Rearrangements of Measurable Functions Citation Day, Peter William (1970) Rearrangements of Measurable Functions. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/6V2Z-F375. https://resolver.caltech.edu/CaltechTHESIS:10292010-132102259 Abstract Let (X, Λ, μ) be a measure space and let M(X, μ) denote the set of all extended real valued measurable functions on X. If (X 1 , Λ 1 , μ 1 ) is also a measure space and f ϵ M(X, μ) and g ϵ M(X 1 , μ 1 ), then f and g are said to be equimeasurable (written f ~ g) iff μ (f -1 [r, s]) = μ 1 (g -1 [r, s]) whenever [r, s] is a bounded interval of the real numbers or [r, s] = {+ ∝} or = {- ∝}. Equimeasurability is investigated systematically and in detail. If (X, Λ, μ) is a finite measure space (i. e. μ (X) < ∝) then for each f ϵ M(X, μ) the decreasing rearrangement δ f of f is defined by δ f (t) = inf {s: μ ({f > s}) ≤ t} 0 ≤ t ≤ μ(X). Then δ f is the unique decreasing right continuous function on [0, μ(X)] such that δ f ~ f. If (X, Λ, μ) is non-atomic, then there is a measure preserving map σ: X → [0, μ(X)] such that δ f (σ) = f μ-a.e. If (X, Λ, μ) is an arbitrary measure space and f ϵ M(X, μ) then f is said to have a decreasing rearrangement iff there is an interval J of the real numbers and a decreasing function δ on J such that f ~ δ. The set D(X, μ} of functions having decreasing rearrangements is characterized, and a particular decreasing rearrangement δ f is defined for each f ϵ D. If ess. inf f ≤ 0 < ess. sup f, then δ f is obtained as the right inverse of a distribution function of f. If ess. inf f < 0 < ess. sup f then formulas relating (δ f ) + to δ f+ , (δ f ) - to δ f- and δ -f to δ f are given. If (X, Λ, μ) is non-atomic and σ-finite and δ is a decreasing rearrangement of f on J, then there is a measure preserving map σ: X → J such that δ(σ) = f μ-a.e. If (X, Λ, μ) and (X 1 , Λ 1 , μ 1 ) are finite measure spaces such that a = μ(X) = μ 1 (X 1 ), if f, g ϵ M(X, μ) ∪ M(X 1 , μ 1 ), and if ∫ o a δ f+ and ∫ o a δ g+ are finite, then g < < f means ∫ o t δ g ≤ ∫ o t δ f for all 0 ≤ t ≤ a, and g < f means g < < f and ∫ o a δ f = ∫ o a δ g . The preorder relations < and < < are investigated in detail. If f ϵ L 1 (X, μ), let Ω(f) = {g ϵ L 1 (X, μ): g < f} and Δ(f) = {g ϵ L 1 (X, μ): g ~ f}. Suppose ρ is a saturated Fatou norm on M(X, μ) such that L ρ is universally rearrangement invariant and L ∝ ⊂ L ρ ⊂ L 1 . If f ϵL ρ then Ω(f) ⊂ L ρ and Ω(f) is convex and σ(L ρ , L ρ' )-compact. If ξ is a locally convex topology on L ρ in which the dual of L ρ is L ρ' , then Ω(f) is the ξ-closed convex hull of Δ(f) for all f ϵ L ρ iff (X, Λ, μ) is adequate. More generally, if f ϵ L 1 (X 1 , μ 1 ) let Ω f (X, μ) = {g ϵ L 1 (X, μ): g < f} and Δ f (X, μ) = {g ϵ L 1 (X, μ): g ~ f}. Theorems for Ω(f) and Δ(f) are generalized to Ω f and Δ f , and a norm ρ 1 on M(X 1 , μ 1 ) is given such that Ω |f| ⊂ L ρ iff f ϵ L ρ 1. A linear map T: L 1 (X 1 , μ 1 ) → L 1 (X, μ) is said to be doubly stochastic iff Tf < f for all f ϵ L 1 (X 1 , μ 1 ). It is shown that g < f iff there is a doubly stochastic T such that g = Tf. If f ϵ L 1 then the members of Δ(f) are always extreme in Ω(f). If (X, Λ, μ) is non-atomic, then Δ(f) is the set of extreme points and the set of exposed points of Ω(f). A mapping Φ: Λ 1 → Λ is called a homomorphism if (i) μ(Φ(A)) = μ 1 (A) for all A ϵ Λ 1 ; (ii) Φ(A ∪ B) = Φ(A) ∪ Φ(B) [μ] whenever A ∩ B = Ø [μ 1 ]; and (iii) Φ(A ∩ B) = Φ(A) ∩ Φ(B)[μ] for all A, B ϵ Λ 1 , where A = B [μ] means C A = C B μ-a.e. If Φ: Λ 1 → Λ is a homomorphism, then there is a unique doubly stochastic operator T Φ : L 1 (X 1 , μ 1 ) → L 1 (X, μ) such that T Φ C E = C Φ(E) for all E. If T: L 1 (X 1 , μ 1 ) → L 1 (X, μ) is linear then Tf ~ f for all f ϵ L 1 (X 1 , μ 1 ) iff T = T Φ for some homomorphism Φ. Let X o be the non-atomic part of X and let A be the union of the atoms of X. If f ϵ L 1 (X, μ) then the σ(L 1 , L ∝ )-closure of Δ(f) is shown to be {g ϵ L 1 : there is an h ~ f such that g|X o < h|X o and g|A = h|A} whenever either (i) X consists only of atoms; (ii) X has only finitely many atoms; or (iii) X is separable. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) ; Decreasing rearrangement, doubly stochastic operator, measure preserving transformation, non-atomic, measure space, extreme point, rearrangement invariant normed space, Muirhead's inequality, doubly stochastic, majorization Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Luxemburg, W. A. J. Thesis Committee: Unknown, Unknown Defense Date: 21 April 1970 Funders: Funding Agency Grant Number NSF UNSPECIFIED Record Number: CaltechTHESIS:10292010-132102259 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:10292010-132102259 DOI: 10.7907/6V2Z-F375 ORCID: Author ORCID Day, Peter William 0000-0002-8076-4048 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 6164 Collection: CaltechTHESIS Deposited By: INVALID USER Deposited On: 29 Oct 2010 21:44 Last Modified: 07 May 2024 21:56 Thesis Files Preview PDF - Final Version See Usage Policy. 45MB Repository Staff Only: item control page

Copyright €) by PETER WILLIAM DAY 1970 REARRANGEMENTS OF MEASURABLE FUNCTIONS Thesis by Pete r William Day In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1970 (Submitted April 21, 1970) 11 ACKNOWLEDGEMENTS The author thanks Professor W. A. J. Luxenlburg for bringing this topic to his attention and for his guidance in its perusal. The author thanks the National Science Foundation for four years of financial support. The author also thanks Mrs. Virginia Conner for her efforts 1n getting this thesis typed quickly. Finally, the author thanks his wife, Susan, for her patience and understanding during the last three years. iii ABSTRACT Let (X, 1\., f,.I.) be a measure space and let M(X, !,.l) denote the set of all extended real valued m.easurable functions on X. If (X l' 1\.1' U 1 ) is also a measure space ann f E M(X, ~d and g E M(X , !.-1 ), then 1 1 and g iff ~t(f-l [r, sJ) are said to be equimeasurable (written f", g) = III (g -1 [ r, s J) whenever numbers or [r, sJ [r, sJ j s f a bounded inte rval of the real = (+ ac} or = [- ac }. Equimeasurability is invesb- gated systematically and in detail. If (X, 1\., u) is a finite measure space (i. e. U(X) < DC) then for each f E M(X, u) the decreasing rearrangement ° t ~ Then Of of f is defined by ~ U(X). Of is the unique decreasing right continuous function on [0, u(X)] such that If (X, 1\, U) is non-atomic, then there is a measure of ...... f. preserving map 0: X .... [0, U(X)] such that ciO) = f \..l. -a. e. If (X, 1\., ~!) is an arbitrary measure space and f E M(X, \..l.) then f is said to have a decreasing rearrangement iff there is an interval J of the real numbers and a decreasing function f", O. 0 on J such that The set D(X, \J.) of functions having decreasing rearrangements is characterized, and a particular decreasing rearrangement defined for eachfED. Ifess.inff~O<ess.supf, then of is Of is obtained as the right inverse of a distribution function of f. If es s. inf f < ° < e s s. sup f then formulas relating (of) + to Cf+' of) and to 0 f- and o-finite 0 -f to 6 f are given. and If (X, A, u) is non-atomic 6 is a decreasing rearrangement of f on J, then there is a measure preserving map 0: X - J such that 6(0) = f u-a. e. IV If (X, 1\, u) and (Xl' 1\1' ~l) are finite Ineasure spaces such that a=u(X)~··Ul(XJl. a c f() (J g 1 ., if f.gl-M(X,~l)UM(XI,fll), and if . are (1111te, g -< f 111eanS and then g -<-< f g <-< f and / IlWanH () & g '. (1"\1 and /1: (,. fOl' ill J 0 . I . it • () [ The p reorder relaliol1H -< and -<-< are investigated in detail. If f E Ll(X, U), let O(f) = £g ELl(X, U'): g -<£} and 6(f) = [g ELI (X, u): g "" f }, Suppose P is a saturated Fatou norm. on M(X, U) such that L P is universally rearrangem.ent invariant and If f EL P then O(f) c L P and LeX': C LP eLI, O(L P, LP')-com.pact. If which the dual of L P is of 6(f) f E L O(f) is convex and £ is a locally convex topology on L P ', then O(f) is the f E L P jff (X, 1\, \1) is adequate, for all L P in s-closed convex hull More generally, if 1, 1 (X I' 1J. ) let (1 f(X, U) = (g E L (X, U): g -< f} and 1 6 (X, U) = [g E L f generalized to I (X, U): g ,..."" f}. Of Theorem.s for O(f) and 6(£) and6.f' and a norm. PI are onM(X ,u.1) is given 1 such that O\f\ C L P iff f E LPI , A linear m.ap T: LI(X I' IJ,I) -+ L1(X, \J.) is said to be doubly stochastic iff Tf-<f for all f E L1(X , UI)' I doubly stochastic If f E L O(f). I T such that It is shown that g -< f iff there is a g = Tf. then the m.em.bers of If (X, 1\. u) is non-atom.ic, then 6.(f) are always extrem.e in 6(f) is the set of extrem.e points and the set of exposed points of O(f), A m.apping ~: 1\1 -+ 1\ is called a hom.om.orphism. if (i) u.(~(A)) = VI (A) for all A E 1\ 1; (ii) ~ (A \J B) = ~(A) U ~ (B) [}.L] whenever A n B = 0 [U.l J; and (iii) ~ (A n B) = ~(A) n ~(B) [u] for all A, B E A!, v where A = B [~J means C A = C B IJ-a. e. If~: AI .... A is a homo- morphism, then there is a unique doubly stochastic operator T~: L 1 (Xl' w. I ) .... L 1 such that T~CE = C~ (E) for all E. (X, u) T: L1(X , ~l) .... Ll(X,~) is linear then 1 iff T = T41 for some homomorphisll1 Let X o If f E L shown to be tg ELI: I Tf- f for all £ E L 1 (X , W,l) 1 ~. be the non-atomic part of X and let A the atoms of X. 1 (X,~) If then the be the union of 1 oc a(L , L ) -closure of 6(£) is there is an h ""f such that glX I o -< hlx 0 and g A = h A} whenever either (i) X consists only of atoms; (ii) X has only finitely many atoms; or (iii) X is separable. -viTABLE OF CONTENTS Chapter Section Title Ac know led gerrwnts 11 Abstract ii j Table of Contents vi 1. Introrluction 1. II. Page viii EQUIMEASURABILITY 2. Spectral Measures 1 3. Spectral Equivalency 3 DECREASING REARRANGEMENTS 4. The Right Inverse of a Decreasing Function 14 5. Decreasing Rearrangements of Functions on Finite m. s. 16 6. Ji)ecreasing Rearrangements on Not Necessarily Finite m. s. III. 29 INEQUALITIES AND REARRANGEMENTS 7. A Theorem of Hardy 42 8. A Preorder Relation 43 9. Embedding of a Finite m. s. in a Non-Atomic Finite m. s. 50 10. An Inequality of Hardy and Littlewood 52 11. The Values of an Integral 55 12. The Decreasing Rearrangements of Sums and Products 13. T f -< f and Some Consequences h1 60 62 vii Chapter IV. Section Title REARRANGEMENT INVARIANT BANACH FUNC TION SPACES 14. Rear rangement Invariant Banach Function Spaces 15. A Representation Theorem V. 65 74 INEQUALITIES OF HARDY, LITTLEWOOD, POLY A AND MUIRHEAD VI. 16. Schur Convex Functions 77 17. The Sets o.(f) = [£1: fl -< f} 83 18. Doubly Stochastic Transformations 92 19. 'Muirhead's Theorem 99 EXTREMAL AND RELATED PROBLEMS 20. Extreme, Exposed and Support Points of 0 f 103 21. Pe rmutator Transformations 108 22. The Weak Closure of tg: g"'-' £} 116 Bibliography 125 viii 1. Introduction. The decreasing rearrangement of a non-negative measurable function has, since its treatment in Inequalities by Hardy, Littlewood and Polya, p lay e d an increasingly important role in analys"is, because of its fundamental part in the structure of nornwcl spaces of n1easurablc functions wh"i ch are rea r rangenlent inva riant. Exalnples of such spaces are the clas sica] LP spaces, the OrU 0'> spaces, and the spaces "introduced by Halperin [l4] and Lorcnt/-; [2~~]. These so-called rearrangenlent invariant Banach function spaces have been shown by Boyd [2], Shimogaki[45], and Lorentz and Shimogaki [25J to be well suited for studying problems related to Fourier analysis and interpolation of operators. Recently Luxemburg [28J gave a general account of the theory of rearrangement invariant Banach spaces for measure spaces with finite total measure. Such spaces provide a natural setting in which to generalize a theorem of Hardy, Littlewood and Polya which gives equivalent conditions that two vectors in R preorder relation -<. n be related by a certain J. V. Ryff [42J has given the generalization for L 1[0, 1], while Luxemburg [28] has given it in part in the rearrangement invariant Banach space setting. We will complete this generalization in Chapter V. With an eye to studying these topics when the total measure of the space is not finite, in Chapter I we investigate the concept of equimeasurability for arbitrary measure spaces, though we include as well results which can only be proved in general for finite measure spaces. ix In Chapter II we define the decreasing rearrangen1.ent for all nleasurable functions if the n1.easure space is finite. If the rneasure spaccis not finite, we characterize the set of rneasurable functions which have decreasing rcarrangernents, and cI('finc olle for each Huch function. Of in,portance is the fact that we can prove a theorClIl relating a function to its decreasing rearrangement by a measure preserving transformation when the measure space is non-atomic and 0 -finite. In Chapter III we introduce the generalization to measurable functions on a finite measure space of the Hardy-Littlewood-Polya preorder relation ties. -<, and investigate it and some associated inequali- In particular we give a new and careful proof of a theorem in [28] about the values taken on by certain integrals, and we characte rize adequate measures. For completenes s we include as Chapter IV Luxemburg's treatment of rearrangement invariant Banach spaces. For the same reason we include his resultsonSchur convex functions in Chapter V, where we give finally a complete account of the generalization of the theorem of Hardy, Littlewood and Polya referred to above. We also extend Ryff's generalization of Muirhead's theorem to finite measure spaces. Finally in Chapter VI we settle some extremal and related problems of some sets which arose in Chapter V. Throughout we us e the following abbreviations: 1"n. s. = measure space; m.p. = measure preserving; pwd = pairwise disjoint. -1- 1. 2. Spectral Measures. EQUIMEASURABIL1TY Let (X, A.,~) be a tneasure space (tn. s.), i. e., X is a non-etnpty point set, A. is a a-algebra of subsets of X, and ~ is a countably additive n'leaSUl'e on f\. Oft~)n we will write Jfdl-l [PI' tIl!' integral of f over X with respcl'j: to }..lwhen X is clear Also we let M (,'oln the c()nt(~xL = M(X, \..I.) denote the set of all extended real valued IJ-tneasurable functions on X, and if E is a set, then C E denotes the characteristic function of E. Let R denote the real nutnbers, let R# denote the extended real nutnbers, and let S = S(R#) denote the Riesz space of all functions s of the fortn s where each Uk E R and each 1k is a bounded interval of R#, i, e., 1k is an interval of R of finite length or 1k =:: [+ao} or 1k =:: {-co}. We call the tnetnbers of S step functions on R Jf , - - For every s E Sand f EMit is easy to see that s (f) is a sitnple tneasurable function on X, for all s ES. If f E M, let 1 :S .... R #- be defined by f l Suppose f has the property that lJ.(f- [u, v]) < oc for every bounded interval [u, v] of RJf. Then 1 (s) is finite for every s E Sand f in fact defines a positive linear functional which is continuous in the sense that if s R Jf n ~ 0 then 1f{s ) ~ O. n Hence there is a tneasure \..I. such that for every s E S we have f on -2- We call1J the spectral measure of f, or sornetinlCs the ~-sp('ctral f In probability theory IJ. is known as the distribution f rneasurc of f. nleasure of f. For many purposes, whether or not f takes the value a on a set of finite or infinite measure is of no interest, so our conditions under which IJ is defined may seem unduly restrictive. f have only to let X, = [x E X:f (x) * a} However, we (called the carrier of f) N =/\ n X, = IJ.!/\' to see that if 1J(f -1 [u,v]) (00 for all bounded intervals [u, vJ of RI - [OJ then 1J'(f\X,-l[u,v])< 00 for all bounded intervals of and IJ' R* and thus the ~-spcctralmeasure IJ.'f\X' is defined. We may in a similar manner ignore whether or not f takes the values +00 or -00 on sets of finite or infinite measure. Observe that if [u, v] is a bounded interval of RI, then kl f( [ u, v]} = J.,l (C 1 [ u, v]). If (X, A, u,) is a fi nit e m. s. ( i. e. hJ. (X) < (0), then IJf is defined for every f E M and can be repre sented by the distribution function d defined for t E R by f df{t} = u,{ (x E X:f(x) > t}) • Letting ef{t) = u.({x E X:f(x) ~ t}) for every t E R we have that df+ef=IJ.(X), d d f is decreasing, e fn t d f and c f Ii m ei t .... -00 t iff d (2. 1) f f is increasing, both d ~ e f whenever fn f and e are right continuous, t f, l~m df{t) = IJ{ U = +oo}), 00 t ) = IJ( U = -co}), df<t-) = IJ.( tx:f(x) ~ t}), n f t and e f is continuous at is continuous at tiff IJ( [f = t}} = O. PROPOSITION. ~~en d f n poi ntwi se a. e. -+ d f If (X, /\, \,t) is a fi nite m. s. and f .... f pointwise -- n at every point of continuity of df' ~ d f ..... d n f -3- Let E = [x:f (x) f f(x}}. PROOF. and let A n lim sup A = Un > t} and A = (f > t}. n Then IJ(E) = O. n cE U AU [f = t}. If d f Let t E R Then AcE U lim inf A cE U n is continuous at t, then IJ( U=t}) =0 so U(A)S:IJ(lim inf A ) ~ lim inf U(A } ~ lim sup \.L(A ) ~1J(lim sup A ) n n n n ~ !-L(A) and hence df{t) = IJ(A) = lim \J(A ) = lim d (t). f n Since d n decreasing on R, d f f is has only countably many discontinuities. If f = g a. e. then s(f) = s(g) a. e. for each s E S. REMARK. Thus for each s f- S we may define a mapping T s :L1(X, 1\, ',.,ij-+ Ll(X, 1\, IJ) by T sf = s(f). Let 51 = fsES:s = C]u, v[' u & v rational}. Then fTs:sES1} separates points of Ll For if f & gEL 1 differ on a set of positive \J - measure, then at least one of {f < g} or fg < f} has positive measure. \= 00 we may assume {f < g} has positive measure. enumeration of all rationals of R, then U By symmetry If ff3 1 is an i co < g} = U < f3. < g} so there . 1 1 is a rational ~ such that \J. (U < v < g}) > O. U 1= 1 Since f E L , u( U= -co}) so there is a rational number ~ s. t. IJ( fu < f < v < g}) > O. =0 Letting s=C] u, v [wehaves(f)=C .u < f< v }sofs(f)~s(g)}:::>[u<f<v<g} . f and thus s(f) and s(g) differ on a set of positive measure. 3. Let (X 1,1\1' IJ ) and (X ' I\Z' iJZ) be measure 1 Z Spectral Equivalency. spaces (m. s. ) with f 1 EM(X l' IJ,1) and fZ EM(X z' IJ )' Z If both m. s. are finite, then we say that f1 and fZ are spectrally equivalent or equimeasurable iff Uf f 1 ..... £Z iff d f = d f 1 = \J.f and in this case we write f 1 --- fZ' Z iff \J.1(f 1 1 Z #= interval [u,v] ofR. 1 Then [u,v]) = \Jz(fz-1[u,v]) for every bounded -4In case one of the m. s. is not finite we write f {,",fZ iff -I [u, vJ) = lJ.Z(fZ-I [u, v]) for every bounded interval [u, vJ of R #:. IJ (f1 1 1 Then f1 ,. .··-[Z iff lJ.1(f1- [B]) = lJ.z(fz-1[B]) for every Berel set Bof R,p,. Observe that if there exist f1 E M(X , lJ,1) and fZE M(X ' IJZ) 1 Z such that f 1 ~ f2. then 1-11 (X 1) =1-12.(X ) in the sense that both are infinite 2 or finite and equal. (3. 1) LEMMA. function. 1£... f'" g and f is a simple function, then g is a simple Two simple functions are equimeasur able iff they take the same value on sets of equal measure. n Let f = ~ a.. C where 0. <.". < a. and tE.} par1 i= 1 1 E i n 1 Suppose g E M(X ' \JZ) and g ....... f. Then lJ.Z( tg = o.i}) = ).ll(E )· i Z PROOF. titions X I' lJ.z(fg ~f0.1'·'" o.n}})=lJ,Z(g-l[-oo,o.l[) n-1 +·L:IlJ,2(g-110..,0.'+1[)+~-J(g-lJo. ,+ooJ) 1= 1 1 <.. n 11.-1 "" lJ. (f - I ]a..,o. I [ )+lJ,l(f - 1 Jo. ,+00 J ) +p 1 1= 1 1 1+ n n Thus if A. = [g= a..} then g = ~ a.. C A a. e. 1 1 i= I I i n Now let f be as above and suppose g = ~ 1= I partitions X z and IJZ(Ai) = iJ. (E ), i=l, .,. , n. 1 i a.. C A 1 L41J 1 (E.):0.. E[u,v]} . 1 1 = 1J.1(UtEi:ui E[u,v]}) = 1-1 (£-1 [u,vJ). 1 where tA.} 1 If [u, v] is a bounded interval of R#:, then ).lz(g-l[u, v]) = UZ(U[Ai:o.iE[u, v]}) ::: L:fIJZ(A.):o.. E [u, v]} = 1 1 i (3.2) -5Let F b(' a Borel measurable function on ;)11 PROPOSITION. interval I , .. R~. ~ G C M(l) ~nd r on ~ I then G ....., F iff G 1 ~ F l' > 0 ~l~ F1(t):::; F(rt) alld G1(t):- G( 1'1) In addition, each of the followi nJ;i functions is equinleasurable with F. r E R, t E 1+ r H(t):::; F(t-r) (i) (ii) H(t):: F(-t) t E- I (iii) H(t) :: F(t+} if F is monotonic (iv) H(t) :: F(t-) if F is monotonic PROOF. Let m denote Lebe sgue measure, and let [ u, v] be a bounded inte rval of R ~ . so m(F r ] 1 m(G -1[ u, v l) m(F 1-1 [u,v)= -Ie u, vl) :: m(G -I ru, v ] ) iff m(F I-I[ u, v ] ) = m(G I-I[ u, v ] ) . (i) ] -I[ u,vJtr)=m(F -I[ u,v~) 1 m(H -1[ u,v)=m(F (ii) m(H-I[u, vJ) = m(_F-I[u, v]) = m(F-1[u, v]) H = F (iii) & (iv) (3. 3) a. e. PROPOSITION. The following are true for all measure space s. (i) f,...., g implies s(f) ,...... s(g) ~or all s E S (ii) f -- g implies rf,.,... rg and ftr '" gtr £~~..!:!. r E R (iii) f", g implies If\ '""' \g I (iv) f,....,g implies ft -g+ and f-""g (v) If a:X 1 -> X 2 is measure preserving and f E M(X ' U ) then Z Z foo""£. (vi) f"""g implies ess.inff= ess.infgandess.supf::ess.supg. (vii)Jif""g and f;<:O ~.then g:2:0~. andthere are sequences [tn } and {g } of simple measurable functions such that f ,..... g and n n n-- o ~ f n t f and 0 ~ g n t g. In addition, if f is essentially bounded ~fl~ g is essentially bounded and fn t f and gn f g uniformly. -6(viii) ~ fl E M(X l' ~1) and fZ E M(X Z' ~Z) and fl' fZ ;;:: 0 then fl ,...., fZ implies Sf I dl-1 1 = Sf2.d~Z in the sense that both arc finite and equal, or both are infinite. (ix) ....!if I (- LI(XI'~I) and fZ EM(X Z ,I-1 Z ) and fZ"'fl then f Z ELI (X 2' IJ 2) and Sf 1 d \J. 1 = Sf Z d IJ. 2 • (x) ..!!:.f EM(X,I-1) andg EM(X',~') and rXi}i~1 and (Xi}i~I are pairwise disjoint measurable subsets of X and X' respectively such f that IJ(X-U X.) = 0 = 1-1'(X'-UX. ) then fIX. ,.... g -- 1 1-- 1 Ix. , i = 1,2,3, ..•. 1 In addition the following ar e true for finite m. s. (xi) .!i. f 1 - g 1 and f Z '" g Z and inf t If 11, IfL.I } = a = inf rig I I, I g zl } the n f I + f Z ....., g I + g Z . (xiii) If f -n .... f and g --n ..... g a. e. and f n "" g n n = 1, Z, 3, .•. , the n f ,..; g. (xiv) If fn' f E L I(X l' 1-1 1 ) ~nd gn' gEL \X Z' I-1 Z ) and fn .... f and gn -- g in L 1 norm and f - (xv) n ...... g n n = 1, Z, 3, ... , then f ,.... g. -- .!i. cp is Borel measurable on R# then f~g implies cp(f) ....., cp(g). n PROOF. (i) Let f '"" g and s =:E ok C I with Uk} pwd intervals n k= I k of R#=. Then s(f) = Ok C f -l[I 1 ,..,~n ok C = s(g) using k-l k~ k-l g Ik LemIna (3. 1). 0 -Ie ] (ii) Let [u, vJ be a bounded interval of RI. ~1 ( (rf) 1J 1 ( (-f) -1 [ -l u,v])=~I(f If r > a then - 1 [u v] -1 u v] -1 r'r)=I-1 Z(g [r'r)=I-1 Z (rg) [u,v]). r·u, vJ) = ~l(f -1[ -v, -uJ) = IJZ(g -1 [-v, -uJ) = ~Z( (-g) -1 [u, vJ). Thus rf ,... rg if r> 0 and -f,.., -g, so rf-rg for all r E R. Let r E R. ] =~Z«g+r) -1[ u , v J). \J.l«f+r) -1 [u,vJ)=U (f -1 [u-r,v-!:J)=iJ.2(g -1 [u-r,v-r) 1 -7(iii) Let f,..., g and let [u, v] be a bounded interval of R:!f,. u :s; 0 ~ v, \J (11 f 1 1 = uz(/gl-I[u, v)). In case [u, v]) = \Jol (f -I[ -v, v]) = uz(g -Ie -v, v J ) If 0 < u, \JI(lf\-1[u,v]) = U (f-l[n,y])+\Jl«-O-I[u,v]) 1 = IJz.(g-I[ u, v])+\Jz.«-g) -1 ell, v]) = Il z( I g ,-I[ u, v J ). The cas e v < 0 is trivial. Let f ~ g and l e t [u, v] be a bounded interval of R#' (iv) W e w i sh to prove that 1J. (f+-l [u, v]) = IlZ(g +-I[u, vJ) which is clearly true jfv<O or 1 \1>0 . Hence supposeu~O~v. Then u (f +-1[ u, v J ) = \Jl(f -lrL -OO , v ] ) 1 -1[ +-1[ _ = IlZ( g -00, vJ) = \JZ( g u, v j). For th e r e st , f ~ g "" -f,..., - g "" f = (-f) + ~(-g) + = g- . (v) -lr vJ) = IJ-z.(f I-Ll((fo a ) -1[ u, v ] ) = 1J.1(a -1 fLU, = inf ft:u( [f > t}) = o} and ess. inf f = (vi) Recall that ess. sup f - ess. sup (-0. -Ie u, vJ) . If f ...... g, then t E R:!f, "" U 1 (U > t}) = IJ- 1 (f-l]t, +ooJ) = IJ)(g -lJt, +ooJ) = IJ ) ( f g > t}) so es s . sup f = ess. sup g. ~ f'" g "" -f ...... -g ~ ~ cs s. inf f = - e ss. sup(-f) = - ess. sup(- g ) = e ss. inf g. Let f rv g and f ~ 0 a. e. (vii) The n ( vi ) abov e "" g ~ 0 a. e . To con- struct [f } and [ g } let n A A 11 , 11 i = n f-l[i-l = f Z -1 n ' _i_ [ Zn B n, i = B [n,ooJ n = - 1 [ i-I n ' Zn Z - 1 g en, 00 J 2-[ g n 2. f g 11 The n Lemma (3. 1) "" f n '" g • n n = "" LJ n i= 1 l~i~nZ.n . 1 1- - In f ..... g "" \ fl ,..., CB . + nCB n,l n 19 I so if f is essentially bounded, then ess. sup! g 1 = ess. sup I f l < 00 so g is essentially bounded. The rest is well known. See [17 , p. 15 9 , (11. 3 5)J . -8- (viii) If f is Cl. sin1ple function and g ..... f then Lemma (L 1) '? g i::; simple and Jfd~I = JgdlJZ' are sequences fn ~gn and fn If f ~ 0 and g"" f then g ? 0 and t:her(~ f£11 } and fg n } of non-negative sin1ple functions s. t. t f and gn t g. Then Jfd 1J 1 = lim SfndiJ1 =lim Sgn d \1 2 = JgdiJZ' 1 (i x) If f E L (X l' iJ 1) and g '""' f the n 1g I ~ \ f \ < ro and thus g E LI(X ' 1.1 ), Z 2 Finally g r-J 50S \g I d Z = SI f I d ~ 1 IJ, f ~ g +...... f+ and g ..... f 50 ('j'g d \J 2 = J+ g d \J. 2. - J g d IJ,Z = S+ f d \J 1 - Sf d \.11 = Sf d \J 1 . I Letting Xo = X-UX. and Xa = XI- UX! we have fIx. '" g IX. (x) I I I I 1 = 0, 1, 2., •.. , and £X.I J~a and £X.' }.rna are partitions of X and XI 1= 1 1= I respectively. Let [r, s] be a bounded interval of R*. Since flx.-g\x. i=O,l,2, ... ,~havelJ(f-I[r, s] n X.) = 1J,'(g-I Cr , s]nx.') . 1 1 1 1 i = 0,1,2, ... , so summing from i = 0 to +co we get iJ(f -I[ r, s])= ~'(g-1[r,5]). . (xi) Hencef....,g . Let fI,fz,gI,gZ be as stated. * Ei = [fi oJ: oJ, Fi = [gi F3 = X z - (F 1 U F Z )' a} Then IJI(X I ) = IlZ{X Z }' Let i= I,Z, and E3= Xl - (EIUE Z)' Then IJ (E ) = UZ(F ) i:= 1,2 and since the m. 5. I i i are finite we may conclude thatIJ1(E3) = IJ,Z(F 3 ). Let [u, v] be a R* bounded interval of and let f = flH ' g = gl+gZ' Then 1J1(f-1[u, v]) Z 3 1 -t='11J1(Einf u,v) -1J1(f u,v])+1J1(f -IrL.U,v_)+1J 1 (E 4 ) I z -;' , -Ie ] - where E4 = £E 3 if 0 E Cu, vJ % if 0 fl [u, v J then fJZ(g-1 [ u, v]' -Ie Hence if F _ 4 - I F3 t if 0 ) [u, v] % if a 'F [u, v] = IJ,Z(gl-1 Cu, v])+UZ(gz-1 Cu, vJ) + Uz (F 4) = \..LI(f -1 [u, v]) (xii) Follows from (iv) and (xi) . -9- (xiii) Using (2. 1) we have = d (xiv) f n so f - g. f n Then g -- g pointwise a. e. ln R f in nonn implies {£ } has a subsequence - pointwise a. e. gn on g n - g in norm so k But then fn -> ~ n } has a n - f k subsequence k f pointwise a. e. so (3. 3)(xiii) mk k implies f'" g. (xv) From the proof (vii) we see that each of cp + and cp - is the n limit of a sequence of simple functions of the form L a. C where i =I I E i the sets [E.} are Borel sets. 1 Hence cp is the limit of a sequence [s } of simple functions of the same form. n and since s (f) ..... cp(f) and s n LEMMA. (3.4) n Hence f"" g ~s (f)- s (g) n n (g) -cp(g) we have cp(f) ...... cp(g). (i) Suppose f E M(X , 1\1' IJ ), g E M(X , 1\2' 1J,2), 1 1 2 El EA 1 , E 2 E1\2' \.11(E 1 ) = 1)'2(E 2 ) < 00 and f and constan t value on E 1 and E2 respectively. g have the same Then f"-' g ~ fiX 1 -E 1 ~gIX2-E2 and fC X -E -gC x -E . 1 1 2 2 (ii) Suppose f, g E M(X, 1\, IJ), E E 1\ has ~(E) < 00 and Then f'" g ~ fl X-E- g IX-E and fC PROOF. = glE \J 1 (f z· -1[ (i) Let fl=flx1-E 1 _ ""' g CX_E. X E and gl =gIX -E ' and let a z 2 Let [r, sJ be a bounded interval of R#. r, s ] _{a~1 (E ) ifa~[r,sJ if a E [r, sJ n E 1) - l fiE = g 1 E. Then = flEl -10- and siITlilarly for g, so iJ. (f -l[ r, sJ n E ) = iJ.Z(g -I[r, s J n E' )' I Z I Hence IJ.I (f -Ie r, sJ) = iJ. I (f -I[ r, sJ n (XI-E » I -l[ -l[ ] =1J.r(f r,s])-IJ.I(f r,s nEll I ~ , .. I[ r , s] ) , :::lJ.t(g -I[ r,s J )-IJ.Z(g -I[r,s ] nl<'Z)=U.Z(g so rJ ~'g 1 . Letf l =fCX1-E ~(El U{(X 1 -E 1 )nf Uz{gI-I{O». l andg 1 =gCXZ-E If I is an interval of R#and -1 1-1 (I»=IJZ(gl-1 (I». (ii) Let f/ = fIX-E and g\ intervalofR#. -1 (0» =: 1-1 (0» = Uz{EZ)+iJ.Z(g 1-1 (0» = -1 (0») =iJ.1(EI)+iJI{f iJ.1(f 1 (I»=UZ{g Z ThenlJ.r(fl ' °~ I, then U (f - 1 {I» I 1 = HencefI"'-'gI' = g\X-E. Since fIE=g\E, Let [r, sJ be a bounded wehave lJ(f-1[r,s]nE) = lJ.{g -1 [ r, s J n E) and the rest is siITlilar to part (i). (3.5) EXAMPLES, 1. The following functions are equiITleasurable with f{x) = x on [0, 1 J and are ITleasure preserving ITlaps of [0, IJ- [0, IJ. (i) (iii) g 1 (xl = 1 -x ~ 0 s; x ~ 1 ------- i-x ~ x: k ={ t 0 g3 (x) ~ 2 -x ~ x ~ 1 I -11(i v) 1 ° I-a SX< a 1-a s x<1-Z 1- a s x s l z ° sa :-:;: 1 Z. Let f = C]_oo, O[-C]Z,oo[ and g = C]_oo, 1[-C J2 ,00[' f + =CJ-oo,O[~CJ-oo,I[=g while m(g -1 + - andf-- g - J butf.,..gsincern(f Then -1 (0»)=2 = l. (O» 3. Letf=C _ ,0[+2C[O,00[" m([O,00[}=m([1,00[) and f J 00 has the same constant value on [0, oo[ and [1,00[ but f1R - [0,00 [ ~ fIR - [1,00[. 4. Let 3' be the collection of all finite subsets F of [0, 1J and let 3' be directed by c. Then fCF}FE3' is a net which converges pointwise to C[O, IJ (since if tE [0, IJ then F CF(t) = 1) but C F ~ ° = ttl E 3' and F 0 c F implies ° for all F E 3' . For i = 1, ... , k let (X., fe., \,.L.) and 11·1 (3.6) PROPOSITION. (Y., 2:., v.) be finite m. s., let 111 Let f. EM{X.,\,.L.) and g. EM(Y.,v.) i=l,. •. , k -- 1 1 1 -- 1 1 and let F(x , ..• , X ) = (f (x L ... , fk(x » for all (xl"'" 1 k I 1 k G(y 1 , · · · , Yk ) = (gl(Yl),"" 1 gk(Yk» for all (Y I ,·'" 1 x ) EX and k Yk) E Y . .!!.. fi '" gi = 1, ... , k then F ....., G. PROOF. rectangle s of R k Let ~k be the semi-algebra of all measurable and let iB k k be the a-algebra of all Borel subsets of R . If A is a set and ~ c ZA let S(~) denote the a-algebra generated by ~; -12- let C(~) denote the monotone class generated by ~; and let R(~) be the algebra generated by~. If f;; is a semi-algebra we recall that R(~) is the set of all finite pairwise disjount [pwd] unions of members of~. Let (J= .2t],' then If Bl X· •• X Bk E: U(F fB ESk:IJ,(F-I[Bl)=v(G-I[B])}. -1 (B1X ... X B »= \J(f -1 [B1JX .... Xf -I[ B ] ) k k k 1 = IJ, 1(f -Ie B 1 ] ) ..• , IJ,k Uk-1[ B k J ) = v 1 (gi 1[B 1])'" 'Vk(gk1rBk J) = v (G - 1 (B 1 x· •. x B k » 0 is clearly closed under pwd unions so R("i) C (J. so"k ctJ. easily seen to be a monotone class, so ~ (J is = SWk ) = S(R("k» = C(R("k» cO. (3.7) Fori == 1, ... ,k let COROLLARY. -- -- (X.,A.,u.) and (Y.,:S.,v.) 1 1 1 -- 1 1 be finite m. s., let f. E M(X., !J.), g. E M(Y., v.), a. E R, and n. -- integers. of If f. - 1 "J 1 1 11 1 1 1 -- 1 1 ? 0 g., i= 1, ... , k then for each of the following definitions 1 - F and G we have F ....... G. n k a. f. (a) F (x l' ... , x ) ~ 1 1 i (x.) 1 k 1= 1 k n. G(Yl""'Yk)=t: a. g.l(y.} 1= 1 1 1 1 =. k (b) F(x , ••. , x ) = 1 k n Tf fi (xi) i.. i= 1 k n. G(Yl""'Yk)=lI gi1(y i ) i::: 1 k '2t PROOF. Use Props. (3.6) and (3.3) (xv) with ['D(t , · · · , t ) = k 1 n. k n. a ti 1 and cp(t , ..• , t ) = .". t. 1 . i 1 k III 1 -13- REMARKS. (3.3) (ii), (1) For finite m. s. (3.3) (xv) (iii), & (iv) by lettingcp(t) = rt, cp(t)::: t+r, cp(t)::: It!, and cr(t) ::: max fO. tl. rp(t) ::: - min (2) gives another proof of [0, t} . For finite Ill. s. if f "-'a f for some number a .s. t. I f I::: 0 or +00. as n ""00. Ia I =F 1, then For we have by induction that If I "" lalnlf\"" 0 or +00 -14II. 4. DECREASING REARRANGEMENTS The Right Inverse of a Decreasi.~Function.. If p is a decreasing function defined on an inte rva1 J of R, we can extend p to a de c reas i ng, function defined on R# by defining [or t ~ J, p(t) = {+ac -oc t ~ inf J t;;;:: sup J if if If P is a decreasing function defined on R#, then its right continuous inve rs e p. is defined by p' (t) = inf (u E R: p(u) :5: t} for each t E R*, where by inf 0 we mean + oc and inf R ::: -ac. • 1 : 1 then p = p If p is -1 It is easy to see that p' is decreasing and right continuous, and for every t E R*, p'(p(t)) st, p'(p(t+)) st, p'((p(t)+€)-) s t, p'(p(t)- 8) ~ t whenever e > 0, and p'(p(t)-) ~ t. Furthermore, for all t E R#, p'(t) = suptu E R:p(u) > t} = sup [u:p(u-) > t} and p'(t-) ::: inf(u E R:p(u) < t} = inf(u:p(u-) < t} . (4,1) PROPOSITION. 1i. p is a decreasing function on R~ then for every t E R we have p' '(t) oc (p')'(t) ::: p(t+). PROOF. Let u E R. inf[t:p' (t) s u} :O::p(u+). Since p'(p(u+)) $ u we have p"(u) = Since p' is decreasing and right continuous we have p'(p"(u)) = p'(inf[t:p'(t) s u}) ::: sup (p'(t):p'(t) s u} s u. Since t .... p(t+) is also decreasing and right continuous we have similarly -15,Ji p(p·(t)+) = suptp(r+):p(r) $ t] :<;: t for all t ER. Hence p(u+) ~;p(p"(p"·(u) )+)$ p··(u). Let 0 < a ER and let p be a decreasing function (4.2) PROPOSITION. defined QQ_ [0, a]. Then for eve ry t E R, p·(t) = m((uE [O,a]:p(u) > t}) = d (t) P where m is Lebesgue measure. PROOF. Now p·(t) = suptu:p(u) > t} so p(u) > t ~ u sp"{t) and thus fu E[O,a]:p{u) > tl c [O,p"{t)] Again, p·(t) = inf {u:p(u):S; t} p·(t) su, i. e., u < p"(t) =:> so p{u):s; t ~ p{u) > t, so JO, p. (t) [c t u E [ 0, a]: p (u) > t} . (4. 3) PROPOSITION. Let p, Pn be decreasing functions on Ri for n = 1, 2, 3, ... and suppose p of R#. n .... p at all but countably many points Then p ..... p. at every point of continuity of p •. n PROOF. Fix t E R. A = [u:p (u) > t} , Let An = {u:Pn (u) > t} , E = (u:p (u) 1+ p(u)} so that p. (t) = sup A , p·(t) = sup A and E is n n countable. An E C Then c lim inf A sup An E C n n C lim sup A n C tu:p (u) ;;:: t} so :s; sup (lim inf A ) :s; sup (lim sup A ) :s; sup (u:p (u) ;;:: t} n n Now the functions p n are decreasing so the sets A n have the forrn [-oc, r [ or [-oc, rJ and thus sup (lim inf A ) = lim inf (sup A ) = n n lim inf p. (t) and similarly sup (lim sup A ) = lim sup p • (t). n n n Since -16E is countable and A is an interval, sup A n E p. (t) $ lim. inf P~ (t) :s:; lim. sup P~ (t) $ p·(t-). t, thenlim.p·(t)=p·(t). n 5. C = sup A = p. (t). Thus if p. is continuous at Also see [47, p. 508, (18.21)]. Decreasing RearrangerrlCnts of Functions on Finitem.. s. f: [0, 1] -> [0, 1 J be Lebesgue measurable. Hence It is Let natural to wonder if the values of f can be rearranged to form. a decreasing function * f : [0, 1] -> [0, 1] such that f * "" £. The affirm.ative answer is well known [15]. We now generalize this idea for a finite m.easure space (X, A, 1-1 ) by showing that if f E M(X, IJ) then there is a decreasing right continuous Lebesgue m.easurable function of on [0, ~ (X)] s. t. of"'" f. For the rest of this section let (X, fl., IJ ) be a finite Ineasure space (m.. s. ). (5. 1) °< DEFINITION. l£. f E M(X, fJ) we define [, f £Y 6 f(t) = d i (t) if t So IJ (X ) . (5.2) THEOREM. (i) l!.. f E M(X, f-l) then of is a decreasing right continuous Lebesgue m.easurable function on [b, IJ(X) ] satisfying Of - £. (ii) Conversely, if p is a decreasing right continuous Lebesgue m.easurable function on [O,I-1(X)] satisfying p ~ f, then p = of PROOF. Now of is by definition decreasing, right continuous and Lebesgue measurable on [0, f..l (X)] and 6 f = di. Lemma (4. 2) ~ -17- = 0; and LeITlITla (4.1) =:) d;· = d£ so do do = d f and thus o£ "-'f. £ Conversely, if p is right continuous decreasing and Lebesgue f measurable on [0, I-l(X)] s. t. p"-' f. then d . P .. = p. H. I} ~ 6 f = d f f = d p = p. using (4.2) 80 == (5.3) PROPOSITION. a (i) where a (ii) £ is integrable if£ Of is integrable in which case Sfdl-l = = I-l(X). J Of o s(£) ...... 8(0£) for all s E S. (iii) fl ~ £2 =:) o£ $; o£ 1 2 (iv) 1£ each o£ £ and g is a ITleasurable function on a finite ITl. S. then - £ "" g -iff o£ = °g . (v) 1i..P is increasing on R# then 0p(f}(t) = P(of(t)-) 0 ~ t ~ I-l(X). (vi) ..!.!. r ~ 0 then orr = rof' while i£ r is real, then 0f+r = of+r. (vii)li £n -. £ a. e. then o£ -+ Of at every p_oint of continuity o£ n (viii)-.!i fn ..... 0 in ITleasure then o£ -+ or 0 uniforITlly on every closed n subinterval o£ JO, fJ (X)[. PROOF. (i) This is an iITlportant special case o£ results o£ § 3. SiITlilarly for (ii). ~ d£ 1 (i v) f '" g=:>o £ = d i °g. = g d =:) ° ° ° =:) £ = df· ~ d£· = £ . 2 1 1 2 2 Conversely, = 0 g =:> d f = 0'f = = d~ = g £ (iii) £1 ~ £2 =:) d £ ° £ '" g. (v) Since f'" o£ we have p(£) '" P{of} = P{of-) a. e. Since t -- P(of{t}-} is decreasing and right continuous, TheoreITl (5.2) (ii) iITlplies -18(vi) These are important special cases of (v) with p (t) = rt and p(t)=ttr. f (vii) n a. e. -- f ~ clfo -- d . at all but cOllntably rnany points 11 Lenllna (4.3) :;.:) of =- eli =: f lirn dfr~ at every point of continuit.y of elf· 0" be (viii) Let [u, vJ c JO, iJ.(X)[ and let € > O. Since u > 0.3 Nl > () j..L({!f l>e})<uwhenevern;:::N1sodf (e)=jJ({£ n n n v < j..L n;::: N2 ~j..L (£If n s. t. >s})<u and thus Of (u) s; of (df (e)) = d/(d f (e)) s; ewhenever n ;:::N . 1 n n n n n 3N2> 0 s.t. 150 Since v < j..L(X) ! > s/2}) < j..L(X) - v so ((! f n ! ~ e /2 }) ~ j..L (Un > - e:}) = d f n (- e) = d i n. (- e) sob fn (v) = d in (v);::: - €: whenever n ;::: N . 2 Since bf is decreasing on [u, v] we have Ibfn(t) I ~ e n for all t E [u, vJ whenever n;::: NI t N . Z Recall that if f E M(X, j..L), then ess. sup f = inf{t: \J({£>t}) = OJ. Writing E C = X -E if E E 1\ and f I E (5.4) PROPOSITION. C for f restricted to E l!.. f E M(X, u) and 0 s; t s;j..L(X), C we have: then c 0f(t) = inf (ess.sup (fIE ): IJ(E) ~t} c 0f(t-) = inf (ess. sup (fIE ): j..L(E) < t} where ess. sup fl 0 = -00. PROOF. Let t E [0, IJ(X)] . Ifu ER:/I:andE u = (£>u} so that df(u) = j..L(E )' then df(u) s; t ~ j..L(Eu) $ t and ess. sup (fIE~) $ u and thus u c inf {ess. sup (fIE ): j..L(E) S; t} s; bf(t). Conversely, if j..L(E) ~ t, let u = ess. sup (fl E C ) so that {£ > u}CE, c df(u) ~ t, and thus bf(t) s; inf (ess. sup (fIE ): IJ(E) st} . -19(5.5) THEOREM. (i) If f E M(X, p) and a = \J(X} then (Of)+ = 0f+ ~~ (of)- = -o_C so that of = 0 f+ + o_C (ii) 6 -f o :-:;: t ~ a. (t) = -0 «a-t)-) f PROOF. (i) of ~ f ~ (O[) + '" f + and (o[) - '" f-. decreasing and right continuous, Ui[)+ = 0f+' Since(o[) + is (o[)- -..f- ~ -(o[)- -- -[- and -(of)- is decreasing and right continuous, so -(o[)- = o_f- and thus (o[)- = -a_C. (ii) f ..... o[ ~ -f,..,. -of'" a where a(t) = -of(a-t)-) using Lemma (3. Z) Since a is decreasing and right continuous, a = (iii). ° -f' From now on we will use 0E to denote 0c ' where E is a E measurable set. Statement (i) of the following lemma seem s to have been first used systematically by F. Riesz ([35], p. 164). n (i) 1L f =i~~l fi C E M(X, u} and 0 < f 1 < ... < f , n n n then of = [I OF +.~ (f. -f. l)OF- where F - = U _ Ek 1 1= 2 1 11 1 k= 1 (5.6) LEMMA. Ei n (ii) If f = ~ f· CEo E M(X, u) with f. > 0 and E 1 ~ ••• ~ E , then i= 1 1 1 -- 1 -n n Of = 1= '~l f.1 0E·1 . PR00F. (i) Let F n+1 = ¢. S :? [ Now n fi ~ s < f i +1 s < f1 i= l, ... ,n-1 -20- n so of = 121 fi CCU(Fi+1LU(F i )[ n n L: f. [) = ~ f. [) i= 1 1 F i i= 1 1 F i+ 1 n = f1 of 1 + 1·r'Z (f. -f. _l}oF . I I i i U Ek and g. = '0 f k , k= i +1 1 k= 1 n (ii) Let G. = E. 1 1 go = O. n Then f = L: g. C and 0 < g 1 < ... < g i= I I G i n so since n F. ::: 1 U G == E. and g. - g. 1 = f. k k==i 1 1 11 E: 0 5: f E M(X, IJ) ~nd EEl\. ~ 0fC PROPOSITION. (5.7) PROOF. Fir st let f = C F 5: of 0E E where F E 1\.. C [ 0, IJ(EnF) r" 5: C [-0, min tU(E),\.L(F) r } [ :: °°. E F n Let f = i=L:If.IC Fi, Fn e · · · C F l' f. > 0, so 1 L I ° fi 0EnF. ~ fi 0E OF. ::: 0E of' If ~ f E M(X, IJ) then :: E 1 1 there is a sequence 0 ~ fn f of simple functions, so of f and 0fC t of C nE t 0fe n and hence 0fe E REMARKS. E t° = lim of C , .c: lim of 0E = ofoE' nE n (1) It follows immediately from Prop. (5. 4) that o£(o) = esse supf and 0f(a-) ::: esse inf f when a = IJ,(X). (2) If £ (x) = xn on [0, IJ then f n n -0 f = 0 in measure but of (0) = esse sup £n = 1 so of (0) f 0f(O) = O. n n -21- (3) We cannot prove Prop. (5. 3)(vii) for nets. Direct the finite subsets of [0, 1J by inclusion and let fE = C E [O,lJ. = CE(t):= 1 so the net fEE} Then t E [0, 1J and ft} c: E;:::) fE(t) = 0 for every E since converges pointwise everywhere to 1 but of fE if E is a finite subset of E = 0 a. l.'. (4) Ifg= 16fl. theno g = 0lfl sincef",of;:::) 1[1-10 [1. (5) Let p be decreasing and right continuous on R#, [ E M(X, U) and a = U(X). f...., -o_f so p(f) "" p(-o -f) which is decreasing and right continuous on [0, aJ. so for 0 ~ t ~ a, 0p(f)(t) = p(of«a-t)-». (6) Let 0': X .... [0. U(X)] be m. p and let a(t) = U(X)-t 0 ~ t ~ IJ(X). Then 0' .-v a and a is decreasing and nght contlllUous so 0O'(t) = U(X)-t. If F E M[O, a] and F 1(t) = F(at) on [0, 1 J then (7) Let a > O. of (t) = 0F(at) on [0, 1]. This follows immedlately from (3.2). 1 EXAMPLES. 1. Let [(x) = x on [a, bJ. Slllce o(x) = b-x IS a m. p. map of [0, b-aJ .... [a, bJ. f a (J.-v [ 0.3) (v). But fa O'(t) =: b-t is decreasing and right continuous on [0, b-aJ so 0f(t) = b-t on [0, b-aJ. 2. Suppose p is increasing on R, a> 0, and f(x) = p(ax) on [a, b]. 3. °( ° b-aJ. p ax )(t) = p(o ax (t)-) = p(a x (t)-) = p«ob-at)-) on ro, ax ab-at Let g(x) = e , a> 0, on [a, bJ. Then 0 (t) = e on Then 0f(t) = g [0, b-aJ using example 2. 4. 011 Let g(x) = log ax, a > 0, on [a, b], a > O. () (t) = log(ab-at) g [0, b-aJ. 5. Suppose p is increasing on R, Q' < 0, and f(x) = p(Q'x) on la, bJ. Since a(x) = x+a is a m. p. map of [0, b-aJ .... [a, bJ, f a a'" f. Thus t .... p(Q'(t +a)-) is decreasing and right continuous and equimeasurable -22- = p ((at + aa) - ). 'Ni th f sao f (t) 6. Let f(x) = sinx on [0, Tr/2}. Then 0f(t) = cost on [0, Tr/2], because cosx::: sin(x+Tr/2) on [0, Tr/2] and thus they are cquimeasurable. 7. Let l' be decreasing on [0, al and extend F to R by periodicity. Let f(x) ::: F(x) on [0, na] where n> ° is an integer. on [0, na]. To prove this, let G(t) = F(..!.+) on [0, naJ. and lei [r, sJ be n Then m(G -1 [r, s J) = m(nF -1 [r, sJ n [0, naJ) a bounded interval of RI. = nm(F -I[ r, s] n [ 0, a]) = m(f -1 [r, s]). Thus G", f and since Gis decreasing and right continuous, G::: or 8. Let f(x) == sinx on [0, Tr]. Then 0f(t) = cos t/2 on [0, Tr]. For if we let g(x) = cosx on [0, Tr/2[ and g(x) == cos(x-Tr/2) on [rr/2, Tr] then g\ [0, Tr/2[ '" fl [0, Tr/2[ and g\ [Tr/2, rr] == f\ [Tr/2., Tr] so g ",f by (3. 3) (x) and thus 0f(t) ::: c5 g (t) == cos t/2 on [0, Tr] using example 7. 9. Let f(x, y) = x-ty on [0, IJ X [0, 1] with product Lebesgue measure. Then t ~ a 1 tt i a~t ~1 o ~. (2 _t)2 1 $ t $ 2 Z $ u $ 1 o t 1 - 1 by inverting 1 - 2"t 10. 2 2 1, ~ $ U $ 1 2 and z(2-t) 2 Let g(x, y) == (I-x)+ (I-y) on [0, 1] X [0, 1]. Then with f as in example 9, 0g = of because g'" f by (3.6). If g(x, y) = Jx+y ° = J6: using (5.3)(v). 11. f g 12. on [0,1] X [0,1] with f as in example 9, The Hilbert Transform of a function f on R is defined by -23- T(x) ::: lim 1 'IT 8-1 0+ S f(t) t - x dt It-xl>€: It is well known that if E has finite Lebesgue measure, then I d Ie" (t) ::: 2 m (E) / sinh 'ITt so L 0l?-! I...~E 1 I(U);::;.L sinh- (2m(E)/It). [46 'IT J At this point it is natural to wonder if, given a right continuous decreasing function Q on [0, ~(X)J, there is an f E M(X, ~) s. t. (5. 7) PROPOSITION. 6 ::: Q f • Every right continuous decreasing function on [0, j..J.(X)] is the decreasing rearrangement of a measurable function on (X, /\,\...1.) iff there is a m.p. mapa: X -+ [O,\...I.(X)]). PROOF. Let 0: X -+ [0, ~(X)J be m.p. and let g be a decreasing right continuous function on [0, ~(X)J. goo'" g so 6 g 0 a = g. Conversely, 6 a Then goa E M(X, \...I.) and let a E M(X, \...I.) s. t 6 (t) ::: j..J.(X) - t,O $t ~\...I. (Xl. a Then and hence a is m. p. Recall that A E A is said to be an atom of (X, /I., IJ) (sometimes an atom of IJ) iff IJ( A) > 0 and B C A, B E /I.=- ~B) = 0 or ~(B) = IJ,(A). If A is an atom of (X, A, IJ.) and f E M(X, U) then ess. sup(fIA) ess. inf(f\A), i. e., f is essentially constant on A. to be non-atomic if it has no atoms. ::: (X, /I., \J.) is said Any Borel subset of R with Lebesgue measure is known to be non-atomic. The following is a fundamental result about non-atomic measure spaces. -24- (5.8) LEMMA. (A. Liapounoff, 1940) If (X, /I., IJ.) is a non-atomic finite m. s. then tlJ.(E): E E /I.} = [0, Il(X)]. For a proof see N. Dinculeanu [5.p. 25J. In order to state Liapounoff's result in another way VIe rnake the following definition. (5. 9) DEFINITION. A mapping r/> : [0, 1 J -0/1. is said to be a Il-resolution of A E /I. if it has the following three properties. (i) r/> (t) C A for all (ii) ° ° t ~ ~ 1. ~ tl ~ t2 implies r/> (t ) C r/>(t )· 1 (iii) p( r/>(t)) = tlJ,(A) for all ° 2 :5: t ~ l. Observe that a IJ.-resolution r/> of a set A E /I. is continuous in the sense that 1J.(~(u) - r/> (t» = (u-t)IJ.(A) - °as u ~ t or as t t u. This is equivalent to saying that r/> is continuous as a mapping of [0, 1] into the ,tnetric space associated with(X, /1.,1,,1). (5. 10) THEOREM. (Also see [26J.) The following four statements are equivalent for the finite m. s. (X, /I., iJ.). (i) (X, /I., IJ) is non-atomic. (ii) There is a IJ.-resolution of X. (iii) There is a measure preserving map of X onto [0, IJ(X)] . (iv) Every right continuous decreasing function on [0, IJ.{X) ] is the decreasing rearrangement of a measurable function on (X, i'. , ;..I.). -Z5PROOF. (i) ~ (ii). Let a = u(X) and let Xo E X. Since (X, A,~) is non-atomic, lJ{{x }) = O. Let i> (0) = [x } and i> (I) = X. If cf; (m/Zn) O O . n m-I m is defmed for each ~ m ~ Z so that ¢ (--) c i> (--) and ° Zn 2;n Zm-l ~li>(-») = - - a, then there are sets i> (-ri-=ry) E A s. t. Zn Zn 2 m m ~ (m-l ) c i> (2m.-l ) c i> (.222:.-) and ~(i>(2m-l Zn Zn+l Zn Zn+ 1 ) = 2m-1 Zn+ 1 a, 0 ~ m -< Zn. ° This defines ¢J on A = £ n~ : n ~ 0, ~ m <' Zn} e [0, 1 J so that for all Z U, v fA we have rI> (u) C rI> (v) when u < v and U(rI>(u» = au. If t E [0, 1 J -A we define i> (t) :-:: n {i>(u): u ~ t}, which also holds for tEA. I-l(i>(t)) = in£[~(i>(u): u ~ t} = t a and if tl < t z there is a v ~B s. t. tl < v < t z and hence rI>(i.) = n(rI>(u): u ~tl} crl>(v) en frl>(u): u ~ t z} = rI>(tz ). Thus i> is a ~ -resolution of X. (ii) ~ (iii). Let i> be a f.l - resolution of X. cr(x) = a· inf {t: x E rI>(t)}. Define Then (x EX: cr(x) > s} = U{rI>{t)c: t>s/a} so dots) = IJ fa> s} = lim IJ (i>(t)c) = lim(a-ta) as t ~ s/a J _oc , O[ + (a-s) C[O, a] so 0a(t) = a-t, 0 ~t sa, and hence a is m.p. =aC Any m. p. nlap is cs sentially onto. (iii) :::) (iv) ~ (iv) (i) This is (5. 7). Let £ E M(X, IJ) be s. t. o£(t) := a-t, 0 ~ t ~a. Then £ is not constant on any subset of X of positive measure so (X, A, IJ) has no atoms. EXA..l.v1PLES. 1. rI> (t) = (0, at] is an m-resolution of [0, a] with Lebesgue measure m. Z. If a: X -+ [O,a}is m.p. then rI>(t) = a-l[O,at] is a \J.-resolution 1 of X, and for all x E X we have a(x) = inf(t: x E a- [0, at]}. -263. cr(x) = b-x and T(X) = x-a are measure preserving n1aps of La, b] onto [0, b-aJ. The following theorem, proved by J. V. Ryff [40J for M[a, 1], shows how each measurable function on a non-atomic m. s. can be related to its decreasing rearrangement. (5. 11) p(X) LEMMA. First a lemma. 1£ the finite m. s. (X, 1\., \.l) is non-atomic and = b-a, then there is am. p. map of X onto Ca, bJ and also one of X onto [a, b[ Let cp: X ... [O,b-a] be rn. p. and let ",(t) = t+a, t E R. PROOF. Then 0 = '" 0 cp: X .... [a, bJ is m. p. and i£ p(t) = 01 = p o t 0 cp: X ..... [a, be. (5.12) Any m. p. map is essentially onto. THEOREM (J. V. Ryff). If the finite m. s. (X, 1\., \.l) is non- atomic and f E M(X, \J) then there is am. p. \..J. - PROOF. cr: X ..... [0, \.l(X)] s.t. a. e. Following Ryff [40, p. 96] let It and At = fx EX: f{x) = t} for each t E R~. £1 = £ u-a. e. {! ~; ~a, be then s. t. = f s E [0, u (X) J: 0 f(s )=t} Since £ ....... of' there is an £1 and o£ have the same range B. Denote fl by £. Also m(I ) = 1••L(A ) for each extended real t, and each It has the form t t [a, bJ or [a, br since Of is decreasing and rt-ctn. t E B there is a m.p. crt: A Then clearly £ = o£ 0 0. many t. onto t --7 It' Hence for each Define crby cr{x) = crt(x) if x EAt' Now IJ,(X) < oc ~ m(I ) > 0 for at most countably t Let F be the set o£ all such t and let I = U tEF It' which is -27measurable since its complement is, and on which of is 1: 1. Let U JnItUJnI so tEF -1 ,,-1 -1 -1 U(o (J)) = LJ 1J(0 (JnI )) + 1J(0 (J nI)). But 1J(0 (J nIt)) = t tEF JC[O,IJ(X)] be measurable. ThenJ= u(ot-1(JnIt)) = m(JflI ) and of is 1:1 on Jnr, so f = of 0 0 ~6f-l of = 0 t there, and hence 1-1(0- m(J n I). Hence 0 REMARKS. 1 (J nI)) = u(f- l = m(of- 1 (Of(J nI)))= (o[(J nI))) is m. p. (1) In general we cannot always find a measure preserving map if>: [0, a] ... X such that f a f/J = or For example, let f: [0, 1] ... [0, 1] be m. p., suppose f/J : [0, 1] ... [0, 1] is measure prese rving s. t. £ a f/J = 0 f' and let 0: [ 0, 1] ... [0, 1] be m. p. such that Of a 0 = f. Then 0f(t) = I-t = fa f/J (t) = 1 - o(f/J(t)) so o(f/J(t)) = t for all t E [0, IJ and hence t1 =+= t z ~ o(f/J(t 1 )) =+= 0(f/J(t 2 )) ~ f/J (t 1 ) 4= f/J(t 2 )· Since f/J is m. p. it is essentially onto and it follows that f/J -1 = 0, so Also o[ 0 0 = f implies f = 1 - 0 so f is a is an invertible m. p. map. nec('ssarHy 1:1. Thus if f is not 1:1, say [(x) =:: 2x mod 1, then there is no Ill. p. f/J: [0,1] .... [0, 1 J such that [ 0 f/J (2 ) (see §20). of" =:: Let (X., fl.., \J..) be finite m. s. with a. = IJ.(X.), i=1, ... , k, 1 1 1 1 1 1 let (X, fI.,lJ) be their product m. s., and let J = [0, a ] x·· ·x[O, akJ 1 with product Lebesgue measure. If o.:X. -+ [0, a.J are m.p. i=l, ... ) k, 1 1 1 and o(x , ••• , x ) = (0 (Xl ), ••• Ok (~)) then 0: X ... J is m. p. 1 k 1 To prove this take £. = o. and g. (t) = t for all t E [0, a.] in Prop. (3.6). 1 1 1 (3) Observe that Prop. in (3. 7) 1 (3. 6) ~ (f , ... , f ) .-..-(of ' ... , o£ ) so as l k 1 k at the end of § 3 we may conclude such things as \f(x) - g(y)1 ,~Io£(u) - 0g(v)\. -28(4) If £ E M[O, 1J, and t E [0, 1J let cp(t) = m(f£ > £(t)}) + m([£ = £(t)} n [0, tJ). Ryff l44] has shown that cp: [0, 1J -[0, 1J is m.p. and £ = o£ 0 cp. (5) Actually Liapounoff proved a Dluch stronger result in 1940 than the one stated above: The range o£ a countab1y additive finite measure taking values in R n is compact, and if the rneasure space is non-atomic, then the range is convex as well [22 J. In 1947 Paul R. Halmo s gave a simplified proof of this result [l1J. In Lemma 7 he shows that a non-atomic m. s. (X, A, ~), with U taking values in Rn, is convex, i. e., for every E E A there is a function c/> : E ... r 0, 1 [ s. t. hl ([ c/> < s}) = s IJ (E) for every s E [0, 1 J. We may define a one- 9imensional vector valued Lebesgue measure A on the line segment [0, U(E)[ = fs U(E): 0 ~ s < I} joining the zero vector and the vector U(E) as follows. Let m be Lebesgue measure on [0, I [. If Be [0, l[ we write B U(E) to denote [tU(E): t E B}, and if B is Lebesgue measurable we define A(BU(E)) = m(B)u(E). measure defined on the a-algebra S = [B U(E): B c [0, I [ is Lebesgue measurable} o£ subsets of [0, IJ(E)[. EE A and let c/> be as above. Thus A is a vector valued Now let (X, A, IJ) be convex, let If we define a(x) = c/> (x) IJ(E) then a is a m. p. map of (X, A, IJ) onto ([0, U(E)[ , s, A) . -296. Decreasing Rearrangem.ents on Not Necessarily Finite m.. s. Let (X, Ii, \.1) be a m.easure space (m.. s. ) and let m. denote Lebesgue m.easure. If f E M(X, IJ) and there is an interval Ie Rand a decreasing function 6 E M(I, rn} such that f ....., 6, ~ decreasing rearrangem.ent ~ f. then we will call 6 In this section we will character- 1ze those functions which have a decreasing rearrangement 1n this sense. (6.1) DEFINITION. We denote by D(X, \..I.) the set of all f E M(X, \..I.) which satisfy (i) fJ.(f (ii) \.l (f -1 -1 whenever [a,bJc] ess.inff, ess.supf[; [a,bJ)<oc ] c, e s s. sup f [) < oc whenever lJ(f -1 (es s. sup f)) > 0 and ess.inff<c<ess.sup f. 1 (iii) \..I.(f- ] ess. inf f, c[) < oc I!SS. inf whenever lJ(f -1 (ess. inf f)) > 0 and f < c < .:!ss. sup f. It is our purpose to show that D(X, \..J) is precisely the set of all those functions which have decreasing rearrangem.ents. If I is an interval of Rand [) E M(I, m.) is m.onotonic, Assum.e 6 is decreasing; the proof when 6 is increas- (6.2) LEMMA. then 0 E D (I, m.). PROOF. ing is sim.ilar. Let J = ] ess. inf 6, ess. sup 6 (i) Let [a, bJ c J. t. so JC]inf 8, sup 8 [. Then b < sup 6 so there is au EI such 1 that b<O(u)andhence6- [a,b]c]u,oc[. Sinceinf 6<athereisavEI such -30t hat 0 (v) < a s 0 0 m(o -1 -1 [a, b ] e ] -0(';, V [ • (ii) Let c E J. [a,b])<oc. 6(v) < c < o(w). If m(6 -1 Hence o-l[a, b] e ]u, vr: so Then there are w, v E I such that (ess. sup 6)) > 0 then there is atE I such that oCt) = ess. sup 6, so o-l]c, ess. sup o[eCt, vJanclthus m(6- 1]c,ess.sup o[)<oc. The proof when m(o-l(ess. inf 6)) > 0 is similar. (6. 3) LEMMA. (i).:!i. f E D(X, U) and fIE M(X l' U 1) ~hcn f 1 ~J f im2li~~ fIE D (X l' 1J. 1 ). (ii) li f E D(X, \.1) then f + r E D(X, 10-1,) for all r E R. PROOF. (ii) (i) use (3.3) (vi) Let f E D(X, \ooL) and r E R. and ess. sup (f+r) = r + ess. sup f. Then ess. inf (f+r) = r + ess. inf If Ca, bJe] ess.inf(f+r), ess.sup(f+r)[ then [a-r, b-rJ e]ess.inf f, ess. sup f[ so u((f+r)-l[a, bJ) U(f -1 [a-r. b-rJ ) < 'JC. f = The rest of the verifications are similar. Thus we see that (6.2) and (6.3) (i) imply that D(X, ~l) contains all measurable functions which have decreasing rearrangements. To prove the convers e we have to construct a decreasing rearrangement for each function in D(X, U). It is convenient to do this first for a special subset of D(X, U), which we now define. (6.4) DEFINITION. Dt (X, IJ) is the set of all functions f E D(X, U) which satisfy (j) ess. inf f ~ 0 < ess. sup f (ii) U(f-l (0)) > 0 if ess. inf f = O. -31If fED' (X, j,l) we define a distribution function by ={ _U(f- 1 JO, tJ) if df(t) ° t ~ ° 1 lJ(f- Jt, OJ) if t < for all t E Rif. Then d 0 (t) f = di(t) f is decreasing and we define l 1 for all tEl = J - p(C ] 0, oc ]), hl(f- [-oc, OJ)[ LEMMA. (6.5) Let £ E D' (X, 1).) and let 1= J -1J(£-lJO,ocJ), 1).(£-1 [-oc, OJ)[. (i) Idf(t) I < oc if ess. inf f < t < ess. sup f. (ii) d f is right continuous 1 (iii) di (t) = +oc iff t < -\..I. (f- J 0, uJ) for all °~ E R u (iv) di(t) ;= -oc iff t ~ IJ.(f-1 J -oc, OJ) . PROOF. If es s. inf f = (i) Since f E D(X, \..I.), this is clear when es s. inf f f O. ° 1 then l).(f- (ess. inf f)) > ° so \..I.(C 1 J 0, tJ) < oc whenever ess.inf £ = 0 < t < ess. sup f. (ii) Now d£ is clearly right continuous at all t < 0, and also at t ~ 0 if ldf(u) Id£(u) I < oc for som.e u > t. I =oc for all u > t. (i) im.plies ess. sup £ ~ u. U(f -1 For such u we have u Hence \.1(£ ° >° Hence let t ;;;: -1 JO, tJ) ;::> ess. in£ £, so = hl(£ -1 JO, ess. sup £]) = JO, uJ). (iii) di (t) := +oc iff fu: d£(u) > t} ::) [0, +oc [ and di(t) = -oc iff {u: df(u) ~ t}::) J -oc, o[ . and suppose -32 - One wouln hope that d£ = d g on R* inlplics f ~ g. example shows this is not the case. X if O:O:;;X:OS::+OC g (x) = { +oc if - 1 :0:;; x < 0 Then d f = d g on R 11= but m(f -1 The following Let £(x) == x on Xl = [0, -tocJa.n<ll<:~t on X (+ oc» 1:- m(g -1 2 (+oc = [-1, + oc J. ». We will be able to prove Of = 0g iff f,...... g, however, by using (6.5) (iii) & (iv) and the following result. Let f E Dr(x ,!-1.) and g E Dr(x , ~2)' 2 1 (6.6) LEMMA. If d£ = d g on R then PI (£-I Ca , bJ) = ~2(g-1[a, bJ) for all intervals [a, bJ of R. PROOF. This is equivalent to proving that £1£-l[R] ""g\g-l[R] so we may assume (and we do) that £ and g are essentially finite. Let d f =d g on R. Then (6.5) (i) says that !df(t) \ < oc and Idg(t) I < ce whenever min {ess. inf f, ess. inf g} < t < max tess. sup £, ess. sup g}, inwhichcase~2(g 1-11 (£-1 ]t, oc [). -1 J-oc,t])=l-ll(f -1 J-oc,t])and~(g -1 [. ]t,oc) = Then es s. in£ £ < t < es s. sup £ implies \J (g -1 Jt, oc J) > 0 2 and I-I.Z(g-l ]-oc, tJ) > 0 so ess. sup g >- ess. sup f and ess. inf g :OS::ess.inf £. The argument is syn1metric in £ and g so we conclude that they have the san1e ess. inf, say u, and the same ess. sup, say v. Now Ca, b J c]u, vr implies \J 1 (£-1 [a, bJ) & \JZ (g -1 [a, bJ) < oc in which case loll (£ -1 [a, b ] ) = df(a-)-d£(b) = dg(a-)-dg(b) = ~Z(g -1 [ a, b J). l [a, bJ c ] -oc, u[ U ]v, +oc [ then IJ,1 (C Ca, b]) = Now f, g E Dr implies u:os:: °< v. ° = I-lZ(g-l [a, b]). 1 If !oJ 1 (f- JO, vD < ce then I-lr (f -1 (v» = df(v-)-df(v) = dg(V-) -dg(v) = biz(g -1 (v». £, g E Dr, 1-1.1 (£-l(v» = 0 = IoL Z (g-l(v». Also if Otherwise, since For the rest, if ~1 (f-1Ju, OJ) < ce, -33- then since 1J.1(f-1(-,;o» = 0 == ~~2(g-1(-oo), we ha.ve 1J.1(f-1(u»). ~1 (C 1 ] -.x., uJ) = 1J. 1 (f -1 ] - 'Y), OJ) - IJ 1 (f - 1 J u, 0:1) = iJ 2 (g-1] -~, OJ) -IJZ(g-l]u, OJ) == bl Z (g-l]-X1, u]) Otherwise, u < 0 and ~l (£ -1 (u» = 0 == blZ(g -1 (u). Let J be an interval of R4F such that inf J (6.7) LEMMA. ~O ~ sup J. ~.! pE: DI(J, m) it> decreasingand pet) ~O iff t;;?; 0, then p' = d PROOF. p ')n R#. From the definition of p' and the condition on p it is easy to see that we have: if t < 0, then [O,pO(t)[cJnp 1]t, OJ C [O,pO(t)] 'i.f t ~ 0, then ]p (t), O[cJnpl]O, tJ c [p (t), 0[, 0 0 Hence pO = d . P (6.8) DEFINITION. , For each fED (X, bl) define b f = 0 if fED' (X, ~) while if f fJ. D (X, \.l) define 'hess. inI f + es s. supf) iI less. inI b f == -1 + eHS. sup 1 + ess. inI f (6.9) LEMMA . esse inI £ < b f I II &1 ess. sup fl <oc Iess.sup fl <00& less.inffl =00 iI 1es s. inf f 1 < 00 & 1es s .sup II = +00 if .!!. f E D(X, IJ) is not essentially constant then < esse sup I and. £ - b f E D' (X, IJ,) . -34- For each £ E D (X, ~) define (6. 10) DEFINITION. if f is not essentially constant if f if essentially constant on I = (6.11) THEOREM. Let f E D(X,~) and let I = ]-~(f -1 Jbf' ~ ]), \.l(f -I (i) r -<X:, b f ][ ) . Then inf I ~ 0 ~ sup I, 6 f E D(I, m.) is decreasing and right 0f'~ f, ,S'ontinuous, and for each tEl, 0f(t) ~ b iff t:? O. f (ii) Suppose J is an interval of R such that inf J ~ 0 ~ sup J, !,uppose p E D(J, m.) is decreasing and right continuous, p,....., f, and for each t E J, P (t) ~ b PROOF. (i) f iff t ~ O. Then I c J and p = 6 f on 1. The result is clearly true if f is essentially constant since m.(I) = U(X). If the result is true for fED' (so b f = 0) and fED is not ess entially constant, then 0 (t) = 0f-bP) -t b ':;: b iff 0f_b (t).:;: 0 iff t ~ 0 for all t E ]u, '\I [ where f f f -1 -1 f u=-~«(f-bf) ]O,oc])=-u,(f ]bf'oc]) and v=\-1«f-b ) f -1 [-ac,OJ)=~(f -1 [-oc,b ] f ) so I = Ju, v (; and f - b f '" 0 f _b f Hence suppose f E D' . If 0 .:;: t E R"", then df(O) = 0.:;: t so di(t) = inffu: df(u) ~ t} :3uppose 0> t E R#. clf(u) For all 0 < u < ess. sup f, df(u) is finite. t 0 as u ~ 0 so there is a di (t) = sup fu: clf(u) > t} ::: U U o > O. o > 0 s. t. t < df(u )' O Then .:;: O. Hence -35- = <I;. Let q Then d = q' := di':: ci f on R. q (Lemma (6.5)):) Ill(o~l [a,b])=rrl(qlr-1[a, b]) for all bounded intervals [a, b] of R. ° f -1 ( - oc ) = [1.1 (f -1 m(oi -1 ] - oC, OJ), 1.1 (f (-oc)) = O. If U(f -1 -1 = m(q-l [a, b]) =\..d{l[a,b]) Now (6.5) implies [ - oc, 0 J ) [ sou (f . (-oc)) > 0 then U(f -1 -1 m(of (-oc)) = l,.L(f (-oc)). But \d[(t)1 <00:) tEl Similarly, ifU(f -1 -1 -1 ( - oc )) = 0 ~ J-oc, OJ) < oc and thus (+oc)) = 0, then I -1 ]O,oc]) asu-oc so(6.5)~m(of-1 (+oc)) = 0. -I.l(f -1] O,u ] ).-I.l(f If U(f -1 (+oc)) > 0, then f..l(f -1 -1-1 JO, oc [) < oc so m(of (+oc)) = I.l(f (+oc )). (ii) Again this is clearly true if p is essentially constant since If the result is true for all then f is) and p '" f implies m(J) = U(X). p ED' (J, m) (so b then p - b f - b f f f = 0), and p E D(J, m) is not essentially constant, E D' (J, m) satisfies all the conditions in (ii) for E D' (X, 1.1) so I e J and p - b Hence suppose fED'. d f = d p m(Jnp m (J n p = p. -1 -1 -1 [ -oc, = lJ(f °J) = IJ. (f -1 f Also, because p "i, ]O,oc]) and [ -oc, 0]), and ]inf J,O [eJ np-1]0,oc] c [inf J, O[ [0, sup J [ c J inf J = -U(f -1 np -1 [ - oc, ]O,ocJ) and i. e. p = 0f_b + b = of on 1. f f = 0f_b Since p ...... f , = p' so di = p" JO,oc]) f °J c [0, sup J J sup J and hence Ie J, and on I, p and so = U(f -1 [-oc, OJ) = di = of . -36- (6.12) THEOREM. _Suppose f E D(x, ~ and ess. in£ £ < 0 < ess. sup f. Let I = J-~(f-l]O, +oc]), a = iJ.(f-l (0» < oe iJ.(f-1 [-0(, OJ) [ and we hav{' on I (i) (ii) (Of)-(t) = o£_((a-t)-) tEl o_f(t) = -of((a-t)-) tEa - I PROOF. F(t) = (of) ((a-t)-) Let Now f,..., of so f- ~ (of)- '" F. of -1 Then (0) c [0, aJ so Now a We prove that 0f- = F on a-I. = m(oi-1(O», &\ (6. II) (i) implies F(t) ~ 0 iff 0f((a-t)-) = 0 F = 0f- on J = J - IJ.((f-) -1 iff t;? 0 and hence 1 JO, 0( J), IJ.((f-)- [-oe, OJ) [= a - I . The rest is siInilar and easier. Vie now show how to obtain an analog of Theorem (5. 10) for general m. s. Observe that unlike the situation for finite m. s., if A is an atom and f E M(X, \J.), it need not be the case that f is constant on A, or even on a subset of A of positive measure. happen, of course, if ~(A) = + oc. This can only The situation is nicer if fED (X, \.J). (6.13) LEMMA. ~ [x E A: f(x) f:. If A is an atom of (X, fl., IJ.) and f es s. inf f, es s. sup fn subset of A of positive measure. = O. f E D(X, iJ.) then Hence f is constant on a -37- LetJ==]ess.inff, ess.supf[. PROOF. IfIJ(A)<octhe n essentially constant on A, so assume hJ.(A) == +oc. f is Since fED , 1 -1[ .- 1[ ]. )nA)<. ·l cr:: lJ(f - I[r,s])<ocwhenever [r,s ] cJso~«(£A) r,sJ)=IJ(t r.s IJ(A) and thuslJ(flA -I rr , sJ) = 0 for all Cr. sJ cJ, since A is an a tom. Since J is a countable union of closed intervals, Utx E A: f(x) ~ tess. inf f, esse sup f}} = lJ(flA -1 (J)) = O. 0< IJ(A) 5; lJ,(f -1 (ess. inf f)) + U(f (6. 14) LEMMA . S. -1 (ess. sup f)) shows the rest. ..!! (X, /I., IJ) is a m. S. and I is an open interval of R t. \J(X) = m(l) and [Xi }i~l is a partition of X by measurable sets s. t. IJ.(X.) < oc 1 i = 1,2, 3, ... , then there are pwd open intervals t}. ,b [} ~ 1 S. t. IJ (X ) = b -a n n 11- - n n n oc m(I - nl.J: Jan' bnU = 0 . I n = 1, 2, 3, ... , and Suppose first that I is bounded below, say a = infI > -oc. 1 n-l oc Let a = a + 6 IJ(X . ) and b = a + iJ. (X). Then U ] a ,b [ = l n i=l 1 n n n n=l n n PROOF. I - fa : n = 1,2,3, ... }. Similarly if I is bounded above. Hence n oc let I = R. Then L U(X.) = \J(X) = m(I) = +oc so there is a partition i=I 1 [B } oc of the positive integers by finite sets Bk such that k k=l L: (IJ(X . ): i E B k } ;:::: 1 for each k = 1,2, 3, . .. . Let {n . } oc and {p o }OC 1 li=I li=l oc oc enulnerate U BZk and U B2k 1 respectively, let rO E R, and let k=I k=loc oc Xl =.ld X , X 2 = . U X , II = ]rO'oc [and 12 = I-oc, r O[ ' We get 1=1 1-1 n Pi i collections of intervals for II and Xl and for 12 and X the union of these collections works for X. 2 as before, and -38(6.15) Let (X, 1\., IJ.) be a m. s. and let I be an open inter- THEOREM. val of R with 0 E I. Then the following four statements are eguiva- lent. (i) There is an f E M(X, IJ.) _so t. t\(t) = -t for all tEL (ii) There is a measure preserving map of X onto_I. (iii) l!'. v E M(I, m) is decreasing and right continuous and for each tEl, v(t),;; b iff t :2 0, then there is an fED (X, IJ.) s. t. v- of = v on I. (iv) (X, 1\., IJ.) is non-atomic and a-finite and ~X) = m(I). PROOF J is (i) ~ (ii) Let 0= -£. Then -0 = f'" of so 0'" -0£. Then m. p. and has the same essential range as -of' namely 1. (ii) ~ (iii). f = v 0 a. Now ess.inf 0= inf I and ess.sup a=supI. Then f ....... v so b f Let = by' I lJ.(f-l] be oc]) =: m(v- ] by' oc]) ::: -inf I \1(f- 1 C-oc, b f ])::: m(v- 1 C-oc, b v ])::: sup 1. implies Hence Theorem (6. 11) ° £ ::: v on I. (iii) ~ (iv) Let f E M(X, IJ.) s.t. 0f(t)::: -t for all tEL Then 1 1 X = UU-I[i, i+l[: i is an integer} and \1(f- [i, i+l[)::: m(of- [i,i+l[) ~ 1 so X is a-finite. Since \.l(£-1 (r)) = m(of- 1 (r)) = 0 for all r E R, £ is not constant on any set of positive measure, so X has no atoms. m(I) = 1 m(of -1(R))= \1(f- (R))::: U(X). oc (iv) ~ (i) Let [Xi }i=1 be a partition of X by sets of finite measure. Since X is non -atomic, so is each X,. ) j )airwise b. -a. disJ'oint intervals []a" b. [}~ I = \.l(X.). I I I 1 1 I:::. Since m (1) = \1 (X) the re are s. t. m(I -i~l ]a , b [) ::: 0, and i i -39Now c.l' (t) = -(t+a ) for 0 <.-. t <: b Let t(t) = t for all t E R. n n -(1 11 n is decreasing and right continuous on [0, \.I. (X )] and hence there is an n f n E M(X . IJ) s.t. n have f n ...... a n ...... v n f n -- - f E M(X, IJ) and fix "'"' a . n t1 [a n,bn [. Let f(x) = f n (x) if x EX. n = f n Now for v (t)=Q' (t-a )= -t (a :S:t < b ) we n Ii n n n n - tl [a ,b [so f - -t n n II. Then Then f ED'(X, \.I.) and Theorem (6.11) irnplies of = - t on 1. 1£ (X, A, \.I.) is non-atornic and a-finite, if f E D(X, \.I.).. (6. 16) THEOREM. 0-nd jf 0 E M(I, m) is a decreasing rearrangement o~ f, then there is a l1wasure preserving rnap 0: X - T such that (J 0 PROOF. Let J t 0=· f u- a. e. = is E I: o(s) = t} for each t ER#. Each J t is an interval, and since the topology of R has a countable base, we have rn(J ) > t ° for at rnost countably rnany t. The rest of the proof is like that of (5. 12). -40REMARKS. 1. If (X, I\,U) is a m.s. and 0 <,f (M(X,U) U({ f ::: ~}) < <L: for some P-, then we may c1l~fjne d.nd [fJ on rO, ';c [ the right continuous inverse of df(t) = U( {f > t}) [H] to IH' This is almost the same as the definition of T given by Hardy, Littlewood [15] and Polya for X an infinite interval of R. Although mn[fJ > S}) = IJnf > s}) for all ~ E R, in general [f] need not be equimeasurable with f. If fED (X, IJ) then a = iJ(r 1] bI' ex: ] ) < ex: and for all 0 ~ u < ex: we have [f](u) 2. = 0f(u-a). If (X , 1\, IJ) is a finite m. s. and f E M(X, IJ) then the decreasing rearrangenlent of f defined in this section is a translation of the decreasing rearrangement defined in § 5 for finite m. s. Letting f>:c denote the one defined in § 5, we have 0f(t) for all tEl = f*(t+a) = J-IJ(f-1]bI'oc]), U(£-l[-oc,b ])[ f where 3. Suppose (X, 1\, IJ) is a finite m. s. and X is the union of a finite number of atoms of equal measure, say X = Al U· ... U An. Then M (X, IJ) is isomorphic with R n unde r the correspondence f - a n = (aI' ... , a n ) E R where a. = riA . . I l For each f E M(X, \,1) let 6 f denote the dec reasing rear rangement defined in § 5. .. k then of is constant on [O'k_l' O'k [ where O'k If f E M(X, IJ) =3:1 Il(A~ k= 1, .... n, and I if a .-. f and a: = 6 f [O'k_l' O'k [ then a * is the point of R n whos e components are the components of ~ arranged in decreasing order. -41- 4. G. F. D. Duff [6J has defined a generally multiple valued decreasing rearrangement for each menlber of M(I, In) where I 1S an infinite interval of R and In is Lebesgue nleasure. The ('on!\itinns (6.1) on f E M(I. In) arc nl:cessary and sufficient that this decreasing rearrangement be single valued. -42- III. 7. INEQUALITIES AND REARRANGEMENTS A TheorelTI of Hardy. The following theorelTI of G. H. Hardy will be used often in what follows. (7.1) THEOREM. (Hardy) f: (i) Suppose fl' f2 EM[a, b], f l , £2 EL'[a,tJ, fl ~ jt f2 for all a S"t :O:;b. 1£ g is a non-negative a decreasing function on [a, b] s. t. fig EL' [a, t] (i = 1,2) for all a :-:; t < b, for all a :-:; t < b and l then ftflg:-:;; f g for all a:-:;; t:-:;; b. a a 2 (ii) Suppose fI' f2 E L1[a, b], / fl ::s; jtf2 for all a ::s; t ::s; b, jb f1 =jbf2 , a a a a and g E M[a, b] s. t. fig EL1[a, b] (i=1, 2). li g is decreasing then Ibflg::s; jb f2g , while if g is increasing then jb flg ;;,: jb f2g . a a a PROOF. (i) Let F. (t) = 1 a It f. (i=I, 2) if a < t :-:;; b. a Then for all 1 jtf2g-flg = jtg d(FZ -FI ) = g(t) (F -F )(t) -l(F -F )dg;;,: O. 2 I 2 I a a a (ii) is silTIilar. a st :-:;;b, (7.2) .!!. f E L 1 [0, a] is decreasing, then; {tf ~ THEOREM . decreasing function of t, while if f is increas ing, then ~ ~t f is an increasing function of t, 0 < t :-:;; a. PROOF. Let 0 < tl < t2 :-:;; a. Then tIt :-:;; t2t for all 0:-:;; t :-:;; a so min [tIt, t t } :-:;;lTIin tt t, t t } and hence I 2 2 l 2 ~ c [0 t ] = ~ lTIin it, t 2 } :-:; ~ lTIin [t, t I } = j t / 02 '2 Z for all 0 :-:; t :-:; a. 1 t z t 10 z f! a = 0 1 a Thus (7. 1) (ii) ilTIplies J1 f if f is decreasing, 1 1 C f<ja_ C f= 1 t[0, t ] t [0 t ] tl a Z a 1 '1 z -43and the reverse inequality if f is increasing. 8. A P reorder Relation. In this and in subsequent sections we will of to denote the decreasing rearrangement defined in § 5, since use the nleasure spaces will always be finite. In addition we will use 0E to denote the decreasing rearrangement of CEo In [15] Hardy, Littlewood and P olya introduced for the first tilne an important preorder relation for n-tuples of real numbers and later for integrable functions on a finite interval. We present this relation for finite m. s. as follows. (8.1) DEFINITION. Suppose (Xl' 1\1' \.lol) and (X, l\,).,l) are finite m. s. with a = ~l(Xl) = ).,l1(X 1 ) and £, g EM(X,IJ,)UM(X1,\J.l) and a a 10 °£+ and. 10 0g+ are finite. t (i) g -< -< f (ii) g -< fmeans t fo 0 g -;:: fo 0 f for all 0 «; t ~ a means a g -<-< £ and a !o 0 g = J0 of' It is obvious that (i) f ""' g iff g -<-< f (ii) g -<-< f iff (iii) g -< f iff -5 g -< 0 £ and f -<-< g iff g -< f and f -< g; 0 g -<-< 0 f (iv) -<-< and -< are reflexive and transitive The reader should cOITlpare the conditions in Definition (8. 1) with the hypotheses of Theoren1 (7. 1). -44The following are some other useful but less obvious properties of these preorder relations. (8.2) LEMMA. (i) 1! g -< -< f then g+r -< -<f+ r for all r ER and rg O. -< -< rf if r·~ (ii) If g -< f then g+r -< f+r for all r E Rand rg -< rf for all 0 -:' r ( R. If}n addition f ELI then -g -< -f. (iii) l.!.. r sf$; s where r, s E R* and g -< f then r $; g $; s. (iv)1!fELl(Xl'~l) then (iJ,l(Xl IX fdhl-l)CX-<f. 1 1 1 (v) If f E L (X l' ~ 1 ) and E E A then f -< C E iff 0 $; f ~ 1 1 JX fd\J.l = \J.(E). 1 In particula~, if f -< C x then f = CX l and \J.l-a. e. (vi) If g + -< £+ and g --<£ then g -< f . (vii) If g -<-< f then g+-<-< £+. If g -< £ELI then in addition g- -<-< C. (viii}.!iO$;£ntf,O$;~1 g, and gn -< in' n=l, 2, 3, ... , then g -< f. PROOF. 1£ g -<-< f and r E R then (i) & (ii) t t t t JOt = I (6 tr) $; f (Of+ r ) = J Of o g r 0 g 0 0+ r If g -< f for all 0 $;t.,:: a then in addition we have equality when t = a. If g -< -< f then ft 0 o rg = ft r 0 0 u~ing (S.3)(vi). Suppose r $; ft r 5 = ft 0 f for all 0 $; t sa. g o fo r then we also have equality when t = a. Now assume g -< f. ~ O. If g -< f To prove the rest it suffices to show that -g -< -f. fto (u)du = / - 0 (a -u) du a-t a-t a-t 0 -g 0 g o = I 0 = J 6 - fa s: J Of - Ja o£ = ft_ o£(a-u)du = fto £(u)du ago gog 0 0 0 0for all 0 $; t s: a with equality when t = a using (5. 5) (ii). (iii) Let g -< £ and suppose first that £ ~ O. set of positive measure then there is a 0 < u < a If g is negative on a s. t. 0 (u) < O. g Then so la Of = lao < jU o s: jll Of $ fa of since g o o gog 0 0 o (t) < 0 for all u $; t s a -456 f ;::: 0, a contrd. f - r Hence g ;::: O. 0 and g-r <: f-r "? f <. s. Then -f f so Now suppos erE Rand r '" L g-r:?: 0, i. e. r s:O'> 0 and -gls .( -Us ;<:: so g. Tlw n Suppose s E R ,~tld g~: -gls:'" O,i. ('. s. (iv) ft 6 o =t g for all 0 ~ t ~ a by TheoreITl (7. 2), and we have equality when t = a. (v) Supposef<:C J CEdU=U(E). X 6 f ~ 1 so E , AssuITle ft 0 f ~ t, and o Then(iii)~O$f$l, and! fdlJ,l ° = Xl Sf sl and fXlf dj..tl = IJ,(E). Then ° S Of so oft Of S rOf = j..t(E). Hence 0 It Of $ min{t, W(E)} = ft 0E for all 0 st S a and we have equality o when t = a. 0 (vi) Now g- -< r- ~ -g- -< -f- t so [ t 0 t o g = =f 0 t f) t ++ J0 0 -g- sf0 of+ g £ o£ for all 0 $ t S a with equality when t = a, using (5. 5)(i) . (vii) AssuITle g EM(X,j..l) & fEM(X1,\.lr)' Let b=j..l([g~O})=ITl({Og~O}) a b b bi a 1 and b =j..tl({f;:::O}). Then f 0 +=/' {) ~f Of$/ 6 =! 6f +· If g -< f E L og og 00 f 0 l then -g -< -f so g - = (-g) + -<-< (-f) + = f . (viii) Use ITlonotone convergence and (5.3). (8,3) LEMMA. Suppose £, gEL 1 [0, a]. f and g are decreasing, ] r, s [ & ] u, v [ C [0, aJ = I, s -r = v-u, and IS f s IV g. l! g -< -< f -- r U then gII-Ju, v[ -<-<fII-Jr, s[ PROOF. The proof is essentially due to Ryff [43 , p. 433J . Let gl = gil - Ju, v[, fl = £\I-Jr, se and let h = s-r = ITlO st at countably many points of I, V-U. Except at -46- og I (x) dnd similarly for f. t We consider fonr cases. t t / °I = / = 0/ O£I g 5 / £ goo o t > m.ax {u, r}. Case 2. t u ° u<x<a-h t < min {ll, r}. Cas e 1. t o S:x ':;U g(x) { g(xlh) = t Now t+h u = / g + / g(x+h)dx = / g + / gI o! o o v u and sirnilarly for f. t t{ 0 g I - / g u t+h f I g <:; / 0 0 t s = / 6£1 -l / f r 0 s v Since Hence t+h v -I g I g - /f~0 u r t t ! 0 I5 o g / 0 o g we obtain s v I + u/ g - / f 5 r t / °fl 0 v r Case 3. t /0 g 1+ u/ s: / o o u /Oyl=/g so r u .:; t c:;: r. u t+h t- / g 5 0 v Thus g I -< -< fl t t+h gs/g+ /g o t t £ = /0£1 t- /£ Case 4. g u s 0 Since g is decreasing, v = / g + / g t t {) u. t v o t+h t 5 5 J° I .:; / 0fl o g 0 t t u t+h I g o t /g + f g = J g 0 lu g 0 as before. t $: I£ = o Then -47f, gEL 1 [0, a J, ] r, s [ & ] u, v [ c (g. 4) L EMMA. S upp 0 S c s-r = v-u, gil - ]u,v[ -<-< flI - Jr, s[. and rO. a. J : .; I, Then gC l _ u, v [taC] u, v [-<-<fC l _ r, s [taC] r, s [forallaER. J J PROOF. Letgl =g\l-]u,v[and fl=fII-]r,s[. sogl-<-<fl, let aE R, let gl :; g C l _ ]u, v [+ a C] u, v [, let f1 = f C1-]r ,s [+ a C] r,s [, and let] w, z[c 0 gl-l(a) & ]p. q[ C 0f-1 (a) s.t. z-w = q-p = h = s-r = v-u. 1 Then 0 g 1 ;,~ a on [0, w J, 6 g 1 ~ a on z. a -h ], and r except at most at countably many points of I, 0 (x) = gl 6 g I(x) o ~x ~ w a w<x~z 0gl (x-h) z<x~a t t and sirrlilarly for f. 10 The inequality $/O o gl exarrlining cases. 0 follows by f1 Rather than go through therrl all we give the details for three representative ones. so Case: w'; p ~ t t w If.> o gl ~ min w t =/f.> 1+/a g 0 Now a tq, z }. / ::: w 0 w t 0 o gl o g w t ! Of I + Ip a = Io 0 f 0 o £1 0 P t I 0f\ + 0 t = I = Now 0fl ~a on[p, w] so w $ t t 7 0 fl t/a ::: P t =I 0 1+ I a p I? +I a Case: p $w $ t $ rrlin tq, z} . 10 I a + w 0 /ofl+ro£1 o w t fa g 1+ t $ ~Ofl on [w, PJ 1 10' w r w t 0f\ +1 a + I a $ 0 p w -48 - Case: w = oI Z ~ . It.cu -_ I W u'- 1 t s g. o t-h ° I + O:'h + 1 g t-h t-h gl o t-h t I 0fl -1- / Of/ 0 t-h t ° = ! ° l+! If t:::;; p, I o t °1= / °1+/0:'. gog t-h w t I 1 glog t-h g s Q' = since Q' ~ 0fl on [t-h, tJ If p < t, then t -; q .c~ t-h sp &a<ofl t t-h / - JOg o 1 0 p 0(,(\+1 Q' Co t-h (8.5) LEMMA. and s -r = v-u. [t-h, PJ so p t IQ' P t-h t fa -1- on I 5', p 0£1 + 0 1 0fl + t-h Suppose£,g ELI [O,a], _. t I o£ o ]r,s[&]u,v[C[O,a]=I s g < f then (ii)!! PROOF. Jr, s [+Q'C] r, S [. I (i) Let £" g be as in the p roof o£ (8.4). a-h a v jOl=/g =/g Ig 1- Ju, v[ u 0 = s -r = v-u. where h we conclude that g (ii) a = Then s I£- I £ = o r Since we already have g I < <£ I from (8.3) I < £\ . Let gl and £1 be as in proof o£ (8.4). v a a I gl Ig J g +Q'h 0 U = u gl I-Ju, v[ < £1 1- ] r, s[ then for all Q' E R, J g r g! 1- ] u, v J < £ 1 1- ] r, s [ . gC I _ u, V [+Q'C] u, v C<£C 1 - o v [J. V. Ryff] If £ and g are decreasing and J£ = Jg and (i) o 1 a = Jf s /f+Q'h 0 r Then = a Ifl 0 -49LEMMA Suppose fELL (Xl' f1.), (8.6) J\, f 1S constant on E1 E IJ I (E )=IJ(E). 1 g is constant on E E 11., flEI ~ glE !!g-<-<f, (i) gIX-E -< -< r/X-E1 (ii) g C X _ E g E Ll (X, ~), and then and + 0' C E -< -< f C x + 0' C E 1 - El for all 0' E R. 1 --- Let (3 = flE , y= glE, a = IJ(X) = iJ (Xl) , l l 1 r = [0, a], ] r, s [ e 0 f- ((3) s. t. s - r = ~ (E 1)' and 1 s v ]u, v[e 0 - (y) s.t. v-u = IJ (E). Then s-r = v-u and! of ~ f 0 . g rug PROOF. If g -< -< f then 0g -< -< of so gl X-E '" 0glr-Ju, v[ -< -< 0flr-Jr, s[-f\X -E I and gC x -E + O'C E""' <> g for all 0' E R S _] u,v[+aC]u, vC-<-< of S-] r, s [+aC] r, s["'" fCXCE/O'CEI using Lermnas (3.3) (xi), (3.4) (i), (8. 7) LEMMA. £ and g have the same cons tant value on E 1 E 11.1 gIX-E -< f\X -E (ii) gC REMARK. 1 _ X E l g and E E II. g -< f then and + 0' C E -< £ C x -E + 0' C E 1 1 for all 0' E R. 1 Let g:f:; £ mean g -< £ and g and f are not equimeasurable, and define g<j;f similarly. f, g E L1 (X, IJ). and g (8.3), and (8.4). Suppose f ELI (X l' 1J ), gEL 1 (X, IJ) and suppos e 1 respectively, where iJ 1 (E 1) = IJ(E). (i) I Then Let (X, /I., IJ) be a non-atomic m. s. and let g:t f ~there is an h ELI (X, IJ) s. t. g;t, h:t f, l' f ~ there is an hELL (X, iJ.) s.t. g<j;h ~ f for by Theorem (S. 10) there is an h E M(X, iJ.) s. t. <>h = ~'[Of+Og] . -50We now paus e to show how a finite m. s. can be embedded in a non-atomic Ill. s. We will use this device to show which results about non-atomic m. s. carryover the general m. s. 9. Embedding of a Finite Ill. s. in a Non-AtoIllic Finite Ill. s. (X, il., 1oL) be a finite m. s. called a u-atom Let We recall that a Illeasurable set A E A is or an atom of (X, /I., I-l) whenever iJ.(A) > 0 and B C A implies U(B) = 0 or U(B) = I-l(A). If A is a IJ-atom and f E M(X, IJ.) then f is essentially constant on A. If A and Bare U - atoIlls, they are equal or disjoint a. e., i. e., \.l(A {:, B) = 0, where A {:, B == A-B U B -A, or U (A n B) == O. A finite (or 0" -finite) m. s. can have at most countably many atoms. Thus X = X number k U U A where P is tl, ... , k} for some natural n(P n is {I, 2, 3, ... } ; each A is alJ,-atom; and n or P o (X ' A n XO' IJ,) is non-atomic. O We may embed (X, /I., IJ.) in a non-atomic m. s. (X II , 1\"" , i-4II ) as Let X If: _ U lCa ,b ] where the l[a ,b ] are disjoint n n n n nEP intervals of R with end -points a & b s. t. b - a == IJ,(A). Then follows. - Xo U n n n (XJl., 1\ Jl., \.l#) is the direct SUIll of (X ' j\nX ' O O n n IJ) and the Lebesgue Ill. s. (I[ a ,b ], m), i. e., E E Aif iff E == EO U U En where EOE ii, EOCXO' n n -nEP and for each n E P, E n and in this case 1oL#(E) is a Lebesgue measurable subset of lCa ,b ], n n = IJ(E O) +:E m(E). nEP n Clearly (X~ f-lIJ.#) is a non-atomic finite m. s. with fJ*(X#) == IJ.{X). Each f E M(X, ii, U) is identified with {'If == f C in M(X# , ) C [ b] x"0 +n r;EP (fiAn ar n ' n il, '~/") and it is easy to see that f,...; # so Of = orlf. Of Illore -51lmportance is the fact that Ll (X, ~) is a retract of Ll (X#, ~#) under the following rn,apping. (9. 1) T \.J f:::£C (9.2) If f ELI (X~ fJ.#) U Mt (X~ ~#) let DEFINITION. "" +0 nEP b Xo THEOREM. n (i) 1-a Jba n (f)C A n n n TfJ.: Ll (Xl, fJ.#) -+ Ll (X, fJ.) . (ii) T~./ = f for all fELL (X, IJ,) UM+(X, IJ,). (iii) TjJ.£ 2 0 for all 0 $f E M(X, jJ.) . (iv) J TtJ,f dlJ = If djJ.:Ik for all f E(LI UM+) (X~ tJ,#) and \\T \Jf\1 1 $ \\f\1 1 for all f ELI (X#, \.l.{l:) • (v) 11 T I-l rll oc s 11 [II oc for all,- f E L oc (X~ ~#) (vi) .!:~r all f E (Ll U M't) (X#, ~:Ik) and g E M(X, jJ.) fg E (Ll UM+) (X#,fJ.Jt.) we have Th e reader s. t. TfJ.(fg)::: (TlJ.f)g who is faITliliar with the concept of a conditional expectation will recognize that TJ is the conditional expectation of f with respect to the a-ring generated by Xo n /.and the intervals l[a ,b n n J. n E P. At the end of § 13 we will prove that TlJ,f <f for all f' .:: '- (L 1 U M+) (X#, ... t,#), type in detail in § 17. W e WI'11'Inves t'19a t e t rans f orma t'Ions 0 f th'IS -52- 10. An Inequality of Hardy and Littlewood. In this section w e prove a n inequality which w as originally proved by Hardy and Littl e wood for non-n e gative fun ctions ([15J, Theor c n,378). This ine q u ality w ill b e fundamental for what is to follow. !! f, g E M(X, 1-1), a = 1-1 (X) <oc, (10.1) THEOREM (Hardy & Littlewood). a nd if 0lflOlgl ELI[O,aJ, then £gELl(X,I-1) and a a a I <\(a-t)o (t)dt = IOf(t)o (a-t)dt ~ I f g dlJ ~ I OfOg o gog PROOF. X First suppose f = C E 0 and g = C F when E , F E A . a + + Then I o£(t) 0 (a-t)dt = (1-1 (E) + I-1(F) - a) = (~(EUF)+~(EnF)-a) ~1-1(EnF)= o g a f fg d~ ~ min tl-1(E), I-1(F)} = I of 0g. o N e xt suppose f and g are non-negativ e simple functions. f and g f = l:: can b e written in the forn> n i=l f. > 0 fi CEo 1 1 i = 1, ... , n and E 1 :=> ••• :=> E n n> l:: g. > 0 g' C F . J j=l J J a Then I 0f(t)6 (a-t)dt = g = o J = 1, ... ,m and F 1 ~ ... :=> F n> a z= f. g. I 0E. (t) 0F.(a-t)dt g i, j 1 J 0 1 J ~ ~. fi gi I CE .C F . dU = / f g du. I, J 1 J a = /Of 0 using (5.6) (ii). o g Then -53If 0 ~- £, g E M(X, \.t) then the re are sequences {r } and n t non-negative sinlple functions such that fn f and gn a Ijnl a Then I o£(t) (a-t)dt = n-ac I o£ (t) g (a-t)dt g o 0 n n ° S lilll n .... ac f g·n } of t g everywhere. ° I £n gn d ~ = I £ g d U a °£n °g n = oI o£ °g o s n-ac lim / Letf, gEM(X,u)bearbitrary s.t. 01£1 0lg\EL Then by what we have already proved /1£ g\ d~ ~ ~a 0 1£ f g E r} (X IJ). , 1 [0,a]. lot g I so Then using Theorem (5.5) -+ -f++ /a 0f-(a-t) 0g_(t)dt s 1£ ++ g dIJ - ! f g dlJ, + / f g dlJ f g d~ o = ! f g d\J, REMARK. Obs crve that the inequalities in (10.1) are true £0 r all ° f, g E M(X,~) even i£ of 0g '1. LI[O, a] . <: (10.2) PROPOSITION. I..!. F and G are llleasurable functions on [0, a] such that O\F\ o\G1EL1[0, a] and G is decreasing, then the convolution F~<G(u) u = / F(t) G(u-t)dt o is a continuous function o£ u ~ [0, a J. -54 - PROOF. u E [0, a]. First we show that \Fl*IG\ (u) is finite for every Let u E [0, a] and let __ {G(U-t) on H(t} G(t) on [0, uJ ]u,a] Then Hlro, u] "-' GI[o, u] and HI ]u, a] "-' GI ]u, a] H "-' G so and thus IF \* IGI (u) I H I - I G \ so u u a ::; flF\ (t) lGI(u-t)cJt~ltF\ (t) \G \(u-t)dt+[IFI \ GI o 0 1.1 a a a : ; LIF I 1H I ~ fo 6 \ F 1°\ H \ = fo 01 F \ 6 IG rex:· Let 0 ~ u < U o :5: a, I F*G(u ) - F*G(u)\ ~ O so u u t. IF(t)G(uo-t)\dt + ~ I (G(u-t)-G(uo-t» F(t)\dt. 0 Let H(t) = G(uo-t) on [u, u ] and O 0 elsewhere. H-G C[O -uJ so (H\ ",IGI CeO _uJand thus ,u o ,uO u a a Then u -u o ~ ~oIF(t)IIG(uo-t)ldt = £ IFIIHI ~ fo 61Fl o IHI ~ foo For the rest, let un on [0, un] elsewhere. and 0 i U o n = 1,2,3, ... , and let Hn (t) = G(u -t)F(t) n elsewhere and H(t) Then H n 61Fl o 1Gr·0 l H pointwise a. e. a G form a set of measure 0, and! IHII o = G('1>-t)F(t) on [0, uJ and 0 since the discontinuities of = \FI*IGI (u l ) < ex: , so -55a U I n o I(G(u -t) -G(tta-t)) F (t) I cit n = 0I IH -H n 1- 0 as 11 .... at by Dominated convergence. The case 0 :s; U 11. o < u :s; a is similar. Let f, gEM (X, \.L). The Values of an Integral. If 0lf 1°1 g I E LI[ 0, a] If' where a = \.L(X) and f'...., [and g' -g then I,.., If I and Ig' I,..., Ig I and we a a know that Ilf'g'ldU ~ 0lf'IOlg'l = loOlflOlgl and a a 10 0r(t) 0g(a-t)dt ~ I fIg' d\.L ~ o Of 5 g . £ J (11. I) THEOREM. where a = \.L(X). PROOF. g Let f, g E M(X, \.L) and 0lf 10 \g I ELI [0, a] , (X, A, \.L) is non-atomic, then We already know that [I: [!f g' dU: g' "-' g} c 0f(t)og(a-t)dt, I: 0fOg]. so it rem.ains to show that all the values are taken on. and u y(u) with y (0) 0 ~u ~ a let a = Jo 0f(t)o g (u-t)dt + Iu 0fo g . a u for = Then y is a continuous function of a !o 0fo g and y(a). = f0 0f(t)o g (a-t)dt, so it suffices to prove that for each 0 ~ u ~ a there is a g' ""'g s.t. Fix u E [0, a]. s.t. IJ(X I ) = u, If g' dIJ= y(u). We prove that there are sets Xl EA and X = Of on 0flx 1 [O,u[ and 2 = X-Xl 0flx (t) = 0f(t+u)onO~t<a-u. 2 -56 To this end, if t ER#: let F = {x EX: f(xl:? t} = [x EX: f(xl > t} t and Now IJ(F _) = 0 ~ u so IJ(F._) = lim IJ(F). t s t s t t ~ so let to = inf[ t E R*: \J.(F ) .:;; u}. Then \.l(Ft ) < \.1, and o IJ(F ) > u, so \.l(Ft.) ~ u ~ IJ (F t _) Since (X, A, IJ.) is nons 0 0 t s < t o ~ atomic, there is an Xl E A s. t. Ft Nows<t o ~df(s)=IJ(F s » u so o c. Xl C Ft - and IJ(X ) = u. 1 o df(s)<u=::)s~ dflxl(S)=U{FsnXI)=IJ(FS)=df(sl. Xl CF to t 0 =::)F dflxl (5) < U ~ S ~ to =::) df(s) = dflX1 as before. we have fs: d f (5) .,; t} = t s: dflxl (s) .,; t} °.:; CFt 0 CX l ~ Conversely, s<to =::) - C Fs ~ dfl Xl (s) = IJ(F s n Xl) = U(X 1 ) = u Now let v(t) = 0f(t+u) on s t < a-u. so so Hence if 0 ~ t < u o/t) = 0flx (t). l Then for each bounded interval [1', sJ of R# we have I -I iJ.(f -1 [ r, s ] n Xl) = 1J«f Xl) -1 [ r, J s ) = m(0tlx ° = m( f -1 [r, sJ Cr, s J ) 1 n [0, u [). 1 Subtracting this equation from \.1(f- [r, s J n X) =rrl(o;l [r, s J n [0, a[ 1 1 we get lJ(f- [r, sJ n (X-Xl)) = m(of- [r, sJ n [u, a[ ), i. e., IJ( fIXz)-l[r, sJ) = m(v-l[r, sJ nco, a-u[)+ u) = m(v- l [r, sJ n [0, a-u[) ° .,; Hence 0f/ X (t) = 0f(t+u) on t < a-u. 2 Recall that if T: (X,!I., IJ) -+ (Y, l: ,'J ) is a measurable trans fo rmation and g E M(Y, L:) then / g(dlJ T -1) = fg oT dll in the sense that either both integrals exist and are equal or neither exist 1 [12, p. 163, Theorem C] where (IJT-I)(E) = IJ(T- (E)) for all EEL: If T is m.easure preserving, then iJ. T -1 =V . -57 Let (11: Xl .... [O,u[ I and 0Z:XZ .... [O,a-u[be m.p. s.t. (\lx o 01 ::: 1 I z() 02 :::: f IX Z· f X 1 and b f X Also 1et G 1 (t) ;..; 6 g (u-tl and ° on u u so G 1 ",-,6 g 1[0, . u[ o <:t<;a-usoGz-6gl[u,a[ letG (t)=6 (t+u) on Z g !o 0f(t)o g (u-t)dt = 10 Then ~,t <: u °I f Xl G 1 dm = ~ (0 f IX 1 0 0 1)( G 1 0 0 1) dlJ 1 Also a-u Jo a-u 0f(t+u)o (ttu)dt = g 10 6 I f X z G 2 theng'-o and y(u) (11. 2) ° ~ f, = dm g "'g 1 f gl dU. COROLLARY !! (X, 1\, fl) is non-atomic then for all g EM(X, U) we have max[ 1 f' g' dU: f' '"'" f, gl -g} ::: max [I f gl dlJ: g' '" g}::: a !o 0fo g , in the sense that both are infinite or are finite and equal. PROOF. As we observed, the inequalities of (10. 1) hold for all ° ~ f, g E M(X, btl even if 0fOg ~ L 1 [0, a]. To show that the sup is attained, let 0: X -- [0, a] be rneasure preserving s. t. of 0 0= f. a 1o Of 6 g = 1 (of 0 0) (0 g 0 o)dIJ = J f gl dlJ, where g' ::: 0 g 0 a '" g. Then -58Many investigato rs have us ed g -< f to mean ([25J for example). Ig I -<-< If I Hence the following theor e m is of i nterpsi. It will also b e v e ry useful. (11 .3 ) THEOREM. ~(X) that Let (XI ' 1\I,\JI) and (X,1\,IJ)befinitem.s . su c h 1 = ~ (Xl) = a, and let f EL (Xl' ~I) and g E M(X, IJ)· g -< f implies gEL 1 (X, IJ) and Then Ig I -<-< \f \ . 1 PROOF . (8. 2)(vii) ~ gE L. From § 9 it follows that (11. 2) implies t 10 0\ g \ = fig I C E dlJ* for some E E 1\# with !Jo#(E) = t = I g sgn(g) C E d',J# a ~ where 10 ° g °h h = sgn(g) C E a s 10 °foh sin c e = f f hI dki 1# for some h I E LI(Xt , !Jot) s.t. hl"'h s 1 lfllhll dI-Lt and = 1 \ f \ C F dlJ~ g -< f lhll,....., lhl =C E some F E1\ # s.t. 1 for all so lhll IJ'*(F) 1 = CF for = ~#(E) = t I o st ::;; a. If (X, 1\, ~ ) has atoms, then Theorem (II. 1) may be REMARKS. 1. false. = Xo U Al where Xo is non-atomic, Al is an atom, and Let X U (A I) > \.L (X 0) > O. Let f == C but gl '" g implies gl = C A 1 xo and g = CA' I Then loo/)g = \J(X O) since U(A ) > ~(XO)' I 1f g I dU = 1 C X 0 CAl dlJ = O. a so -59If (X, J\, ~) is a finite m. s. consisting only of atoms of equal 2. measure, then it is said to be discrete, and the set of values of the corresponding sums do not fill lip the whole interval, although the endpoinls are attained. (Sec [15]. Theorem .)()H). (11. 4) DEFINITION. for all 0 ~ f, g E M(X, \.l) we have A finite m. s. (X, fI, \.l) is called adequate if a !o 0fo g , a = IJ(X). max { / f g' dbl: g' '" g } = (11.5) THEOREM. The following are equivalents for the finite m. s. (X. A, U) (i) (X, l\, ~ is adequate. (ii) (X, fI, ~) is non-atomic or consists only of atoms of equal Ineasure. (iii) For all A, BEll. we have a sup [f C A C E dU: CE~CB} = fo 0A 0B PROOF. Now we know that (ii) ~ (i) ~ (iii). to prove (iii) :~ (ii). Suppose (ii) is not true. Hence it remains Then either X has at least two atoms A, B with 0 < !J(B) < ~(A), or X has an atom A and a non-atomic part Xo of positive measure, in which case (5.8) implies there is aBC Xo such that 0 < \.l(B) < 1-1 (A). for all E E A. with C E '" C B I-1(A n E) ~ \.l(E) = bl(B) < bl(A) for all E E A. with C E ,.." C B In either case, we have \.l(E) = !J(B) and hence so U(A nE) = O. Thus / C A C dlJ. = 0 E a but /0 0AoB = IJ.(B) > O. - 60lZ. The Decreasing Rearrangements of Sums and Products. f, gEL 1 (X, ~) then in general If °f+g t of + g' For example, let E 0 and F be disjoint sets of equal positive measure. Then We can, however, prove the following. (IZ. I) THEOREM. t t n t t ~) .) f. ! .~;. ~) ~\. (a -u )du <: fo 0 f ! . . . . 1 f J lIT I S n .2= fa °f. -- 1, .•. , n J =1 1 with equality on both sides when t = a. for_ all 0 S t <: a PROOF. 1 From § 9 it follows that Theorem (11.1) implies t t = max{ fE(fl+fZ)d~,l": E E /I."" & \..l"(E) = t} s fa 0f + fa ofZ 1 t and Ja t :? t f of + f of (a-u)du . lnax a a 1 Z The general case follows by induction. t t .!! f ELI (X,~) then fa 10fl ~; fa 0 If I fa r all o ~ t sa with equality when t = a. (1Z. Z) COROLLARY. t PROOF. t t t folofl = fa of+ t fa 0c using Theorems (5. 5) and (IZ. 1). Along with (ii) of (7. I), (a-u)du:;;; fa O£ttr We have equality when t = a (IZ.I) also yields the following. -61 (12. 3) THEOREM. !! (X, A, IJ) is a finite :m. s. with a = IJ, (X) and is a bounded decreasing function on [0, aJ. then the function g a 10 0f(u)g(u)du is sublinear on Ll (X, 1).). p(f) = I gEL (X, IJ,) then (12.4) LEMMA. l! f, I~ I~ 0 g I for all of - PROOF. t t <, <, For all ° so 10 I 0 f _g - (12.5) <t S:a . t t ~ t fo ag + f0 al f -g I and f0 t ° t !o ag + !0 0 f-a t ~ a, ° g t t '0 10 t b r 0 g- f ~ r of + !o °lof -.-g I <, t t I ~ 10 0 f - 10 &g <, 10 c. If _ g I . COROLLARY. and gn -<fn' n 1 I Suppose f, fn EL (Xl' fJ ) and g, gnEL (X, IJ) 1 = 1,2,3 . . . .. !! "fn -fill -> 0 and IIgn -gill -> 0 then g -< f . PROOF. \ J~ ofn - ~ of I fo °Ifn -£1 <, <, J: 0lf -fl = Ilfn -flh for n all 0 ~ t ~ a, and si:milarly for g. As in the case of su:ms, if f, gEL 1 (X, IJ) then in general, ofg f:, 0 fO g even if 0 <, f, g. For exa:mple, let E and F be disjoint sets of equal positive :measure and let f = <;:+2C F and g=2<;:+C fg = 2 C EUF ' and of = 0g = 2 C[O,U(E)[+ C[IJ(E), 2U(E)[ so OfOg = 4C CO , ~(E)C + CC\.l(E), 21J.{E) [ ~ 2C CO , 2~(E)C = Ofg' Recall, however, that if 0 <, f E M(X, \.I.) and E E A then 0fC <, E of 0E' More generally, we have the following. F . Then -62(12.6) t THEOREM. t (i)..!£ 0~f1'" ., f E M (X, IJ) then n - - loof .... f s;lo of ... of for all Os; t s; a, so in particular, 1 n a 1 n I fl'" fn dlJ, s; 10 Of l' .. of . n 1 (ii) !f..f 1,···, fn E M (X,IJ) and 0l f 0lfnl E L [O,a] 1 where a :;: iJ(X), then for all Os; t S; a we have 1'" t t t 10 lOf 1 ···fn I ~/oO'f 1 .. ·fn I s;loo lf 1 , ... O\f n \. PROOF. t I (i) There is an E E A>Il with IJ* (E) :;: t #: .. IE s. c. a of f :;: flf2 d IJ :;:_&(£1 CE)(£ZC E ) dlJ S; 10 of C of C a 1 2 X" 1 E ·2 E a t ~ f of ~ of 0E = f of of' The general case follow s by induction a 1 2 a 1 2 using (7. 1) (i). 13. T f < f and Some Consequences. II (13. 1) f (ii) follows from (i) and (12. 2). THEOREM. Let (X, A, IJ) be a finite m. s. Then for all ELl(X~,)..L#) UM+ (X*,iJ#)we have T f<f. IJ PROOF. Let E EA*. and T C E = C x n E + ~ \.l a nEP hence TIJ, C E < C E I TIJ C E dlJ = I C E dlJ+ = IJ+ (E) by (9.~) (iv) m(E n l[a ,b n n J) m{Ila n ,bn J) by (8.2) (v). Actually, for all f E L l(X*, iJ >Il) we have already that a :;: = = fa Of using (9. 2)(iv). Let f be a non-negative simple measurable function. n can be written in the form f = '6 i= 1 f. C 1 Ei Then f with f. > 0 i = 1, ... , n 1 -63For all 0 ~t ~a, and E1::::J .... ::::JEn' t r Jo n 0 '6 T f"; i= 1 !J f. 1 t n t fa 0T \..L C E. .,; i='61 f.1 fa 0E i t 10 = Then there is a sequence (f n J of non-negative simple functions s. t. f n t f. U so T f -< f. IJ. 1 Let a : :; f E L l(x#, IJ~) be arbitrary. of T Of' we see that a : :; T f t T f. \.tn \..L Since we have T from (8. 2)(viii) that T f -< f. \.t If f E L1(X~, \..L JI:) then TUf+ -< t+ and Tu,f- -< fall a ~ t :5: a From the definition f -< f it follows un n so -TIJ C -< -C. For we then have + so T f -< f • U (13.2) THEOREM. Let (X,fI.\J) and (X ,fl. ,1-4 )befinitem.s. such 1 1 1 that a = IJ,(X) = 1J1(X ), and let f E M(X,IJ) and g E M(X1,1J,1)' 1 (i) ~ 0lfl 0lgl E LIea, a] then J. f!f g' dlJ,: g' E M(X,IJ) & g' -< g} = [ / a 0f(u)o (a-u)du, fa Of 0 ago g a (ii) If a ~ f, g then max{ f f g' dlJ: g' -< g} = I of 6 in the -0 g sense that both are infinite or are finite anc:I equal. PROOF. g a E M(X*, \J (i) According to Theorem (5. 10) there is a '*) such that 6 go = °g . Let I be the interval in question. Then I::::J U f gl dlJ: g' -< g}::::J [/ f TUg'dw,: g'", g, g'E M(X~w,#)} = £ffg'du.#:g'",g} = I using (7.1), (13.1), (9.2) and (11.1). a (ii) Follows from (11. 2) in a manner similar to (i). -64EXAMPLE. sup[ !0 1 x n If n ~ 0 is an integer and 1° ff 0 -;; f ~: A and a ~ f ~A and 1 f C 1 I" -<'[O'IJ' sup t! o find 1 f(x)dx: 10 f(x)dx = 1 } . 1 Now t.. ~ 1, 1 xn f{x}dx: . 10 usmg f -. 1 a iff ( 8 2) ( ) f -< C [0 using (13.2) and (5.3) (v). . I I --f < 1 and A ~} ~f A 1 A Thus the sup is the saIne as v. 1.. ] } = ! , A ~. 0 1 I 6 xn 6, C I\. I [0, I ] 2 n (I-x) dx, o = A! -65 IV. REARRANGEMENT INVARIANT BANACH FUNCTION SPACES studies havl' been nlactc of Banach function spaces ane! Illor(~ genl·rally narrned Riesz spaces (see [27Jand [29J). In thjs chapter we review saIne rearrangeInent invariant properties of such spaces when the underlying Ineasure space is finite. LuxeInburg in [28J. The treatInent closely parallels RearrangeInent invariant Banach spaces will be the setting in which we will work in future sections. (14. l) DEFINITION. Let (X, 1\, ~) be a finite In. s. and let M +=M + (X,~) denote the sct of all non-negative extended real valued Ineasurable ~ I.! ) . _f unClons on (X ,n, t· A Inapplng . p: M+ R# IS . ca 11 e d a f t 'Ion norIn unc -0' if it has the following two properties. (i) o ~ p(£) ~ +oe for all f E M + and p(f) = 0 iff f = 0 1-4 - a. e. (ii) p(£+g) s p(f) + p(g) for all f, g EM+ p(af) = ap(f) f s g ~ forallfEM + and p(f) ~p(g) a~ O. + for all f, gEM. In addition, p is said to have the sequential Fatou property and is called a Fatou norIn if it also satisfies. (iii) 0 $£ n t fpointwise everywhere iInplies p(f n ) jp(£). We extend the dOInain of definition of a function norIn p by defining p(f) = p(/fl) for all f E M(X, U), and we denote by LP=LP(X, u,) the set of all f E M(X, IJ) such that p(f) < oe. If U-a. e equal IneInbers -66 of L P are identified as usual, then L P is a normed linear space with the nornl Ilfll :: IIfll p = p(f). These spaces arc clearly generaliz.ations of the classical Lebesgue and Orlicz. space';. Unfortunately the hypotheses we have placed on p so far do not preclude the existence of a p-purely infinite set, i. e., a set which p(C B ) :: +oc for all Be A such that U(B) > O. f E M(X, U) and A B.3]. oc hi, - a. e. Fortunately it can be shown that X has a maximal p-purely infinite set X infinite and X -X for Of course, if is p-purely infinite, then f C A :: 0 [29, NoteIV, p. 251, Lemma 8.2]' A oc , i. e., X oc is p-purely has no p -purely infinite subsets [Note IV, TheoreITl We will aSSUITle in this and in subsequent sections that X has oc been renlOved frOITl X so that our ITl. s. has no p-purely infinite subsets. In this case p is said to be saturated. Given a function norITl p we define the first and second associates of p as follows. For all f E M (14.2) DEFINITION. + :: sup £ ! Ifg IdlJ: P II (f) :: sup [ ! Ifg Idu.: pi (g) 51} pi (f) p(g) 51} It is not hard to show that pi and pI! have the sequential Fatou property [Note IV, p. 254, Theorem 9.2]' and that p is saturated implies pi is a norITl [Note IV, TheoreITl 9. 7J. to show that in addition, 11. 6J. It is true but harder p! is saturated [Note IV. p. 261, Corollary Hence p" is also a saturated Fatou norITl. The spaces L and L P " are called the first and second associate spaces of L P pI -67 respectively. We have the following lIc)lder type inequality [NoteIV, p. 2£>1, Corollary 11. 7J. (14.3) THEOREM. I If g dU! .0; If f E L P and (14.4) then I \fg\dIJ::::; p"(f)pl(g)::::; p{f)p'(g) . Note that p" ::::; p. shown. g E LP' In the other direction the following has been [Note IV, p. 259, Lemma 11. 3]. THEOREM. (Lorentz and Luxemburg) p" = p if and only if p has the sequential Fatou property. The following converse of the Holder inequality also holds. [Note V, p. 499, Corollary 14. 2J. (14.5) THEOREM. completeand Suppose L P is (Lorentz and Wertheim.). gEM(X,U). ThengEL P' iff IlfgldhJ.<oo forall f E L P. It has been shown that L P is complete iff P has the following property called the Riesz-Fischer Property: P(L If n I) < 00. L p(fn ) < 00 implies [Note I, p. 143, Theorem 4. 8J. is complete whenever p is Fatou. Thus LP' and LP" are complete. In particular, L P [Note II, p. 149, Theorem 5. 3]. -68:> If L P = L I , 1 ~ P <; oc, then L p' = L q where p -1 I q - 1 :::. In general L P' is a closed normal subspace of the Banach Dual (L P )':< , = (L P )* iff p(f ) l ° whenever f l O [30, p. 153]. We have L P everywhere. G E (LPt is in L P' n iff G(f ) l n n ° whenever f n ~ 0, Since the m. s. is finite we will, in accordance with the Lebesgue and Orlicz space situation, assume in this and in subsequent sections that L oc C L P, L P' c L 1. L·quivalent to p(C LocCLPCL L oc C L p' (14.6) l c L iff 1 X ) < oc and p'(C p(CX)<oc and X ) < oc. p'(C This is easily seen to be If P is Fatou, then x ) <,.x:, in which case also. DEFINITION. A function norm invariant (r. i. ) if f1 --f 2 implies p is called rearrangement p(f ) = p(f ). 1 2 L p is called rearrangement invariant if f1 ..... £2 E L P irnp1ies f 1 E L P. If P is rearrangement invariant then so is L P , but the converse 1 IS not true. For example, let p(f) = L P is r. i. since g"-' f E L P implies 1 2 2 3 J~ 0 f ~ 3 fo 6 3 fo If \ < oc . 1I If 1= 2 Jo lf(t)ldtt2 f1If(t)ldt if fEM[0,2]. 121 fo Ig I + 2 fl 1 Ig I ~ fo O\g \ + 2 fo O\g I = However p(C[O, 1J) = 1 and P (C [ 1, 2]) = 2 even though C [0, 1] '" C [1, 2 J . The following lemma is fundamental to what follows. -69(14.7) LEMMA. atomic) and Suppose (X, 11., fJ) is adequate (i. e. discrete or 110n- = U(X). is a Fatou norm and let a p Then L P is re---- arrangement invariant iff 0fOg E LI[O, a] for all 0 ~f EL P and o ~ gELPI, in which case L P' and L P " are rearrangement invariant iff L P is. PROOF. Suppose L P is rearrangement invariant. and 0 ~ g E LP' Hence If f' '" f then Let 0 ~ f E L P ! f' g dlJ:::; p(f') pI (g) < ex:.. f' E L P so 0/) gEL 1 [0, a] since \J is adequate. Suppose 0fOg E Ll [0, a] for all 0 :::; f EL P and 0 ~ g E L If a o ~f' "-' f E L P, ! f' g dw. ~ fo 0llg < oc so f' EL P " P' = L P using (13.4) and (13.5) since pis Fatou. The rest follows by replacing P by P I and L p" by L p. (14.8) (i) THEOREM. Suppose (X, 11., w.) is adequate. If P is a Fatou norm and if L P is rearrangement invariant then 0 ~ g -<-< f E L P implies g EL P and hence g -<f E L P implies g EL P. (ii) Y L P is rearrangement invariant, then 0 ~ g -<-< f E L P' jJl1plies g E L P' and g -< f E L P' implies g ELP'. (i) Suppose 0 ~g -<-<f EL P . Then for all 0 ~ hELP' a we have! g h dlJ :::; fo 0gOh s 0foh < oc using (10.1) and (i) of (7.1), PROOF. r: so gEL P" = L P so If g -< f E L PeL 1 then (11. 3) implie s Ig I -<-< If 1 EL Ig I E L P. (ii) Follows from (i) since pI and P" are Fatou norms, and If (r. i. ) implies LP' and L P " are (r. i. ). P - 70- A functi~n_ norrn. p is called universally re- (14. 9) DEFINITION. arrangement invaria_nt (u. r. i. ) if fl ",f ~ 0 i.mplies p(Tp [I) ~ p(O for a.1I E M(X-IJ, IJ.I). [I L P is called universally rearrangement invariant (u. r . i.) if fl"", £ E L P implies TIJ £1 E L P for all fl E M(X~ I.l). If (X, /\, U) is adequate (i. e. dis crete or non-atomic) then we will see that p is (u. r. i.) iff it is (r. i. ) and L P is (u. r. i. ) iff it is (r. i.). (14.10) Our previous results generalize as follows. THEOREM. o s f EL P and (i) If L P is (u. r. i. ) then Of 0 - -- 0 s g EL P ' where a = IJ.(X). Y then the converse holds, and in that case L g ELI [0, a] for all P is a Fatou norm, pip" and L . are (u. r. 1. ). iff L P is (u. r. i.) (ii) Y p is a Fatou norm and L P iUu. r. i. ) then 0 s g « f E LP implies g E L P and g -< f E L P implies g E L P. (iii) Y LPis (u. r. i. ) then 0 sg <-<f EL I g -< f E L P implies gEL PROOF. (i) pi . Similarly for PI L implies g EL P" PI and . I For all fl "" f, £1 EM(X , IJ#)' we have ! fl g dlJ.* = ! (T fA. fl)g dlJ :5: p(T V fl)pl(g) < DC, . (X df. ,~ #). ' Slnce ' 1S non-a t omlC. Jf Suppose P is Fatou and let 0 s fl ....., f E L P, £1 E M(X#, v ). a I /(TI-/')g dIJ = J fl g cl s OfOg < ex for all O:5:g E L P so J TIJf l EL P " L P " by L P = L P. Then 10 The rest follows by replacing L P by L P I and -71 (ii) and (iii) Follow as in (14.8). We now investigate rearrangement invariant (and (u. r. i)) norlns. (14. 11) Suppos e (X, 1\., ~ ) is adequate and p is a re- THEOREM. arrangement invariant norm and let a = bJ, (X). Then for all o sf E M(X, IJ) we have p I (f) a = sup t J p(g) s of °lg 1 : 0 a p ll(f) = sup [ J of 0 1} 01 g j : p'(g)sl} so pI and pl! are rearrangement invariant. g p is a Fatou norm, then In addition, a p (f) for all = I I: sup { fo 6 fO g pI (g) s I} o 5 f E M (X, IJ ) • PROOF. Since (X, 1\., U) is adequate, for each g with p(g) :::;; 1 there is a gl ~J g such that JoaOfOlgj=Jf!g'ld\J.. p(gl) = p(g) 5 1 since a p is (r. i.) f J0 0 fO 1g !: p (g) s I} C gl~g implies so t J fIg I d},l: p (g) :::;; I}. a But f fig j dW 5 fo of 0lg ISO a pl(f) = sup{J flgjdlJ: p(g) <:: 1] -. sup{ fo 0lllgl: p(g) 51 ] We get the formula for p l! by replacing p by pl. p = pl!, giving the last formula. If P is Fatou, then -72- The following is a kind of converse which will be useful later. (14.1Z) Let (X, 11., ~) and (X', 1\', IJ.') be finite m. s. with LEMMA. IJ.(X) = U'(X') = a. y A C M(X', U') with rCx, EA for some r :J: 0, then the mapping 0 $ [ ..... p (f) defined on M + (X, ~) EY a p (f) = sup { fa 0[0\ g I: g E A } is a Fatou norm which is universally rearrangement in variant. f = 0 U - a. e. implies p(f) = 0 is clear. PROOF. a we take g = rCX' and obtain fo of = 0, since f ~ o. If u ~ 0, then 0uf so If p(f) = 0 of and hence f = 0 a. e. = u of so p(uf) = u p(f). If 0 $ fl $ fZ then of $ of so p(f ) $ p(fZ)· To prove the triangle inequality we 1 1 Z t t have from (1Z. 1) that f of +f $ (of + of ) for all 0 $t $a, so o 1 2 0 1 Z 1 if p(f ) & p(fZ) < ~ 1 a then (7.1) implies that for each gEA we have a a f 0 6 [ 1 +f Z I) Ig I ~ fa 6 flo Ig I + fo 0 f Z6 Ig I a Since 0 $ fn a t f everywhere implies fa Of 01 g I t 10 0/) Ig I for each n g EA, we have 5. 4J. p(f ) n t p(f) as in [Z9, Note II, p.149, Theorem Hence p is a Fatou norm. To prove tha1 p is (u. r. i.), let o $ f E M(X, U) and let fl '" f where fl E M(X#-, ~#). a a I I solo 0 T f' 0 g \J p(T\.,/') $ p(f). 10 of 0lgl for each Then 0 ~ T g E A and hence IJ f' -<-< f -7.)- (14.13) THEOREM. Suppose (X, /I., u.) is adequate and rearrangement invariant norm. (i) 0 ~ g « f implies (ii) g < f E L Whether p If p p(g} ~ p(f). \X, u.} implies p(g) ~ p(f}. is Fatou or not, p' and p" satisfy (i) and (ii). Theorem (14.13) may not For example, suppose X = Xo U A where A Xo is llon-atornic, 0 < u.(X ) < u.(A), and u,(A) > 1. O p(h) is a is al so Fatou, then If (X, A, u,) is not adequate, then Iw trut'. p = IX \h\du. + \heAI. o and let f = C A and g = C is an atom, Let Pick E: > 0 such thatO<l-e:<\J.(XO)~(l-E:)',J,(A), x o + E: C AO Then p is a Fatou rearrange- ment invariant norm and 0 ~ g -<-< f, but p(f) = 1 < \J(X ) + 8 = p(g). O For general m. s. our previous results take the following form. (14. [4) THEOR EM. p i s a univer sally rearrang(~ment invariant If no/·m, then the following four statements are true. (i) For all 0 ~ f E M(X, IJ) we have a p'(f) = sup f fo Of 0lgl: p(g) ~ I} p"(f} = suPt fo Of 0lgl: p'(g} ~ I}, and a a = \J(X). If p a p(f) for all is in addition Fatou, then = supt fo 0fOlg,:pl(g)~ I} 0 ~ f E M(X, U) where -74(ii) p' and pIt are universally rearrangement invariant. (iii) 0 ~ g -<-< f implies similarly for p p'(g) ~ p'(f) and p and if p is Fatou. (iv) g -< f E L l(X, IJ} implie~ p'(g) ~ p'(f) similarly for p"(g} ~ p"(f), ~ p and p"(g) <:;; p"(f), and is Fatou. (i) For each g with p(g) ~ 1 there is a g' E M(xi, \J.*) a suchthatg'''''gand 10 0fOlgl = Iflg'ld\J.* = IfT~lg'ldIJ.. PROOF. gf,.., g implies Ig'l -\g\ so p(TIJ.\gl\) ~ p(lg\) ~ 1 and thus a flo of 0lgl: p(g) ~ I} c { I flgld\.J.: p(g) ~ I} . Then a pf(f) = supt I flgldU: p(g) ~ I} = supt 10 of 0lgl: p(g) ~ I} as in (14. 11). (ii) T IJ This follows from (i) since 0 ~ ['" f' E M(X~ IJ #=) implies f' -< f im p Ii esT [' >- 0 (8. 2), IJ (iii) & (iv) are immediate using the representations in (i). 15. A Repre sentation Theorem. The rear rangement invariant spaces such as the classical Lebesgue spaces, the Orlicz spaces, and the spaces introduced by Halperin and Lorentz (see (14 J, [23J, [24]), Boyd [2] and Shimogaki (45] are all of the following kind. Let (X, /I., U) be a finite m. s. with a= IJ(X) and let M+ [0, aJ denote the set of all non-negative extended real valued Lebesgue measurable [unctions on [0, aJ. -75- (15. 1) LEMMA. . defmed on M o sf -> g A is a Fatou rearrangement invariant norm +e 0, a] such that L 00 c L A c L 1 , then the mapping p(f) == A(of) is a Fatou norm which is universally rearrange- ment invariant, and p '(f) = A'(0 f) for all ° s f E M(X, U). Further- ~~Loo C L P , LP'C L 1 Since Lebesgue measure is non-atomic, (14.11) PROOF. implies that for all Os F EM[O,a] we have a = supf 10 ~ 0 I GI: A'(G) s I} X(F) so a p(f) -- To show that sup( 10 of 0IGI "A'(G) s I} p is a (u. r. i. ) Fatou norm we have by (14. 12) only to show that \'(r CeO,a]) s 1 for some r take r = 1. Otherwise, let r = '* 0. 1£ \1(CeO,a]) = 0, 1/\' (CeO, a]). For each G with A(G) s 1 a there is a g' E M(X~,u*) such that g'"" lGl and OfO\Gl = Ifg'du~. Finally we show that p'(f) = \ '(of). 10 But I fg' rilJ~ == 1fT g' dU and p(T g') IJ u sinceOsT U g'«lGI, = \(oT g ,) s \(IG1)s 1 U so a t fo of 0lGl: A(G) s I} C [ I flgldU: p(g) s I} a For the rest, if p(g)::;; 1, then IflgldU s p'(C x ) == )..'(C eO , aJ) shows the rest. and thus \'(of)::;; p'(f). 10 of O\gl and -76(15.2) THEOREM . .Y' (X, 1\, \J) is a finite m. s. and p is a Fatou norm, then p is univer sally rearrangement invariant (r earrangement invariant if (X, 1\, IJ,) is adequate) if and only if there is a Fatou rearrangement invariant norm 'A. on M+[O, a], where a = IJ,(X), such that p(f) = 'A.(o£) for all 0 ~ f E M(X,IJ,}. PROOF. It only remains to prove the existence of 'A.. By (14. 14) a we have that for all 0 ~ f E M(X,IJ,) p(f) = sup[ loolilgl: p'(g) ~ I} . For every 0 ::;; F E M[O, a] we define 'A.(F} = sup[ 1:%01 gl: p'(g) ~ I} • (14. I2) implies that 'A. is a (r. i.) Fatou norm.. Clearly p(f) = 'A.(of) for all 0 ::;; f E M(X, IJ,). -77I V. INEQUALITIES OF HARDY, LITTLEWOOD, POLYA AND MUIRHEAD 16. Schur Convex If Functions. (X, fI, u,) is a finite ITl. s. consisting of n atoms of equal nleasure, then each ITlember of M(X, IJ) may be identified with a point ;; := (x l' ... , x ) of R n, in which case its decreasing rear rang eITlent, n -*, is the point obtained by rearranging the components denoted by x -+ of x in decreasing order. --+ -iii- If x, y E R n -+ -It then the definition of y -< x as sume s the form k k :Ie ~ x. i= 1 1 * ~ y.1 ::; i= 1 with equality when k = n. It was in this form that the relation -< was first introduced by Hardy, Littlewood, and Pblya. Let S denote the symmetric group of all permutations of n [1,. ") n}. For each;; ERn let 6(x)= [yERn:y,,-,x} = Recall that an nXn matrix A = [a .. ] 1J is said to be doubly stochastic (d. s. ) if a .. 1J n ~ 0 and n :0 a .. = 1J i= 1 (16.1) =~ 1 a .. for all i, j = 1, ... , n. 1J j= 1 THEOREM. (Hardy, Littlewood and Pblya). --.... The following n ar e equivalent for x, y E R . (i) Y -< -x -> n n (ii) ~ cp(y.) i= 1 1 (iii) Y 1S ~ ~ cp(x.} i= 1 1 for all convex functions CO in the convex hull of ti;) on R. -78--t (iv) There is a doubly stochastic matrix A such that y ...... = Ax. An interesting discussion of this result and of the theory of doubly stochastic transfornlations can be found ill a paper by L. Mirsky [31J. (16.2) THEOREM. iff for all pas itive " LJ ...... --t n --t --t Then y -< x (R. Muirhead) Let x, y E R • a ERn we have y( 1) ao(l) o E Sn a y ( n) < " - ) o ( noES --t Equality holds iff a .... is constant or x( 1) a o( 1) L.J a x{n) o(n) n .... y"-' x . It will be our purpose in the next few sections to show in what sense these two theorelTIS can be extended to functions in LP(X, A, U" where (X, A, U) is a finite m. s., L 00 C LP, L P ' eLI, P is a saturated Fatou norm, and LP is universally rearrangement invariant. The gene ralization to L 1 [0, 1 J of the equivalenoe of (i), (iii) and (iv) and Theorem (16. Z) was given by Ryff ([39J, [40J, [42J). The generalization of the equivalence of (i) & (ii), was given for L oo(X, A, 1.1) by Grothendieck [10J. The generalization of the equivalence of (i), (ii) and (iii) was given for LP(X, U) independently by Luxemburg [28]. -79(16.3) DEFINITION. A mapping ~: LP - R* which satisfies the following propertie s is called a Schur Convex function. - ao < iP (f) ~ +00 for all f E L P , and iP (f) < +00 for some f E L P . (i) is convex, i.e., ~(rf1+(1-r)f2)s r 4i(fl)+(l-r)4i(f ) for all Z f l' f 2 E L P and 0 s: r s: 1. (ii) '"1' I (iii) iP is a (L P , L P ) - lower serrli-continuous, i. e. , (£: iP(f) s: r} is o(L P , L P ') - closed for all real r. (iv) iP is rearrangement invariant, i. e., fl ,...... f2 irrlplies iP(f ) = ~(f2) for all f , f2 E L P . 1 1 If iHs convex, then for all real r, the set , convex. Hence it is [f: ~(f) s r} is o(LP, LP ) closed iff it is closed in each locally , convex topology on LP in which LP , P is the dual of L P • In particular 101 (LP, L ) generated by the serrlinorrrls , P (f) = I I fg IdlJ, f E LP, g E L P is a locally convex topology on LP g the topology , in which L P is the dual of L P , and thus a satisfies (iii) iff: Jlfn-f! ~ (f) s lim inf iP (f ). n - 00 I gldlJ - 0 as n - ~ satisfying (i) and (ii) t 00 for all g E L P implies In particular, such a iP satisfies (iii) whenever: n inf iP (f ). If n I s: f o E LP and f n - f IJ-a. e. irrlplies ~ (f) s lirrl n-oo n We would like to prove that g -< f Convex functions ~. iff qi(g) s iP (f) for all Schur However it is clear that we cannot prove this for gene ral m. s. because any rearrangement invariant Fatou norm PI is Schur convex, and we have seen that unles s (X, fl., IJ) is adequate, we cannot guarantee that g -<f irrlplies Pl(g) s P (f). 1 -80- In order to alleviate this problern we introduced the concept of a universally rearrangement invariant norm. The saIne idea works here also. (16.4) A mapping DEFINITION. ~: LP -+ R* is said to be a universal Schurconvex function (u. s. c. ) if it satisfies properties (i), (ii) ~nd (iii) of (l6. 3) and (iV)1 ~ is universally rearrangement invariant, i. c., H T fl) ~ ~ (f) whenever f E LP, f' E L P (X*, \J*) and f' ~ f. ~! If ip is universal Schur convex then it is Schur convex. The next lemma win imply (among other things) that if (X, !\,IJ) is ~ is u. s. c. whenever it is Schur convex. EXAMPLES. (i) If cp is a real convex function on R adequate, then q; (f) = / q:l(f) dlJ is Schur convex on LaO. on L 00. /0 cP(of) = fo 0(1)(f) is, for each 0 ~ t ~ iJ(X),Schur convex If lim inf r~(u)/u is finite, then ip and ip U -+ (ii) If in addition cp is increasing t t then ip t(f) = then ip 1 (f) = are Schur convex t -O(l ! f dlJ and ip2(f) = !( -f)dlJ are u. s. c. on LP. t (iii) For each 0 ~t :S:\J(X), ipt(f) = fo Of is u.s.c. , a ° (iv) If g E L P and a = IJ,(X) then q; (f) = ! of + b is universal gog Schur convex on L P where b ER. (v) [RyffJ q; (f) = a t a !o !0 0f(s)dsdt =!0 0f(s)(a-s)ds is u. s. c. on LP and q; (f) = q; (g) iff f ""' g. -81(vi) If ~ is a Fatou rearrangcll1ent invariant function norm., tl1<'ll cj) is Schur convex on L P . (~;('e [30J, Theorem 3. t, p. !(,Z). (vii) The sup rC111un1 of a Ltlnily of Schur convex [Llni versa} Schu r Convex] functions is Schur Convex [Universal Schur Convex]. (16. 5) LEMMA. adequate and If cP IS universal Schur convex, or if (X, /I., U) is is Schur convex, then there are functions g. E L cj) Pf 1 and real nUInbers b. such that 1 a ~(f)=sup£f Of 0 i If in addition t 0 1 is inc rea sing, then cj)(f) = s~p{ PROOF. gi +b.} for allf EL P . a 10 of 0 Igi l + bi} for all f EL P . It is well known that if ~ IS convex and lower semiL. and real continuous, then there are continuous linear functionals 1 numbers b. such that ip (f) = sup {L.(£) + b.}, the L. being non-negative 1 . if ~ is increasing [1]. 1 1 1 1 1 For each i, there is a g. E LP s.t. L.(f) = 1 1 f f g.1 clu, for all f E L P, and g. ~ 0 if L. is non-negative, so 1 1 a cj) (f) = sup (ffg. du. + b.} ~ {! Of 0 i l l 0 Suppose cj) is u. s. c. and let f E LP. fl "-' f with £1 E LP(X~ '1-)") such that cj)(f) ~cj)(T ff):? I g. T U. IIJ gi + b.} 1 For each i there is an a I fl g.1 clJ = !0 Of 0 gi . f' dk/. + b. = Iff g. du.#-+ b. 1 1 and this holds for all i 1 so Then -82- a ~(f)~ supff 0[0 + b.} o gi 1 . The proof when (X, f\., I.J.) is adequate and ~ is Schur Convex is similar. REMARK. Y(g) Luxem.burg has observed that if we let = supf /fgd\.J.-iP(f):fELP}thenthe conclusion of the lemma assumes the form a ~ (f) (I6.6) ::: sup I r /o of 0 g - Y(g); g E LP } THEOREM. (i) Suppose (X, fl., \.J,) is any finite m. s. and f l' f2 E LP(X, IJ). 1. f1 ~~ f2 iff ~(fI) ~ iP(fZ} for all increasing universal Schur convex functions iP on L P . 2. fl -< f2 iff iP(fl)~~(f2) for all u. s.c. functions ~ on LP. (ii) Suppose (X, fL, u,) is adequate and fI' f2 E LP. 1. f1 -<-< f2 iff iPUI)"S: iP(f2)for all increasing Schur convex functions <jl on L P . 2. f 1 -< f 2. iff ~(f 1) :-s: iP(fZ) for all Schur convex functions iP on L P . PROOF. Lemma (16. 5) in conjunction with the Hardy inequalities and CP is increasing. ~ f - iPU } for all increasing Schur convex iP, then since Z increasing and u. s. c. for each 0 :S t ,;; \.J.(X), we have -83- is universalSchurconvexwe have f1 <f · 2 Reca 11 that if p i ti a Fatolt It. r. i. Bonn, then thc .J"t.· is ,\ Fatoll 1'.1. norm (16.7) A on M + [0, a] such that p(f) = A(of)' Suppose p(f) = Ho ) where f THEOREM. arrangement invariant norm. Then ~ A is a Fatou r e- is a universalSchur convex function on LP iff there is aSchur convex function I is u. ti. c. on L P then there are g. E L P and b. E R If ~ PROOF. ~ 0on LA[O, a] - 1 a 1 ~(f)= suPt J 0fO + b.}. Since p'(g)=AI(O) . 0 g. 1 g 1 AI 1 for all g E LP we see that [) E L [0, a] for each i. Hence define gi a ~ (F) = sup f J of + b . } for F E L A[ 0, a]. Then ~ o is Schur convex o i 0 gi 1 such that for allf EL P , I ° and <Po(of) = ~(f). Conversely, if ~ Gi E L AI [0, a] and b A. F E L CO,aJ. o is Schur convex on LA[O, a] then there are E R s. t. i a ~o(F) = s~pf fo of 0G. + b i } for all A l l But f E L P iff of E L [O,a] so for all f E L P , a ~(f) = ~0(6f) = stpf fo of 0G + bi} and thus ~ is universal Schur i convex. 17. The Sets O(f) = (£': f' -< £}. Recall that we are assuming that p is a saturated Fatou norrn on M+(X, U), L 00 LP(X,~) is u. r. i., P pI 1 (u) C L (u), L (U) C L (\.I.), and a = \J(X). -84- The equivijl.lence r--to -+ l.y: y -<x} y-<;; iff YE the convex hull of li~) can be reformulated; -+.-. -+ = the closed convex hull of [y: y'" x}. 1 let D(f) = [[I EL (X,1-41: £1-< f}, and 6(£):::; ££1 If f E L p (X,~) we ELl(X,~): fl ~f}. Since L P is u. r. i., fl -< £ E L P implies £1 E LP and hence both O(£)& Mf)e L P . n(£) has a smallest element T = (\J~X) for all fl E nit) iff h = T. ! £ dfJ) eX in the sense that h -< fl This follows from (8. 2)(vi). Also O(f) is contained in the hyperplane [£, ELI: ! fl dIJ = / f d~}, and T is L~quldistant (in the L 1 norm) from the members of ~f). Any a(L P, LP')-bounded set ACL P is (17.1) PROPOSITION. p-bounded. Since A is a(L P, LP')-bounded, for each g E L P' we have PROOF. For each fl EA, Lf,(g) = ! £' g dfJ defines sup(! /f' g dfJl: f' E A}< oc. a pi-continuous linear functional on L P ' with IILfl1i = p"(fl) = p(fl) Then for each g EL P ', sup { ILf,(g) [29, Note IV, p. 257J. I: £1 E A} <oc, I and since L P is complete, the Banach Steinhaus Theorem implies sup{IILf,ll: f' E A} = sUp{p(fl): £' E A}< oc. (17.2) COROLLARY. (17.3) THEOREM. 1£ {£ } is a net in LP which is a(L P, L - Q pl ) For all £ E L P, ["2 (f) is p-bounded. I PROOF. We have only to show that 0 (f) is a{L P , L P ) bounded. This follows at once since for each gEL pi we have sup { 1 I fl g dlJ I: f' E O(f)} $ sup [ ! 1£1 g 1dlJ: £' E O(£)} $ a 10 0 I £ I [) I g I < oc -H5(17.4) Y £ E L P then 0(£) is a convex and a(L P, L PI ) THEOREM. compact subset o£ L P . Let f E L P, let fI' £2 E O(f) and let 0 s r~· 1. PROOF. Then t 10 0rf +(I-r)f2 (12.1) and (5. 3)(vi) imply that for all 0 s t sa, t ,;; fo rO t fl + f (1-r)Of o t s fo o£ so 2 I rf + (l-r)f2EO(f). I Hence O(f) is convex. Let [£ } be a net in O(£). For each 0', F 0' a pi-continuous linear functional on L [29, Note IV, p. 257]. PI 0' (g) :: f f with IIF 0' 0' g dfJ defines II :: p"(f0' ) :: p(£0' ) Since O(f) is p-bounded, there is a nUITlber M > 0 such that p(fl) s M for all fl E 1(£), and thus lIF 11 :: p(£ ) s M 0' 0' for all 0'. Hence {F } is a net in a a{(L pi 0' * pi ), L )-coITlpact set [Alaogluls TheoreITl, 7, p. 424J, so jt has a convergent subnet F \3 -> F 0 say. To show that F sufficient to show that F o o E L P ::: L P " it is necessary and (g ) ~ 0 whenever g ~ 0 [30, p. 155]. n n Now for each gEL P I and for each 0', IF 0' (g) I :: I f fO' g du.1 sfl fO'g 1 d~ a ~ fo 0lfl 0lgl because fO' -< fELl implies 1£) -<-< If I . Since the a. bound is independent of 0' we have as 19 I ~ 0. F 0 (g) :: Hence F o IF 0 (g) I s fo °\f \ 0\ g \ and this ~ 0 EL P, i. e., there is an f Jfo g du. for all g E L PI 0 Since F f3 -. F 0 we have fff3gdf,l=Ff3(g) .... Fo(g)=f£ogd~forallgEL O(L P, L PI ) convergent to £. o EL P such that pi ,i.e., {£!3} is Thus 0(£) is o(L P, L P ') compact. Let V be a locally convex linear topological space and let V denote the dual of V, i. e. V * * is the collection of all continuous linear -86functionals F F on V. For each A C V and for each linear fUllctional on V let F[A] = [F(v): v E A} and let cov(A) denote the convex hull of A. As a corollary of the Hahn-Banach Separation Theorern we have the following characterization of closed convex sets. (17.5) PROPOSITION. Let K Then for each v E V, v E K PROOF. 1£ v E K, be a closed convex subset of V. * iff F(v) ~ sup F[K] for all F E V . * then clearly F(v) ~ sup F[K] for a1l F E V . Conversely, if v ({ K, then the Hahn-Banach Separation Theorem [36, p. 30, Cor 1] implies there is an F E V * such that F (v) rI:. F [K] :::J [inf F [KJ, sup F [KJ ] so either sup F [K] < F (v) or sup -F [KJ < -F (v). (17.6) THEOREM. Letf EL P . Then fl EO(£) a iff Iflgd~ ~/06f6g fo r all gEL P , . PROOF. The continuous linear functionals on L P with the ()(L P , L P ') topology are Fg (f) = f f g dUo where f EL P, For a each g E LP' it follows from (13.2) that sup Fg [O(f)J = Io 6 f 6 g . Now use (17.5). We may similarly give a criterion for deciding when a closed convex set is the closed convex hull of another set. -87- (17.7) PROPOSITION. and D C K. Then K Suppose K is a closed convex subset of V is the closed convex hull of D iff ~( sup F[DJ = sup F[K] for all F E: V . PROOF. Suppose the criterion holds. Then * sup F[K] = sup F[D] ~ sup F[cov(D)] for all F E V . is a closed convex set, since D c K and K (17.5) implies K ccov(D). is closed and convex. Suppos e now K = cov(D). in cov{D) such that w Let v E K. Since cov(D) But cov{D) cK Hence K = cov{D). Then there is a net t w Of} Of -v, so for each F E V * we have F(w Q ) - F{v) anc1 thus sup F[DJ :::: sup F[cov(D)J:?: sup Q F(w ) :?: F{v) Q since F[DJ :.:: F[cov(D)] . In view of (17.7) we have immediately: (17.8) THEOREM. Let f E L P and let in which L P' is the dual of L P . r, be a topology on L P Then O(f) is the; -closed convex hull of a set D c O(f) iff supt If' g d~: f' ED} = (17.9) THEOREM. dual of L P . In particular fn l 0 the~ S = the (i) S be a topology on L P in which L P ' is the S = o(L P, L P ') or if p(f ) -- , n a whenever p - topology are such topologies. O(f) is the S -closed convex hull of 6 (f) = [£, ELI: fl ""' £} fo r all f E L P iff (ii) Let a !o Of 0 g for all g E L P . (X. A, w.) is adequate. For any finite m. s. (X, 11., IJ) and f E L P, O(f) is the ;-dosed convex hull of f.T hi- fl: £I ELI (X#, \.l) and £1 '" f } . -R8- PR.OOF. If (X, A, f1) .is a.dequa.te then the re~.,uH follow::; from (i) (17.8) and (11. 1). If (X, A, iJ) is not adequa.te, the condition in (17. 8) fails. (ii) then (11. 5) shows that Follows from (17.8) and (9.1) (iv) & (vi). As Luxemburg has pointed out, Theorem (17. 9) answers the following question asked recently by Z.Nehari [32J: Let (X, A,IJ) be a non-atom.ic finite m. s. and let E E A have positive measure. What is the smallest closed convex set A cL 1 (X, IJ) which contains all the functions C F such that W,(F) = ~(E)7 dosed convex hull of .6(C and E ) so A = O(C E Obviously A is the ) = {£ ELI (X, bJ,): 0 $ f $; 1 J f dlJ = IJ(E)} using (8. 2)(v}. We can defin~ the sets 0 in a slightly more general way. Let (Xl' AI' IJ I ) and (X, A, 14) be finite m. s. such that 141 (Xl) = IJ(X) = a. 1 1 IffEL (X1,IJI)letOf(X,u)= [hEL (X,\J.):h-<f} and L\f{X, iJ) = [h E L 1 (X, 14): h "'-'f}. let Observe that Of is never empty because it contains I = (_(IX ) f f diJ )CX' but it may happen that i IJ I 1 .6£ = 0. This is the only interesting case, because if 6 =1= (/;, then f w~~ are doing nothing new, since for each fa E 6 we have Of = O(fo)' f / Thcorenl (5.10) implies that .6 (X . iJ#-)=I= 0 for all f ELI(X , U ), so if I I f fa E lIf(IJ#) we then have Of (X, IJ) CO(fo}' Now let p be a saturated Fatou norm on M + (X, IJ) such that L'X:; (14) C L P, L P ! C L 1 (u) and L P is u. r. i. I w ha t f EL (X I' iJ I) is (2 f C L P (X, )J,) ? The question arises: For -89(17.10) PROPOSITION. LetfELl(Xl,~.J.1). DlflDlgl EL1[O,a]forallgEL PROOF. PI ThcnOlflcI..JP , in which case rtC-LP. IfC2lflC L P then fl -< If\, fl E M(X,U) implies o S£I E L 1 (X, U) so £1 EOlfl c L P and thus for all g E L that jff PI we have 1 fl Igi dlJ is finite, and hence (13.2) says 0lfl 0lgl E Ll[O, aJ. Conversely, if 0lflDlgl E L1[0, a] for all g EL 1£11 -<-< Pf then fl -< f implies 1£\ so ('lfllOlgl E L 1 [0,a]for all g E L PI and hence £1 E L P by (13. 2) & (14. 5). Observe that Theorems (17.3), (17.4), (17.6), (17.8) and (17.9) are true for C2 £ under the hypothesis that 01£1 c LP, because in view of (17. 10), the proofs are practically the same. I I cL P it is natural to wonde r: In view of the condition [2 f p Is there a norm PIon M+(X , U ) such that f ELI implies 01£1 C L P ? I 1 a = supt 10 0lfIDlgl: pl(g) -:; I}. Pl(f) Then Pl is a Fatou u. r. i. norm satisfying p (i) pi L ~ (IJ 1 ) C L 1, L 1 C L 1 (U 1 ) (ii) fl -< £ E L PI implies p(£I) -:; PI (f) (iii) 0\£1 C L P iff f E L PI . -90a PROOF. P1(C (i) X1 = supf / Igldu: p'(g) ~ I} 1 gl = p'(C <sup{ t X CX' ) ) = 8Up[ fa algi: p'(g) <11 = p(C x ) < DC. Since 0 < p'(C x ) < DC let 1 a Then - - flfldu 1 = 1 °lflol I p'(C ) a gl X 0lf!Olgl: p'(g) ~ I} = Pl(f) so f E L P1 implies f E Ll(U l )· p' 1 Since PI is Fatou, L oc: (U ) C L 1 C L (U ) also. 1 1 (ii) In order to apply the theorem of Hardy (7.1) we need to know that 0lf' 10 Ig f' -< f E L P1 C I E L 1 [0, a] for all g E L P ' with P' (g) < 1. LI(U ) implies Of' -< Of E LI[ 0, a] so I a 01£' 10 1 g I ELI [0, a] for all gEL oc (U), so for all gEL OC $ gn a fa 0' f' ,0, g' ~ fa 01 f ,01 g 1 ( lJ). Let g E L P ' (U). o Now Then there is a sequence [g } CLoc(hl) such that n t Ig I so {) gn t a, g I and hence the monotone convergence theorem implies a a a <oc fa °If'IOlgl = lim fa °If'IOg n < lim /0 °l£! °gn = a ~ sup [ /0 01 £ 10 1g I: P' (g) ~ I} = PI (f). (iii) Part (ii) shows that f E L Then 01 f! 01 g 1 PI implies o'lfl C L P . I E L [0, a] for all g E L P I Suppose -91a somaxf ffllgldlJ: fl -< If I} = fo 0lflOlgl <x forallg ELPI and PI(f) ~- Sup [rllax[ ff'lgld~: £1 -<If\}: p'{g) <: I} hence -Sllp[Sllp[ / j"lgl<lfJ: f"{g).; l~: [' ...: If I } :: suptp(£I): 1" E Cl f }<oc because 01£\ cL P implies O£ is p-bounded. If (as in ?IS) A is a L ac A cL, L A' cL 1 Fato~ r. i. norm on M+[O, a] such that and p(£)::A(of) for + fEM (X,I,J.) then the natural norm to choose onM+(X ,1J ) is PZ(f):: A(of)' 1 1 In this connection we have the following. (17.12) PROPOSITION. (i) L ac (btl) c L (iii) £1 -< £ EL PROOF. PZ P2 , L 1 c L (U I) (i) ·See (15.1) . PI (f) :: sup [/0 ° °Ihl: A'(oh) ~ a < sup PZ ' implies p(£I) $;PZ(£) a (iii Let A, P, PI' Pz be as above. If! t J I) If I I:i IG I: A (G) < 1 } == P2 (f) . 0 I (iii) Use (17. lZ) (ii). 1, h E M(X, IJ)] -92Doubl~tochastic 18. Transformations. irn.plies that an nXn rnatrix A n -+ Observe that (16. I) is doubly stochastic iff A-;' ,;;Z for Let (Xl' AI' ~tl) and (X, fl., U) be finite nWaSllr(' spaces all x E R. with ~l (Xl) = IJ(X) = a. (18. 1) DEFINITION. A linear mapping T: L 1 (X l' U ) .... L is called doubly stochastic (d. s.) iff Tf -< f for all f E L EXAMPLES. a J 1 (X, U) (Xl' IJ ). I If (X, A, U) is a non-atomic m. s. and 1. a: X .... [0, a] is measure preserving (m. p. ) let T f = f 0 T 1 1 o. Then : L I [0, a] -+ L I (X, IJ) is d. s. 2. I # JI, 1 . TIJ:L (X,U ) .... L (X,V)lsd.s. 3. Tf (18.2) (i) 1 = (\.il (Xl) PROPOSITION. T X = C I I I (\1) -+L (IJ') be d. s. t L f E L ac 1 (Tf) (Xl' k-i )'" L l ~ 0. 1 ITfl <;Tlfl for all f EL (U ) 1 <; Tf-, c<: (X, U) (U ) implies /lTf1l 1 ° implies Tf Then x (iii) (Tf)+ <; Tf , (iv) T: Let T: L is non-negative, i. e., f ~ (ii) TC (v) J f dw.I)C X defines ad. s. T: L I (Xl' f.l l ) .... L I (X,u). 1 ~lIflll f E Loc(IJ 1 ) implies IITfik <; Ilfllac -93- PROOF. (i) Follows from (8.2)(iii). Follows from (8. 2)(v). (ii) (i i i) Sin c e T f = T f and (Tf) s; (TO-. (iv) & (v) Hence ITfl Suppose Ii Tf + -< f + and (ii) T = (T£)+ + (T£)- < Tf+ + Tf- = Tlfl. Follow from ITfl s T If I -< Ifl. (18.3) LEMMA. (i) + - T f - and T f +, T f I f :? 0 we have (T f) :--: T ( Tf l l T:L (lJ ) ..... L (lJ)islinear. l --<f - then Tf -< f. l is d. s. iff Tf -< f for all 0 $f EL (1J ). 1 PROOF. for all 0 st $ a with equality when t = a since Tf- -< f implies -Tf- -<-f (ii) Tf-<f foraH f:?O implies Tf+-</ and TC-<f- soTf -< f for all f. The following Theorem, first proved by J. V. Ryff =39J for . f un d amenta 1. L 1 [.0, 1 ] ,IS (18.4) THEOREM [J. V. RyffJ. A linear transformation T rnapping the simple functions of (Xl' 1\1' V- ) into Ll(X, 14) has a l unique extension to a doubly stochastic transformation of L 1 (1J ) .... L \4) 1 iff for all E E 1\1' we have 0 s T C PROOF. T C E -< C E E $C x and!X T C E dlJ = 1J 1 (E). (8. 2}{v) shows that these conditions are equivalent to for all E E 1\1' and are thus necessary. is given satisfying the above conditions. Hence suppose Let 0 s; f E M(X l' V- ) be 1 T - 94n simple. whereE c · · · · c E and£.> 0, Then f = ~ f.e . l I E .1 n I l 1= so n Of = L: f. 0E' i::: I I i Then (12.1) implies for all 0 ~ t <a, t n t Jo0' T f " "'J"! t. i:::l .I n 5 lTC () "'E. ' t'J' < . i:.::l t t !"°r ,. ! 0 f 1 () "j 0 f' with l'quality will'" 1 t = a, so Tf -< f for all non-negative siluple functions and hence for all simple functions. Thus T is a contraction in the L 1 and Lac norms on the simple functions, so it extends uniquely to L I T o show that T is d. s. Then there is a sequence (f } of simple functions s. t. 0 ~ f n Tf n n i f. Then Tf n is increasing so t Tf and since Tfn -< f n , (8. 2)(vii) implies Tf -<f. Hence (18. 3)(ii) implies T is d. s. If T:L 1 (U ) - L 1 (U,) IS linear, let T>:' denote the adjoint of T, 1 >:< 1 ac defined by g Tf du,::: X f T g dW,1 for all f E L (U- ), gEL (u). 1 X >.~ 1 It follows that T: L1x,\J) - Lac (Xl' U ) and hence T is weakly contin1 I ac 1 ac uous under the topologies a(L (U,l)' L (~)) and a(L (u), L (u)) ! f ([36, p.38, Prop. 12J or use nets and the defining equation. ) If T: L1(U ) - L1(1-l) is d. s. then T 1 1 1 extensiontoad.s. mapofL (u,)-L (U ). 1 (18.5) THEOREM. PROOF. This follows from (l8. 4) as follows. >:< has a unique Let E E A. - 95(18.6) PROPOSITION. A linear mappin~ I I T: L (~ll) --- L (1.1) is doubly stochastic iff (i) T:;:: (ii) TC ° - C X1 X * x ::: C x {iii} T C PROOF. 1 This follows from (18.4) since for all E E Al Observe that i£ (X ' A ' /J.Z) is also a finite m. s. with Z Z hJ (X )::: a, and T : L 2 2 1 1 (WI) d. s. then TIT 2: L 1 (W ) Z -0 1 -0 1 L (fd,) and T : L (fd,Z) .... L Z 1 (IJ ) are both I L 1 (U) is d. s., since for all £ ELI (1J 2) we ha ve TIT 2 £ -< T 2 f -< f. (18.7) LEMMA. Let £ ELI (X, IJ) and let c;:x* -0 [0, aJ be measure preserving s. t. PROOF. For all measurable J c [0, aJ we have /0a C J Tc;* £ dm::: /fTc;C J dlJ'* =[(6 f a a)(C J a a)dlJ'¥ =/0a o£ C J dm. Let stochastic}. ~(Xl' X) ::: {T: LI(X 1 , 1J 1 ) -oL1(X, \.J,) such that T is doubly If f E L 1 (fd,l) let ~f(Xl' X) = [Tf: T E ~(Xl' X)} recall that (1 fiX, IJ) ::: {g E L 1 (X, IJ ): g -< f} and and we -966f(X,~) = [g E L 1 (X.U); g,....,f} As we indicated in § 17, if 6 that ~f C Of for all f E L (18.8) THEOREM. 1 f :\= J6, then Of = 0(£0) for all foE!::.C Notl' (~j 1)' ~(X l' X) is convex and compact in the weak operator topology determined by the linear functionals 1 00 f E L (IJ)' gEL (IJ )· 1 T ... /£ Tg dIJ PROOF. To show 19 is convex, let T l' T Z E ~ and 0 S; r S; 1. 1 Then for all f E L (IJ ) we have Tlf, TZf E Of so r Tlf t (l-r)T f Eel , Z f 1 i.e., (r Tl t (l-r)T )f= r TIft (l-r)T f.-<f, sor Tl t (l-r) T E~. Z Z Z Let S 1 be the unit ball of L 00 (IJ 1) and let S be the unit ball in L 00 (u.). 1 Since Sis a(L oo (IJ), L (1J» compact, the set of all linear operators which map S 1 ... S is compact [18]. that ~ is closed. Let T a .... T. and T a C weakly. TC E E TC -i> E 0 C . E Hence it only remains to show For each E E Al we have T a C E E 0C 1 Since liCE is a(L , L (0) closed we get E Hence (18.4) implies T E ~. E (l8.9) THEOREM. 1 Let f E L (X l' U ). 1 !! gEM (X, IJ) ~g -< f iff there is a doubly stochastic T: L I(IJ 1) ..... L I(IJ) such that g PROOF. TIJ 6 f (IJ#) C We have to show that ~f = Of' = Tf. It suffices to show that ~f (where, of course, TIJ 6f(lJ~ = [TlJf':f'ELI(X~IJ#)&f'--f}) and that 19 is a(L I(IJ)' L 00(1J)) closed and convex, because we know f from (17. 11) that Of is the closed convex hull of TU 6f(IJ~) in this topology. 19 is convex and closed in this topology, because it is the f -97ilTlage of the convex, cOI11.pact (in tllt' weak operator topology described >\'C in (18.8)) set .\9(X, Xl) under the continuous linear map T -- T f. the rest, let f' E L\(~*). For There are lTleasure preserving tran!:lforma- tions 0: Xl -+ [0, a] and ~: xt Since ft -- f, of' = of so T,,/' = TIJ. TaT¢' f E ~f" -+ [0, a] such that To of' = f' and T ~o f= f. >\'C THEOREM. (18.10) g -< fl + fZ then there g 1 -< f 1 and g Z -< f Z· PROOF. There is aTE .I9(X ' X) s.t. g = T(f +f )' so let 1 Z 1 g. = Tf., i= I,Z. 1 1 A good exalTlple of a class of doubly stochastic operator s is provided by conditional expectations. 1 let f E L (X, iJ.). It follows frolTl the Radon-Nikodyn theorelTl that there is a unique !\-lTleasurable E E /\'. Let 1\' be a a-subalgebra of 1\ and function Tf such that IEf dlJ = IE Tf dU for all Using (18.4) it is clear that T: Ll(X, 1\, U)->Ll(X, 1\',1.1) is doubly stochastic. 1\' conditional expectation. T is called the As a special case of this process let X = X U U X. be the o iEP 1 union of an at IllOst countable nUlTlber of sets of positive lTleasure and let (I.' be the a-sub-algebra generated by the sets A n X Then the A' conditional expectation has the form Tf = f C x o +6 iEP 1 \J{X.} 1 (Ix. f du)C x .· 1 1 o and fX.}. r:::p. 1 1, -9BOf course T ~ is of this type. Now let P be a saturated. Fatou norrn on M(X, IJ) such that LP(\.J) is u. r. i. andL11J.) C L P , L P ' C L1(\.J)' and define the u. r. i. Fatou norm PIon M(X , \.JI) by I a Pl(f) = supt 10 blfl blgl: p'(g) ~ l}. and (ii) p(Tf) ~ PI (f) for all f E L PI T has a unique extension to a doubly stochastic T: Ll(\.Jl)~Ll(IJ). I (iii) T * : LP' (IJ) ..... L PI (loll) and PI (T*g) ~ p'(g) for all g E LP ' PROOF. (i) We already know that Tf -< f E L p(Tf) ~ Pl(f) so Tf E LP(\.J). P1 (\.JI) implies P1 oo This follows from (lB. 4) since L (\.Jl) C L (\.Jl)' pi >!< (iii) Let g EL (IJ.). Since we have! f T g dIJ. =! gTf d~ is finite for all , 1 P I P1 I f E L (u ), (14.15) implies T*g E L (loll)' Let g E LP(U). 1 (ii) pll (T*g) = suPt ! If T*gld~l: p (f) ~ l} 1 ~ sup f I IfiT * Ig I dU 1: P 1 (f) ~ I} = sup(! Igl T If I dlJ: PI (f) ~ 1 } ~ supf ! Igi If'ldu: p(f') ~ 1 } = p'(g). -99(18.12) s. t. COROLLARY. 1 r PROOF. Let gl -< g E LP (u). g 1 = Tg. Then T * so applying (IB.l1)toT * 1 There is ad. s. T: L (IJ)-+L (1J 1 ) extends uniquely to a d. s. map of L 1 (IJ 1) --L 1(U) * wehaveg 1 ** =(T)gEL PI ( 1J 1) and >,'C pll (gI) = pll «T ) g) ~ pl(g). REMARKS. (i) If (Xl' fir 1J ) = (X, fl., IJ) and 1 p is u. r. i. then PI = p. (ii) If "- is a r. i. Fatou norm on M+ [0, aJ and p(f) = "-(of) for f E M+(X, IJ) as in §15, and PZ(f) = "-(of) for f E M+(X l' U ) then I (18.11) holds with PI 19. replaced by PZ' J. V. Ryff has given a generalization of Muirhead',s Theorem. Muirhead's Theoren-l for bounded measurable functions on [0, 1 J (see [42J). In thi s section we will show when this generalization is valid for arbitrary finite measure spaces. Let (X, fI.,U) and (Xl' fl.1'lJ ) and (X , fl.Z,IJZ) be finite m. s. such 2 l If u E M(Xz.IJZ} is positive (i. e. that U(X) = 1J1(X ) = IJZ(X ) = a. l Z u(x) > ° IJZ-a. e. ) and f E M(X, [f;uJ = I log (I X X \J) let u(x/(Y) dIJZ(x} )dlJ(Y) z and similarly for g E M(X , 1J }' l 1 -100(19.1) LEMMA. [f;uJ = [Of;o u J - u E M(X ' ~Z) is positive Z If and f E M(X,~) thet::.. in the sense that both are finite and equal or both are infinite with the same sign. PROOF. Let pER and let \jr(t) = t P for all t E R. Since u'" ° u and the measure spaces involved are finite, (3. 3)(xv) says \jr(u)~ \jr(o ) a so /,X Ci'J(P) Z \jr(u)dU = ! \jr(o ), i. e. i-X Z 0 a II u a uP dlJ,Z = ! (0 )P. Let lOU = log(/X.,uPdUZ) = log( fo(ou)p). Again,sincef",-,6fwehavcCj:~f)~ct(6f) a L. so[f;uJ=J X cp(f)dlJ.= JoCP(Of) = (19.l ) THEOREM. (i) [of;ouJ· 0) . LetfEL 00 (X,IJ.)andgEL0 (Xl,U l Il g < f then [g;uJ ~ [f;uJ and both are finite for all positive u E M(X , 1J. ) such that uP E L l(Xz,lJ.z) whenever l 2 ess. inf (min(f, g» ~p s: ess. sup (max(f, g» [we call such u admissible for f and g. J (ii) f '" g ~E. g [g;uJ = Cf;uJ ~nd u is admissible for f and g then is constant uz -a. e. U (iii) If (Xl' "'Z,)..I2) is non-atomic and [g;uJ s: [f;uJ for all i>ositiv~ u E M(X Z' )..12) such that uP E L 1 (X 2 , )..IZ) whenever pER [ we call such u PROOF. X = Xl = X admissible for L ooJ, then g < f. Ryff has shown that this theorem is true when z = [O,1J with Lebesgue measure ([4Z ,p. 596J and [43 , p. 436J). Our fir st step is to show that the theorem is true when X = Xl = X z = [0, aJ with Lebesgue measure. and let u E M[n, aJ b e admissible for F Let F, G E M[O, aJ and G. Then -101[F;u] = J: log( 1 =aJ log( o :.- a u(s)F(t) dS)dt 1 10 u(as)F(at) dS)dtl-aloga [1' 1; u 1 ]Ia Log a where F 1 (t) = F(at) and u (s) = u(as) on [0, IJ. 1 Similarly a[G1;u1J + a loga where GI(t) = G(at) on [0, I). = [G;u] I: Since a> 0, (3.2) implie s G 1 -< F 1 iff G -< F, and thus it is easy to see that the theorem is true when X = Xl = X z = [0, a]. 0) Now let f E L 00 (X, u,) and gEL 0 (Xl' U . 1 (i) Suppose g -< f and let u E M(X ' u,Z) be admissible for f and g. Z Then 0g -< of and it is easy to see that 0u is admissible for of and 0g so = [0 g ;6 u J ~ [of;o u ] = [f;uJ. rg;uJ - (ii) Suppose [f;uJ = [g ;u]. 0f~ ° g (so f,.., g) or 0 u Then [of;o ] = [0 ;0 ] so either u g u is essentially constant (so u is also). (iii) Suppose (X ' I\Z' U,Z) is non-atomic and [g;uJ~ [f;u] for all Z u E M(X ' ~Z) admissible for L 00. Z LOO[O, aJ. Let v E M[O, a] be admissible for Then there is a u E M(X ,U,Z) such that 0u = 0v' in which 2 case II is admissible for L 00 and [0 ;v] g = [0 g ;0 u ] = [g;u] ~ [f;uJ =[of;v], so [Og;v] ::; [of;v] for all admissible v and hence 0g-< of so g-<f REMARKS. l.Ryff has also shown that (iii) is not in general true if Wv intt.' n~hang~~ the ord('r of integration in the definition of [f;u]. 2.. (iii) mayalso fail if (X ,A ,I-4 ) has atoms and (X, 1-10) and(Xl'~) are Z Z Z more cOITlplex than (X z,t-Ioi. Forexample.if X Z is an atom and both X and Xl are not atoms, then (iii) fails. The assumption that (X ' liZ' U,Z) Z -102is non-atomic is sufficient to insure that there are enough admissible u E M(X!" P ) to di stinguish when g -< f, no matter how complex (X, j.l) Z and (X1,f,l1) are. -103- VI. 20. EXTREMAL AND RELATED PROBLEMS ExtrelTIe, Exposed and Support Points of Of" Let locally convex topological vector space and let K subsd of V. be a be a convex A point v E K is said to be all cxtl"t::me point of K v ~ ivl+ivz & vI' V point of K V z E K iUlplies v if there is an F E V :f< = vz." l v is said to be an exposed (the continuous linear functionals on V) such that F(w) < F(v) whenever w E K, w '*' v. in a hyperplane, then a point v E K if there is an F E V for SOlTIe w E K. if If K is contained is called a support point of K '* such that F(w) ~ F(v) for all w E K and F(w) < F(v) It is clear that every exposed point is both an extreme point and a support point. It would be de sir able to characterize the extrelTIe points of K '* in terms of the sets F[K], F E V. For example, it is clear that * if F(v} is extreme in F[K] for all F E V , then v is extreme in K. The converse is not true as can be seen by considering the closed unit disk. is an F E V However, it is true that if v ~ in K, then v is extreme in such that F(v} is extreme in F[K]. K, then there For if v is extreme is a boundary point of K, and since the interior of K is convex, the Hahn-Banach Separation theorem [36, p. Z9J gives the required F. It is we II known that v E K is not extreme iff there is a 0 such that v+u and v-u E K. Suppose K is closed and convex. easy to see using (17.5) that this condition become s: extreme iff there is a 0 '* u E V such that both F(u) ~ sup F[K] - F(v) F(u) ~ F(v} - inf F[K] '* u E V It is v E K is not -104>1< for all F E V. This condition has been given for O(f) by Luxemburg [Z8, p. 141J. It does not appear likely that a useful characterization of this type is possible. Let K be an ice-cream cone in the plane fonned by intersecting tangents to a circle. The points of tangency arc extrCITIC but not exposed, and there seems to be no way to distinguish them using closed hyperplanes from the points on the sides of the cone. Let (X, fl., I-L) and (X l' fl. , I-L 1) be finite measure spaces such 1 that U(X)::: 1-L (X ) = a. 1 Of (X, I-L) ::: 1 1 Recall that if f E L (X ,1-L ) we let 1 1 [g E L l(X, U): g ~ f} 1 {g E L (X, I-L): g "'" f}. The reader is referred to § 1 7 for a detailed discussion of the se sets. The extreme, exposed and support points of Of have been determined by J. V. Ryff when X = Xl = [0, IJ (see [41J). The proof that the functions in 6 are all extreme in Of is due to J. L. Doob f [41J. We present it in the following way. The proof of (20.2) is different from Doob's. (ZO.I) LEMMA. f1 fZ ~ a (ii) (J. L. Doob). (i) If -;t(f1+ f 2) '" f1 -- f2 ~ I-L - a. e. Let g E Of' PROOF. (i) .!!.g::: -;tf I +#2 & fl' f2 E Of implies fl -- f2 "'" g fI'"-'f2 implies !\I-I£zl so Also !f I + fzl --.-zlf I ' so fifl + izldl-L= !(I f 1 1 + Ifzl)dlJ triangle inequality implies I fl + fzl = I fll + 1fZ! flfZ = Iflfzl ~ a u-a.e. !!fll dlJ = !1£zldl-L' U-a. e. sothe and thus -1051 1 (ii) Suppose g, fl' fl E Of and g == a f l + afl' we have to show fl = fl' By symmetry it suffices to show that [x: f (x) <f (x)} has measure zero. 2 1 (f1-r) ""' (fZ-r) ....... (g-r) Let fr For all r E R, = i(fl-r) + i(fl-r) so (fl-r)(fl-r) :2: 0 \.l-a. c. } be an l'llUIl"leration of all the rationals of R. oc f (x) < fZ(x)} U [x: f (x) < rn < fZ(x)} . fx: f Then g"'" f 1 ....., land n = 1 Then 1 n= 1 0() U [x: (fl(x)-r )(f (x)-r ) < o} 2 n=l n n zero. C (ZO.2) THEOREM. PROOF. each of these sets has measure If g E Of and g""'" f then g a t Jo 10 Let ~(h) == 1 1 hE L (X,\J) U L (X ,\J1)' 1 h} "" h ' Z and ° h is extreme. a = 10 0h(s){a-s)ds if ~ is Shur convex and ~(hl) == ~(hZ) iff 1 1 1 1 Suppose gEL (X,U) and g ....... f E L (X ,U )· 1 1 If g == Pl+af Z where fl,f2 E Of then ~(f) = Hg) == ! Hf l ) + t~(fZ) ~t~(fl) + !qj{f) so ~(f)~~(fl)-O;~(f) and thus ~(f)=~(f1) Hence g ~ f 1 ~ f 2. so g so f ...... f • l Similarlyf......,f Z · is extreme. Using Theorems (5.10), (5. lZ) and (10.1) it is easy to see that Ryff' s proofs are valid when (X, 1\,).).) is non-atomic. Hence we have the following. (lO.3) THEOREM. 1 If (X,I\,IJ)isnon-atomicandfEL (X ,U ) 1 1 then the set of extreme points and the set of exposed points of Of are identical with the set 6 C A function g E Of is a support point of -106t t Of iff there is a 0 < t < a such that fa 0g = fa Of' If (X, fl., IJ) is not adequate, then there will be a function f E L l(X, IJ) such that O(f} = [g E L 1(1J}: g -< f} is not the closed convex 1 1 hullof,Mf}= [gEL (1J):g ..... f}.lffoEL (X1,1J ) suchthatfo~fthen 1 Of will not be the closed convex hull of 6£ ' and hence 6£ 0 0 cannot 0 contain all the ext reme points of Of . o It is a good conjecture that every extreme point of Of has the form T fl where fl E L l(X*, IJ#) and £1 ..... f, and that every function in ~ Of equimeasurable with an extreme point is an extreme point. For example, if £ is extreme and either (i) g a a = £ where a: X ...... X is l a = fl X a and g' A,.." fl A (where X a is measure preserving or (ii) g X the non-atomic part and A is the union of the atoms of X) then g is extreme. The following example shows that not every function TlJf l , ft E L 1 (X"", IJ#)' fl ..... f is an extreme point. of two atoms A and B with IJ(A} < IJ(B}. has the form g = xC A + yC B · Then every g E M(X, IJ) Let f = 2C A + C B x ~ 2, Y IJ(B} ~ 21J(A) + (IJ(B}-IJ(A}) = IJ(A) + IJ(B} xU(A) + Y IJ(B) = 21J(A) + I-l(B}. ::p(xC A + yC B Let X be the union Then g E O(f) iff . and If we define cp: M(X, IJ) -+ R by ) = (x, y) then cp is linear, 1:1, and onto the line segment joining the points (2, 1) and (1, 1 + ~~~~ ). of D(f} are f = 2C A + C B Hence the extreme points and g = C A + (1 + ~~~ ) C 6(£) = [f}, 6(g) = (g}, and g = Tlffl where £1 = Ceo '"" ",f. 2 B ,I-l . Observe that (B)[+2C r-IJ (B) ,IJ (X)i ,~ It is clear in this case that [TlJf': f' ...... £, f' E Ll(~,IJ:I(.)} = 0(£). -107The following exatnple shows how extretne points not in lI(£) Inay arise when X has an atotn A of positive tneasure. Let B C X (5.10) there is an f on f to be 0 B Let l[a, rJ on the rest of X. + IJ(B}-t g' "",f so 1 2 C Using Define Define g' ELI (XJI:, U ') + IJ,(B} and g' = 0 elsewhere. g~£. ::: TIJ g' -< f and A o be the interval of as in §9. A if a < t < a g::: Z f,.L(B) be s. t. 0 < IJ(B) :s; IJ(A). o 0fIB(t) = ~4B)-t on [0, IJ(B)[. s.t. (X#=, !I. ~ \.I. #) co rresponding to g'(t) = a and a non-atotnic part X by Then To show that g is extren"le, suppose g ±u -< f (here ± means + and -) where u E M(X, IJ}. Then! u dlJ ::: 0 and since f:;:: 0, we have g ±u ~ O. O:s;g±ulAc=±uIA itnplies u IA = 0 c so ulA so u::: O. c =O. Hence Thus ThenO=!udu,=(ulA)bJ,(A) g is extretne. In each of the above exatnples extretne points were obtained in the following way. WriteX:::X U U X where P o iEP 1 p = [1,2,3, ... }, X are the atotnS of X. o For the intervals 1 0 k : :1 1 Let f E L (X, \..l. ). l[a., b. ] defining X# 1 1 l\: 0 L, k= 1 IJ(X k )[ 1 1 1 1 of the interval 1 Now for each tneasure preserving , n, cr ::: " 0 cr)C (fn X take . For each partition n ::: [[a.,~. [}. EP 1 iEP, i i, bJ,(X ) + such that [J.-a. = IJ(X.) define [0, f,.L(X) [ f . is the non-atotnic part of X, and X, i i-I l[a., b. ] ::: [IJ,(X ) + 6 iJ(X 1 = fl. ... , n} or o a: X " C _ + fn X X o . 0 -t [0, bJ, (X ) [ 0 let -108- , # ~ # ), f E L 1 (X, Then f , IT, 0 T )..l 21. f is c:x-trcn1e in IT, 0 , IT, 0 ~ f, and it is a good conjectu re that Of" Pern1utator Transformations. lI(f) = £h: h -- f} the sets Becaus e of the importance of it seems natural to investigate the neces- sarily doubly stochastic operators T such that Tf,..., f for all f. Such an operator is called a permutator, because for discrete lueasures they correspond to the permutation matrices. Let (X, fI, U) be a finite m. s. with u(X) = a. write A = B [U] to mean C A = C where A /::, B = A-B U B-A. CAS C B IJ - a. e., B We also write A cB [~] to i. e., U(A-B) = O. B E fI we IJ.-a. e., i. e., IJ.(A 6. B) = 0 mean Note that A c B [IJ] implies U(A) $; U(B) and A = B [14] implies bJ,(A) = IJ (B). A If A, If A c B [IJ.], then = B [u] iff U(A) = IJ(B). Now let (Xl' AI' btl) be a finite m. s. with 141 (Xl) = IJ,(X) = a. (21.1) DEFINITION. A mapping ~: Al .... Ais said to be a homolUor- phism of fll into fI if it satisfies (i) lJ(qi (A» = 1J 1 (A) for all A E Al (ii) ~ (An B) = ~ (A) n~ (B) [U] for all A, B E Al (iii) ip (AUB) = ~ (A) Uip (B) [IJ] whenever A n B = ~ [1.),1]. We call ip an isomorphism if in addition its range is A [IJ,], i. e., for every E E A there is an A E Al such that ip (A) = E [kJ.]. In this case there is an isomorphism ip is the identity [U] on A. -1 : 1\ - Al such that ip 0 ip -1 -109- (21.2) PROPOSITION. (i) q; I( q;: Al .... /\ is a homorphism, then disjoint sets, i. e., AnB = 0 [IJ] implies preserves i!i(A)flq;(B) = 9J [IJ]. (ii) if? is monotone, i. e., AC B [~J implies q;(A)c~ (B) ijJ]. (iii) i!i preserves differences, i. e., ip (B -A) = ip(B)-i!i (A) [~J whenever ACB[IJ ]· I oc (iv) iii is countably additive, i. e., if [Ai }i=l c /\1 are pwd [1J 1 ] then -- ip (U A.) = Uili (A.) [~]. 1 (v) 1 if? is 1:1, i.e., q;(A) = <!i(B) [IJ] implies PROOF. (i) AnB = f/J [1J ] 1 U(ili (A) n ili (B)) = U(ip (An B» (ii) & (iii) If AcB A=B [1J iff fJ (AnB) = 0 1 = U (AnB) = 0 l 1 J. so implies <!itA) n(B) = 9J [U]. [1J1J then B = B-AUA [1J ] 1 is a pairwise disjoint union so <!i (B) = ip (B -A) U <!i (A) [uJ is a pwd union (U] and hence ip(A)cif?(B) [IJ] and ip(B) -ip(A) =ip(B-A)[uJ (iv) Let tA.} c Al be pwd [WI]' 1 Now A. c UA. for all j J 1 implies i!i (A.) c q; (U A.) for all j and hence Uq;(A.) cq;(UA.). Since 1 t>C nIl n U(q;(U A.»=U (U A.)= lim 1J (UA.)=lim u,(ip(U A.» 1 i=1 1 i=1 1 n .... oc 1 i=l 1 n .... oc i=l 1 n oc :: lim IJ( U ip(A.» = IJ(U ip(A.» we have ip(UA·)=Uili(A.) [uJ n ..... oc i=l 1 i=1 1 1 1 J oc (v) Suppose ili (A) = ili (B) [IJJ. 1J (E - AnB)=\.l(ili (E-AnB» 1 Let E = A or B. = lJ(ip (E)-ip (An B» = lJ,(ili(E)-<p (A)nili (B» = 0 so A :: B [u]. If (X, A, IJ) is a product of a possibly unc ountable number of copies of [0, 1] with Lebes gue measure, then every homomorphism -110~: A -> A is induced by a measure preserving a: that ip (E) = a -1 (E) for all E EA. X ...... X in the sense If 0: X -> [0, a] is measure preserving 1 then HE) := 0- (E) is a homorphism of the Borel subsets of [0, a] into Recall that in this case T f = f 0 0 defines a doubly stochastic o transformation of L 1 [0, a] into L 1 (X, \.I.). 1\. PROPOSITION. (21. 3) To each homomorphism ip: 1\1 -. 1\ there corresponds a unique linear transformation Tip: L I (Xl' ~1) -> L 1 (X, u) such that Tip C E = Ccp{E) for all E E AI' In addition we have the following: (i) T cP f ~ f for all f ELI (X l' ~ 1 ) (i1) Tcp(fh)::: (T ipf){Tiph) whenever f, h, fh ELI (Xl' 1.11) then ip = ip I [~J. (iii) If ip I : 1\('" A and Tip ::: Tip I n PROOF. If f = I: 0'. C A is a simple function with fA.} a partition i=l of X 1 [IJ 1] we define i 1 1 n Tf="'O'C cP L.; i ip(A.) i= 1 1 n It follows from (21. 2) that [i[>(A ) }i::: 1 is a partition of X, so i Tf and f are simple functions taking the same values on sets of equal n1.easure and hence Tf-f. m To see that T~ f is well defined, suppose also f = 6 1". C B ' 'It i=l J j m where {B ]j:::l j f = Then I: I: O'i C A . nB . ::: ~ ~ f3 j CA. n B. j i so is a partition of X l' 0'. 1 ::: 1 J 1 J (3. whenever A.n B. =1=0 J 1 J J 1 [u. ] and thus '-1 -1110'. 1 = 13. wheneverp (A.nB.)::\=0 [u.]. J 1 J Hence L: L: O' i C~(A. n B.) = ~ ~ f3 j Cq?(A.nB.) j i J 1 1 J 1 J L: L. O' i C~(A.)np(B.)= L: L: f. Cp(A.)n~(B.) lJ J 1 ij.J - ~~ "i 1 C<r(B.) J 1 It is easy to see that Tip(rf) = J rTipf for all r E R. To show that m Tip is linear, let g = L: ",C where [B.} is a partition of Xl' i =1 J B j J £+g = L:.I:.(O'.+I3,)C nB and fA.nB.} is a partition of Xl so A'. 1J1J 1 J 1 Then J Tp(£+g) =L:L(O'i+~j) Cq?(A.nB.) =LL:(O'i+l3j)C~(A.)nip(B.) = Tq?f + Tipg 1 J 1 J since [ip(Ai)nip (B )} is a partition of X j [1I.J. = Similarly, Tip(fh) (Tipf) (Tiph) for all simple functions f, h. It follows from (18.4) that Tip extends uniquely to a doubly stochastic trans formation of L 1 (U 1) .... L I (u). Since the simple functions 1 oc (U ) and Tp is continuous in the L and Ll norms, it I 1 follows from (3.3) (xiii) that Tipf""f for all f EL (kJ ). Similarly, l 1 Tq? (fh) = (Tipf)(Tq?h) whenever f, h, fh EL (VI)' arc dcns<' in L 1 (U ) .... L (U) is also linear such that TC = Cq?(E) for all 1 E n and f = I: a. C A where fA.} is a partition of X 1 then i=l 1 i 1 If T: L E E 1\.1 1 Tf = L:a i TC A. = L:a i Cip (A.) = Tip£. 1 It follows that to a d. s. transformation of L 1 (1J,1) .... L 1 (\.l). functions, T T extends uniquely Since T = Tip on the simple 1 = T ~ on L 1 (1J 1 ). FillctllY. if~': 1\.1 -+ 1\ is a hornornorphism for which T~, = T C.p'(E) -;" "'.p,C E ' T<j\CI~'::' C~(E) for all E ( 1\1 so ip'"' <P iu]. f then -112- It is easy to see that if (X 2' 1\.2' bi ) is a finite m. s. 2 1J (X ) = U(X) = a 2 Z IS and 'J!: 1\.2 ..... 1\.1 is a homomorphism, then ~ 0 'l': /\2 -+ /\ a hOJnornorphism and T~ Let T: L 2. (~),l) ..... L 2 0 'J! = T~ 0 T,+,' (boL) be linear. T (fh) = (Tf) (Th) whenever f, h E L JTf Th dlJ = J f h dlJ l if both T and T with 2 (~1)' whenever f, h E L T is called multiplicative if Tis called isometric if 2 T (1:4 ), 1 is called unitary * are isometric. Observe that if T is multiplicative and isometric then (18.4) irnplies that T has a unique extension to a d. s. If T: L 1 (1J ) .... L 1 1 (1J) is d.s., then (17.13) (or Remark (ii) 2 2 after (18.12)) implies that T: L (1:41) .... L (IJ). It is easy to see that it is multiplicative iff T(fh) = (Tf)(Th) whenever f, h, fh ELI (14 ) iff 1 T(fh) = (Tf)(Th) whenever f, h E L oc (U ); similarly for I (Z1. 4) THEOREM. Let be linear. T isometric. Then the following are equivalent: (i) T£ ,...., £ for all f E L (ii) T 1 (14 ), 1 is induced by a homomorphism ip: /\1 ..... I\. • (iii) T is d. s. and multiplicative. (iv) T is d. s. and T*T is the identity function on L 1 (\J )' I (v) is d. s. and isometric. T We have already proved (ii) ~ (i) in (21. 3). PROOF. If E E J\1 then TC E and lJ(p (E)) = 1-11 (E). + 2 C AnB , ,....., C E so there is a ~(E) (i) ~ (ii): E I\. such that TS: = C HE Let A, B E 1\.1 and f = C A +CB=C A-AnB+CB-An B Then Tf:: Cp(A)+Cp(B)::C~(A-AnB)+C4'(B_AnB)+ 2C~(AnB) ) -113so if? (An B) C iji (A) n iji (B) and thus to show equality it suffices to "how Now iji (A) n q;(B) = [Tf = 2} and AnB = (£ = 2} they have equal ITleasure. so since Tf", f, = ~1 (AnB) = lJ.(iji(A) n <I>(B)) and thus U(~(AnB)) if?(A) n iji(B) = iji (A n B) [uJ . If An B = '/1 [jJ 1J then C A UB = CAt C get Ciji (ALJB) = Ciji (A)+ Ciji (B) = Ciji (A) Uiji (B) f-i-a. e. B so applying T we since IJ(~(A) niJi(B))= lJt(AnB)=O. Henet' <r (AUB) = <P (A)Uiji(B). (ii) ~ (iii). This is (21. 3). (iii) ~ (iv). so 1 (1-L ). 1 For all gEL oC (~1) 1 we have f gEL (101. ) 1 ffg d~1 = f T(fg) d~ = f Tf Tg dU = Jg T *Tf d~I and thus f = T *Tf. (iv) ~ (v). = Let f E L Let f, h, fh E L 1 (U ). I * Then f fh dl-Lr = Jh T Tf dU 1 ITfThdbl. (v)~(ii) C3,p.22]. LetA,BEA I · JTCATCBdU=lcACBd\J.I = Ic AnB dU I = I TC AnB dU· Since T is d. s., 0 S;TC A s; 1 so o s; (TC A) 2 s; TC A s; 1, and for A = B, f TC A - (TC A) 2 dlJ = 0 (TC A) 2 = TC A and thus TC A takes only the values 0 and 1. q;(A) = fTC A = I} so 0 S;TC AilB we have TC A = Cq; (A)" so Letting Now TC AnB~ITlin[TC A' TC B } = (TC AnB ) 2 s; (TCA)(TC B ) and since ITCAnBdlJ = I TC A TC B dU we have When AnB = 0, apply TC AnB = TC A TC T to C AUB B so iji (AnB) = iji(A) niji(B). = CAtCB to deduce iji(AUB)=iji(A)Uiji(B). Finally, lJ(iji(A)) = ITC A dU = I C A dIJ- 1 = 1J (A) 1 since T is d. s. -114(21. 5) COROLLARY. (i) Tip (ii) Tip* maps oc L oc (u.) onto L oc (~) and L 1 (\.l) onto L * = T<p Tip (~1) .... L (21. 6) Then is 1:l. PROOF. Tip: L Let ip: 1\1 -+ 1\ be a homorphism. oc 1 (14 ), 1 the identity mapping on L 1 (\.lI) and (14)· THEOREM. Let if?: 1\1 .... 1\ be a homorphism. Then the following are equivalent. * is 1 = 1. (i) Tip (ii) T <p T iJ?* is the identity mapping on L 1 (fJ.) (iii) The range of Tip (iv) if? (v) is all of L 1 (14) is an isomorphism. * is induced by a homorphism of 1\ .... 1\1' T~ (vi) Tip is unitary. PROOF. (i) ~ (ii). * * * Then Tip (TiJ? TiJ? f) = Til? f so )'( T iJ? TiJ?' f = f. (ii) ~ (iii). Obvious. (iii) ~ (iv). Let A E 1\. TiJ?f A = CA' Then there is an fA E L * 1 (fJ. ) such that 1 For all g EL oc (14) we have !fA2 Tipg dlJl = ! g TipfA2 dlJ Since TiJ?* maps Loc(IJ) onto L 2 have fA = fA so fA = C B soiJ?(B) =A [u.]. oc (I-!l) (this is implied by (Zl. 5)(ii» we where B = [fA = I}. Then CiJ?(B) = TiJ?f A = C A -115(iv) ~ (v). 11 -1 : 1\ ..... 1\1 is a hom.orphism., and * = T~* (Tq, Tq,_I) = (Tqi* Tgi)T~_1 = Tqi_l . * (v) (vi). If Tq, == T,¥ then Tq, is isom.etric so Tq, is unitary. T~ ~, =:> (vi) ~ (i). * If T<l? is unitary, then Tq, is isometric and hence induced by a hornonlOrphisrn, and it follows from. (21.5) that it is 1: 1. REMARKS. Tn 1. Let 0: X ..... X I be measure preserving and let bedefinedonM(XI,\JI) by all E E Al so To = Tq, 2. Tof::::£(o). ThenT C a E ==c _l(E) for 0 where 1> (E) = a-I (E). The relationship between m.ultiplicative unitary operators Td where a: X .... X (which are necessarily d. s. ) and the operators is m.easure preserving, was studied as early as 1932 by von Neum.ann [33, p. 6I8J, who assumed that X is a complete separable metric space with a finite Borel measure kJ, such that spheres have positive measure and every measurable set is contained in a Go with the same measure. 3. With (X, 1\,1../.) and a as in 2, Paul R. Halmos has used the operators To to find necessary and sufficient conditions for the existenceofa square root of a. [13J 4. It follows from (20.2) that the operators Tq, are extreme points of ~(XI' X) (another proof can be gleaned from [34, p. 269, Theorem 1. 4 J ). is extreme in It is easy to see that T E ~(Xl' X) is extrem.e iff T fi.X, Xl)' * * Hence the operators T ~ such that ~: A. ..... 11.1 is a homomorphism are also extreme in fi.X l' X). A complete characterization of the extreme points of .19 (X l' X) does not seem to be known. -11622. The Weak Closure of fg: g "-' f}. As always let (X, fl. IJ) and (Xl' Al,U ) be finite measure spaces with IJ(X),:: 1J1(X ):::" <l. 1 I Hccal1 that if f ELI (Xl' U ) then Of (X, IJ) = fg ELI (X, IJ): g -< f} and 1 l 6 (X, U) = [g ELI (U): g ....., £}. The problem is to determine the o(L , L:X:) f closure of 6 The case 6 r f = 0 is easy, while if 6 then Of = O(f ) and 6 = 6(f )' f o o f and fo E 6 ::j: C/J f Hence from the beginning we work only I with the sets 0 (f) and 6(f) where f EL (X, IJ). J. V. Ryff has shown the following result which we state as a lemma. For the p roof see [43, p. 432 J . .!! FE Ll[O, IJ then O(F) = fGE Ll[O, 1J: G -< F} (22.1) LEMMA . is the I o(L , Lex:) closure of 6(F) = [G E Ll[O, 1]: G,....., F}. (22. 2) THEOREM. f ELl (X, u) iff PROOF. 0: O(f) is the 1 o(L , L oc) closure of 6(f) (X, A, bJ.) is non-atomic or X is an atom. Suppose (X, 1\, \,1) is non-atomic. Let g E O(f) and let X -+ [0, a] be measure preserving such that 6 F(t) = 0f(at) and G(t) = 0g(at) on [0, 1J. for all g 0 G Then G <: F = g. Let so there is a net [Hale 6(F) such that HO! -- G in the O(Ll[O, IJ, Loc[O, 1J) topology. oc Letting h (t) = H (t/a) on [0, aJwe have h ,....., 6 , and for all v EL [0, a], f a O! 1 a a 1 a ! OO! h v cit = a ! H (t) v(at)dt -+ a ! G(t) v(at )dt = f 6 v dt, i. e. , OO! 0 og 1 h --6 intheo(L [0,a], LOC[O,a]) topology. SinceT isweakly O! g continuous, T dens e in 0(£). G h GO! .... T 6 og = g and TGO! h E 6(£). Hence 6(f) is weakly -117- is an atom, then 6. (f) = {f} = O(f). If X For the converse, suppose X is not an atom and (X, A, 1-1) is not Then X = Al U A2 where Al and A2 are disjoint sets non-atomic. of positive H1Casure, and X has an atOTIl B. Let f = 2e A + C A ' 1 1 2 let g o = -a f f dl-J. = 1 + ~(Al)/f,.L(X) so I < g 0 < 2, and let g = g 0 eX' If h E 6(f) then h = 2C so g E O(f). so !(g-h)C u,(B) since B B BI + C B where \.I. (B. ) = w.(A.) i = 1, 2, 2 1 1 dw. = go U(B) - 2u(B nB) - U(B nB) and u,(BnB ) = 0 or 2 I i is an atom. Hence for E: = IJ(B) min (g -1,2-g ) there o 0 is no h E 6.(f) such that I ! (g-h)C B dU 1< E: so lI(£) is not dense in 0 (f). (22. 3 ) COROLLARY. 1 For any finite m. s. (X, fl., IJ) and £ E L (X, U) £1 "'£} . PROOF. Let g E (2 (f). Since (X#,I,i) is non-atomic, there is a net [£1 } cL I (X~IJ*) such that £1 -- g in a(L 1(14#), L ~ (J)). a a is weakly continuous (see ~I7 or use (9.2)), T 1 a(L (\.I.), L ~ £1 U a Since Til ~ -- T . .e: = g in ~ (\.I.)). 1£ (X, A, IJ) consists only o£ atoms o£ equal measure then lI(£) is finite and hence weakly closed. We will now determine the weak closure o£ 6(£) when X consists only o£ atoms, or X has only finitely many atoms, or when X is separable. Let Xo be the non-atomic part o£ X, let (An}nEP be the atoms o£ X and let A = U A; also let a = \J.(X ) (see ~ 9). nEP n 0 0 -118If f EL (X, ~) let I (22.4) DEFINITION. Z (f) be the sct to which g I b~~longs iff there is an h ...... f such that g Xo ~ h Ixo and (22.5) LEMMA. PROOF. For all £ ELI 6. l.~ Z is easy. h 0( -, glX is a weakly. 0 net For the other inclusion, let g E Z (h } eLl (X 0( 0 Thenh oc 0( -h-f (X ,~) 0 g v dj..l, but since ~hO( v dj..l::: fA gv d~ we have 0 finally that I X hO( v dj..l -+ ! X g v dj..l, i. e., ha .... g weakly. (22. 6) THEOREM. f()r all Since Xo is 0( Extend eachh to X byh IA:::hIA. a 0( hO( v dj..l-+!X o so ,~) with h ...... h I X 0 and (see (3. 3)(x)) and for each vE Loc(X,~) we have vlx EL o so Ix IA. ~(f) CZ(f) eMf). we have tll\~n.~ is an h...., f Ruch that glXo -<hlxo and glA::.: hiA. non -atomic the re g IA:::h If Hence g EX. (X, 1\, j..l) consists only of atoms, then ~(f) ::: Z (£) f ELI (X, ~). PI~OOF. We have onLy to show that D.(O C Z(O. Then there is a net {h } c 6(f) with h .... g weakly. a a Let g E ~(f) . Let B be an atom. Now ~(B) S;~(h -l(h IB))::: ~(f-l(h IB)) and f-1(h IB)) nf-1(hAIB)::: cP a a 0( whenever hO( I B =1= h~ I B so \J.(X) < oc implies there are only finitely rnany different values ha I B. implies h that a 2a a n I I B ::: g B. implies h ~ 0( I But ha B -+ g I B, so for some a ' a ~aO O Hence there is an increasing sequence a a IA k ::: glAk' k ::: 1, " ' , n. . g ·~f and hence g EMf) ::: Z (f) in this case. I can now prove 6(f) = Z(f) in general. Then IIha n such -gill -+ 0 so n -119- (22. 7) THEOREM. (X, fI, IJ) has only a finite nUITlber of atoms, If t.(f) = Z (f). then PROOF. We have only to show that 6(f) C Z(f). It is easy to see that the condition hilA = h21A defines an equivalence relation on t.(f). Since there are only finitely ITlany atoITls, there are only finitely ITlany t.(f) == H U·.·. UHn 1 equivalence classes, HI'····' Hn say. t.(f) == HI U· ... U Hm' so there is an h E Hk and a net [h } CH with h - g weakly. a a k o atoms X B, so thus g IB = h 0 I g A =h o IA. Since be an atOITl of (X, fI, u,) Let B so (gl B )IJ(B)= IgCBdU=liITlafhaCBd\.L g is constant on B (h o IB)U(B) and . == Then g EHk for SOITle 1 ~k ~ ITl, Let g E 6(f). h ,h EH , h IA = h IA for all a. a 0 1'a 0 (so B cAl. so IB since k!(B) > O. This holds for all Let v E Lac (X ,IJ) and extend v 0 by vIA == 0, so v E Loc(X, fJ)· Then Ix to all of Ix g v du, g v dfJ = o == liITl h (}' a tx h a v dU = liITl /,X h a vd~ so h a IX -+g Ix weakly. QI IA == h 0 IA and h Hence glX 10 0' "" f '" h E t.(h Ix ) 00 (22.8) PROPOSITION. 0 o 0 so (3. 4)(ii) iITlplies h = n(h Ix ), i. e., glX 00 If f, gEL 1 (X, u,) But 0 and ~ h a IX -h Ix . 0 0 0 Ix . 000 g ~ f and g IA == flA I 0 ~ fIX 0 . then g X ---- PROOF. Define F n' G n inductively by G G n +1 == G n C X _A ' F n +l == Fn C X _A ' n n l == g, F 1 == f, Since g -< fwe have G I -< Fl' Since flA == glA we have by induction using (8.7) that G n -< F , n EP. n -120Ii X has only a finite number of atoms, = G G n ~ g Cx ~F n = f C n = f £ C ~ A. d~ .... i=n Xo and (8.7) implies glX 0 ~ fix. 0 and F n~ f C x o then for some n E P, ° as n 0 and IIF -fCX II = n 01 -+ oc 0 1 so g Ix o If f & geL REMARK. 1 0 A LEMMA. o ~ fix. 0 then (22.8) may fail since we could have flo ! £ dlJ = J g d~ = +oc with J (22. 9) IF n -f C x dIJ. since f ELI. Hence (12. 5) implies g eX A Otherwise, ° I < Jao ° I x goo If f ELI (X, IJ) then f X Z (f) is 0 o(L 1, L oc} sequentially closed. PROOF. g n IA -+ Let the sequence [g } c Z (f) with g n g I A pointwise, and there are functions h gnlxo~ hn\XoandgnlA =hnIA, n=1,2,3, ... n n -+ g weakly. Then "" £ such that Lettcan' bn[}nEP be pairwise disjoint intervals such that U [a , b [::: [a ' a [ and O nEP n n b -a = U(A }, n EP(see §9}. For each n = 1,2,3, .. , let n n n = 0h Iv C [0 H n n r 0 ' a °[+ z::; kEP (h n IAk ) C rLa k' b k [ Then tEn} c [-oc, oc J[ 0, a [ which is compact in the product topology, oc so there is a subsequence IHn} which converges pointwise l' k k= 1 everywhere to a function H. HI [0, a [ O is the limit of a sequence of decreasing functions so it is decreasing, and X there is an h E M(X ' IJ} such that 0h o h to X by h IA = g IA. Now o is non-atomic, so Ix = HI [0, a O[ a. e.. Extend o H "" h "" f and Hn .... H pointwise so n n k -121- H"" f. Also HI [0, a O[ = b h IX ""' hi X o O' and since 1 .H I [a O' a [ pointwise, we have HI [a ' a[-- h!A. O 1 Hence h", H ,..... f ELand thus Since O(f) is weakly closed, g E O(f) g -< h. so g -< f...., h SincegIA=h!A, we have using (22.8) that glX o and hence -<hIX. 0 Thus g E Z (f). Recall that the metric space associated with a finite rn. s. (X, A, \,j) is (A(\.I), d) where A (IJ) is A modulo the sets of measure zero and d(A, B) = \.dA-B) + IJ(B-A). A(IJ) the equality A = B[~J ~ -a. e. iff C A = C B A with will be viewed as A finite m. s. is said to be separable if its as sociated metric space is separable. Note that Lebesgue measure on bounded subsets of Rk and StieltjesLebesgue rneasure on bounded subsets of Rare separable [47, p. 69J. (i) (X , A nx ,\.I) is separable o 0 (22. 10) PROPOSITION. iff (X, 1\., IJ) is separable. 1 (ii) If (X, /I., \.I) is separable then for each f E L (X, 1..1) the relative 1 oc a(L ,L ) topology on O(f) is rnetrizable. PROOF. (i) (A n X (\.I), d) is a subspace of (A(IJ), d) o separable implies (A n X (lJ), d) is separable. o (I\. so (!I. (1-/.), d) Conversely, if n X o (\.I), d) is separable, then the union of the atoms of X and a countable dense subset of (A n X (U), d) is countable and dense in o -122(/I (~), el). 1 Let a be a countable dense subset of (1\(\J.), d) an! let f E 1.. . (ii) Then 0 (f) is weakly compact. so according to [20, p.143, TheoreU1 16. 7J we have only to show there is a countable subset of 1..oc which separates points of L 1 Let f; = teE: E E d3}, so f; is countable. show that f) separates points of L 1 let g, h ELI for all C Ef) then E If f (g -h)C E To db!, = 0 JE(g-h)db!, = 0 on a dense subset of (A(~), d). I Since g-h EL , E -> JE(g-h)du, is continuous on (A(~, d) and hence we conclude that JE(g-h)dw, = 0 for all E E II(bt) (22. 11) f EL 1 If (X , THEOREM. (X, ~ we have PROOF. - 0 so g = h. An x 0 ,bt) is separable then for every - 6(f) = Z (f). Now the weak topology on O(f) is metrizable so is closed and thus 6(f)c Z(f)C6(f) iU1plies Z (f) 6(f) = Z(f). Now suppose p is a saturated Fatou norm on M(X, IJ) such that L oc cL P, L P ' eLI and L P is u. r. i. If f E L P then 6(f) cL P and the problem is to deterU1ine the a(L P, LP')-closure of 6(f). A c L P we denote its o(L P , L P ') closure by P A. o(LP,Loc)ca(LP,LP')we see that If Since P A cA. By examining the proof of (22.10)we see that if £ E L P and (X ,1\ nx ,uJ is separable, then the relative a(L P, L P ') topology on o 0 0(£) is metrizable. -123- , (22. 12) PROPOSITION. If the simple functions are dense in then for every f E L P and Ac O(f), the a(L P, L PI Lp ) closure of A ac 1 equals the o(L ,L ) closure of A. PROOF. Now O(f) is p-bounded so there is an M> 0 such that We have only to show that A c Px . p(g) < M for all g E O(f). p- . A c 0(0 by (J 7.4). 1 in o(L ,L Let go EA. ac ). Ifh E L tinuous linear functional on LP. for all hELP'. an 0' 0' 20' a , (22. 13) , Fh(g) = J g h dU defines a conIt suffices to ",how that F h (gQl) ..... F h (go) such that Then there is a pI (h-v) < e, so for all g E O(f), I J (h-v)g d~ I ~ p(g)pl(h-v) ~ M 8. Now there is such that 0' 2 o Then there is a net {gO'} C A with Hence let hELP' and let IS > O. simple function v IFh(g)-F)g)1 = pI Now 0' IFh(g )-Fh(g 0' THEOREM. I < e. a implies IF (g ) - F (g ) vO' va Hence for 0 )1 ~ l F h (g 0' )-F VO' (g )1 + IF (g )-F (g )1 vO'vo If the simple functions are dense in L P ' and 1 f E LP then the a(L P , L P ') closure of (I.(£) equals the a(L , LO())~losure of !\(f). REMARK. The intuitive idea behind the definition of Z every member of 6(f) can be reached by a net in which eventually the rearrangements of f are formed by rearrangements on X rearrangements on A. This means that if g E tJ(f) (h } ctJ(f) and an index 0' 0' is that 0 such that for 0', p 2 QI a a and there is a net we have -124 - h IX 0 '" hAlx t-' 0 Q' lmpliesglX o and -<b Hencedefiningh h ,....., f, I I Q' o h Ix 0 by g Xo -< h X 0 Q' IA,.....,bAIA. t' and h Q' In this vase h 0:' IX 0 . . . glX . 0 IA ..... gIAimplicsgIA"'-,h IA. Q' 0 hlX o =h and Q' IX 0 o g I A = hi A. and hlA=glA we have -125BIBLIOGRAPHY 1. Bourbaki, N., Espaces Vectoriels Topologiques, 2nd E(l. , 1967, Chaps. I and 2. 2. Boyd, David, The I-Iilbert Transform on Rearrangement Invariant Spaces, Canadian J. of Math. 19 (1967), 599-616. 3. Brown, James. R., Approximation Theorems for Markov Operators, Pacific J. Math. 16 (1966), 13-23. 4. 1 oc Spaces Between Land L and the Calderon, A. P., Theorem of Marcinkiewicz, Studia Math. 26 (1966), 273 -299. 5. Dineuleanu, N., Vector Measures, Pergamon Press, New York, 1967. 6. Duff, G. F. D., Differences, Derivatives, and Decreasing Rearrangements, Canadian J. 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