeneous only if the normed linear space is abstract Euclidean. Thus Pythagorean orthogonality is additive if it is homogeneous. The converse of this is also true, and it therefore follows that such orthogonality can be additive in a normed linear space only if the space is abstract Euclidean. 1. The space used for Example 4.1. 40. Theorem 5.2. If Pythagorean orthogonality is homogeneous in a normed linear space T, then T is an abstract Euclidean space. Proof: It will be shown that the condition given by Jordan and Neuraant for the existence of an inner product in a normed linear space is satisfied, namely: }/x+y 112+ l/x.-yJ/2 = 2 }J1xJJ2+JJyJI~ for all elements x and y. Because of Theorem 5.1, it is known that for any elements x and y of T a number a exists such 2 2 that //xl/2+.}Jax+y// = //x - (ax+y)J/ • Assuming homogeneity, it 2 2 2 follows that k2 //xJ/ +Hax+y// = }/kx - (ax+y)// for all numbers k. 2 Set k equal to a±l. This gives /lx7Fy// 2 = (a±l)2/Jxl/ +//ax+yJJ 2 , and: 2 }Jx+yl} 2+/Jx-y// 2 = 2J/ax+yJ/ + 2(a 2+1))/x/J2. But using homogeneity again, it follows ' that ax...Lax+y, and hence 2 2 2 2 2 2 JJax+y// + a.2 /JxJ/ =//Y// • Therefore }Jx+yJJ2+ }/x-yJ/ = 2 f]JxJ/ +J/yJJ J. It has been shown that the condition for the existence of an inner product is satisfied for all elements x conditions of Definition 1.9 are satisfied and and y. T is an Thus the abstract Euclidean space. • The inner product (x,y) known to exist by Theorem 5.2 2 2 is equal to ¼[lfx+y// - //x-y// ~ • It is interes_ting to note that, 2 2 2 2 by using })x71=yl} = (a±l) /JxJI + Uax+yJ/ , as shown in the proof of 2 Theorem 5.2, this reduces to (x,y) = ¼[Ja-1) 2 //xJ} - {a+l) 2 )1xl/~ , or {x,y) = -a J)x/j2. This is the definition of the inner product used in establishing 1. See Jordan and Neumann (IV). 2. See Theorem 1.2. 41. Theorem 4.8, which is the analogy of Theorem 5.2 for isosceles orthogonality~ It was shown that isosceles orthogonality is homogeneous in 1 any normed linear space in which it is additive. This can be strengthened for Pythagorean orthogonality: Theorem 5.J. The properties of homogeneity and additivity of Pythagorean orthogonality are equivalent for normed linear spaces. Froof: If Pythagorean orthogonality is homogeneous in a normed linear space T, then T is an abstract Euclidean space because of Theorem 5.2. The orthogonality is then additive because of Theorem J.2. Conversely, suppose and y x .Ly, are arbitrary elements of a normed linear space. Theorem 5.1 gives the existence of a number x .J..ax-y. hence where a if x -,: o. Thus x ...L -y. Also, Then such that If the orthogonality is additive, then a = 0 x x -1.. ax, and y ..L x, because of 2 the nature of the condition for Pythagorean orthogonality. Using additivity, it now follows that nx.J.. my for all integers 2 2 2 m and n. Thus llnxll + llmyll = Unx -myll , or = ltx - ~ t ft • 3 2 2 2 Since the norm is continuous' it follows that llxh + lltyll = IJ x-tyn 2 2 t, or x .J...ty llxll + I(~ y/1 for all numbers 2 for all t. Thus Pythagorean orthogonality is homogeneous if it is additive. 1. Lemm.a 4.9. ~ ~ 2. That 11:x:ll.a. +lfy H Ux - y ll • J. As follows from Condition (2) of Definition 1.6. Clearly Theorem 5.2 is still valid if -y throughout. y is replaced by This and the combination of Theorems 5.2 and 5.3 give the following Corollaries: Corollary 5.2. If two elements x and y of a normed linear space are said to be ortho~onal if and only if l!xJl2+ Jlyl\2 = llx+yJ2, and if such ortho linear s ace T, is homo eneous or additive in a normed then Corollary 5.3. is an abstract Euclidean space. If Pythagorean orthogonality is additive in a normed linear S;PfiCe space. T T, then T is an abstract Euclidean 43. 6. SPHERICAL ORTHOGON.ALITY IN GENERAL NORMED LINEAR SPACES. It is clear from the forms of their definitions that both isosceles and Pythagorean orthogonality are symmetric, and that spherical orthogonality is homogeneous. l It has been shown that a normed linear space in which either _isosceles or Pythagorean orthogonality is additive is an abstract Euclidean space, 2 while it will be shown that a normed linear space in which spherical orthogonality is symmetric and additive is an abstract Euclidean sp,a ceo 3 .As with isosceles and Pythagorean orthogonality, it follows that for any elements exists a number a x and y of~ normed linear space there such that x is orthogonal to ax+y. 4 Actually, a stronger theorem than this can be established by relating spherical orthogonality and the theory of linear functionals. 5 Thus for any element maximal linear subset x of a normed linear space there is a H such that x .L h for all This "hyperplane" is not necessarily unique. h in H. 6 In fact, it will be shown in the next section that its uniqueness is equivalent to the existence of a differential of the norm, as well as to additivity of the orthogonality. 7 In Section to the relations between spherical 1. See Definitions 2.i-2.3 .. Homogeneity, symmetry, and additivity of orthogonality are discussed on page 13. 2. Theorem 4. 9 and Corollary ·$ . 3, 3. See Section 8. 4. Whenever "orthogonality" is used in this section, it will mean the "spherical orthogonality'' of Definition 2. J. 5. See Definition 1.8 and the following discussion. 6. Theorem 6.20 7. See Corollaries 7.3 and 7.40 orthogonality and linear functionals will be used in finding an evaluation of linear functionals for certain types of normed linear spaces. The present section will deal primarily with showing the existence of this hyperplane and a dis cussion of the resulting number a for which x ..L ax+y, and with showing the existence, for general normed linear spaces, of limits whose equality would give the existence of a t ype of differential of the norm. 1 An element Definition 6.1. is orthogonal to a set U (x ..LU} x of a normed linear space T if and only if x ..1.. u for all, -u u. E: Definition 6.2. A maximal linear subset of a normed linear "'fVl s-'tC-tl~) space is a11 linear subset linear subset. Such a subset called a hyperplane. . u which is not Vcontained in any other 1-'¾rA U, or any translation x+U, is 2 The properties of linear functionals which will be needed here are discussed on page 8 of this thesis. the modulus Vt JI of the linear functional In particular, if r is 1, then ?(* for all is as near to x, and elements vcan be found for which Will as desired. The equality may not actually hold for any element, the following theorem showing that it will 'X 1 hold if and only if there is an element orthogonal to the hyperplane consisting of elements/\ for which t = o. 3 This theorem also hET l,, See Corollary 6.6 and Theorem 6.7. 2. Hyperplanes are dis·c ussed by Mazur (IX}, pg. 71. 3. That the elements form a hyperplane isshown implicitly in the proof of Theorem 6.1. 45. enables one to show the existence, for any element hyperplane H f or which Theorem 6.1.. If x ..LH. for all elements Proof: elements IJ :t'fl = l ,~ is a linear functi onal f, with on a normed linear space x ..L h x, of a }f (x)/ = Ux l( T, t hen h of T if and only if f {h) = o. such that Let H be the linear subset consisting of all 1 such that f(h) - o. This linear subset is maximal, h since, for any two elements x and y not in H, f(x+ky) =O if k - ~ ~~~.,. -Thus x+ky £ H, and hence any linear set1containing x and y H contains }f (x)/ = llx ll • )f(x+h)k:,,-\_ T. Ut ll = 1, it follows that Since o and therefore all of lf(x) I~ Nx+hll . Hence llxll~ 1/x+hll Suppose f {h) = O implies for all h<c'...H. ~-------~--✓ Since H is closed under ·multiplicati on, this gives x ..L.h if h £-H. Conversely, suppose p :::: l llt ll = 1. since. x ..LH. Then Jr (x) J = p lfxl/ , where Take Uxll ~ llx+h tl if h £-H, and }t(x+h) J = lt(x) I = P UXII !::: p /jx+hl(. Since H is maximal, every element -for which of the form k(x+h), where y £.T. }/t i/~ p Then Theorem 6.2. and If x h &. H. p = 1, or Thus f is not zero is )t (y) I~ p \l y fl for all It (x)f = Rxll . is any element of a normed linear space, then there is a maximal linear subset H such that Proof: space Let T, and x f x J...H. be an arbitrary element of a normed linear be a linear functional such that f(x) = lfxll l. H is closed under addition because of the linearity off (see Definition 1~8), and is closed under multiplication by real numbers because of the homogeneity of f which results from its additivity and continuity.,. and lltll =1. 1 Take H as the maxim.al linear subset consisting of all elements h such that f(h) = o. Then by Theorem 6.1, x..LH. The/ converse of: Theorem 6.2 is not true. The most obvious 2 H is dense in its space. This is the failure is the case when case for Hilbert space 3 when the maximal linear subset contains the set S H of elements with only a finite number of non-zero components, since any element x = {x1 ,x2 ,x3,x4 ,•••) (x ,o,o,•••), {x1 ,x ,GQ·i;··) 1 2 The converse of Theorem 6.2 will be further is the limit of the elements • • •• investigated in SectionJD. It is closely related to the problem of finding an evaluation of linear functionals for normed linear spaces. Theorem 6.2 is stronger than the analogous theorems proven for isosceles and Pythagorean orthogonality. 4 However, while for any elements number a x such that and y of a normed linear space there is a x..Lax+y, 5 or }lx+k(ax+y}ij ~ lfxl( for all k, it is not always possible to choose a such that Ux+k{ax+y)J/ > I/xi for all k TO -as can be done for abstract Euclidean spaces. This can be shown by using the normed linear space consisting or all pairs of real numbers (a,b}, with ll(a,b}ll as the larger of 1. The existence of such a fU]lctional f is given by Banach {I), pg. 55, Theorem .3. 2. H is dense in T if H T• .3. The space of all sequences x (x1,x2,x.3,•••) for which 2 n 1 is convergent, with Uxll = ~~{xi) 2 • 4. Theorems 4.5 and 5.lo 5. Theorem 6• .3, below. = = \a\ and lb} • 1 For let x = {l,O) and y = (0,1). Then #x +k(ax+y) II= /l(l+ka,k)II, and is less _than and a Thus _ llx+k(ax+y)II~ I/xii -2<ka<O. = o. Thus x..Lax+y only if a = o. /Jx+k(ax+y)lf-=> llxll Theorem 6.J. x a Proof: a. only if Ux+kyll = llx ti if y are any two elements of a normed and for all k. a such that That is, such that x = 0, then clearly If k Jkl< 1 such that linear spaee, then there exists a number llx+k(ax+y)U > llxU it k :j: o. for all If for all But Jkj ~ I, and hence there is no number /lxJ/ = l x.J.. ax+y x..Lax+y. 2 for all values of x =, o, then Theorem 6.2 gives the existence of a maximal · If ,=i,t- ,, , r,; linear subset H such that ax+y must be in H for some number subset gotten by adding x = by+h x.LH . . . Since y to x . , . ,., /"I . -1, is not in H, For if not, the linear H would not contain implies - ; x+y =-l h t-H. b a. ,:, , Thus x, since H would be properly ~, -( .1:>kci:,,wi.c. ~-l- ~e,J ""-"1 Cv-\.-"t.-\.-;, X) contained in another linear subset and hence would not be maximal. If a is taken so that ax+y E.H, then x..Lax~. Jor ordinary Euclidean space, it ha~ been.shown that I aJ:s; f&1, if x.Lax+y. 3 It was shown that this inequality was valid for isosceles orthogonality in normed linear spaces if 1/yH :S. JJxH, but had to be weakened for )lyll > Vxll . 4 The following Corol lary shows its validity for spherical orthogonality, without any assumption about the relative sizes of 1. The space of Example 4.1, pg. 21. 2. In the sense of Definition 2.J. J. See the discussion and figure on page 29. 4. Theorems 4.6 and 4.7 and Example 4.6. IJylJ and I/xii. Corollary 6.1. If x (# 0) normed linear space, and x...1..ax+y, then By definition, i · x-1..ax+y Proof: y are elements of a and lal:::S. ~~J,• if and only if llx + k ( ax+y )// ~ I/xii for all k. If I/; Yil ~ llxll , or la l:S t~i~\ • 1 this gives k =--, a Unlike isosceles and Pythagorean orthogonality, spherical orthogonality is not symmetric in general normed linear spaces. That is, x..Ly does not imply y ..1..x. This lack of symmetry is shown by the normed linear space consisting of number pairs (a,b), 2 with //{a,b)ll=lal+lbl. Then l/x+kyll = II (l+k,k)II = Ji+kJ + lkl ~ 1 = llxll. for all k, and llyll = 2, to x. ( x..Ly. //y+kx 1<-1/yll But if Forlet and y:(1,1). Thus llx+kyil;::,,. llxll IIY+kxll = ll{l+k,l )II= µ.+kl + 1. Since -2 "'k< o. Thus y is not orthogonal Because of this lack of symmetry, Theorem 6.3 does not give the existence of a number shown hy Theorem 6.4. such that that x=(l,O} x .1.. ax+y bx+y .1..x. a such that ax+y.L x. This is Likewise, the uniqueness of the number is independent of ·that of the number b a such These types of uniqueness will be studied in Section 7, and will be shown to be equivalent, respectively, to Gateaux differentiability of the norm and to the condition for a norpied linear space to be strictly normed. 3 1. Definition 2.3. 2. The space of Example 4.2, page 24. 3. See Definitions 7.3 and 7.4, and Theorems 7.3 and 7.8. Theorem 6.4. If x and y are any two elements of a normed linear sp½l.ce, then there exists a number ax-+y...Lx. This number a I is a value of a such t.hat I for which //kx+yll k takes on its absolute minimum. By definitioh, 1 ax+y ...Lx Proof: if and only if 1/( ax+y) + kxll ~ llax+y ll for all ax+y ...L x Thus k. if and only if ll ax+yll is the smallest ~6 'v~~ \<./t,v,,l 1 value of -p~iJv..~ II kx+ylf . Since {M 1/ kx-tyll I<.~ becomes vinf ini te ,,a:t' +o1> and is a continuous function'--'and ;;t;, ~ . - =, it must take on its minimum. I The number a can then be any value of ,. :,. ,._, to _.'i'c I k for which Jlkx+yll takes on its absolute minimum. Because of the difference in the methods of evaluation a nd interpretation of the numbers for which x -1.. ax+ y, and for which ax+y ..L x, it is interesting to consider the effect of assuming symmetry of spherical orthogonality. The following corollaries follow immediately from Theorem 6.4, using Theorems 6.1 and 6.2. A further result of this type is given in the next section by Corollary 7.,5. Corollary 6.2. If a normed linear space is such that spherical orthogonality is SYplilletric, and with llr ll = 1 and f ( kx+y) = 0, ~ If ( x) k =- ~~ Corollary 6.3. I - llxU , then f is a linear functional ll kx+y H is minimum if i? • If a normed linear space is such that spher- ical orthogonality is symmetric, then for any element exists a maximal linear subset 1.· Definition 2.3. H such that x ....LH and x there H..Lx. 50. / Symmetry also aids in a further investigation of the relation between the numbers y ..1.. by+x. a and b, where x ..L ax+y and Such a relation was very important for isosceles orthogonality, it having been shown that b //y/[ 2 = a /lxll 2 if isosceles orthogonality is homogeneous and additive. 1 This was the key t.o the proof of Theorem 4. 8, which showed that isosceles orthogonality can be homogeneous and additive in a normed linear space only if the space is abstract Euclidean. "'l\,(Y"'\ O"./ -1} Theorem 6.5~ If x normed linear .space, and y are any two Yelements of a and I . x..L ax+y and y ...Lby+x, then Jab j~ l. If the space is such that spherical orthogonality is symmetric, then 0 ~ ab ~ l. By Corollary 6 01, }a l.::s 11~~1~ and Proof: labl;:$ 1. If symmetric o x ..L ax+y, then Hence y J.. by+x, then ax+y..L x lbl~ :J~/[. Hence if the orthogonality is ll(ax+y) + k1 x ii~ llax+y\\ )l y + k 2 (by+x) ll~ JlyU for all for all orthogonality is symmetric. The validity of Theorem 6.5 for ordinary Euclidean space can be easily shown. From the figure, i-t is evident that a and b are both neg- ative or both positive 1. Lemma 4.8•, page 330 1 If k • If kl= -a, we 2 ll ax+y n ~ 11-abl ltyll . II YI ~ /lax+y ll . If k2 = l-~b' we get Hence IIYJl > Jl-ab l IIYII , and 1 ~ }1-ab I. Thus get k • ab ~ O, if spherical 51. sih 2 cx. = \ab ! , and Then san;i.i ,s,1ign, ~ . '_ - .._ ~~, I Also, sino<.= 1}~~1 = e is acute or obtuse. according as lab I~ l. Since a and b l\~1. 1 are of the 0 ~ ab ~ l. .'., Suppose x ...Lax+y .and y .1- by+x. By assigning particular k1 and k 2 in )lx + k1 (ax+ y )II ~ Uxli and \\ Y + k 2 ( by+x) H~ II y ll , a number of interesting inequalities can . l be derived: values to If k = ± ½: l) (a±b)x + y l\;?; lblllxU, if 1 (1). l1 2ax + y/f ~ laJUxll, 1 = ;: 1 =~ ±1 : k1 b . 1/ by + x ll ~ l1-abj Uxll , = 1-ab· k k kl k 2 = ± ;: Jl x+ (b±a) yfl ~ \al Jlyll , k2 = • ½: IJx+ 2by II~ lbl /lyll, k 2 = l ~ b: IJx _± yfl ~ fl + b) llyM, /I x ± y fl ~ ' 1 +- al llxll , k2 = l-~b: Uax+y)I > Jl-abf fl YII, k 2 = a :: b: ffay+x ll ~ la - bl #YII • l . = b-a'· /l bx + y lJ ~ )a - bl ltxll , If spherical orthogonality is symmetric, it also tollows that and llx+k1 (ax+y)\I ~ l~I llax+yll Jly+k 2 (by+x)II~ lk 2 \ llby+xll, and other inequalities can be ax+y .1.. x and by+x--Ly. Thus gotten by giving values to k1 and k 2 • But because of Theorem 6.4, these are nothing more than consequences ot Jlax+yll and llby+xll being minimum. Thus by making the same substitutions as above, one gets the following inequalities--all or which are obvious results of this minimum condition, but are interesting in that they are the variations in equations (1) above made possible by the assumption of symmetry. /l(a±b)x+yll ~uax+yll, llx + (b ± a)yll.:::,.. l/by+xll, ll2ax + yll ~ »ax+yfJ, Ux+2byll> lfby+xll, Jlx±ylJ ~ Jlax+Yll, /Ix-+- YII ?:!:. llby+x II, 1. k 1 = --;- (0r analogously, k2 Corollary 6.1. =-½) was used in proving 52. (2) (Cont.) If }lby+x)I~ )bl /lax+y /1 , lax+yl > \al )by+xll, llbx+yl ~ llax+ Yll , II ay+ x ii> I/by+ x j. and X and x.1-y, then all n. are e leme.n ts of a normed linear space, y Ux+kyll ~ llx ll for all >"-"-- "'t'ftr:o..l..._ k, or It n:x:+yll ~ 1/nx II for It thus seems interesting to investigate the limit of llnx+y/l - llnx/1 as n becomes infinite, both when orthogonal and when they are not. and X y are This limit will be shown to exist, and to give a condition for orthogonality, as well as~ determination of all numbers a such that be related to the limits of /lx+hr!- llx /1 x.1-ax+y. as h 4 It can then approaches zero from the right and left, and to the Gateaux differential of the l norm if these limits are equal. Theorem 6.6. If x normed linear space,~ Proof: r and s infinite. y are any two elements of a lim IJnx-t-yll - llnxll exists. m~oo Because of the triangular inequality of the norm, 2 J llnx+y/1 - llnxll f ~ )/Yll • infinite. and Thus Assuming that llnx+yll - llnxll Hnx+yl - lnxll llnx+yll- llnxll becomes as n becomes is a continuous function of there must be arbitrarily large values of llnx+yl/-llnxl = n t~- \lnx+yll- Unxll does not exi:st, let be two limit points of Since is bounded as n n 3 such that ' T~ If such a value of n is greater than lrtsl , then, for that n, 1/nx+yll = U(n+~)xj/ , or l\ITn+z;w)x+½~+I: ~-;x+½illl =J}un+ ~M)x+½i) - [- · !tiili x+½fj ll ~ 1. See Definition 7.3. 2. Condition(!) of Definition 1.6. J. The continuity of the norm follows from Condition (2) or Definition 1.6. 4. The existence of this limit also follows from results of Ascoli (XV), pp. 53-55. 5.3. Theorem 4.1 then gives rtx: r~: f;~i )X+½i] + k[x+½ill/?: }](n+ )x + ½Y II for lk l ~ l., If k is replaced by 2m and this inequality divided by 1ml , 1 it gives }l[]n+ )/(~ +~ )x+ ¾Y - 21~tt x+ ylf ~I/(~ + ::i&,>x+ ¾YI/ . Now take any number m p, and let n become infinite, keeping =~. This gives p ~~~ l)(p+ PL~fx11)Jx+ -!nY- 2r~1X+YII ~ ¼~ Jl(p + t~n~) )x + -fn YII , or ll(P- 2 x + yJI~ VPxll , ;rx:) for all p~ From Lemma 4.5, it follows that all limit points o~ llPx+yll - 1/pxll as p ➔ oo are greater than or equal to this is impossible if the assumption that Corollary 6.6. r and s are both limit points. lim Unx+yll - llnxll ltl ➔ OO If x and normed 1 inear • space, _then ..h~+o 2~~. But y Hence does not exist is false. are any two elements of a lim llx+hhl/ - Jlxlf .;_.._ and Alim Dx+hy)I ~-o h - llxll exist~ Theorem 6.6 gives the existence of Proof: Setting n = ~~ this becomes lim ljx-f.hfill - 1/ x ll . J.➔ + O ll lim /l nx-y#- ll nxll "'➔ "° y lim !lnx+yll - Unxll . 11\➔ 00 The existence of is also given by Theorem 6.6, since the element was arbitrary and can ,therefore be replaced by equal to lim Ux-h.hll -llxll , -"-++O or -y. This is -lim IJ!x+hi_ll - Uxll . i➔- 0 ' Theorem 6.6 and its Corollary are essentially the same, but the proof is e5=sier to follow if the limits are kept as in the theorem. The two forms will therefore be carried together. 1. Using Condition (3) of Definition 1.6. Conditions under which the llmits of Corollary 6.6 are equal 1 will be studied in Section 7! The fol1owing theorem gives an evaluation of the limits known to exist because of Theorem 6.6. The proof is very much like that of Theorem 6.6, and it could have been revised so as not to make use of that theorem and to give it as a coroliary. However, the proof is much more complicated in that case and it therefore seems advisable to include both. Theorem 6.7. If x normed linear space, with where and y are any two elements of ' a x ± O, -the.n Z:: lim Hnx+yJ/-J/nxtl = -Al}xJI, m ➔ o0 A is the algebraically smallest number such that x..LAx+y. ~~~ la:rgest number such that Proof: xJ_Bx+y. The existence of Theorem 6.6. B is the -------- lim Jlnx-yil-Unxll = Bllxff, where ~' lim /lnx+y# - llnxll M+OC> But, for any number is given by a, this limit is equal to lim ll(n-a}x+ax+yll-Dnxll, or because of Lemma 4.5, to .m ➔ ao lim Inx + (ax+y) JI - Unx/1-a IJxll •• A➔~ If llnx + (ax~)II ~ Nnxll, and hence x.1.ax+y, then lim llnx+yl/ -Unxll::::::.-a I/xii. /1\~oo of all numbers a Thus if A is the greatest lower bound such that. x ..Lax+y, then lim IJnx+y(I -Hnxll = -r Dxll. lim llnx+yJf-Mnxll~ -AjlxJI. Suppose be any positive number. Then there must exist arbitrarily large f'\➔ 00 values of n Let~ fh~CO such that \llnx+ytl-Jlnxl/+rllxl/1<- E.#x//. If n > )rl, this can be written Illnx+yl/- ll(n-r )xii f ~£I/xii. There is then a number )el<€. and e such t .h at l}nx+yll- ll(n-r)xll-el/xll = o, l. See Theorems 7.2, 7.3, and Corollary 7.4. or = ll (n-r+e)xl\ . llnx+yll But this can be written: })fin-½rt½e )x+½Y] + fiir-½,e )x+½iJ}f = /lnn-½r+½e )x+½i.] - Il½r-½e )x-+½ill\. Theorem 4.1 then gives: J)nn-½r+ ½e)x+ ½;j + kU½r-½e)x + ½ilJI > l~n-½r+½e)x + ½Y/J lk \ ~ l. for If k 1/H:-¼+ ';Silx+ l Now take any number m= .!!.. p /<~- ,&+ °;Si}x+ ~ y/1 • p, and let be c ome infinite, n This gives lf(p+r-e )x. + y Q~ Upxl( p. Since le I --< E. Jfpx + ( rx+ y )}/ -=>- //pxf/ has thus been sh.own that if A YII~ \~ }J(p -¥n +?n ):x:+ /n yJI , f or all was arbitrary, this gives x ..L rx+y. and the inequality Y] + (r-e }x+y// ~ l~oo IRP - ~ + ¥n)x + fn Y] + (r-e )x + or 2m lml , 1 this gives divided by keeping is replaced by lim #nx+yU- llnxJI hl~OO But it was ahown t hat It now foll ows that x ..LA:x:+y • If -B B r or all £ p. It = -r llxll , then •· lim lln.x+yff - lnx/1~ -A llxll , where m ➔ co is the greatest l owe r bound of all _numbers x ..L ax+y. and a such that lim IJnx+yll - lnxll = -A lf:x:lt , and lh,oo is the largest number such that is the smallest number such that lim llnx.;.yj/ - \lnxll = B 11:x:II , where B :x: ..L-Bx-y. 2 X...L.Bx+y, then Hence is the largest number such "'...,. oO that .:x: -L B:x:+y. Cor ollary 6.7. If normed linear space, and and x and y are any two elements of a x :/= O, then i¼1E'o ll:x:+hhl/ -llx lf =-B llxll , where smallest and largest of the numbers lim H:x:+hhll - llxll ..lt.~t-o A and a B = -A llxll , are the algebraically for which 1. Using Condition (3) of Definition 1.6. 2. Since spherical orthogonality is homogeneous. and Definition 2.3. x..Laxty. See page 13 Proof: This follows immediately by replacing n by ½ in i~ llnx+ytj- l/nxll = -A IJxll , and n . by - ½ in ,¾~ J/nx-yl/ - llnxJI = B I/xii , both of which are gi veh by Theorem 6. 7. For elements x and y of a normed linear space, Theorem 6. 7 gi ve.s the upper and lower bounds of all numbers for which :x .J..ax~. for all numbers The following results show that a a :x.J...ax+ y between these bounds, and hence give an eval- uation of all such numbers. This gives a necessary and sufficient condition for th~orthogonality of elements of a normed linear /,}r~e,,.J) space. Lemma 6.8. If linear space, and x and y are any two elements of a normed x..L Ax+y and x -L Bx;;r, then is a nUl,liber between Proof: A and x .L ax+ y then llx + k(Ax+y)/1 ~ llxU for all k. Likewise, if then llx + k(Bx+ y) j ~ llx ll k. Let that A~a~B. )Ix + k (ax+y) II I.f for all Jlx + k(ax+y)// . a If x ..J.. Ax+y, x .L Bx+y, be any number such k ~ O, then = IJ D- + k ( a-A[\ X + k (Ax+ y )\I ~JfJ. +-k ( a-An x )l ?!!. \lx ll. k :S: 0, then /Jx + k ( ax+y )(( = }}fJ. + k ( a-B)] x + k (Bx+y )I\ ~ I\D- + k (a-B)] Hence \\x + k(ax+y) }I ~ U xll a B. A< IS · For definiteness, assume -; ~~....t::.'B.,. If Consider if for all k, and therefore x\l~ \\x l\. ~ax+y" Theorem 6.8. If x y are any two elements of a and normed linear space, then x ..L ax+y if' _a nd only if , -lim llx+hyll - llxJJ ~ a Uxll~-Iim lx+hyll -hx ll J.➔+o Proof: :K➔ -o h It follows from Corollary 6.7 that the inequality of Theorem 6.8 is satisfied if x -LAx+y B = h and x ...L Bx+y where , a A is such that x ..1.. ax+y. = -lim Jlx+hyU - IJxll h uxll and .;{-++O - ~0}Jx+~~ii /lx ll . Thus by Lemma 6.8, x ..Lax+y Corollary 6.8. If x and y if Also A~a ~ B. are any two elements of a normed linear space, _t_h_e_n_x_.L_,Y.___1._·f_a_n_d__o_nl___..y_i_f and lim Dx+hylJ- llxll ~ 0.1 ,,K-} -0 li It follows from Theorem 6. 7 that lim 'hixll -lfnx-yll], or ~~=~ lim fjnx+yll - llnxlij~ ni➔ "° ~ lim ' llnx+yJl + llnx-yJf -2 JlnxJil~ -~ o. This is also M➔ ~ ~ 2 evident rrom the triangular inequality of the norm, which gives 2 llnx//:::; /lnxt-yll-t- llnx-yH. This limit · was investigated before. 3 It is zero in ordinary ~uclidean space, and because of Theorem 6-.7 (or 6.8) it being zero is necessary and sufficient for the uniqueness of the number a for ·which x .!.. ax+y and for the existence of the Gateaux differential of the norm. 4 For any elements x and y of a normed linear space, Theorem 6.8 enables one to determine all numbers x .1. ax+y. a such that How this can be done is illustrated by the following example: 1. Also see Corollary 7.2. 2. Condition (2) of Definition 1.6. 3. Page 27 of this thesis. 4. See Definition 7.3. • 58. Example 6 •.1. Let T be the normed linear space consisting , of all pairs____ of (a,b) _._______,;;_,;;;._.;r;;;..e.;...a..;;;;l;;;..._n.;...um_b_e.;;;.r..;..s, with JI (a, b )/l as the larger 1 of Let x = ( 1, 1 ) and y = ( 0, 1 ) • Then la ! and l bl • Ux-thy II = I}( 1, 1 +h )I/ h '<( 0 and smali. ll x+h~ - J/xl/ = 1 and 1 im Nx+hyll - }lxll Thus n ~~+o Then by Theorem 6. 8, x..L ax+y is, (1,1) -1... (a,a+l) = l if and only if h > O and zero if 1 im IJx+hyfl - llxll = O. ~~-c 1i -1~ a ~ o. That and if and only if -l ~ a ~ o. This conclusion 2 By definition, x -L ax+y if and can also be verified directly. only i .f if llx + k(ax+y) /1::,,.. llxl/, or }Jl}.+ka,l+k(a+lfl)} ~ IJ(1,1)U ,- for all k. Clearly this is true for a a+l are not. of the same sign; that is, if and only if and jk l 1. small if and only if The two limits of Theorem 6.7, and of Corollary 6.7, can be added. h ➔ O, This has the advantage of making it possible to let since tne right and left limits are then equal. rt will als o be valuable in interpreting the following results. Theorem 6.9. If x and normed linear space, . then y are any two elements of a !~1! /lnx-t-y/t - Unx-yll = -(A+B) llx ll , or 1~ Hx+hYII hIJx-hyU = -(A+B) llx ll , where A and smallest and largest of the numbers Proof: Theorem 6. 7 gave lim llnx-yll - Unx/1 = B I/xii . ~ ~ oO a B for which are the x _t_ ax+ y. lim /Jnxt-yM-fnxU = -Allx/I , and /1\ +ao Subtracting these gives lim »nx+y/1 - llnx-yll = - (A+B) R:x:11 . /)\ ~ CO 1 •. The space of Example 4.1. 2. Definition 2.3. If n is replaced by 1 n' this 59. bee omes l im flx+hyjJ - nx-hyjl which is obviously equal to h ' im IJx+hyfj - 1/x-hyll • ~::++o J-+-o h • The limits of Theorem 6.9 are twice the differential of the norm.1 if it exists. For ordinary Euclidean space, these limits being zero is clearly ne c essary and sufficient for the orthogonality of x and y. The following theorem shows that even for normed linear spaces there is always a unique number for which x ...L ax+y lim Jfnx + (ax+y) ll - llnx- (ax+y) fl = o. and Theorem 6.10. ~ ~~ If x normed linear space, and largest of the numbers and • y A and a are any two elements of a B are the smallest and for which x .l. ax+y, then ,~ .. ¾~ llnx + (ax+y)D- llnx- (ax+y )II = 0 Proof: But since largest of the numbers a' x ...1.. ½(A+B)x+y if and only if a= ½(A+B). It follows immediately from Lemma 6.8 that x ..L ½(A+B)x+ y. ½(A-B) and ½(B-A) for which a A and B are the smallest and for which x ...1.. ax+y, it follows that are the smallest and largest of the numbers x .J...a•x + ½(A+B)x + y • Theorem 6.9 that Hence it follows from lim //nx+ [½(A+B)x + iJ II - J/ nx-[½(A+B)x + yJ fl = 11\➔00 -[½(A-B) + ·½(B-A)] = o. From this and Lemma 4o5, it is clear that lim //nx + (ax+y)Jl - //nx - (ax+y )Jj = 0 ll'l4-"° if and only it a = ½(A+B). It is now interesting to look at Example 6.1 again. the x and is -1, and a y of this example, the number B = O. Thus 1. See Definition 7.3. ½(A+B) For A of Theorem 6.10 Hence Theorem 6.10 shows 60. that we must have o, or lim Jlnx + (-½x+y)Jl - llnx - (-½x+y) II m 4-°"' = lim jj(n-½ ,n+½)JI - U(n+½,n-½)II lim O = o. m+oo ~ ~= 1 While isosceles and spherical orthogonality are not equivalent in general normed linear spaces, it is possible to establish a relation between them. ½(A+ B) Thus the unique number of Theorem 6.10 can be shown to be nx j_anx + y in the isosceles sense. lim an, where 111 +<::IO The following Lemma used in establishing this is a strengthened form of Lemma 4.5. Lemma 6.11. For any normed linear space numbers E elements x la l <:: E and Proof: N T and any positive and £ there e,xis ts a number such that for all and y of 'T fJJ( n+a)x + yJl - llnx+yU-a Hxllf <£; it n > N [ ljyjl + :iJ . ..!!.. n+a + ~n+a = 1 identically, ll (n+a)x + ylJ can ll nx + (n!a)Yll+ a}lx + (n}a)YII , if n is positive Since be written as and large enough that n+a > o. Thus II (n+a) x + yff - llnx+y/1 =JJ!nx + (n~a) y/J - Jlnx+ yl/] + a Ux+ (n!a_) YI/ . But from' the triangular inequality of the norm., J llnx + (n!a) ylf - 1/nx+yflf ~ ln~J IIYII , ja llx + ( n!a) Yll -a llxll J~ In~\ lfyff • and If N is large enough that ln!a\ IJy~I< IN l/~ :~+a\ <:: and g_ , choose \al< E. Thus Then N N+a > o, then for . n > N ~ yll +JJ , lN\ for all y. For positive numbers E large enough that l n!a\ II YII"" ¼ if and N+ a > 0 iJ and \a l< E. n > N UIYU+ /ll{n+a)x + y)l - llnx+ yd -a llx ll} ~ £ 1. See Definitions 2.1 and 2.3. J;J< t if Ja l< E and if n > N [!IYN + 1] • 61. Theorem 6.11. If x and y are any two elements of a • normed linear space, and the numbers an Unx+ (anx+y) fl = llnx - (anx+ y) IJ , then lim an M-+00 such that x _1_ ax+y. Proof: n = a exists and is #~ llnx+yU-Nnx-yJ/ =-2a nx11 . 1 Also, The existence of lim Unx+yl( - llnx-yJI ,., ..,.00 is known. 2 But =ijnx+y/l - ll nx + (8l.nx+y)ILJ + IJ!nx - (anx+y )JI - llnx-yli]. llnx+y ll - llnx-yJI If are such that II YII ~ llnxll , then Theorem 4. 6 gives is large enough that J:n/ n~1,, or lanl~ ~ - This, with a double application of !:$ Lemma 6.11, gives the existence, f or any positive number ~, of a number such that N n> N implies J11 ( n-tan) x + ylt - )lnx+y)I -an UxUJ--=-:: ){ £ • and JlH n-an) x - YII - JJnx-y/l+an llx ll J-< K.i. I Hence • ll nx+y/l - llnx-yl/ + 2an llxll lim )Jnx+yJI - Unx-yJI m +~ exists, and that )< £ if = -2a llxll . Fr.om Theorem 6. 9, a= ½(A+ B), where A b such that x and ax+y Corollary 6.11. If x normed linear space, then and and B are the smallest :x: _1_ bx+y. The now follows from Lemma 6.8. y are any two elements of a lim llx+hyJ\ -llx-hyll exists and is h lim -2an 1Jxll , where Jlnx + (9.nx+y)JI = llnx- (anx+y)fl. - equal to lim an = a ~ -)-«I lim \lnx+yl/ - ffnx-yfl and largest of the numbers orthogonality of Since exists, it how follows that M ~OO it follows that n .:::. N. - .:l+o m ➔ co It is interesting to note that while the number such that x J_ax+y a is not unique in general, the particular 1. For any value ·of n, the existence of such a number given by Theorem 4.5. 2. See Theorem 6.9. an a is 62. of Theorem 6.11 is unique. lim Ux+hyf - jl x-hy lf li Theorem 6.11, and ~~o x ...L ax+y In any normed linear space, = -2a )Jxll ' where a is the unique number of x ..L ax+y. If the number -a llx ll . for which lim Hx+hhll - H xH exists is assumed to be unique , then and is equal to a ,R~ 0 1 Furthermore, Theorem 6.8, or Theorem 6,7 and Lemma 6.8, gives a simple evaluation of all numbers The largest such number is the smallest is I /I XII ' xB or for which x ..Lax+y. -lim llx+hy/1 - /Jx ll • ~~-o h //x ll ' -lim llx+h~II - Ux /( . The lim llnx-f// - 1/nxll m-+ 00 -lim llnx+ y /l - /lnxl/ m+ oo a ' ,).~+o or hi XII numbers between these two bounds are the totality of other numbers a for which . x ..L ax+y. the unique number a for which The mean of these limits is lim l/nx+ylJ -Unx-y/J /tl+ oo = o, or lim llx+h.yl/ -llx-hyJJ o, which are obvious conditions for li orthogonality if x and y are elements of ordinary Euclidean Jt-=,.o space.· 1. See Theorems 7.2 and 7.3. 63. 7. SPHERIC.AL ORTHOGONALITY, DIFFERENTIABILITY OF THE NORM, MTD STRICTLY NORMED SPACES. In the previous section, spherical orthogonality was investigated for general normed linear spaces. This section will consider the effect on normed linear spaces of assuming certain types of uniqueness of orthogonal elements. X For elements I and y, these types of uniqueness will be defined in terms of the number 1 a for which x..1- ax+y. Since spherical orthog2 onality is not syrametric, x ..1..ax+y does not imply ax-+y ..L X, and this uniqueness can take the following forms: Definition 7.1. elements x (i 0) such that x ..L ax+y. Orthogonality is right-unique if for any and Definition 7.2. elements x (i 0) such that ax+y ...1... x. and y :there exists at most one number a 0-rthogonality is left-unique if for any y there exists at most one number a For a type of orthogonality which is homogene~us~ 3 these two concepts of uniqueness are equivalent, respectively, to: 1. The existence of this number a is given by Theorem 6.3. 2. This follows from the independence of the types of uniqueness given by Definitions 7.1 and 7.2, as shown by Theorems 7.3 and 7.8, taken with Examples 7.2 and 7.3. A simple example could also be given to show this. • J. See page 13 of this thesis. Spherical orthogonality is homogeneous by the nature of its definition (x-1..y if and only if Ux+kyll~ llxll for all k) and property (3) of Definition 1.6. "If 1 l con t a1n1ng • • x ..1. F O · , then any pane an elemen t at most one element y such that x ...Ly", and: l cont a1n1ng • • then any pane an e 1 emen t 1 element y such that x "If x contains x ¥= O, contains at most one y ..Lx". If the orthogonality is symmetric, then right and leftuniqueness are equivalent. This was the case in normed linear spaces for the other two types of orthogonality which have been investigated, as is clear from their definitions. 2 Right and left-uniqueness of spherical orthogonality are not equivalent, and will be related, respectively, to the concepts of Gateaux differentiability of the norm and strictly normed spaces as given in the following definitions. For isosceles and Pythagorean orthogonality, additivity implies uniqueness (left and right). 3 For normed linear spaces, it will be shown that spherical orthogonality is right-unique if and only if it is additive, and additive if and only if t ,h e Gateaux differential of the norm exists at all non-zero points. 4 ilso, spherical orthogonality is left-unique if and only if the normed linear space is strictly normed. 5 This gives a relation between these and other well known concepts applicable to normed linear spaces, and a means of investigating them by use of spheric~l orthogonality. 1. "Plane" here means a two-dimensional linear subset. Definitions 2.1 and 2.2. J. Lemma 4.8 and Corollary 5.J. The uniqueness in abstract Euclidean spaces follows from Theorem J.2. 4. Theorems 7.J and 7.4. 5. Theorem 7.8. 2. Definition 7 •.3. l .inear space T 1 defined on a normed is Gateaux diff e~entiab-le 2 at a point x if li~ f(x+hyl-f(x) A f'unctional f exists for all elements y limit exists it is the Gateaux differential at increment of T. x If this with y, and is written f(x;y). Definition 7.4. A normed ' linear space is strictly normed 3 if from the equality I/xii+ IIYI! = ll x+yll there- is a number for which t x (y ;t; 0) it follows that = ty. An abstract Euclidean space is strictly normed. For if 2 2 2 Ux ll + IIYII = // x+yll , then /lx+y l/ • - I/xii - lly/1 = 2 llxll • UYH• If Y :j: 0, then a positive number t can be chosen such that Then (2) of r.lI1heorem 1.2 gives 11:xff = lltyl/ . IJxll • l/ylf = (x,y), and 11:x:JJ • JftylJ = (x, ty). 4- Applying ( 2) of Theorem 1 .• 2 now gives 2 11:x:-tyll 2 = ll xU2 + lftylf - 2 1/xll • IJty/1 . Thus lJx-tyl/ = 0 and x = ty c> Therefore any abstract Euclidean space is strictly normed. That the norm of such a space is Gateaux differentiable will be shown by Corollary 7.3,. Right-uniqueness of spherical orthogonality can be readily applied to the results of the previous section. The following theorem is not as important as some later ones, but is interesting in that it involves limits which exist and have been evaluated without assuming right-uniqueness of spherical orthogonality. 1. See page 7. 2~ Sometimes called weakl¥ differentiable, as by Mazur (VI). 3. Called strictly normalized by Smulian (Il), and strictly convex by Clarkson (XIT). _ 4. See Condition (1) of Definition 1.9. 66. Theorem 7.1. If a normed linear space is such that spherical orthogonality is right-unique, then, for any elements lim lnx-wll - lfnx-y /1 = -2a m~00 x and y, ,,xn ' and --1.7limo llx +hy/1 h //x-hy// = -2a /JxJ/ , - where a x .J..ax+y. is the number such that Proof: This is a restatement of Theorem 6.9, making use of the assumption of right-uniqueness of the orthogonality. Corollary 7 •.1. If a normed linear space is such that spherical orthogonality is right-unique, and the isosceles sense, then and only if Proof: lim an= a. x -L ax+y nx ..L anx + Y in in the spherical sense if 1 m~00 This follows easily from Theorem 6.11. Corollary 7.1 gives an interesting relation between 1 spherical and isosceles orthogonality. The number a of this Corollary is the mean of the largest and smallest of the numbers b for which x -L. bx+y in the spherical sense when such orthog2 onality is not right-unique, but is the only such number if the orthogonality is ri ght-unique. limit of the numbers sense. 3· Theorem 7.2. a n In any case, it is the unique for which nx .La x + y n in the isosceles If a normed linear snace is such that spherical orthogonality is ri ght-unique, then, for any elements 1. See Definitions 2.1 and 2.3. 2. This follows from Theorems 6.9 and 6.11. J. That is, Unx + .( anx+y) IJ = Unx - (anx+y) II existence of such a number an for ea ch for all no The n is given by Theorem 4.5 x (:p O) lim llnx+yll - llnxll = -a Jlxll M~c:0 y, and l ~ l/x+hh/1 - llxll -a llxtl , where a and is the number such that x ..Lax+y. This is merely a restatement of Theorem 6.7 and Proof: Corollary 6.7, making use of right-uniqueness. using That is, A= B = a. Corollary 7.2. If a normed linear space is such that spherical orthogonality is right-unique, then necessary and x (# 0) sufficient conditions that two elements and y be orthogonal are: lim Vnx+y l/ - Unxll = o, hl~OO o. lim Ux+hr21 - llx JI and ~~"' If spherical orthogonality is right-unique, it is then possible to define orthogonality in terms of the conditions of Corollary 7.2. Spherical ortho gonality is homogeneous, 1 and symmetry and additivity can then be restated as follows: Symmetry or 1 im ~70 (x.l y implies Ix+hh/1 - llxll = O J then y .L.x) : "If lim Jlnx+y /1 - ll nxll == O, /h~oO lim ffny+x ll- /lny// = 0 l)t~CIO and Jy+hxll-/Jy/1 - 0 h - Addi ti vi ti • (x.L y and X ..L Z imply X.L y+z): "If - 1/xU = o and l¾1g llx+hhJl-llxll = O, then l~ llxl)x+hhll + h(;y:+z )11-llxlf lim ,,f.+o h =o. 1. By virtue of its definition (Definition 2.3) and Condition (3) of Definition 1.6. 68. This statement of additivity is equivalent to the additivity of the Gateaux differentia1 1 of the norm, which follows from the existence of this differential. 2 This and the preceding theorems can be neatly restated by use of this concept. Theorem 7 • .3. A normed linear space is such that spherical orthogonal~ty is right-unique 3 if and only if the norm is Gateaux 1 differentiable at each non-zero poi_nt. If the norm is Gateaux differentiable, then the differential, f(x;y), is equal to where a is the number such that Proof: -a UxJl , x .J..ax+y:. If spherical orthogonality is right-unique, then Theorem 7.2 gives the existence and evaluation of the Gateaux differential of the norm. Conversely, if the Gateaux differential of the norm exists at each non-zero point, then the numbers A and B of Corollary 6.7 must be equal. x (t 0) and y For any elements there can then be only one number a such that x .Lax+y ,. and hence s,pherical orthogonality is right-unique. Corollary 7.3. The norm of a normed linear space is Gateaux differentiable at each non-zero point if and only if for each non-zero element such that Proof: x there is a unique maximal linear subset x ...1.. H. For any non-zero element existence of a maximal linear subset if y H x, Theorem 6.2 gives the H such that is any element not a multiple of x ..J...H. But x, then any maximal 1. See Definition 7.3. 2. This is evident from Theorems 7.3 and 7.4, or see Maxur (VI), pages 129-130. • 3. See Definition 7.1. linear subset not containing number a. 1 x must contain ax+y for some Thus if spherical orthogonality is right-unique, then the hyperplane H is unique. 2 The Corollary now foll ows from Theorem 7.3. The norm of an abstract Euclidean space 3 is Corollary 7.3'. Gateaux differentiable at each non-zer o point, and the Gateaux . I differential x f(x;y) is orthogonal to An element of the norm is equal to ax+y if and only if a= Proof: The ratio ll x+h~l - flx ll can be written as 2 2 t +hyJJ - »x11 Using Definition 1.9, this becomes (ffx+hy~+ llxJI ) • ~x+hy,x+hy)-(x,x) 2(x,y) th(~ y) otx+hy/1 + llxll J 1/X-+hyJI + I I~ Letting h approach zero, it is seen that the Gateaux differential = of the norm at to .' x with increment y {J-.¾~ llx+hhl/ - Jlx lf } is equal ¥xlr1• The rest of the Corollary follows from Theorem 7. 3. Corollary 7 .3n. If the Gateaux differential of the norm exists at each non-zero point of a normed linear space an element x of T is orthogonal to an element ohly if the Gateaux differential Proof: f(x;y) T, then y if and is zero. This ·is a restatement of Corollary 7.2, and is also immediate from Theorem 7.3. 1. See Definition 6.2. 2. See Definition 7.1. J. See Definition lo9e 70. Theorem 7.4. In a normed linear space, spherical orthog- onality is additive if and only if it is right-unique. L€{ Xro Proof: Suppose spherical orthogonality is additive and that there exist numbers a and b such that x .1.. ax+y and x ...L bx+y. Then from the homogeneity or spherical orthogonality, x ..1- -bx..;.y • .. . Hence -additivity gives all This is clearly not t rue if k. ll x + k(a-b)xll~ ltxll x -L (a-b)x, or 1 k =-~• a-o Thus for a= b and spherical orthogonal ity is right-unique if it is additive. Convers ely, if spherical orthogonality i s right-unique, I The orem 7.3 gives the existence of the Gateaux differential of the norm. If x ..L y and x ..1.. z and the Gateaux differential of the norm at x is denoted by f (x;y), then f (x;y) + f (x; z) = o. 1 2 But Mazur has shown that the Gateaux differential of the norm is linear if it exists. Hence f(x;y+z) =o, and by Corollary 7.3" Thus right-uniqueness implies additiv~ty. x -L y+z. That right-uniqueness of spherical orthogonality implies its additivity can also be proved nicely directly, and the proof seems interesting enough to be included. Let x be any non-zero element. Then there exists a linear functi onal f such that f (x) = //xU and llt ll 1. 3 Let y be any element such that x = Jr (x+ky )f ~ l}x+ky/1 . Yo is taken so that kf(y) Dividing by Jr (y)J ~ IJtx+yJI-JI! xJI • lkl , Since #r/1 = 1, is orthogonal to is positive, then If k )f(ky) J~ Jlx+kyfl - llxll . Letting k spproach zero and assuming uniqueness, it follows from Corollary 7.2 and the assumption x ....Ly f(y) = O, and likewise that x ...1.. z f (y) = o. implies Thus x ...L y f(z) = o. implies Since f is 1. See Corollary 7.J". 2. Mazur (V:t), pages 129-130. }. Banach (I), pg~ 55, Theorem 3. Also see page 8 of this thesis. 71. linear, it follows that f [ax + (y+z)J f (y+z) = o. If • x..L ax + (y+z), then =o. But f (y+z) ...: o. Hence a = 0 and x ..L {y+z). Thus additivity has been shown to follow from right-uniqueness. Corollary 7.4. Spherical orthogonality is additive in a normed linear space if and only if the Gateaux differential of the norm exists at each non-zero point. Proof: This follows immediately by using Theorems 7.3 • and 7.4 together. By making use of the Gateaux differential f(x;y), the statements of symmetry and additivity of spherical orthogonality when right-unique, as given prior to Theorem 7.3, can now be stated simply: Symmetry: • Addi ti vi t y: "If "If f(x;y) = O, then f' ( x; y) It has been shown 2 f(y;x) = o. 1 = 0 and f (x; z) = 0, then f (x; y+z) 0. that additivity of spherical orthogonality follows from the assumption of right-uniqueness or of Gateaux differentiability of the norm, but symmetry does not. This added assumption of symmetry will be studied in Section 8. Any normed linear space of more than two dimensions will be shown to be an abstract Euclidean space if spherical ortho gonality is symmetric and unique. It is interesting t .o verify the evaluation of the Gateaux differential for a two-dimensional Euclidean space. Suppose x 1. See Corollary 7. 5'. Also, Theorem 8 . .f shows that a strengthening of this condition can imply that the space is abstract Euclidean. 2. Theorems 7.3 and 7.4. 72. and y are situated as in the figure, with x ...1. ax+y. In this position, is negative, and a for simplicity h will be taken as positive. Then !lx+hyl/ = 1/x/l sec e+ llahx/lsec e = Jlxl/ sec e -ah //.xll sece. Hence IJx+hhll - lJxJ/ = //xi (sece- 1) _ a //xl/ sec 0 • But h and 0 are clearly That is, lim sec B - l ,J..:,. o h = O • of the same order of magnitude. lim IJx+hyJ/ - llxll = - a lfxll . Therefore ~+ o h - It has been shown that right-uni.queness of spherical orthogonality is equivalent to Gateaux differentiability of the 2 norm. 1 It also follows that such orthogonality is left-unique if and only if the normed linear space is strictly normed. 3 It will first be shown that the concepts of left-uniqueness and strictly normed spaces can be interpreted in terms of functions of the form f(n) = Ux+nytl .. These concepts will then be investi- gated_ for spaces of continuous functions, and their independence shown by means of such examples. Theorem 7.5. If x and y are a:ril two elements of a normed linear s J?!:l!Ce, then the curve f (n) = Jlx+nylJ is concave up except for possible straight line sections. Proof: To prove the theorem, it is sufficient to show that for any straight line intersecting the curve in two points, the segment of the curve cut off by these t wo points lies below or 1. Theorem 7.3. 2. See Definition 7.2. 3. Se-e Definition 7 .4 and Theorem 7. 8. 73. on the line. If the points of intersection are for values i ~ c./ n, then the ordinate of the ..,.,,..,.., / of and n1 I I point having abscissa an + bn 1 2 and lying on the ~...~t-il-P,11 I straight line joining these points is af ( n ) + bf (n ) • 1 2 To establish the the orem, it is then sufficient t o show that f or all numbers a and b such that f(an + bn 1 ) ~ a+b =1, it follows that at(n ) + bf{n ) , 2 1 2 IJx + (an1 + bn2)Yll ~ a Jlx + n1Yll+b lfx + n2yn. or IJ x+ (an1+bn 2 )yll • /} a(x+n1 y) + b(x+n 2 y) ff , this follows from Since the triangular inequality of the norm. Theorem 7.6. 1 If a normed linear space is such that the Gateaµx differential of the norm exists at all non-zero points, then d llx+nyl is a continuous monotonic increasing function of -dn _n__ e_qu_a_l_t_o__-_an..ll:iJL , where Proof: d l~~nyll It follows immediately from Theorem 7.5 that is a monotonic increasing function of To show its continuity, let f(n) = llx+nyll . differential x+ny f(x+ny;y) at n if it exists. Then the Gateaux with increment y is lim /I ( x+ny) -+- hy /1 - lx+nyfl or f' (n). Thus by Theorem 7. 3, -1:-+o h ' f' ( n) = -an »YII , where (x+ny) _L an ( x+ny) + y. Hence f' (n) is 1. Cmndition (2) of Definition 1.6. 2. Also sere Corollary 7.5". The existence of such a number for each value of n is given by Theorem 6.J. an 74. continuous if and only if Suppose an an is a continuous function of is not continuous for n = n1 • n. By Corollary 6.1, -< DYil la\ n - Hx+riylf' and hence an is bounded and must have at least one limit point a~ an1 as n approaches n1 . Since the Gateaux differential of the norm exists, spherical orthogonalit y 1 is right-unique and it is impossible for x+n y to be orthogonal 1 2 to a(x+n y) + y. Therefore there exists a number k such that 1 IJ(x + n1 y) + k[a(x + n 1 y) + yJ J}<IJx + n1 y/l . But since a is a limit point of an as n it would then be poss ible to select a number ln 2 -n1 ) and approaches n 2 Jan - .a f are small enough that 3 2 /J(x + n 2 y) + k [an2 (x + n 2 y) + y] JI~ Ux+ n 2 yJJ , which is impossible if (x+n 2y) ..L an (x + n 2y) + y. 2 is a continuous function of n. f'(n) n , 1 such that Therefore Theorems 7.5 and 7.6 raise the question of whether it is possible for a curve section. 4 f(n) = Ux+nyll to have a straight line The following example shows that this is possible even when the norm is Gateaux differentiable. d 1/x~yl! dn In other words, can be constant over an interval. Example 7.1. pairs (a,b), with Let T be the space consisting of all number (a,b) + (c,d) = (a+ c,b+d), k(a,b) = {ka,kb), and lal tt li a + a1 if ff (a' b)ff = /}_b)f ::S I lbI , lal ~ lbl . Clearly Conditions (1) and (3) of Definition 1.6 are satisfied. 1. Theorem 7.3. 2 Since x + n1y -L an1 (x + n1y) + y. 0 }. It is possible to do this beaause of the continuity of the norm, which follows from Condition (2) of Definition 1.6. 4. That this is possible was implied by Theorem 4.4, and shown by Example 4.3. However, the norm of. Ex. 4.3 is not Gateaux a_ifferentiable. 75. Thus T i:s a normed linear space if Condition (2) is satisfied. That this "triangular inequality" is satisfied can be shown by considering the following curves: (\b I lbl ~ l, if II (l,b) II = "l3(l+b~) or lbl~ 1, if )/(0.,1)1\ I -I - f---+ 1 - 1 - I- +- I I I 1 -, ===--t--+--~~ 1- -t1 I I ..--t----+--,...--t-~t---+---t---t- (l... Clearly the curve 11( 1, b )II is never concave downward, and hence r II (1 , b ~I~ s II (1 , c )JI ~ lJ ( 1 , r ~~~ c )JI for all r and greater than 1 s. The equality holds if or less than r' and sc by If IJx+yJI ~ llxll+llylt normed linear space. rb is replaced s', it follows from the above 1.nequai'ity that JI ( r+s , r '+ s ' )II ~ II( r , r ' )/I + I/ (s ,Js ' )II . Thus are both c -1; i.e. if only values of 1/(1,b)II are considered which lie on a straight line. by and b for all x and y . of • T, •and T is a The Gateaux differential of the norm of exists at each non-zero point. This follows from the following relations, which are derived from the definition of H(a,b)ff: .: if lal ~ lbl and if la\ ~ lbJ and if Jal ~ lb I. if la J ~ lb I , ½(1 - ab: ) if Ja J~ \b I i b::t. -z(l - a!-) if Jal'=> lb J 0 d 11( a, b )II da and a~ o, and a~o. T 76. By using these relations (or more simply by direct calculation): + B d IJ(abb) II lim IJ (a,b) + h(A,B) JI - ll( a,b) II _ Ad ll(a,b)II h da d ,t+o =; if ja 1~ lbl , 2 bB aA 11 b A = faf -,.t-a jal + I{.laf This is the Gateaux differential of the norm at increment if Jal :?:Jbl • (a,b) with (A,B). /l (a,1) 11 = 1 Since for ' }al~ 1 if x if la J~ l, it follows that ax+y .Lx = (1,0) and y = (0,1). Thus spherical orthogonality is not left-unique for the space of Example 7.1, even though it is right-unique by virtue of the existence of the Gateaux differential of the norm. 1 Another illustration of this will be given later, 2 as well as an example 3 to show the converse. Before doing this it is to advantage to further investi gat e left-uniqueness of spherical orthogonality. The following theorem is different from Lemma 608 because of the lack of symm.etry of spherical orthogonality. that Ux+ky ll The most interesting difference is has the sarn.e value for all That this is not true for all k for whi ch x ...1.. kx+y x+ky _L y. k for whi ch can be y are elements of a normed linear seen from Example 6.1. Theorem 7.7. space T, then If x Ux+kyll and has the s ame value for all numbers betwe :en or equal to the smallest and larges t numbers x+ay_i_ y. Als o, Proof: Hence l}x-tay lf If x+ky...L y for all numbers x+ay_t_ y, then llx+kyll for which in this interval. )f( x+ay) + ky ll~IIYII is the minimum value of l. _See Theorem 7.J. 2. Example 7.2. J. Example 7.J. k a k for all k. if and only if 77. x+ay ..L y. Therefore if IJx + a yJI = JJx + a 2 ylJ . 1 x+a 1 y ...1.. y x+a 2y ...J_y , t hen Ohviously the set of numbe rs a From the continuity1 is minimum is bounded. 1 Jlx+ayJI and f or whi ch it follows that there exist smallest and largest numbers B of this set. Then minimum values of t hat Ux+ky l\ all k 1/~ J1·t.ke.. of A and ll x+Ay ll = )lx+By JI , and these are both It now f ollows from Theorem 7.5 llx+ky\l . is constant for A~ k ~ B, and that x+ky ...L y fo r in this interval. Theorem 7. 7 es.sentially st a tes that spherical orthogonality ·\IYt J X, c,,,.,J ~ E t"...!,f'·t'-~ f (n) = ll x-i-nyll is left-unique if and only if vno curve of the form has a "flat bottom" like that of 1/ (a,l) JI of Example 7.1. But if any such curve has a strai ght line segment, it can be shown that there is another with a "flat bottom". This follows from Theorem ·7.8, and is also the basic idea of the proof of that Thus it is shown that 2 llxll +llytl = Jlx+yll theorem. curve /1( 1-k)x + ky/j below). of has a strai ght line se gment (as illustrated It then follows that elements such that the curve ~{a,1) 11 implies the /lx'+ ky•JI x' and Y' can be found has a "flat bottom" similar to that of Example 7.1, and ,thus that spherical ortho gonality is not left-uniqueo Theorem 7.80 A necessary and sufficient condition that a normed linear space be strictly normed 3 is that spherical orthogonality be left-unique. Proof: Suppose spherical orthogonality is not left-unique. There then exist elements x and y {y #, 0) and unequal 1. This follows from Condition {2) of Definition 1.6. 2. The following assumes the space is not strictly normed (see Def. 7.4). 3. See Definitions 7.2 and 7.4. 78. numbers A arid B such that x+Ay...L y follows from Theorem 7.7 that and x+By .L y. x + ½(A+B)y_L y rt then and llx+AyJI = llx+Byll = IJx + ½(A+B) yJI . )lx+Ay/1+ IJx+Byll = /l(x+Ay) + (x-tBy}IJ . Thus If the space is strictly normed, there then exists a number =t(x+By). 1 t If t = 1, then either y = 0 or A= B, contrary to assumption. If t i 1, then such that x+Ay x is either zer9 or a multiple of y, which is impossible since it would then be necessary that =x+By =o, and hence that A= B. Hence x+Ay 2 a normed linear space in which spherical orthogonality is not left-unique cannot be strictly normed, and spherical orthogonality is left-unique in any normed linear space which is strictly normed. Conversely, suppose the normed linear space strictly normed~ There then exist elements such that x =/: ty number such that k for any t O~ k ~ l. and 11:xll x T is not y (y-:;: 0) and + IIYII - Ux+yU. Take any Then 11(1-k)x + ky /l~ (l-k) 11-xJI + k llyll because of the triangular inequality of the norm.3. But from x+y = D_l-k)x + ky] + kx + (1-k)y, it also follows that jµ:+yll~}l (l-k)x + kyll+ k llxJl+ (l-k)j/yJI . llx U+ UYII Since it was assumed that = l/x+ylf, this gives 1/(1-k)x+kyJI:::.. (1-k) UxJH-k llylJ . Thus II( 1-k) x+ky II = (1-k) llxll+k llyfl for O~ k ~ l. . Therefore if 0 1. See Definition 7.4. 2. Since ky ..J.. y if and only if k x+Ay .L y and x+By .L y. J. Condition (2) of Definition 1.6. '/,._ 1. = 0 (see Def. 2.3), and 79. _ k - llxll Hxll+ Jlyfl and x• = (1-k)x + ky and y• = (1-k)x - ky, then Jlx•II = 21:t1f6U, and Theorem 7 .5, 1 \k ) ~ 1. for »x~+y•II llx•+ky•II = Hx'-Y'II '.'"- Jlx:•11 . = llx'II JI~+¾'» _,,, / This with ...... ..._ ..;..I ' Theorem 4.1 gives fix'+ ky •JI ~ }Ix' for all Hence from Thus k. 1/x' I/ be the minimum of JI I must · I 1/(x'+ ay') + ky'II ~ lJx'+ay'II Thus X if not be zero because x i ty 1(-,..'JI -1 )Jx' +ky' IJ ' and '+ ay ' .J.. y t ---- -+ :1. = Jlx'H for all Since Y' for any t. = k, if IJyll x +UxllY llxll+llyll ' lal ~ 1. Y' can Therefore spherical orthogonality is not left-unique, and a normed linear space for which ·spherical orthogonality is left-unique is strictly normed. Corollary 7.5. If a normed linear space is such that spherical orthogonality is symmetric, then the space is strictly normed if and only if the· norm is Gateaux differentiable at each non-zero point. If spherical orthogonality is symmetric, then such orthogonality is clearly left-unique if and only if it is right-unique. 2 The above corollary then follows from the equivalence of leftuniqueness to the space being strictly normed and of rightuniqueness to Gateaux differentiability of the norm. 3 But it has also been shown that if the Gateaux differential of the norm, f(x;y), at to y x with increment if and only if f(x;y) y = exists, then x is orthogonal o. 4 Thus in this case the 1. Or from Theorem 4.4, which is now a corollary of Theorem 7.5. 2. See Definitions 7.1 and 7.2. 3. Theorems 7.3 and 7.8. 4. Corollary 7.3". 80. orthogonality f(y;x) = o. is symmetric if and only if O i mplies This gives t he foll owi ng corollary. Corollary 7.5'. If the norm of a normed linear space Gateaux dif fer entiable, then i mpl ies f (x ;y) f(y;x) =0 Corollary 7.5". T is strictly normed if for all non-zero elements x a nd T Proof: y, T is is strictly normed if and only if the Gatea~x differential of the norm, f(x+ny;y), at increment is f(x;y) = O 1 y of T. If the norm of a normed linear space Gateaux differentiable, then T is an increasing function of n. x+ny with 2 As in the proof of Theorem 7.6, it is clear that f(x+ny;y) = d ll~~yJ/ . It also follows from Theorems 7.7 and 7.8, or directly from the condition for a normed linear s pace to be strictly normed, 3 that such a space is strictly normed if and only if no curve any f (n) = 1/x+ nyJI (y # 0), for which x :f: ty for t, contains a straight line segment. · The conclusion of the corollary then follows from Theorem 7.6. It is interesting to investi gate the effect on spaces of continuous functions of the assumption of right- or left-uni queness of spherical orthogonality. This will be seen to give an easy means of establishing the independence of these types of uniqueness of orthogonality, and hence also the independence of Gateaux differentiability and strict normedness. 1. The ~stronger relation f(x;$) = f(y;~) holds in abstract Euclidean spaces, but by Theorem _8 . K ~ in such spaces. 2. As follows from- Theorem 7.6, this difference is continuous in if it exists. 3. As in the proof of Theorem 7.8. n 81. Definition 7.5. The space (C) of continuous functions consists of all functi ons which are continuous in the closed (9, 1), with /lf /l interval Theorem 7.9. defined as max. o~x -"'l A subspace 2 T Jr (x) J . 1 of the space of continuous functions is strictly normed if and only if all non-zero functions of T whose absolute values take on their maximum at a common point are multiples of each other. Proof: such that Suppose there exist non-zero functions ffl is the maximum of Jt +g j , or Then either )t(a)-g(a) I )t(a)+ g(a)J is the maximum of Ht ll+IJgH = llt-g ll . T is strictly normed, there then exists a non-zero 3 number lf-g l • f or If t = tg. Conversely, if T is not strictly normed, there exist non- and g such that )lf ll+ )lgll = llt +gll for any number t. Suppose }If )) It (a )I , zero functions f :f. tg llf ll+)g l/ = ll f+g/1 Hence either such that g )gJ are both maximum at .some number a. and I/ti/ = jf(a) / and Ugj/ = )g(a) / . That is, f , and // g /1 = )g(b)J, f and IJf + gll = Jr(c)+g(c)J. and But It ( c ) + g ( c ) I ~ Jr ( c )) + Jg ( c )I . Also, 1r(c)l!S )r(a)I and Jg(c)/~ Jg(b)}. \ f ( C ) -t-g ( C ) \ = }f (a ) ft follows from this that ~hat is, both \fl and }t(c) J By assumption, l + Jg ( b) ' • = }f(a)j and )g(c)j = lg(b)~. jg\ take on their absolute maximum at c. 1. See Ban~ch (I), page 11. 2. A subspace here means a subset which is also a normed linear space. J. Non-zero since f , o. 82. The above theorem enables one to tell readily whether a space of continuous functions is strictly normed~ In order to find a simple condition that the norm of such a space be Gateaux differentiable, the limits of Corollary 6.7 will first be evaluated. Conditions for differentiability of the norm can then be deduced from these limits. Theorem 7.10. If uous functions, and T is a subspace of the space of contin- f(;c 0) and g are elements of lim llf+kgll- llf l/ - max.(g.f) and Jim /Jf+kgl/-Jlfjf k --'-')-tO where - U Orr/ - ➔-0 k U is the set of all numbers for which Proof: for which number Because of the continuity of T, then = min{g)f{\ U /ti -; is maximum. 1 f, the set of numbers ltl a' is maximum is closed, and hence there exists a 2 of this set for which takes on its maximum. 1#- Because of the way a' was chosen, it is a member of the set U of num- a... I bers for which lt l Since is maximum. • gfx)f(x) f {x)J is maximum for x =at, it follows that lf(a)+kg(a)J for If (a' )+kg(a' )f a ~U, provided k is the maximum value of is positive and small 1. The only function of the factor½ is in determining the sign (,;fhJe. fi· gure, at • . d.ica t e d poin . t s; U 2.of Int can be ei· th er o f th e in consists of the points marked a' and the points between a] and a 2 ; and x.J.. is the point at which /jf+¼:g J/ takes · on its maximum. 4- 83. lkg{a) I~ It {a)/ enough that number for which as k a fr u. If xk is maximum, then since lt +kg/ bounded and the norm of for all T is a g(x} is is continuous, 1 all limit points ot u. approach zero belong to Then for small k and positive: Ut +kg ll = It (xk)+kg ( xk) I = Jr (xk ))+ kg (x~) f (xk) -<..Y {a')/+ kg (xk )f ( xk) Jf xk}J Jf (xk)I ) lim Uf +kgft - Uf ll ~ lim kg (x~ )f (xk) _ g (a )f (a) k __, --&++o kj(xk)f )f{a)J' for the number a of U which is a limit point of xk and ~ ~+o approaches zero. But as k ur + kg ll~ lt(a') + kg(a')) , 2 and hence lim llf+k~l/ - 1/f l} !::,.. lim)r(a')+kg(a'}l-)f(a'}j _ gfa')f~a') ~ g(a)f~a) . - - --h -+o . k f (a ' f jf (a I for all a £. U, because of the way a' was chosen. Hence .&➔-to 1 im l!f +kg II - llf ti _ ,• a ' ) f ( a ' ) _ max. g ~ a ) f ( a ) • k f a' u f (a) J From this, -i ~ +<> limVf+k~A- ll t U _ lim Uf-kgll - Jlf lf _ -max. -g(a)f ~a) _ min.g(a)f (a). ~ .+-o - .Jt+-t-o -k - u jf (a I - u jf {a )I It is interesting that Corollary 6.6 (and therefore also Theorem 6.6) could be given as a corollary of Theorem 7.10. This follows from the fact that any separable Banach space can be represented as a subspace of the space of continuous functions, 3 in effect only only two-dimensional spaces being involved in the proof of this type of theorem,i. The following corollary is Theorem 7.10 restated with norms replaced by their defined values. 4 Corollary 7.7 is gotten by applying Theorem 7.10 to Corollary 6.7. 1, g(x) is bounded since it is continuous, and the norm of T is continuous because of Condition (2) of .Definition 1.6. 2. See Definition 7.5. 3. Banach (I), page 185, Theorem 9. A separable space is one which contains a countable subset whose closure is the whole space. Clearly any finite dimensional normed linear space is separable. 4. See Definition 7.5. Corollary 7.6. If f and g are continuous functions defined over a closed interval, then lim max. Jf +kg /-max. !f f = max. ~ u k ~~+o and JfT ' lim max. jf +kgJ-max. )f J _ min. ~ ~~ -o where k - If I ' U fr l U is the set of numbers for which Corollary 7.7 . . If T tinuous functions, and f is a subsuace of the space of conand smallest and largest nu.mbers f ..L Bf+ g are A= -m~x. is maximum. g are elements of A and T, then the B such that . f --L. Af+ g and -f- and B = -min. f, where U is the jf j set of numbers for which is maximum. Corollary 7.7 gives a simple means of finding all numbers a for which f ..L af+g, if f and g are elements of a space of continuous functions. If the orthogonality is ri ght-unique, then 1 the norm is Gateaux differentiable at all non-zero points. It is possible to use Theorem 7.10 to find a condition for the existence of the Gateaux differential of the norm, and a means of evaluating this differential if it exists. This leads to simple conditions for orthogonality of elements of a space of continuous functions. Theorem 7.11. If T continuous functions, and is a subspace of the space of f (# 0) and g are elements of then the Gateaux differential, lim Ht+kgll - llf ll ~~ o if Jg(a)/ , has the same value for all nurnbers lo Theorem 7.30 T, exists if and only a for which Jfl g (a ) is maximum , and and f(a) are either of the s ame sign for all such a, or of opposite sign for all. such a. If it exists, this limit eq_uals g(a)f(a) /f(a) / , where a is any number for which Jr/ is maximum. Proof: From Theorem 7.10, this Gateaux differenti al exists if and only if mux{,;f)= m~n•e:rf) , where U is the set of numbers is maximum. has the same value for for which 1t l all Hence for this eg_uality to hold, a e;U. a ~u, the same si gn for all value f or al!. But (f(a) ) and /g(a) / If T T is Gateaux differentiable at all non-zero points , then for elements f + g , where Proof: must have the same is a subspace of the s pace of con- tinuous functions, and tAe norm of =~ must have a £- U. Corollary 7.8. f -L -~f g(a)•f(a) a f and g of T, is any number for which jf j is maximum. This follo"WS immediately from Corollary 7.7, since spherical orthogonality is right-unig_ue if the n orm is Gateaux differentiable 1 . It also follows from Theorems 7.3 and 7.11. Theorem 7.12. If T is a subspace of the space of contin- uous. functions and the norm of T is Gateaux differentiable at all non-zero points, then an element onal to an element for which jf / g if and only if f is spherically orthogg(a) = 0 for numbers a is maximum . This theorem is really a weak form of Corollary 7.8, but provides a very simple means of testing the orthogonality of 1. See Theorem 7.3. 86. elements of a space of continuous functions in which the norm is Gateaux differentiable. If the orthogoµality is also symmetric, then two elements are orthogonal if and only if each is zero at every point where the absolute value of the other is maximum. will be used extensively in the next section~ This Its application is made possible by the fact that any separable Banach sp,ace 1 is equivalent 2 to a subspace of the space of continuous functions. 3 The above results enable one to ~onstruct simple examples to show the independence of Gateaux differentiability of the norm and the condition for strictly normed spaces--and hence of right- and left-uniqueness of spherical orthogonality. Example 7.1 showed that Gateaux differentiability of the norm does not imply the space is strictly normed. This is shovm more simply by Example 7.2, while the converse is given by Example 7.3. Examule 7.2. A normed linear space whose norm is Gateaux differentiable, but which is not strictly normed. Let T be t ,he space of all functions of the form f = a sin x+ b(x-1t) in the interval (0, 1Y ). Since ft = a cos x+ b, the slope of the curve f is of the same sign throughout la/ ~ /bl. 0 (o, ,rr ) if Hence in this case 1. See Definition 1.7. A space T is separable if it contains a countable subset U such that U = T. 2. Two normed linear spaces are equivalent if there is a 1-1 correspondence between them which preserves the norm and the operations of addition and multiplication by real numbers. 3. Banach (I), page 185, Theorem 9. the maximum of lf l Theorem 7.9, T is not strictly normed. is attained at la}> )b I , then f' If O and 1, . Thus by is of opposite signs at the ends of /t i must take on a larger value than that or the interval and '1tl b/ which it has at O and 1r. la/:::.. lbJ, Ir ) Thus if takes on its maximum at the unique point for which la)> lbf , then · It I Hence if point. 1,. lal~ lb) , If jf f takes on its maximum at only one takes , on its maximum at both sin O = sin 1Y' = o, the values at Since other function Asin x +B(x- ~ ) values of are f the norm of T b cos x=- -• a are +~• O and O and 1T' of any Since the corresponding +~b, it now f ollows from Theorem 7.11 that is Gateaux differentiabl~· Example 7. 3. A normed linear spac e which is strictly normed, 1 but whose norm is not Gateaux differentiable. Let T be the space of all functions of the form f = a sin x + b sin 2x in the interval ( O, 1Y' ). Then sin 2x has a maximum of at ~ and ~ , 1 being of opposite signs at these points. But sin x has the same sign at both points. Hence by Theorem 7.11, im //sin 2x + k sin xJ/ - // sin 2xll ~o k does not exist, and the norm of T is not Gateaux differentiable 1- at the "point" sin 2x. 88. f = a sin x+b sin 2x The function is zero at both ends of the interval (0, 1r ), and hence the maximum of must be taken on for a number x0 Jasinx+bsin2xf such that a cos x 0 +2b cos 2x 0 = o. If the absolute value of another function takes on its maximum at the same point F = Asin x + B sin 2x x 0 , then A cos x 0 + 2B cos 2x 0 = O! Since cos XO follows that and cos 2x 0 Ab= Ba cannot be zero simult.aneously, it and that any functions f and F are multiples of each other if they take on their absolute maxima at the same point. It then follows from Theorem 7.9 that T is strictly normed. Theorem 7.lJ. A normed linear space can be strictly normed without its norm being Gateaux differentiable, and conversely. Theorem 7.14. There are normed linear spaces for which spherical orthogonality is right-unique and not left-unique,~ others for which such orthogonality is left-uni que and not ri ghtuniqueo These theorems are immediate consequences of the above Examples 7~2 and 7.3, the second making use of the equivalence of Gateaux differentiability of the norm and ri ght-uniq_ueness of spherical orthogonality, and of the equivalence of strict normedness and left-uniqueness of spherical orthogonality. 1 1. Theorems 7.J and 7.8. 89. 8. SPIIER IDAL ORTHOGONALTI1Y AND T.dE EXIBTENCE OF mmR PRODUCTS It has been shown that if either isosceles or Pythagorean orthogonality is either additive or homogeneous in a normed linear space T, then T is an abstract Euclidean space. 1 These types of orthogonality are symmetric in any normed linear space, 2 while spherical ortho~onality is -homogeneous, but neither symmew-ic nor additive in general normed linear spaces. Two conditions for the existence of an inner product 3 in normed linear spaces have been used in this thesis as an aid in proving the existence of inner products from assumptions on the orthogonality. Dl. connection with Pythagorean orthogonality, it was convenient to use the condition given by Jordan and Neumann4 that a normed linear space is abstract Euclidean if and only if 2 ltx+y Jj2+ U:x:"'."ylf 2 = 2[ 1J:x:1l + fly1f 2] for all elements x and y. With isosceles orthogonality, a condition given by Ficken was found useful: 5 That a normed linear space is abstract Euclidean if and only if II a:x:-ttr II = fl x+ayJI for all numbers _~ and elements :x: and y for which Jl :x:11 =JI YII . It will be seen that it is sufficient to assume only that 6 • fh.➔ lim \flnx+y lf - ll •+nY'ft] = o if llxll = IIYll. l!!. OD I. Theorems 4.9 and 5.2 and Corollary 5.3. Also see page 13 of this thesis. 2~ See Definitions 2.1-2. 3. 3. The existence of an inner product is equivalent to the space -being abstract Euclidean-. · See Definition 1.9. 4~ See page 40 of this thesis~ and Jordan and Neumann (IV). 5; See page 35 ·of this thesis, and Ficken {XIV)• 6. Theorem 8 _.4. 90. Garrett Birkhoff has shown that a normed linear space of three. or more dimensions is abstract Euclidean if and only if spherical orthogonalitf is symmetric and uriique, and that this is .!!21 true for spa?es of two dim.ensions. 1 It has been shown that right-uniqueness of spherical orthogonality is equivalent to additivity of the orthogonality and to Gateaux differentiability of the norm of the space, and that left-uniqueness is equivalent to the space being strictly normed. 2 Thus Birkhoff's theorem can ,.be stated in any of the following forms: Theorem. 8.1, A normed linear space of three or more dimen- sions is an abstract Euclidean space if ' and only i:e it is such~tll;~t spherical orthogonality is synnnetric and additive. Theorem s.2. A normed linear space of three or more dime~- sions is an abstract Euclidean space if and onlt if it is strictJ,y normed and spherical orthogonality is synnnetric.,. Theorem 8.3. A normed linear space of three or more dimen- sions is an abstract Euclidean space if and only if its norm is Gateaux differentiable at each non-zero point and spherical orthogonality is symme$ric. Examples have been given 3 of normed linear spaces which sJ:nw the independence of the condition for strict normedness and the 1. Birkhoff (XIII). Condit ions which are sufficient for the existence of an inner product for two-d:ilnensional spaces are given by Theorems 8.4 and s. 5. The uniqueness assumed by Birkhoff can be either left or right, because of the assumption of symmetry (See Definitions 7.1 and 7.2.). 2. Theorems 7.3, 7.4, and 7.8. 3. Examples 7.1, 7.2, and 7.3. 91. existence of the Gateaux differential of the norm at non-zero points, making use of the equivalence of these to left- and rightuniqueness of spherical orthogonality. Since syrametry of spher- ical orthogonality clearly implies the,::equivalence of these properties, 1 it ,is interesting to consider whether the assumption of both left- and right-uniqueness :implies symmetry of spherical orthogonality. That is, whether spherical orthogonality is symmet- ric in a strictly normed space whose norm is Gateaux differentiable, at all non-zero points. Tha~ this is not true is shown by Example 8.1, and also by the following analysis: Suppose a par- ticular linear representation of a normed linear space T taken and the unit pseudo-sphere the points whose norms are unity-2. S is constructed as .representing Then it can be seen that T strictly normed, or spherical orthogonality is left-unique in if and only if S ~- T, does not contain a straight line segment. Also, the norm of 'f is Gateaux differentiable at~.; non-zero points, or spherical orthogonality is right-unique in T, if and only if 3 S has a tangent hyper-plane at each point. Clearly these two conditions can be easily met without satisfying the condition for symmetDy: namely, that a diameter drawn through the origin parallel to a tangential hyperplane at any point S p ot s cut in points at which the tangential hy:perJ.)lane is parallel to the diameter through Example s.1. p. A normedJ.linear s;paee which is strictly normed and whose norm is Gateaux differentiable at all non-zero ;points, 1. See Corollary 7.5. 2. See Birkhoff (XIII), page 169, The pseudo-sphere s can be in an Euclidean space .of any number of di!p.ensions, while Birkhoft uses a circle. 3. See Definition 9.2. This will be discussed in more detail in the next section. 92. but for which spherical orthogonality is not syrn;metric. Let be the normed linear space of all functions of the form F =x(2-:x:) The function f about the line in b (1,2) and =af+bg 1/Fn = ~J· IF) and in the interval {0,2), where f 2 ( 1-x g = for O~ x ~ l, l _(x-2} 2 -1 for 1s x ~. is positive throughout x = 1, while g (0,2) and symmetric is positive in and symmetric about the point (l,O). (O,l) and negative Thus if are of the same sign, then )F / has its maximum in a and (0,1)--at kt,• Likewise, if a and b are of opposite the point x = sign, then the maximum is in (l,2)--at the point 2b-a x = ·t, .. 8 • Clearly then the absolute values of two functions F = af+bg G = Af-t-Bg T and can take on their maxima at the same point only if they are proportional, and the absolute value of a function can take on its maximum at only one point. norm of T Thus T is strictly normed and tm is Gateaux differentiable at every non-zero point. Spherical orthogonality is therefore symmetric in · T 1 if and only if the absolute value of the function G which is zero where IF) is maximum, is maximum where F is zero. 2 If a and b are of = the same sign, then IF l has its maximum at :x: ab and F...LG 2 a _ [ a a \\ r ., a ~+ e,nd only if G(a+bl -A~(2-a.+b)J+B~ -(a+b)2...J O, or if = (1). Aa(a+2b)+Bb(2a+b) = 0~ But then A and B are of opposite sign and IGt has its ma:x:mum 2B-A 2B-A at X = 'ii"=A• If G.1-.F, then F (i3=A) = O, or (2). Aa(2B-A)+Bb(B-2A) = o. But (1) and (2) are both satisfied if and only if Ab= -aB, and hence (1) and (2) are not equivalent in general. orthogonality is not symmetric in 1. See Theorems ?.9 and ?.11. 2. See Theorem 7.12. T. Thus spherical 93. It has been shoWJ;l; that for the three types of orthogonality of Definitions 2.1-2.3, and any elements x and y of a normed linear space, there exists a number a for which x..1.ax+y. 1 For isosceles and Pythagoren orthogonality it has been shown that the assumption of either homogeneity or additivity of the orthogonality implies the normed linear space is an abstract Euclidean space. 2 The assumption of both homogeneity and additivity of spherical orthogonality in normed linear spaces of three or more dimensions :tm.Plies the space is a~stract Euclidean, but this is not true for two dimensional spaces. 3 However, several conditions can be given which are valid for two-dimensional spaces. The first of these is similar to the condition of Fioke~'s, 4 and makes use of orthogonality concepts only in the method of ..proof. Lermn.a. 8.4. implies If for a normed linear space lim. \l nx-ty~-· llx-+nYI/ M700 = O, then spherical orthogonality is • symmetric, additive, and unique in T, and illlply ak:x:/1 2 Proof: T 1J x ii= J/y/1 x ..1-ax+y y .1._by+x = b fty/1 2 • Let x and y be elements of T, A and algebraically smallest and largest of the numbers x .L a:x:+y, and and A' a B be the for which and B' be the algebraically SIIlE!.llest and largest •of the numbers b for which y..L by..1.x. If llx ll = II y/1 , then lim Unx+yJI - bx-tnylU= ~➔~ lim [ nx+y "- llnxli1 - ~➔= lim rux-+nylllJnY!l = o, ~ ~ ~➔~ since these limits are known to exist. Their existence follows from Theorem 6.7 , which also gives 1. That is, there is a number a for each type of orthogonality, but not necessarily the same number. 2. Theorems 4.9, 5.2 , and Cor. 5.3. Also see page 13 of this thesis . 3. See Theorem 8 •.1 and the preceding discussion. 4. That ll ax+y~ ==- /I x+ayd if llxlt = Uy1{ . Ficken (XIV). 94. =-(A-A') llx l\ == o, and hence that A = A'. Likewise, ~xii = Hy lt implies lim fll nx+yll - 11 :x:+nytl] /)1 ➔ 111:1 U! lim I!: r,,nx-yfl - lrx-nyl[! m -)-ao =~~lim L!.f)nx-Y1I - UnxJ~ - lim I..!!fTl:x:-ny/1 - fl nyJQ = o. oo I?\.+o0 .Another application of Theorem 6.? then gives lim U, nx-yl/ - 1/x-nylTI I>\.~ and = (B-B•) lfx ll = o, =B'. Now suppose :x: _1_ y and 11:x:II = IIYII . Then clearly B A~ O!::i:: B, and from the above A'~ 0 ~ B'. from Lemma 6.8 and the definitions of That y ..1.. x now follows and B'. Since A' s~herical orthogonality is homogeneous, there ,was no loss of generality in assuming Ox " = /l y/1. It has thus been shown that spherical orthogonality is symmetric in a normed linear space for llxll = lty n implies 1'1-?"'° lim l/nx+y11- l/x-,.nyll = o. How suppose which spherical orthogonality is not unique in such a normed linear space. There are then elements e for which I/x ii x and y and a positive number =J/y}/ and x ..L ax+y for la )<£ . 1 From the symmetry that has been proved, it now follows that )a }<£ , and hence from Theorem ? • ? that )a }<.£ . IJax+yJ/ for = II Yil = llx /f if Then it follows from the original assumption that lim [inx + (ax+y) ll - ll x+ n(ax+ y )lt] = O /1\➔ oo But ax+y_t_x llx+n (ax+ y) /J if la l<£ . = II (l+ na) x+ny II = ti nyf\ if n is large enough tl:lat ,a t-1/n 1<£ o Also, by Lemma 4.5, l-i! ffinx-+(ax-+y) ll - llnx+ylLJ = a \lx ll. Thus l:im D_lnx+{ax+y)ll - 1\:x:+n(ax+y-) lij ~➔~ = l:im (}! nx+yll - \l nyli]+ a Ux ll , ~➔~ and therefore can clearly not be zero for all values of a with )a \<£ . The assumption that spherical orthogonality was not right-unique in T then right-unique in is therefore false. such orthogonality is T, and it follows from Theorem ?.4 that it i. That this is true if spherical orthogonality is not right- unique follows from Lemma 6.8 and the homogeneity of spherical orthogonality. The symmetry makes left- and right-uniqueness equivalent (see Definitions 7.1 and 7.2). 95. is also additive. It now follows that if only does and A= A' just pr0ved, A= B if llxll = I YII . If x ...L ax+y = UYII , then not =B', but because of the uniqueness = A' = B'. Thus l/xH=F IIYU, take If and B llx lf x -1.axt-Y r implies such that y ..Lay+x llrx fl = IIY/1. y_t_by+x, it follows from the homogeneity of spherical orthogonality that rx..1._ !.(rx)+y and 2 r 2 Then = br, or a ll x ll = b l)y ll • y_1_ (br)y+rx. F Theorem 8.4. A normed linear space .e an -space if and only if is an abstract Euclid1 lim )lnx+ylf - llx+nyll = O whenever Uxll= IIYll • "'-➔ oo Proof: Define the inner product x ..LaXff• T (x,y) as -a l!x11 2 , where This value of the inner product is unique because of Lemma 8.4, and it is only necessary to show that it satisfies the conditions of Definition 1.9: (1). (t:x:,y} =t(x y). If x L ax+y, and t # o, then 1 tx.Ll(tx) + Y because of the homogeneity of spherical orthogonality. 2 Thus {tx,y) -flltxlt -at U:x:11 2 • Hence (tx,y) ~ t(:x:,y). If = = = o, the proof is trivial. (2). (:x:,y) = (y,:x:). If x ..L a:x:t-y, and y J_ by..1.X, then 2 {x,y) -a ll:x:11 and (y ,:x:) = -b \lylf. These are equal by Lemma a.~ t = (3). (:x:,y} + Cxaz) Then oc J...{a+b):x:+ (y+z) a.4. Hence (4). = (x,y+ z) ~ Suppose x ..Lax+y and :x:..L bx+z. because of the additivity given by Lemma 2 (x,y) = -a llx 1' , (x,z) = -b Uy /1 2 , and (x,y+z) = -((a+b) \\ x \l ~ (x,x) = )lx\\ 2 • This is immediate from :x: ..L (-l)x+x. Conversely, it follows from the definition of an inner product 2 that-cfor abstract Euclidean spaces and elements x and y l) nx+yH- llx+nyl\ = o for all for which Ux ff = Jlyfl or I. See Theorem. 9.3 for an equivalent statement. 2. Definition 1.9. (x,x) = (y,y). n It has been pointed out that if the norm of a normed linear space is Gateaux differentiable at each non-zero point, then this differential gives a convenient statement of properties of spherical orthogonality. 1 It was seen that the existence of this differential of the norm is a necessary and sufficient condition for spherical orthogonality to be additive. of the norm at means that x f(x;y) with increment y =o and f (:x:;z) = If the differential is denoted by f(x;y), this O imply f(x;y+z) = o. 2 Thus spherical orthogonality is additive and symmetric and the normed linear space is strictly normed if and only if the norm is Gateaux differentiable at all non-zero points and f(x;y) = o 1m,p1ies f(y;x) =o. This is a necessary and sufficient condition for the existence of an inner product in a normed linear space of three or more dim.ensions. 3 The following theorem shows how this can be strengthened to give a like condition for two-dimensional spaces. Theorem a.5. A necessary and sufficient condition for a normed linear space to be an abstract Euclidean space is that the Gateaux differential of the norm exist at each ·non-zero point, ~ f(x;in) = f(y; 1~,) for all elements is the Gateaux differential of the norm at x and x y, where f(x;y) with increment y. 4 Proof: (x,y) To prove ~he sufficiency, define the inner product of Definition 1.9 as -a}l x lf 2 , where x .Jl.ax:+y. If the norm is Gateaux differentiable, then the uniqueness of given by Theorem ? • 3. (x,y) is It will be shovm.t t hat it satisfies the 1. See page 71 and Corollary ?.5•. 2. Corollary ? .4. That linearity of the differential of the norm follows from its existence was shovm. by Mazur (VI), pp. 129-130. 3. Theorem s. 3. 4. Also see Theorem 9.3. 97. conditions of Definition 1. 9. (1). =t{x,y). If :x:.J..ax+y, and t =p o, then (tx,y) tx.J..t(tx)+y Thus because of the homogeneity of spherical orthogonality. (t:x:,y} = -~Ht:x:11 2 -atllx/1 2 = t(x,y). If t = o, the proof t . = is trivial. (2). {x~y) = (y,x). Theorem 7.3 gave f(:x:;y) =-all:x:N, where :x:..Lax+y. Likewise, f(y;:x:} = -b//yll, where y..L by:r:x:. Then 1 ~ ' x.L )JY11 ( ax+y) and y_t_ /J:x:lt(by+:x:), because of the homogeneity of spherical orthogonality~ Therefore f(:x:;~) ~ f(y; ~ } becomes 11 -a -b • 2 2 I -ilYil lbcll = -mi IIYII , or -a IJ:x:l1 -b llYlf • This is the same as = (x,y} = (y,:x:) • (3). {:x:,y)+ (x,z) - (:x:,y+z). Suppose :x:..La:x:+y and :x:..Lb:x:+z. It follows from the existence of the Gateaux differential of the norm that spherical orthogonality is additive, 1 and that :x:..L(a+b):x:+(y+z). That is, (x,y) = -a\l:x:11 2 ; (:x:,z) = -b llxfl 2 ; and 2 (:x:, Y+z) = - (a-+b) II :x: 11 • (4). IJxff"'2 =. (:x:,:x:J. This . is immediate from :x: ..L(-1):x:+:x:. conversely, if a normed linear space is an abstract EUclidean space, its norm can be defined by a (bilinear and inner product (:x:~y) • 2 Then: ~~~-fri~) !1:x:~-tbd( _ • - 2 2 ):x:+h II· - 1:x:\I · ~ (:x:+hy,:x:+ht)-{x.:x:) ~ 2(:x:~+h()ifi). :x: Yl + :x1[1 h (!tx:+hy +IIX,0 tlx II+ ( Hence the Gateaux differential f(x;y) of the norm exists and is (:x:.y) L) (xfl XI ~, = -i:x:,y) ~ But also equal to ~ • Then f ( x;j;yll ,1 · IIYU. :x: -- --,ryr, ~ -_ JJilFTiYff ~• f(y; J'.flf) 1. Corollary .7 .4. 2. See Definition 1.9. = Therefore . :x: • f(:x:,.L llY1\) -- f(y ,16ctJ). 98. 9. SFHER ICAL ORTHOGONALITY'i HYPERPLANES, AND FUNCT IONAIS It has been shown that for any elements x and y of a normed linear space there can be found at least one number . a wh ieh x ..Lax+-Y, and a number b for which tion of all such numbers- a and b boc+yix. 1 for An evalua- has also been given, and their uniqueness has been shown to be equivalent to Gateaux differentiability of the norm and to strict normedness, respectiveJ.y. 2 Spherical orthogonality is homogeneous, a.nd the proper- ties of additivity and symmetry were also studied and related to strict normedness and to Gateaux differentiability of the norm and used to get conditions for the existence of an inner product. 3 In this section, some of these results will be interpreted and extended by use of functionals and the concepts of conjugate spaces and supporting and tangential hyperplanes. I1' a linear functional sphere at a po int f takes on its maximum in the unit x~ then x is orthogonal to the maximal linear subset consisting of all elements for which f is zero. 4 Thus while spherical orthogonality is not additive in general normed linear spaces, there is {for each element x) always at least one 5 The problem of finding an maximal linear subset H with X..LH. x . orthogonal to a given linear set H will be studied in the next section, and will be related to the problem of finding elements on the unit sphere for which linear functionals take on their 1. Theorems 6.3 and 6.4. In this section, "orthogonalitytt is the spherical orthogonality of Definition 2.3. 2. Theorems 6.8 and 7~7; 7.3 and 7.8, 3. Corollaries 7 .4 and 7.5 and Theorems s.1-s. 3. 4. Theorem 6.1. 5. Theorem 6~2. maximum values. If the Gateaux differential of the norm exists, then it is linear. i'Uilctional f If for an element t'or which f(:x:) x there is a unique 1-inear = llxll an.4··· ll f lt == l, then f can be - 1 evaluated in terms of the differential of the norm and hence in terms of spherical orthogonality. Spherical orthogonality and linear functionals have been related by showing that Otll = 1 x.J....H if a linear functional f, with f(:x:) = ll x ll , is zero for all elements of and H. 1 The following theorem establishes a reciprocal relationship. Theorem 9.1. If an element :x:0 of a normed linear space is orthogonal to a linear subset H of 1inear functional f (x0 ) if h £ H. ~ f for which T, then there exists a = IJ'fl{J x l( and f(h) = O 0 There is also a maximal linear subset M of -T such :x:0 .L M and He M. Proof: and _H. Let G be the linear subset of Define the :f'unctional F h e. H. Then F llax0 11 and ll a:x:0 H~Uaxc,t-hll , since Since F(:x:0 ) T generated by by F(axcth) is clearly linear over fF{ax0+h) J~ lfa:x:ot-hll . \IFft T G. Xo-1-h x0 =allx 1f , where 0 But also, \F(a:x:dh) if h e- H. I= Thus = ff:x: \1 , it follows that 0 = 1. But there then exists a linear functional f defined = over all of T and such ~hat llf ll 1 :x: t:- G. 2 Then f(:x: ) )\:x: 11 and f(h) 0 linear subset = 0 M of all elements H and is such that =f{:x:) if and F (x) =O if h e.. H. The maximal m for which f = 0 contains x.1-M4' 1. That is~ if at least one such functional is zero on H. Theorem 6.1 • . 2. This is shovm by Banach (I), pg. 55, Theorem 2. S,ee 100. The following concept~ of tangential and supporting hyperplanes of the unit sphere of a normed- ,linear space will be found to be closely related to spherical orthogonality._1 Definition 9.1. T . A hyperplane H of a normed linear space is a supporting hyperplane of the unit sphere S distance between H and S points of if the is zero and H contains no interior s. A hyperplane 2 Definition 9.2. T 2 H of a normed linear space is the tangential hyperplane of the unit sphere S boundary point x0 of S ;plane containing the point at the if H is the only supporting hyper~ :x:0 • Theorem 9.1 and the preceding remark state that any element x of a normed linear space is orthogonal to some hyperplane through the origin, and that any hyperplane H through the origin for which there exists an element for which and :x:0 is the set of elements f = o, for some linear functional f(x0 ) = )/x 0 l1 . through x 0 ..LH But then f (x) = l f with IJtll = 1 defines a hyperplane which contains no interior points of the unit sphere. This hyperplane is thus a supporting hyperplane of the unit sphere. Hence there is a supporting hyperplane through every point of the unit sphere. ~ 1. See Mazur {Dt), pp. 71-73 and 77, for a more general discussion of supporting hyperplanes (Stutzhyperebene) and tangential hyperplanes. 2. See Definition 6.2. 3. This also follows from Mazur ( JX), page 73, Theorem 1. 101. Theorem 9.2. S .If x 0 is a boundary point of the unit sphere of a normed linear space and y is any element, ~ x0 ..Lx0 -y if and only if s has a supporting hyPerpmane containing x0 Y• and Proof: and y. Suppose the supporting hyperplane H L = H-x0 Then the set of all elements of the form contains is linear, where H-x . 0 h-x0 with h E. H. 1 x0 consists Therefore = H-x0 for all k, and x0+k(x0 S'ince contains no interior points of s, ffx0 -t-k(x0 ~y) IJ~l/x0 11 • 1 and k(:X:0 - y } £. L - y ) x 0 .L x 0 -y. Conversely, if E H . H x 0 ..Lx0 -y, then by Theorem 9.1 there exists a linear functional f for which f (x 0 ) = llf1Hlx 0 II and s, then )lx0 t/ .:_ 1 am f(x 0 -y) = o, If x 0 is a boundary point of :r(y) = )/t ll . Thus is in the supporting hyperplane defined by f (x) y = J/ f ll ~ It is known that if spherical orthogonality is right-unique or additive, then the Gateaux differentigl of the norm exists at all non-zero points. 2 However . elements x and y .J.im llxt-hylf- l)xll , Ai~+o h of a general normed linear space and can be evaluated in terms of spherical orthogonality. and this limit is denoted by Thus if x ~ 0 f+Cx;y), then: A = _ fx(x;-y) (l}. nxlt where A and for which exists for all B- are the smallest and largest of the numbers 3 If spherical orthogonality is additive, x .la:x:-y • then this function a f+ is the Gateaux differential of the norm 1. See Def. 6.2. His a coset of the subgroup L of the abstrac t Abelian group of the normed linear space. 2. Theorems 7. 3 and 7 .4. 3. Corollary 6. 7. 102. I ' It is not linear in'.;general, but does satisfy the following weakened linearity conditions: 2 and is ·h ence linear. then 1 (2). f+(x;y+z)~ f'+(x;y)+f+Cx;z), (3). f+Cx;ty) = t•f+ Cx;y) <4) • t+Cx;y)~ )Jyll. If B B+T is the largeE!t of the numbers for is the largest of the numbers . x..l.bx-{n+s.y) • Using this, and b t~O, a for which :x:J_ax-sy, for which f+(x;rx) = rllxtl, it follows trom (1) that: :f+ {x;rx+sy) = f+Cx;rx) +f+Cx;sy), ( 5). = rllxll+s•f+Cx;y), for s~ O and all r. It is known 3 that, for a linear functional F(x) = IJxll and for all y ,:;. o. F with lfJrtl = 1, it follows that -f+Cx;-y)~ F(y)~ f+Cx;y) .Also, if a linear functional F -f+Cx;,-y).:::. a ~f+(x;y), then there is such for which F(y) = a. The following is a 4 statement of this in terms of spherical orthogonality, but follows more easily directly from the evaluation of and f+{x;y) (6) • f+(x;-y) given by (1) above: x..t.ax-y if and only if -f+(x;-y)~ altxl,~ f'+(x;y) • In the previous section, it was possible to establish ' the existence of an inner product by assuming Gateaux differentiability 1. That \~ ( IJi-iliyn- llx\l) /h is linear if it exists has been noted by Mazur (VI), pp. 129-130, but also follows from the additivity of spherical orthogonality resulting from its right-uniqueness or from Gateaux differentiability of the norm (see Theorems 7.3 and 7.4) • • 2. These three relations were given by Ascoli (XV), pp. 53-55. The first follows trivially from the triangular inequality of the non:n., while the others can be pr0ved ' easily from the theory of spherical orthogonality. Thus ( 3) follows immediately from f+ (x;ty) /llxl1 and f+Cx;y)/IIXII . being .the largest of the numbers a and b, respectively, for which x.J...ax-tJ!: and xJ_ b:x:-y. The inequality Ct) follows from x..L fr+ (x;y) /uxN] x -y and Corollary 6.1. 3. Mazur (IX), pg. 75, statement 7. 4. The equivalence follows from Theorems 6.1 and 9.1. 103. of the norm and a type of symmetry of this differential. 1 This is also possible .without explicitly assuming Gateaux differentiability . of the norm. Theorem 9.3. A normed linear space Euclidean space i:f' and only i:f' non-zero elements Proof: x y. 2 and T is an abstract f+Cx; 1~) ~ f+CY;iji) :f'or all f +(x;y), 2 it follows that: Using the definition of _1!:_) - f + ( y; ±-) llx+h,1;;11- llxll -~im lt y+h,wll - -llYII f+ ( :X:; a•df f'flf --iim ....,_,_,.,.._..i.-i.;....... ----, ,. =-Hh ➔ +<> h 0 = }~ 11:x:+ tift:~11 -IJxll - }➔~ ffy+\ryfl - JJyJf' - lim llw, + hQ - ~ -tt0 11-lld,- +h w, II , h But by Theorem 8.4 this is zero if and only if the space is abstract Euclidean. The above theorem does not explicitly assume Gateaux differ:entiability of the norm. of f+(x; i~I) and However, it must follow from the equality f + (Y; ,~ 1) for all non-zero elements :x: and y~ since the norm of an abstract Euclidean space is Gateaux 4 differentiable. 3 This gives the following corollary. Corollary 9.1. ments x and 1_ X l!: f+ Cx;ffi) = f+ (Y;(Jxi) for all non-zero ele- y of a normed linear space T, 2 then the norm of T is Gateaux differentiable at all non-zero points and f(y;~}, where f(:x:;y) is the differential at Theorem s.5. 2. Where f+(x;y) = ¼~0 V!_xt-hy A- 11:xi\] /h. 3. Corollary 7.3'. 4. Compare with Theorem s. 5o 1 .. f(x;~) = x with increment y. 1040 If spherical orthogonality is symmetric in a normed linear space T, then the orthogonality is rightf unique if and only if it . is left-unique. 1 The space T is then strictly normed if and only . if the norm is Gateaux differentiable at all non-zero points. 2 It was also seen that spherical orthogonality is synrrnetric in a normed linear space whose norm is Gateaux differentiable if and only if f(x;y) = 0 differential at•: . x implies f(y;x) = O, where with increment y. If f(x;y) f(x;y) is the is replaced by f+(x;y), this condition implies symmetry and the existence of f(x;y), Theorem 9.4. Spherical orthogonality is symmetric and additive in a normed linear space f+(x;y) = o, 3 where Proof: If x x :¢ 0 T f+(y;x) = 0 whenever if and y are any non-zero elements. and x.-1. y, then all the numbers a for which x..L ax-y must be .in the interval (A,B), where A~ O::!::: B d B _ f+(x~y) 4 Then the largest number and A _- - f+(x;-y} llxll an • /lxJ • a for which x..L (A-a)x-y, or x..Lax-(Ax-y), must be zero. Hence f+{x;Ax-y} = o. 4 the smallest nuinber must be zero. f+(Ax-y;x) = O and Ax-y..Lx. Then a for which Therefore f+(-(Bx-y) ;x) = o and x..1-(B-a)x-y, or x..1-ax-(Bx-y), f+(x;-(Bx-y)) o. 4 Then Bx-yi x. it follows from Theorem ? .? and = Since A:i.-y..1. x A~ 0~ B that spherical orthogonality is symmetric if f+(y;x) Likewise, and Bx-y..L x, y-Lx, Thus f+(x;y) = O implies = 0 for all non-zero elements x and y, For elements 1. See Definitions ?.1 and ?.2. 2. Corollary ?.5. 3. Where f+(x;y) = \i¥1 Ulx+hyH-Nx!}/h. 4, See (1), page 101 of this thesis. 0 105.• x (=;t O) and y, suppose that x.1-ax-y A'~ a-6B'. if Then \fax-yl\ because of symmetry, ax-y..1-x and (from Theorem 7.7) is minimum and constant for A'~ a~B'. If A'< a'<-B', then ll(a'x-y) +hxlf-lla'x-ylf = O if \h\< a'-A' and fhl~ B'-a'. Thus f+'a'x-y;x) = f+(-(a'x-y) ;x) .:.."o; and hence f+Cx;-(a'x-y)) A! . = B', f+(x;a'x-y) = o. But with (6) of page 102 this implies that and that there is a unique number a for which x.1- ax-y, The additivity of spherical orthogonality follows from Theorem 7.4. Corollary 9.2. A normed linear space of three or more dimensions is an abstract Euclidean space if and only if f+(x;y) = 0 implies f~(y;x) = O for all non-zero elements x and y. 1 This corollary follows easily from Theorem 8.1 and Theorem 9.4. It was shown that spherical orthogonality is additive, or right-unique, in a normed linear space T if and only. if the norm of T is Gateaux d,ifferentiable. 2 This differential is a linearr functional if it exists, and hence furnishes a link between spherical orthogonality and the theory of linear functionals. 3 Thus the fact that for every element x of a normed 0 linear space there is a linear functional f with f(x 0 ) - 11:t.1l·\\x0 \\ was used to prove the existence of a hyperplane H such that x ...LH, H consisting of all elements for which f o. 4 = 0 = .limo l])x+hyll - llxU/h. 2. See Theorems 7.3 an~ 7.4. 3.• The linearity of this differential has been noted by Mazur (VI), pp. 129-130. It also follows from Theorem 7.3 and the additivity of -spherical orthogonality given by Corollary 7.4. 4. Theorem 6.2. That f exists is shown by Banach (I), pg~ 55, Theorem 3~ 1, Where f'+(X ;y) 106. llx 0 11 = 1, then the hyperplane defined by f ( x) = 1/f )I can not contain interior points of the unit sphere l)x}I ~ 1 and is If x0 • 1 therefore a supporting hyperplane at This hyperplane is unique, or also a tangential hyperplane, if and only if there is a unique lihear functional f with /lf)I = 1 and f(x 0 ) - Ux Jf= 1. 0 The tangential hyperplane then consists of all elements satisfying • f(y) = 1. 2 But f there is a unique hyperplane y is then unique if and only if H for which x ..L H, 3 and hence · if 0 and:' only if spherical orthogonality is right-unique or the norm is Gateaux differentiable at defined by x0 • f ( x 0 ;y) = 1Jx 0 11 , where This tangent hyperplane is then f(x •y) 0' = lim ltx:e+ hy}/ftx lf • 4 h 0 JC➔ o These results can be stated in terms of spherical orthogonality to give the following: Theorem 9.5 . Spherical orthogonality is right-unigue in a normed linear space T if and only if for any element x0 /lfll = 1 and there is a uniaue linear functional f(x 0 ) = llx 0 //. where This functional with is then defined by T f(y} - -allxJ\, x 0 .lax0+ Y• Proof: If the number unique for each element not unique. a f f of If f y x 0 _i_ ax + y 0 is not y, then Theorem 9.1 shows that f is is not unique, then Theorem 6.1 shows that is not unique for a lly. functional of a . for which Tha t f(y) = -aVx JI is a linear 0 follows from its being equal to the Gateaux f(x 0 ) = 1/fll , the elements x satisfying f(x) = //f II are of the form x 0+h, where f(h) = ~ and the elements h form a maximal linear subset. See Def. 9.1 and the discussion on page 100 of this thesis. 2. This has been shov1m by Mazur (VI) , page 130. J .. See Theorem 6.1. 4. Also shown by Mazur (VI), page 130. 1. Since 107. differential of the norm at y, the linearity of this differential of the norm following from its existence. 1 Since x0 with increment x 0 _J_ (-l)x 0 + x 0 , it follows that f(y) = -a1Jx0 /I for ally, where Corollary 9.3. f(x 0 J = llx 0 fl x 0 L ax 0 + y. Spherical orthogonality is right-unique in a normed linear space T if and only if there is a tangent hyper- plane at each boundary point of the unit sphere of Theorem 9.6. if T. There is a hyperplane tangent to the uhit sphere of a normed linear space T at the boundary point x0 if and only if spherical orthogonality is additive at y for Such a tangential hyperplane consists of all elements which x0 J.._ x 0 -y. Proof: Since Theorem 9.2 gives is in a supporting hyperplane at x 0 _t_x 0 -y if and only if y x 0 , it follows that if the tangent hyperplane exists it must consist of ally for which x 0 ..L x 0 -y. But by definition, there is a tangent hyperplane if and only if there is a unique supporting hyperplane, or by Theorem 9.2 if and only if the totality of elements x 0 J_ x -y is a hyperplane. elements xo-Y 0 y satisfying That is, if and only if the set of forms a linear set. 3 1. See Theorem 7.3. The linearity of this differential has been noted by Mazur (VI), pp. 129-130. It also follows from Theorem 7.3 and the additivity of spherical orthogonality given by Corollary 7.4. 2. Orthogonality is additive at x 0 if Xo..L y and Xo..l._Z imply X0 .Ly+z for all elements y and z. See III, page 13 of of this thesis. 3. Such a set is necessarily maximal. See Th. 6.3 and Def. 6.2. 108. In the above, the equivalence of Gateaux differen~iability of the norm to right-uniqueness of spherical orthogonality in normed linear spaces was used. Analogous results can be gotten by using the equivalence of left-uniqueness and strict normedness. Thus the norm of a normed linear space is Gateaux differentiable if and only if for each element functional f with /lfll = 1 x there is a unique linear f ( x) = /lxtl, while it can be and shown that a normed linear space is strictly normed if and only if for no linear functional than one point f x (with llxll = 1) with 111'11 =1 for which is there more f(x) = llxfl -that is, no linear functional takes on its maximum. on the unit sphere l)xll~l at more than one point. This has been shown by Smulian, 1 but also follows easily from the theory of spherical orthogonality: =1) takes on its maximum in the unit sphere at the points x1 and x 2 • Then f(x 1 ) = lJx and 1 For suppose f {:-·1/:ftl \I f(x 2 ) = llx 2 II , where f(x 1 +k(x 2-x1 )) llx1 11 = /lx 2 11 = 1. = 1. If all k. that x1 +k( x 2-x1 )__t_ x 2-x1 f(x 1 -x 2 ) = 0 and l~ llx 1 +k(x 2 -x1 )11 for Thus llf ll = 1, then But it also follows from the triangular inequality of the norm2 that llx1 +k(x 2-x )1\ ~l if O~k~l. Thus 1 Ux1 +k(x 2-x1 )JI = 1 if O~k~l, and it follows from Theorem 6.1 if 0~ k~l. Thus spherical orthog- onality is not left-unique, and hence the space is not strictly normed. Conversely, if a normed linear space is not strictly normed , then spherical orthogonality is not left-unique and for some 1. Smulian (XI) Theorem 6. 2. Condition (2f of Definition 1.6. 109. elements x and y and number a ' y_1_x and ax+y-'--x. Theorem 9.1 gives the ·existence of a linear functional l f = O, and ll f ll = 1. But it then follows that f( ax+y) = l) ax+y/1. 2 Thus f takes on with f(y) = /ly//, f(x) from ll ax+y H = IIYII its maximum in the unit sphere at the two points ,,;,, I/:~:~,,. and For every hyperplane H there is a linear functional and a number c such that H consists of all x f satisfying = c, and conversely all x satisfying an equation of the form f(x) = c form a hyperplane. 3 If lltll = 1 and r(x = f(x) 0 llxo !l = 1, then the hyperplane points of the unit sphere hyperplane. f ( x) l!xll ~ 1 =1 ) contains no interior and is therefore a supporting Thus the above can be restated: "A normed linear space is strictly normed if no supporting hyperplane of the unit sphere contains more than one boundary point of the unit sphere." Using the equivalence of strict normedness and left-uniqueness of spherical orthogonality gives the following theorem. Theorem 9.7. Spherical orthogonality is left-unique in a normed linear space if and only if no supporting hyperplane of the unit sphere S contains more than one boundary point of s. Whi~e right- and left-uniqueness of spherical orthogonality have been related to Gateaux differentiability of the norm and to strict normedness, no direct relationship has been developed between these concepts except the analogy between Corollary 9.3 and Theorem 9.7, and the equivalence which results l. See Theorem 7.8 and Definition 7.2. 2. llaxt-yll - \l y l\ because of 'rheorem 7. 7. IIA.x+yll if o~A~a. 3. Mazur (IX), page 71. Actually, f (Ax+y )= 110. from the assumption of symmetry of the orthogonality. 1 However, it has been shown by Smulian that if every linear functional defined in a normed linear space T attains llxll~ 1, then the conjugate space its maximum in the unit sphere T' is strictly normed if and only if the norm of T is Gateaux differentiable. 2 With this condition, it can also be shown that T is strictly normed if the norm of T' is Gateaux differentiable. These statements are equivalent to the following theorem, 1 which will be proved by using the theory of spherical orthogonality. Theorem 9.8. linear space If every linear functional defined on a normed T attains its maximum in the unit sphere then spherical orthogonality is left-unique in if it is right-unique in is right-unique in Proof: T'. T' I/xii~ l, if and only T, and it is left-unique in T if it T'. Suppose spherical orthogonality is not left-unique in Then using Theorem 7.7 it follows that there are non-zero elements f and af+g..Lf if \af <.e, and that l/gll = 1. Since is an element g g x0 of T' and a positive number /laf+gll = Ilg/I for which faJ<e. Take attains its maximum in the unit sphere, there for which llx 0 1f = 1 and g(x 0 } = 1. But there are both positive and negative values of • llaf+g# = l. for e Therefore f(x 0 ) = O, and in the unit sphere at the point x0 if. a af+g for which attains its maximum \al<e. Because of is orthogonal to all elements for which g = 0 Theorem 6.1, x0 and also ( if \al< e) , to all elements for which af+g = o. Since 1. As a result of Theorems 7.3 and 7.8. 2~ Smulian (XI}, Theorem 8. 111. the maximal linear subsets of which f and g T consisting of elements for are zero, respectively, do not coincide, it follows that spherical orthogonality is not right-unique in Thus spherical orthogonality is left-unique in right-unique in T1 T. if it is T. Suppose spherical orthogonality is not right-unique in T. Then it follows from Theorem 9.5 that there is an element x0 •r with for which there are two linear functionals l)r}I = /lglf = 1 f(x 0 ) = g(x0 ) = }/x 0 1f. and f and g of But then f(x 0 ) + k ~ (x 0 )-g(x 0 )] = ll:x: 0 11 , and therefore ll t+k(f-g) I ~ 1 = lltlf for all k. Thus g-1...f-g. Hence f..Lf-g. Likewise, f+k(f-g).L f-g if k llg+k(f-g) fl~tl gll = 0 or k = -1, and spherical orthogonality is not left-unique in cal orthogonality is right-unique in in and T T1 • Thus spheri- if it is left-unique T1 • Suppose spherical orthogonality is not left-unique in T. Then it follows from Theorem 7.7 that there are non-zero elements x and y and a positive number la l< e, and that IJ ax+y ff = fl YII e for there is then a linear functional and f(x} _ o. and g(x) ~ o. Let 1 g for which ax+y .Lx \a f< e. f with if From Theorem 9.1, f(y) = I/YI( , llrff = 1, be a linear functional with g(y} =0 Then f ( ax-t-y) + k [pr (ax+y) + g{ ax+y[) = f ( y)+ k l},f (y) + ag ( x)] • Thus if lbl~ Je:~;: J , then a can be chosen so that Jr(ax+y} + k [ bf (ax+y) + g(ax+y j] , ~ lf(ax+ y) / = l) ax+yl\ . Therefore tlf + k(bf+g) ll ~urn= 1 if lbl~ \ e(g(x)J I -L f(y) -,- O, and 1. The exist·ence of g follows from Banach (I), pg. 57, or from Theorem -9.1 and the existence of a non-zero nlLmber a for which ay+x J.. y, as given by Theorem 6. 3. 112. spherical orthogonality is not right-unique in spherical orthogonality is left-unique in uniq_ue in T T'. Hence if it is right- T' • The fol~owing theorem serves to complete Theorem 9.8. It makes use . of th!.e concepts of weak convergence and weak compactness, both of which will be used extensively in the next section. Definition 9.3. linear space A sequence (xn} of elements of a normed is weakly convergent if T all linear functionals f verges weakly to an element _ _ _. . . __ _ _- _ _ _ _ _ __ linear functionals x if ______ f Definition 9.4. defined in subset T. Such a se quence con- lim f(xn} - f(x) for all />1.7~ T. defined in A exists fo-r lim f(xn) 1'1.700 S of a normed linear space is weakly compact if every sequence of elements of S a sequence vmic~ converges weakly to an element of T. Theorem 9.9. T contains If the unit sphere of a normed linear space T is weakly compact, then spherical orthogonality is right-unique in T' if it is left-uniaue in Proof: in T' • T. Suppose spherical orthogonality is not right-unique Then spherical orthogonality is not additive in and there therefore exist linear functionals defined in T, orthogonal to for which f.Lg g+h. J)f+kg ll ?=:.ll t i\ 1. See Theorem 7.4. Thus and f, g, and f _Lh, but with and f 8f+kh ll? ll f fl T' 1 h, not ' 113. for all k, but there is a value k1 of for which k ll t+kl {g-t:h)I/-< 1/f' II . Wi th6ut loss of generality·, ll t ll can be taken as unity and k • -1 1 to be positive. Then ll f + k{g+-h) fl<ll f ll for 0-<k~k . 1 Now let [ :x:n~ {n 1,2,3,•••) be a sequence of elements of = T = 1, f(Xu)~O, and fr(:x:n)+ !g(:x:n) f ~ 1/ f ll = 1 for which for each Jf:x:n fl n, and let [ Yn1 be such that )JYnl\, =1, f{yn) ~ O, and If ( Yn )+ !h ( Yn} / ~ II t ll = 1 for each n. Since from O~ f(:i:cn_)~ 1 and n >-/l hll . and h{Xu) fg(xn) J~ 1/g lf that fh(yn }J~ 11 h 11 , it follows and h(yn)~ 0 if n> lf g(f 1 n=S kl' then g(:x:n) is this large and n If g {:X:n )'.:?:- 0 and must be of opposite sign,since otherwise / f(xn)+ ; f§ (xn) + h(~ )]/ llf ll would not be less than .• = 1. Likewise, g{yn) and h{yn) would also be of opposite sign. Since T there are subse quences of [xn1 and of t Yns weakly to elements y f{JSi) -+ l 1/ f It and x and f{Yn) ~ l. = l, it follows that Thus II F II which converge T, respectively. f{:x:) 1 -=:; 11:x: lf linear functional for which \F(:x:n) of is weakly compact, Clearly = f(y) = 1~ Since and 1~ ffy ff . If F is a = 1 and F(x) = fl x lf , then I~ IJx:n U =1. But since F{xn) ~ F{x), it follows that and thus that ll x ll~ l. 11:x: tl £ \ly II = 1, and f takes on its maximum in the unit sphere at both x g(y) ~ 0 and and y. But h(y) ~ O. Likewise fl yN~ l. F(:x:)-< l g(:x:) ~ O and Thus x- y Hence h(x) ~ O, while is possible only if 1. This foliows · trom the triangular ine quality of the norm, or from Theorem 7.5. 114. g(:x:) = g(y) = h(x} =h(y) o, and hence only if \t(:x:)+k1 Tu(:x:)+h(xil\= ltrlf . But this would contradict the assumption that flt+k 1 (g+h)l\ < \\t~. Since :x::; y, f takes on its maximum ~ the unit sphere at two distin.ct points and spherical 1 orthogonality is not left-unique in T. Therefore spherical orthogonalltyy is right-unique in T' if it is left-unique in T. If the unit sphere of a normed linear space is weakly compact, then each linear functional attains its maximum in the unit sphere. 9.9, 2 It is then possible to combine Theorems 9.8 and However, the concepts of uniform convexity and regularity 3 are also related to weak compactness,and hence to the question of whether a linear functional attains its maximum. It is known that a uniformly convex Banach space is regular, 4 and that a regular Banach space is weakly compact. 5 Thus any of . these three concepts can be used for the following theorem. Theorem 9.10. If a Banach space T is uniformly convex, regular, or 'i'ts2~:unit sphere is weakly compact, then spherical orthogonality is left-unique in unig,ue in T' if and only if it is right- T, and it is right-unique in T' if and only if it is left-unique in T. 1. See the discussion on page 108 of this thesis. 2. This is implicitlJ shown in the proof of Theorem 9.9. 3, These concepts are defined and used in the next section. See Definitions 10,1 and 10.3. 4. Milman (XVI) , Theorem · 2. Also see Pettis (XDC}. 5. Gani,;makher and Smulian (XVIII) , Theorem 1. Also, see Milman (XVI}, page 244. 115. 10 • Fl ALUAT IONS OF LINEAR FUNCT IONAIS • An essential step in finding an evaluation of a linear f'Unctional F defined in a normed linear space an element for which F (x) x Theorem 10.1. is finding = Jf]t\ ·lbd\. This is equivalent to the problem of finding an element maxim.al linear subset T x orthogonal to a given 1 H of T with ff'#, T : A necessary and sufficient condition that there exist an element orthogonal to each closed linear subset H of a normed linear space T is that for each linear functiona.D. F defined in T there is an element x with F (x) =lfFIHtxH. \)Fll i/:. o, then the set H of elements for which F = O is a linear subset with H ;; T. It follows from Theorem Proof: If 6.1 that any element F (x) x orthogonal to this set is such that = )j:E1t· )fx:IJ,, Conversely, suppose H is any closed linear subDefine the functional f as being zero for elements of H and unity for some element x0 not in H. set of T. This fun.ct ional is clearly additive over the space gotten by adjoining and its continuity follows from H being closed. to H,. It is then to all of T,· 2 Theorem 6 1 1 then showing possible to extend f that an element is orthogonal to H if x x0 f(x) = ))fl)•l!Kt\. A.1 normed linear space has the properties of Theorem 10,,1 if and only if each supporting hyperplane of the unit sphere S 1, See Def. I.a and the following discussion, and Def. 6.2. In this section, "orthogonality" is "spherical orthogonality" of Definition 2.3. 2, Banach (I), page 55, Theorem 2. 116. contains a boundary point of s. It is known that there is a plane of support e.t each boundary point of equation F(x) = )IF\\ defines a supporting hyperplane. hy:perplane contains the boundary point only if s, and that the x0 { l(x0 1 U= 1) This if and F(Xo) ~ JIFff . It is known that for any element linear subset H with x there is a maximal x.J..H, a linear functional F with F(x•) .:.. ltFlf•llxll , 2 and a supporting hyperplane to the unit sphere at the point W~ The problem of this section is an investiga- tion of the conditions under which the converse of this is true and of the resulting evaluations of linear functionals. It shoula. be recalled that a normed linear space is strictly normed if and on1y if no supporting hyperplane of the unit sphere S contains more than one boundary point of S, or if and only if for no linear functional F one element 3 with F(x) · = IIFfl•llxO • x is there more than Likewise·, the norm is Gate~ux differentiable if and only if there is a tangent hyperplane at each boundary point of element F(x) s, or if and only if for every x there is a unique linear functional F with =JIFfl·lbd\ . I 4 The latter will make it possible to evaluate linear functionals in terms of Gateaux differentials and spherical orthogonality for normed linear spaces in •which the norm is Gateaux differentiable and 1:inear functionals take on their maximum :in the unit sphere. 1. See Definition 9,1, and the discussions following Definition 9.2 and on page 109. 2. Theorem 6.2 and Banach (I), page 55, Theorem 3. 3. See Theorem 9.7 and the discussion of page 108, remembering that spherical orthogonality is left-unique if and only if the space is strictly normed (Theorem 7.8). 4. Corollary 9.3 and Theorem 9.5, using the equivalence of Gateaux: differentiability of the norm and right-uniqueness of spherical orthogonality (Theorem 7. 3) • Also, Mazur (VI) , page 130 • 117. The evaluation of linear functionals has been done by LBwig for complete abstract Euclideansspaces. F He showed that if is a linear functional defined in a complete abstract Euclidean space, then there is a unique element where = (x0 ,y), x0 such that F{y) l x 0 e.nd y. Then (Xo,Y} is the inner product of 2 2 F(x0 } = (x0 ,x0 } = ffx 0 \1 = lfE'f·ltx0 1J. The functional F thus Xo takes on its max:imum in the unit sphere only at the point lr.x:on ' and is the only functional te.king on its maximum at that point. As follo~~ from the above discussion, 3 x is orthogonal to the 0 satisfying {x0 ,~) = o. It follows that for every maximal linee.r subset H (with H ::fi T} set H of all h abstract Euclidean space which Xo.L H. 4 and for is the Gateaux differential of the norm at x0 It is also worthwhile to recall that f(x 0 ;y) x 0 ..Lax0+y. l!Fll = 1) Uxoll , or 5 (x ,y} j§ II = -a llxo II , 1 0 Thus for any linear functional F{y) {with in a complete abstract Euclidean space, there is e. unique element {xof,y ) T there is a unique element x0 f ( ~ ; y) = where in a complete x0 -a )lx0 1f . such that F{y) is equal to f(x0 ;y), .All but the form using the inner product can be extended to certain normed linear spaces. There are a number of restrictions which can be put on normed linear spaces e.hd which are suf'ficient to assure linear functionals takihg on their maximum in the unit sphere, or to give an element orthogonal to any ·given closed linear subset. shown that for any linear functional F Mazur has in a normed linear space 1. Low.ig (V), page 11, Theorem 11; {x,y) is defined br Def'. 1.9. 2~ See (4) of Def. 1.9. That )\f tl = Nx0 D, or 1{x0 ,y) J~ )\x0 "·\Jytl , follows from (1) of Theorem 1.2 and {2J of Def. i.6. 3. Also see Theorem 3.1. 4. This is given by Lowig {V), page 15, -Theorem 16. 5. Theorem 7.3 and Corollary 7-.3,. 118. with a weakly compact 1 unit sphere there is an element = \\Fll· )Jx:I\ , and equivalently that every supporting hyper- which F(:x:) plane of the unit sphere s. point of for X S in such a spijce contains a boundary This is equivalent to the following theorem. 2 Theorem 10.2. If the unit sphere of a normed linear spaee T is weakly compact, then for each closed linear subset T there is at least one element H of x.J..H. The following conditions are more restrictive than that used above, but are worthwhile in that they are of quite different f0rm and still give the desired conclusion of Theorem 10.,2. D;finition 10.1. A normed linear space and only if for every linear functional F conjugate space F(f) = f(x 0 ) T' there is an element for all elements Definition 10.2. A functional f there is a S> O of T such that 3 defined on a normed linear space T possesses the Frechet differential for any £ > 0 is regular if defined -over the x0 of T'. f T f(x;y) at :x: if for which ) f(x+h~)-f(x) - f{x;y) J< s\\Yll if lhl-< & ; number s5 _f__{_x....;_y__) __is_._a_,_,_'un_1_·f_o_rm __F_r_e_c_h_e_t_d_i_f_f_er_en_t_ia_l_'_'_1_·f_t_h_e 4 can be chosen independently of :x: for lxlt = 1 . 1. Mazur (VI), pp. 129-130. Also see Definition 9.4. 2 •. See Theorem 10.1. 3. Some authors use "reflexive" in place of "regular". Conjugate spaces are defined on page 8 of this thesis. 4. These are Gateaux differentials (Def. ?.3) for which the convergence to f (x;y) is uniform in y when llYll.:!S 1 , and uniform in x,y when 11:x:M= 1 and lfyH~ 1. 119. It is known that the unit sphere of a regular normed linear space is weakly compact, 1 and that a Banach space is regular if its norm is uniformly Frechet differentiable at each non-zero point. 2 Thus either of these c.oD'.ditions can be used in place of the weak compactness of Theorem 10.2, if completeness is assumed with the differentiability. 3 Uniform Frechet differentiability of the norm has the added advantage of implying Gateaux differentiability, which will be either directly or implicitly needed in the evaluations of linear funetionals. Theorem 10.3. or if T If the normed linear space is complete and the norm of T is regular, T is uniformly Fr~chet differentiable at non-zero :points, then for each closed linear subset H of T there is at least one element X..LH .• If F is a linear functional defined in T, then there is at least one element X for which F{x} - IIFll·llxll • Definit:i.Qn 10.3. A normed linear space is uniformly convex if for every £. > 0 there is a number sS > O for which llx\l = lfyll = 1 and llx-yH> £ imply 11:x: + Yll < 2- S • 4 It has been shown by Milman that a uniformly convex Banach space is regular. 5 It therefore follows that the regularity of 1. Gantmakher and Smulian (XVIII) , Theorem 1. Also see Milman (XVI}, page 244. Using these and knowing that the conjugate space of a normed linear space T is the same as the conjugate space of the Banach space gotten by completing T, it follows that a regular normed linear space is a Banach space and• its unit sphere is weakly compact. 2. Smulian (XX} , page 648. . 3~ See Theorem 10.2 and the preceding discussion. 4. Originally given by Clarkson (XII), Definition 1. 5. Mil.man (XVI), Theorem. 2. Also see Pettis (XIX). 120. Theorem 10.3 can be replaced by uniform convexity and completeness. Furthermore, if F is a linear functional defined in a uniformly convex Banach space x for which F ( x) ~ )IF~· ltxl\ T,then there is a unique element and there is one and only one element orthogonal to a given maximal linear subset of T. Theorem 10.4. If a Banach space T is uniformly convex, then for each maximal linear subset H with I unique element xJ_H. If F 1 H~ T there is ,a is a linear functional defined in !, then there is a unique element x for which F(x) = IJFQ·ll;lf. If the Gateaux differential of the norm at a point x0 , f(Xo ;y) 1-¥% Uxo+hr!- liXoll , exists in a normed linear space T, then it is a linear functional of y. 2 For this linear functional F(y) = f(x ;y), it is seen = 0 that F (x 0 ) ~ lfx0 1 and that IIFI( = 1. 3 linear space is Gateaux differentiable at But the norm of a normed x0 if and only if there is a unique linear functional F with liF II = 1 and 4 F (x0 ) , llxo ll • Hence if the Gateaux differential of the norm exists at each non-zero point, then any linear functional F, with ))FU = 1 and for which there is an element x such that F(x} = l x ll , is equal to the Gateaux differential f(x;y). 1. The first statement is given by Pettis (XIX:), Lemma 1. The two statements can be shown to be equivalent by the same reasoning as used to prove Theorem 10.1. They then followa from the equivalence of left-uniqueness of spherical orthogonality and strict normedness, a uniformly convex space being strictly normed. 2. That f(Xo;Y) is linear if it exists has been noted by Mazur (VI), pp. 129-130, but also follows from the additivity of spherical orthogonality resulting from its right-uniqueness or from the existence of f(x ;y) See Theorems 7.3 and 7.4. 3. Since \Jlxo+hylf - ~lxollr~ llhYII , because of (2) of D,efinition 1.6. 4, Mazur (VIJ, page 130. This also follows from Theorem. 6.1 and the eq~ivalence of Gateaux differentiability of the norm and right-uniqueness of spherical orthogonality (Theorem 7.3). 121. Using Theorem 7.3, it is seen that such a functional is also equal to -aJJxll, where xJ..ax+y. The following theorem now follows easily from Theorems 10.2 and 10.3. Theorem 10.5. T .2!. If the norm of a normed linear space is Gateaux differentiable at each non-zero point and: (1) the unit sphere of T ( 2) T is weakly compact, is regular, then for a clinear functional F least one element x with IIFi( = 1 there is at for which F{y) = f(x;y) = -alixlf for all elements the norm at x y, where and f(x;y) • is the differential of x...Lax+y. However, it was shown that a normed linear space is Gateaux differentiable at each non-zero point if and only if spherical orthogonality is additive. 1 This gives the following corollary. Corollary 10.1. If spherical orthogonality is additive j_n a normed linear space or T and: (l) the unit sphere of ( 2) T x is weakly compact, is regular, then for a linear functional F one element T with llFI) such that it follows from F (y) = =1 there is at least x...Lax+y that,for ally, -allxll • If the norm of a normed linear space is uniformly Frechet differentiable, then it is clearly Gateaux differentiable. 1. Corollary 7.4. It 122. therefore follows from Theorem 10.3 that: 1 Theorem 10.6. If the norm of a Banach space is uniformly Frechet differentiable, then for a linear functional F with . l(:Efl = 1 there is at least one element • x for which F(y) = f(x;y) = -anx H for all elements at x and y, where f(x;y) is the differential of the norm x .J...ax+y. In Theorem 10,5, 10,6, and Corollary 10.1 it was not possible to say that there is a unique element F (y) = f(x;y) = -a/1:xf/ . be more than one such x for which However, it is known that there can not for any linear functional F if, and only if, the space is strictly normed. 2 Therefore if this conx dition were added to either of these theorems, it would be possible to conclude the existence of one and only one element for which F(y) = f(x;y) = -a ft:xif for all y. x Rather than assuming the space to be strictly normed, this can be included in uniform convexity and the following gotten from Theorem 10.4. 3 Theorem 10.'7, If F is a uniformly convex Banach space whose norm is Gateaux differentiable at all non-zero points, then for a linear functional F unique element x with llF M = 1 such that for all elements there is a l F(y) = f(x;y) = -a lj:x.U, where f(x;y) is the differential of the norm at I. By the same reasoning as preceded Theorem 10.5. x and xJ_ax::7. 2. Smulian XI, Theorem 6. Also see page 108 of this thesis. 3, See Definitions '7.4 and 10.3. The reasoning is the same as that preceding Theorem 10.5. 123. Corollary 10.2. If spherical orthogonality is additive in a uniformly convex Banach space~,,. then for a linear functional F with flFI\ = l there is a unique element F ( y) for all elements x for which = -a lf:x:11 y, where x..1..ax+y. This theorem would also give the result of Low:ig 1 as a corollary, since an abstract Euclidean space is uniformly convex and its norm is Gateaux differentiable--the inner product being equal to llxtl •f (x;y). 2 (x,y} However, the conditions of Theorem 10.7 are rather severe. Gateaux differentiability of the norm implies that spherical orthogonality is additive, 3 while uniform convexity is a stronger condition than strict normedness. 4 It is therefore seen from Theorems 8.1-8,3 that requiring spherical orthogonality to be symmetric in a space of three or more dimensions satisfying the conditions of Theorem 10.7 makes the space abstract Euclidean. Because of this, it seems worthwhile to consider what kind of a generalized inner product can be defined in normed linear spaces'• It was seen that for abstract Euclidean spaces the inner' 5 product can be evaluated as (x,y) = -allx11 2 , where xJ_ax+y. This can also be stated, equivalently, that if H linear subset for which - y = ax-h with x..1..H h £H, then and a is the maximal is the number for which 2 (x,y) = -alJxlt • 6 1. Lowig (V), Theorem 11. See page 117 of this thesis. 2. See Corollary 7.3'• 3. Corollary 7.4. 4. See Definitions 7.4 and 10.3. 5; See page 19 of this thesis. 6. The existence of H is given by Theorem 6.2; it must be unique since the number a is unique. 124. Theorem 10.s~ elements It is 12ossible to a:ssjgn to each pair of x,y of a normed linear space T a real number oc,yJ satisfyin_s: (1). [ tx,YJ = t [x ,YJ (2). [e ,Y+zl = a:,iJ+ Cr,zl , ( 3). Bc,il is continuous in ( 4). ~ 1 :x:J = lb:U2 and Bc,tyj = t [x ,y] , Y, I[x ai1 [ .:$ HxH •Jl,vU• and There is a unique possible def'inition of [i 1 y] satisfying 1 (1)-(4) if and only if orthogonality is additive. In that case: Proof: ( 5). [;x: 1y] is continuous in (6). :x:..Ly if and only if For each element a linear functional F:x: with :x: of IIF:x:\\ :x:, u ,y] T = o. for which lfxll = l choose = 1 and F:x:(x) = 1, and let 2 Rx be the corresponding maximal linear subset with x ..LRx• Define [kx, ~ as k \lx ll •Fx(y) or as -ak /bcH 2 , where a:x:+y £.H.K . Since F:x:(a:x:1-y) = o, 3 these definitions are the same and clearly satisfy (1)-(4), the second part of (4) following either from )IF:x:lf 1 or from ~ ,~1• 4 = fal 1 Spherical orthogonality is additive in T if and only if for any elements which x (# O) and y there is a unique number a for 5 :x:J...a:x:+y ~ From the way llc ,:ij was defined , it is clear that i. Other conditions equivalent to additivity of spherical orthog- onality are: (1) Right-uniqueness of spherical orthogonality; (2) Gateaux differentiability of the norm at non-zero points; (3) Uniqueness for each element :x: of the lihear functional f with llf l\ = 1 and f(x) = llxll ; (4) The existence of' a tangent hyperplana-eat each point of the unit sphere. Any of' these could replace sph erical orthogonality in Theorem 10.a. (See Theorems 7.4, 7. 3, 9.5, and Corollary 9.3). 2; See Theorem 6.2 and its proof. 3. Theorem 6 .1. 4. The latter is given by Corollary 6. 1 . 5. Theorem 7 ._4. 125. the uniqueness of such definitions of ~,Y] ness of the number a for which implies the unique- :x:..l..ax+y, and therefore also implies additivity of spherical orthogonality. @c,y] Also, if (4) gives 11xu2 .:$ u:x:q·ltx-tkYll =o, then ~ ,x+ki] = ~,x] and \l:x:+kylf > 11:x:ft • Thus [ :x:,y] = TI O implies :X:.L y, and 2 ~ ,Y] -a ll:x: Q implies :x:..La:x:+y. Hence if s:pher ical orthog- or = onality is additive, then the only value [lc ,y] -a U:x:li 2 , and :x: _Ly if and only if 8C , ~ 0. = m.a ins to establish (6): Suppose ~ ,y] can have is It now only re- is not continuous in :x:, =-a ltxll 2 • Then for some sequence .f?c11 with 2 11:x:ill =1 and :x:r~:x:, it isn •t true that ~ i , ~ --r -a ,~n • Since (4) shows that l_x1,Y] is bounded, there is a -subsequence [ :x:j 1 of and a number A such that [ :x:j ,Y] = -Aj U:x: 11 2 and and that p_c ,Y] A.-+A. Then, as above, for all k. Sitice>tt.he norm is continuous, it follows that J a U:x:j+k(Aj:x:j+y) lf>-..}l:x:jll = 1 ~\\:x:H= 1 for all k for which :x: ..L a:x:+y- is unique if spherical orthogonality \\:x:+ k(Axw)'\1 number :x:j..LA:X:j+Y, or is additive. Hence a and that =A and [ x,yJ :x:..1-A:x:+y. But the is continuous in :x: if spherical orthogonality is additi~e. This result gives another form in which to express • Th eorems lo • 5 - lo • 7 , b y rep 1 acmg f ( x;y ) = -a )Jx,, b Y ~'11:x:Hy] . 11 If in a ddit i on to additivity of spherical orthogonality its symmetry is also assumed, then [;, yJ = O :implies fr ,x] = o. All normed linear spaces of three or more dimensions satisfying 1 these conditions are abstract Euclidean. For a two dimensional space it is necessary to , make the assumption I. Theorem s.1. ~ ,Y] = fr ,x] , and 126. hence have all the conditions of the definition of an inner 1 product. In the following examples, the generalized inner product [x,y] is expressed in terms of the general form of the linear functionals defined on these spaces ( for p > 1) , and interpreted in terms of spherical orthogonality (for p-;;:::1). Example 10.1. Consider the Banach space lp (p ~l) of all = (x1 ,x2 ,• • • ) for which ~fxil p is convergent, 2 where llxll = ~ fxil ~ rt, • By definition, xJ...y if and only if sequences x llx+kYi/-lfxll~O 10 [~ Jxi+kyiJ1 if for all k. If y = (y1 ,y2 , • • • ) , this becomes -}j, lxil~Yt>~o, or ~Uxi+kyi JP- lxi l~ ~o. Thus x...Ly, then d> lim -¼➔+O h IJ p . pl Jxj_+ky j I - Fil :J ~ o. k But this clearly implies that Jim llx+kyll- lb: I\ ~ O and that ➔ +o K 3 }lx+kyll ~ Jlx lt :f.'or k~O. Using the convergence of the seri,es tft Jxi i P and ffl; Jy 1 1P, it is not difficult to show that r~;i~:t ;r,:i~r: lim J;, ➔+o and ~ Jxi+kYitp - II x-tky /l ~ \he ff :te::: a:1 f:r ::::h lid ~ 1. ~ 1 fxi l p Thus = Z. lim !xi+kyilk P_ lx1 lp ' ~:::.I ,k-')--+O for · k ~ O if and only if neither of these is less than zero. As k ➔ +o, lx1+1cy1l P_ lx1 1P approaches k 2 p : l if or and xi:; o, and it p> l P lx1 l p- (xiyi), . approaches )yif if p = 1 and xi= o. Hence )Jx+ky Jl ~ ltKH for k ~ O if and only if 1. Definition 1.9. 2~ This space is defined by Banach (I), page 12. 3. See Theorem 7.5. 127. or Putting (1). ii lxi\ P- (x1y (2}. ~ ,~~r ax+y and -(ax+y) 2 1} ~ o, if + (~) fyi \ ~ O, if in p = 1. in place of 2 Jxi\ p- (xiy 1 (l) y shows that / if and only if la.J/xJ/ P+?; .fl'J =o, and that r oO p-2 11 i: 1 -ch lx1 J P (XiYi_jx + y. From Theorem 7. 3, it follows that x...Lax+y L- • x _1_ p>l; nxll f (x·y} = ( 3}. where ' f(x;y) ~ I ·l p-2 . . . .:. -,xi • ( iiYJ,) )lxll 15 ' is the Gateaux differential of the norm. at Thus the norm of a space lp with p >l x. is Gateaux differentiable and spherical orthogonality is right-unique and additive in hence if ¾,• 2 A space lp (p >l} LJC,~ is defined by (4) • Ix y7 = ~' ~ #, lxi lp-2(XiYi) UxffP-2 then for any linear functional unique element x0 is uniformly convex, and ' F . defined in for which F(y) =[ x ,y] 0 lp there is an 3 for all y. Also , this "generalized inner product" satisfies conditions (1)-(6) of Theorem 10.s. For p = 1, putting (2) shows that or • (5) • x .1-ax- y ax- y and -(ax-y) in place of y in [ {f..~ x •Y ftJ-, :o) if and only if ±LllxU- ~ ~il/+~I Y1 \~0 , i1 ~ ~ r - ~) IY1 I S allxll ,; ~ x~~f + ~ IY1I • Thus spherical orthogonality is right-unique, or additive, at an element if x = (x1 ,x 2 , • • •) xi :;p O for any i. of the space 1 if and only 1 This condition is also necessary and 1 . There must be at least one number a for which x J:..ax+y. See Theorem 6. 3. 2. These three concepts are equivalent in any normed linear space. See Theorems 7.3 and 7.4. 3. The uniform. convexity of (p > l) was shown by Clarkson {XII). The rest follows from Theorem 10.7, and has been known before , e.g. as a direct consequence of Pettis (XII) ~ Lemma 1 , and the satisfying of (1)-(4) of Theorem 10.s by ~,~ . . :I> 128. sufficient for the existence of the Gateaux differential at and for the unique evaluation of the linear functional IJFU = l and F(x) = 11:xfl . Thus if x 1 ::,. 0 for any with i, then the F ( y) = EC,Y] differential of the norm is given by ( 3) and . ally, where F x for E,~ is given by (4). Comparing (5) above with (6) of page 102 gives: Example 10.2. ·functions x{t) Consider the Banach ~pace (0 1 1) in · Jlxll = [ [fx(t) f Pd~Yt> • for which [ lx(t) JPdt ¾, .An 0 x 12c,Y] defined on for which F(y) = [x,Y] 1 ( exists, where 2 1'.P (p > l) there is a for all = t \x(t) \ P- x(t) •y(t) dt • • of all space i; uniformly convex if p >1, and for a linear::·functional F unique element 1p (p ~ l) 2 y, where 3 This "generalized · inner product" satisfies (1)-(6) o'f Theorem 10.s. For the space L1 the following is analogous to the corresponding equations for the space 1 1 , U being the set of all numbers t in (O,l) for which f + (x;y) = ( x(l!)nt >at+ )4 Since x(t)-:/, 0 x J_ ax+y and 1-U the complement of f I Y(t) j dt; t-U if and only if -f+ (x;y) 5 a \lxl\~ -f_(x;y), spherical orthogonality is right-unique, or additive, at Gateaux differential of the norm at numbers t 'for which u. x(t) ~ 0 x and there is a x, if and only if the set of is of measure zero. 1. Where 'f± (x;y} = } ~ 0 f] x+hyjJ- ltx lQ/h. 2. Clarkson (XII)• c""t3. This follows from Theorem 10.7 and the evaluation of the differential of the norm given by Mazur (VI), pg. 132. It is also immediate from Pettis (X:CX:), Lemma 1, and the satisfying of (1)-(4) of Theorem 10.8 by [x,i] . 129. B IBLIOGRAP.HY I, s. Banach, Theor~e des Operations Lineaires, Warsaw (1932). II. M. 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