A Counterexample in the Theory of Fourier Transforms in the Complex Domain Citation Delaney, William Kenneth (1975) A Counterexample in the Theory of Fourier Transforms in the Complex Domain. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/8s0b-ck03. https://resolver.caltech.edu/CaltechTHESIS:08312021-161900257 Abstract The Borel transform of an entire function of exponential type is defined outside a closed bounded convex set D. Paley and Wiener have given a necessary and sufficient condition on the entire function F(z) such that φ(w), the Borel transform of F(z), is contained in E 2 (ℂ\D) for the case when D is a line segment. Kacnel'son has shown that the natural extension of this result provides a necessary condition for a general closed bounded convex set D. Here, by counterexample, we show that the natural extension does not provide a sufficient condition. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Luxemburg, W. A. J. Thesis Committee: Unknown, Unknown Defense Date: 23 May 1975 Record Number: CaltechTHESIS:08312021-161900257 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:08312021-161900257 DOI: 10.7907/8s0b-ck03 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 14348 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 31 Aug 2021 17:17 Last Modified: 02 Aug 2024 22:57 Thesis Files PDF - Final Version See Usage Policy. 7MB Repository Staff Only: item control page

A COUNTEREXAMPLE IN THE THEORY OF FOURIER TRANSFORMS IN THE COMPLEX DOMAIN Thesis by William Kenneth Delaney · In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1975 (Submitted May 23, 1975) ii ad majorem ])ft gloriam iii ACKNOWLEDGEMENTS . I wish to express my thanks first to my Lord, Jesus Christ, who provided the courage and the strength to complete this thesis. I wish· to thank especia.l.ly my advisor Professor W. A. J. Luxemburg for his patience, encouragement, and constant help throughout the research and writing of' this thesis. I wish also to express thanks for the support I received while a graduate student in the form of a fellowship received under the National Defense Education Act and a teaching assistantship received from the California Institute of Technology. A special thanks goes to Professor Emeritus H.F. Bohnenblust whose insight and encouragement led me to seek the counterexample contained in this thesis. I also wish to thank my family and friends who encouraged me during times of frustration and shared the joy of each step forward. Finally I must thank Mrs. Frances Williams for her excellent typing of the manuscript. iv AOOTRACT The Borel transform of an entire function of exponential type is defined outside a closed bounded convex set D. Paley and Wiener have given a necessary and sufficient condition on the entire function F(z) 2 such that ~(w), the Borel transform of F(z), is contained in E (t\D) for the case when Dis a line segment. Kacnel'son has shown that the nat~al extension of this result provides a necessary condition for a general closed bounded convex set D. Here, by counterexample, we show that the natural extension does not provide a sufficient condition. V TABLE OF CONTENTS Page ACK;NOWLEOOEMENI'S. iii AffiTRACT • iv CHA.Pl'ER I. INrRODUCTION 1 CHAPrER II. A COUNI'EREXAMPLE 6 REFERE:OCES. .. 23 1 CHAPrER I INTRODUCTION 1• For a given entire function of exponential type, a • • • + ~ zn + • • •, the Borel transform cp(w) of n! +···+aw n -n-1 + •••• Let D be the conjugate indicator diagram of F(z) and let h(8) be the in- A discussion of the indicator function and dicator function of F(z). the conjugate indicator diagram is contained in Polya [11, p. 571-597), and Boas [ 2, p. 66- 77]. Then the following integral transforms relate F(z) and q:,-(w): where r is a rectifiable Jordan curve homotopic to a positively oriented circle with Din its interior, and 00 ~(w) = ·e )exp(-wre1·e )e1·edr J F(re 1· e) > h(8). 1 0 where z = re Our i8 and Re(we (11, p. 578-585], (2, p. 73-75]. work here begins with a consideration of the classical Paley- Wiener Theorem. following: The theorem as formulated by Paley and Wiener is the 2 THEOREM. (Paley-Wiener) [9, p. 1-13], [10, p. 224-234], (1.1) [7, p. 386-389]. The two following classes of entire functions are identical: (1) the class of all entire functions F(z) such that F(z) = O(eAI z I) and F(z) E ½/-ex,, co) (2) the class of all entire functions of the form A F(z) = J-Af(u)eiuzdu, such that f(u) E ½(-A,A) Note that from Plancherel's theorem we have the result co J 2 IF(x)I dx = 2~ -oo A · J-A jf(u)I 2 du. In order to restate (2) in a more convenient fonn, we will need the following definition. DEFINITION. [ 4, p. 168]. least two boundary points. Let D be a simply connected domain with at A function f analytic in Dis said to be of class F!(n) if there exists a sequence of rectifiable Jordan curves, c1, e 2 , ••• , en, ••• in D, tending to the boundary in the sense that en eventually surrounds each compact subdomain of D, such that 3 Restating (2), we have (2~) the class of all entire :functions F(z) of exponential type such that the conjugate indicator diagram Dis contained in the line segment [-iA,iA], and ~(w), the Borel transform of F(z), is contained 2 in E («: \ D). [1 O, p. 224-234]. We can extend the result of Paley and Wiener so that a conjugate indicator diagram different f'rom a line segment along the imaginary axis can be considered. The first result is an extension to the case of any line segment, and the second result is an extension to the case of a convex polygon. (1.2) THEOREM. The following two classes of entire functions are iden- tical: (1) The class of all entire functions F(z) such that e -OzF () z = O( eAlzl) , and e -arei& F ( re i0) E L ( -oo, oo ) • 2 (2) The class of all entire functions F(z) such that .e a+iAe - i F(z) = J a-iAe -iS f(u)euzdu,where f(u) is in L of the line segment 2 For the next theorem, let D be a convex polygon with corners a , 1 a2, ... , at numbered clockwise. Let Al.J.. = la.l. - a.J I be the length of the side from a. to a. (j = i + 1 or j = l. J . ,, i = -t). Let e.. be defined l.J 4 such that 8ij E [ 0,2~ ) and a 1 - aj = A1 je -i9•1 · J. We will also let a= ½(ai + aj), A = ½Aij' and 9 = - ½~ + 9ij (1.3) THEOREM. (7, p. 389-391] The following three classes of entire functions are identical: (1) The class of all F(z) such that F(z) is entire, Dis the con- jugate indicator diagram of F(z), and F(z) ~ F12 (z) + F23 (z) + ••• + Ftl (z), where Fij(z) is in class (1) of Theorem t.2, with a, A, 8 as above. (2) The class of a11 F(z) such that•F(z) is entire, Dis the con- jugate indicator diagram of F(z), and F(z) = F12 (z) + F23 (z) + ••• + Ft 1 (z), where Fij(z) is in class (2) of Theorem 1.2 with a, A, 9 as above. (3) The class of all F(z) such that F(z} is entire, Dis the con8 18 jugate indicator diagram of F(z), and exp(-Orei )F{re ) E ½{O,oo), with a, 8 as above for i and j as above. The Paley-Wiener theorem has been extended by Kacnel'son [6, p. 106] in one direction to the case of a general conjugate indicator diagram D. The result is: (1.4) THEOREM. Let F(z) be an entire function whose conjugate indica- tor diagram is the bounded closed convex set D. Let cp(w) be the Borel 2 trans form or F ( z ) where cp( w) is contained in E ( t \ D), and let 6 be the 5 boundary of D traced counterclockwise. Then :for all 8 in (0,2,r), where C~ is a constant which depends only on~. The question which we will answer here is whether the condition: 00 J-IF(rei 8) 2exp(-2rh(8))dr ~ M for all O < 8 2,c, given in Theorem 1.4 0 1 . <_5_ 2 as a necessary condition :for ~(w) to be in E (~ \ D),is also a su:fficient cond.i tion. 6 CHAPrER II A COUNTEREXAMPLE 2. The question we ask here is whether the necessary condition given in Theorem 1 .4 is also a sufficient condition. This would give us an extension of the classical. Paley-Wiener Theorem for the case of a segment and for the case of a polygon. (Theorems 1.2 and 1.3). The con- jecture would be: (2.1) CONJECTURE. Let F(z) be an entire function whose conjugate in- dicator diagram is the closed bounded convex set D and whose indicator function is h(t). If 00 for all v in [0,2~), then where ~(w), the Borel transform of F(z), is contained in½(~) and~ is the boundary of D traced counterclockwise, or equivalently, ~(w) is in 2 E (a: \ D). The following counterexample will show that the conjecture is false for the case of a circle, even with the additional hypothesis that 00 J 0 IF(rei~)l 2exp(-2rh(*))dr is continuous in,. 7 [4, p. 2). A function f'(z) a.na.1.ytic in the unit disc is DEFINITION. said to be of class Hp(l '::_p < m) if ln jf(rei 8 )jPd8 is bounded as r 0 increases to one. NarE. HP=> Hq if O < p < q ~ 00; and if A is the unit disc, [4, p. 2]. then for all p > 0 we have that #(A) = Hp(A). (2.2) THEOREM. [4, p. 17]. Given f(z) in class HP, then f'or almost a.11 8, there exist non-tangential limits for f. We will ca.11 these 8 limits f{ei ). (2.3) THEOREM. If(rei8) j2d8 = l [4, p. 8]. 2 2n Ia n 1 r • Hence n=O 2Jt 8 2 jf(rei )j d8 increases to J lf'(ei 8)1 d8 as r increases to one. 2 0 Hence f(z) is in -if if and only if ln 00 8 2 jf(ei )j d8 = 2n I janj 2 is n=O finite. (2.4) THEOREM. [4, p. 20]. Every function f'(z) r O of class Hp can be factored in the form f(z) = B(z)g(z), where B(z) is a Blaschke product and g(z) is an Hp function which does not vanish in NOTE: [4, p. 19]. lzl < 1. If B(z) is a Blaschke product, then IB(z)I < 1 in Iz I < 1 and IB(z) I = 1 almost everywhere for Iz I = 1. 8 We shall now turn to the construction of the counterexample. (2.5) COUNI'EREXA.MPLE. The main idea used in the construction of the counterexample is the fact that for each O ~ t < 2~ the function 1/(1 - e-itz} is of class HP, 0 < p < 1, in the unit disc D = {z : Iz I ~ 1) and not of class H1 in D. This will allow the con- struction of a function analytic in the unit disc D with the property 1 that for all O ~ ; < 2,t it belongs to the class H (D( ;)), where D(t) = {z: lz - ½e 1'1 <½},but not to the class H1 (D). To accomplish this we sha.ll first prove a series of lemmas. In the remainder of' this chapter we shall denote the circle {z: lz - ½ei ♦ I = ½), the boundary of D( ♦ ), by r(,). Furthermore, for each O <A< 1 we de- fine ¾(,t) = d-- s ,t r( ~) 1, - z , -A Idz I, 0 ~ t < 2,c, where the function (1 - z}-A denotes that branch in lzl < 1 which for z = 0 is equal to 1 • The power series expansion of (1 - z)-l/2 in the neighborhood. of the point z = a, lal < 1, is of the form 00 (1 - z)-l/2 = l (- 12)(-l)n (z - a)n/(1 - a)n+A/2, n=O which holds for all Iz - al < 1 - I~I• In particular, for .,., i8 . "' z = (e 11 + e )/2, O ~ ;, 8 < 2,t, and a= e 11 /2 we obtain the expansion (1 - z)-A/2 = f (n=O 12)(-l)n ein8/2n (1 - ei~/2)n+A/2. 9 (2. 6) If O < A < 1, then for all O ~ ♦ < 21t we have LEMMA. ¾(0) ~ 1/(1 - A). PROOF. If O < A < 1, by Isin }I :::_ 1!1 for !el < ,c, we obtain 11_ (0) = ½ (1 - A)-l < 1/(1 - 11.) (2.7) LEMMA. For a.l.l O <A< 1, HA(v) is a 2~-periodic even function of f which attains its maximum a.t t = 0 and its minimum at PROOF. From Parseval's relation it follows innnedia.tely that I\(~)= 2 L (- ;/2) /4n(f - cos ,)n+A/2 00 (2.8) ~ n=O = n. In 10 Since(~ - cos ♦) is a 2~-periodic continuous even :runction attaining its minimum. at t = 0 and its maximum at += ~, the result follows easi- ly from Lemma. 2.6. If cos ♦ > }, then I\ (♦) is increasing in A, 0 < ). < 1. (2. 9) LEMMA. PROOF. If O < a < 1, then a 0 < X < 1• For cos ♦ > X is a decreasing function in A for 1 5 4 we have that (4 - cos ,v) < 1, and so 2 (,f - cos •>-A is increasing in A. If we then observe that (- ;(2 ) is increasing in)., the required result follows from (2.8). (2.10) LEMMA. If ~-/ 0 and j wl < 1, then for all O < A < 1 we have z E r(\f,) we have that 1, - z1-1 ~ c✓ ~ - cos ' - ½)- 1 • Hence, if ,ir -/ 0, J\ (,!, ) ~ H, (,) ~ ½ (J f - cos ~ - ½) -l • Fram the elePROOF. For •l that J f - cos 2 ,1, - 2 tain l¾_(w) ~ 1/(, ~4 2 + 4! and Iit I < 1 , it follows easily mentary inequality cos • ~ 1 - 4 ½~ ~ - ~ , and so, finally, since I+ I < 1, we ob2 - 2* /3) ~3/l; and the proof is finished. We are now in a position to show that there exist functions in the co unit disc of the type F(z) = \ l n=1 . -A a f (z) where f (z) = (1 - ze-itn) n n n n 11 with O < ). n < 1 , O < ♦ n < 2,t, and f n {0) = 1 for all n = 1 , 2, • • • ·s uch · that F EH, (D( ♦)) for all O ~ ,t < 2,r and Fi_ H, (D). we will see that it For this purpose, is necessary that the sequence {A. } increase sufn ficiently rapidly to 1 and the sequence[ ♦} decrease sufficiently n slowly to O. From Ifn (z)I <_ (1 - IZD-1 it follows immediately that, if 00 an > 0 for all n and l an < 00, then F exists a.nd is analytic in the n=l unit disc. In view of this fa.ct we shall assume from now on that We begin with the following theorem. (2.11) THEOREM. 00 00 If \L. an< 00 and \l n=l n=l 2 a/tn < oo, where 'n is a se- quence strictly decreasing to O, then H, ( .,F) = J--,t Jr(,t) IF(z) I Idz I is a 2,r-periodic f'unction of t, continuous for all O < ; < 2,r and contin- uous from below at ,t = 2n: • PROOF. For each 6 > 0 let S~ = {+: I ♦ - tnl > 6, n = 1, 2, ••• }. Then from lrn(z)I ~ c✓ f - cos(, - •n> - ½)-1 on r(.,), n = ,, 2, ••• , it follows that IF(z)I is bounded on the set U(D(t) : t E S ). 0 Hence, from the bounded convergence theorem it follows immediately that H,(w,F) is finite and continuous in, for a.11 ♦ except possibly at the points tr= ,itn(n = 1, 2, .•• ) and i' = o. To show that H, (,t,,F) is 12 continuous at ♦ =· • n (n = 1, 2, ••• ), let On be a sma.11 neighborhood of ♦ n such that I♦k - , I > 6 for all k I n and for all ♦ E O • - n neighborhood exists since~ Then for ♦ E O n l ak lfk(z)I n is a sequence strictly decreasing to zero. and z Er( ,t) we have that + anlfn(z)I. Such a IF{z) I < - For the same reason as above the func- k;ln tion l akffk(z)I is bounded on U(n(,) k#n ♦ E on ) , and If n ( z) I E L1 (r ( v)) with the L norm uniformly bounded for t E O by the L norm for ♦ 1 1 n Lemma 2.7. n by Hence it :follows again fran the bounded convergence theorem that H,(t,F) is continuous at the points~= tn(n = 1, 2, .•• ). For it= 2n:, we first note that it is clear from (2.8) that 1\(4') increases as 1,1 decreases to zero. Hence, I Jr(2~) I F(z)I fa.zl - Jr(e) fF(z)I la.zll < J IF(½ ½eH) + ,c -1( . 00 + 4:n: I n=m+l F(½ ei 8 + ½eit) I dt 13 00 The last step is :rrom Lemma 2.10. Since the second term converges to zero a.s m tends to infinity by hypothesis, and for any fixed m the first sum clearly goes to zero as 8 increases to 2~, H, ( ♦,F) is continuous f'rom below at 2~, completing the proof. Note that if we prove H, (i, F) is continuous from above at o, we then will have H, ( ♦,F) continuous for all ♦ and hence uniforml.y bounded in ,. So far, no particular hypotheses on the sequences {a}, (X }, and n n 00 { ,i, } have been placed other than a n n > O, \l a n < co, 0 < Xn t 1, n=l 00 0 < vn t o, a.nd \l a /♦ n 2 n < oo. We shall now show that there a.re se- n=l 1 quences {a}, {A.} and {11.} such that the resulting function FiH (n), n n n but H, ( "' F) is continuous for all t. a lemma. To this end, we shall first prove 14 (2.12) LEMMA. , o, 0 <Ant 1, Assume that ; 1 > ~2 > •·• > ~n > ··• co and an > o, n = 1, 2, • • • with l an < 00. Let p 1 > ~l and pk be the n=1 midpoint between ♦k-l and tk (k = 2, ••• ) • JI I= I ( ) I I I ~ 1 2,r 1 F z z dz Then 1¾ (D, F) = 00 1 L °it ;-:x:- (.k - Pk+1 ) -Ak - l 2,t \ k.=:1 k 00 - t" l (pk - Pk+l) l an(d(1it, ♦n))_).n, where d(~,,J,n) is the distance lQ:=1 PROOF. ~k With ½t a.s defined above we have J IF(z) I la.zl ~ lzl=l 00 00 00 00 00 p > - k s I ~1 8 - "' , f "k ♦ -A k d~ > - J k ~1 -A I& - t k I k dA = It - p k , , -A k k+ l k 1 - A • Now 15 L an J If (z) I I I• For this case observe that, ~k 7t consider the sum \ n L dz where d(1it, ;n) is the distance of 'n to the interval 1it(k / n); and the re- quired result fol.lows. We sba.ll now make the following selections: a 2 and An = 1 - (n + 1 )- , n = 1, 2, • • • • Then n 3 = n- , l an < oo, ♦n = 1/ Jn, a.nd so F n=1 00 exists; l ajtn2 n=l = \ L. n -2 < 00, and so Theorem 2. 11 holds. What is n=l 1 left to show is that F I. H (D) and ~ { t,F) is continuous from above at t = O. 1 To prove that F f. H (D) we shall use the result contained in Lemma 2.12. (2. 13) THEOREM. (n = 1, 2, ••• ), then Fis a.nalytic in lzl < 1 satisfying the following properties: (ii) (i) II, (w,F) is a continuous function in~, 0 ~ * ~ 2~, 1 F f. H (D), and, (iii) F E Hp(D) for all O < p < 1. 16 PROOF. From Theorem 2.11 it follows that to establish (i) we need only show that H, ( ♦,F) is continuous from above at ,t = O. To this end, let 8 E (O,o), Then f 6 > O. Jr(e) IF(z)I ldzl - fr(o) IF(z)I l<izl f 00 n=L+1 00 l n=L+l an HA ( • n - e) = T, + T2 + T3. n For any fixed L, T1 converges to zero as 8 goes to zero. From Lemma 2.10 it follows immediately that T converges to zero as L tends to in2 3 4 finity. Let N be defined such that 8 E (tN+l' ♦N], a.nd let A= [N-N / 1 Then we write T 3 = n=L+l B + 00 l an 1¾_n( tn - 8) = s, (t - 9} + L a n H.. -"-n n \' n=A+1 + s2 + s3 . n=B+l To estimate the parts T , T , 2 1 and T we observe first that, by Lem3 1 2 mas 2.7 and 2.10, HA(t) ~min ((1 - AJ, 3/t) and that .!. 3 4 2 (k - k / )- - k-½ ~ ½ k / then HA ( ~n - 9) ~ 12 n n 3 4 . k- 3 / 2 = ~- k- 3 / 4 fork~ 1. Hence, if n ~A, 3 312 , and if n ~ B + 1, then HA ( •n - 8) --s_ 24(N+ 1 ) /2, 2 ( ♦n - 8) '.:;_ (N + N3/ 4 ) • Observing that and for A+ 1 < n< B we have that H,. n 17 A \ L k -3 lB • k-3 ~ 2(N - ~/4) -3 • Ji314, and k=A+l k=L+l fL k -3 3/4 -2 ~ (N + N ) we obtain, finally, that s 1 + s2 + s3 ~ lo=B+-1 3 24(L + 1)-½ + 2(N - N3/ 4 )- • 2 i 14 . (N + N3/ 4 / +24(N+ 1) 3/ 2 (N+N3/ 4 )For a given e > o, first which tends to zero as N ➔ oo and L ➔ oo. choose L so large that T < e/3 a.nd s < e/6. 1 2 Then with this L fixed choose 6 > O so small that s + s < e/6 and T < €/3. 2 3 1 continuous :from above at t = 0 and (i) holds. ak oo n, (,t,F) is For (ii) we use Lemma Observe that 2.12. \ Hence l 1 - Ak (♦ _ k ' = loo ( k + 1 ) 2 • 1 ( _1 _ 1 ) ( k+ 1 ) -2 ) 1 -1\k 2 Pk+l k3 Jk_ Jk + 1 > k;::1 lc=l Observing that for k I- n, d( Iit, 'n) ~ > ( ,t - ·n - > 1 p n+ 1 ) = ½ (-1 --=---===-:,~:-----:=:=:..... Jn _Jn 1+ 1 ) = ½Jn Jn + 1 (Jn + Jn + 1 ) 1 - 4 (n + 1 )Jn + 1 k=1 00 00 00 \ L , we obtain the result that (pk - :Pk+l) l an(d(~, ♦n))-An <- l n/k k:=1 ~ (pk - :Pk+l) l n=l 4 ( n+1 ).Jn➔T n 3 < - 18 l (n+l ) ./ii+f < ""· 3 1 Hence, F f. H (D). In order to show that n n=l F E HP(D) for a.11 0 < p < 1 we observe that 00 From Lemma. 2. 7 it :follows that Jlzl=l I n ( Ip I I ~ f z) dz co latter stnn converges for all 1 > p > 1/3. 2rc/ ( 1 - pA n), and 00 Hence, :rrom the Note (before Theorem 2.2) it follows that F(z) E Hp for all p satisfying O < p < 1. This completes the proo:f of the theorem. REMARK. It was shown by R. M. Gabriel and F. Carlson (see [5],[3]) that, i:f :f is analytic in the unit disc Dandy is a. cl.osed rectifiable curve inside D, then for a.11 p > 0 when V(z) = J I~ arg(x - z)!a.x. If vis the circler(~), then V V(z) ~ 21t for all Iz I = 1, and so, (*) 19 The :function F of Theorem 2.15 shows that the inequality in(*) cannot be reversed, that is, even if the left-hand side of(*) is uniformly bounded in ♦, the right-hand side need not be finite. We return now to the construction of the counterexample. From F E If {O < p < 1) it follows from Theorem 2.4 that F(z) = B(z) • G(z}, lzl < 1, where Bis a Blaschke product and GE Hp for a.11. 0 < p < 1 and G(z) IO for all lzl < 1. Let g{z) = B(z)•(G(z))½, where (G{z))½ is a well-defined branch of the square root Then IBl ~ 1 shows that g E function. Jr(,) I I I I <- Jr(,t) IF(z) I I I for g 2 dz since IBI ~ 1, the :fact that dz J r( ♦) If ( D ( ♦)) for all ♦ and all O ~ ♦ < 2,r. Furthermore, 2 I g 1 I dz I is continuous in ♦ can be shown in exactly the same way as in the proof of Theorems 2. 1 1 and 2. 15. 1 Since Fi H (D) it follows from IBI = 1 almost everywhere on lzl = 1, that g f lr(D). Now let q>(w) = !.. g ( !..) • Then cp is analytic for lwl > 1, Ice(w) I ➔ O w w 2 as lwl ➔ oo, and q> E L 'a long every line in the complex plane touching the unit circle {for each ♦, 1/w maps r( ♦) into a line t(t) tangent to norm continuous as the point of tangency is varied. However, 20 Let f{z) = 1 2re 1 J V q>{w)ezwdw, where 'Vis a. simple closed contour containing the unit circle in its interior, be the entire function of exponential type whose Borel transform is equal to cp. Then by Plancherel's theorem we have that for all O ~ • < 2rc 2 1 2rcJ e- rlr(re ;)1 ar = (2.14) J l~{e-i ♦(1+it))l dt cc, 2 0 2 -co Hence, we have obtained the following resuJ.t disproving the Conjecture 2.1. (2.15) THEOREM. There exists an entire function f of exponential type 2 such that the function p(8) = [ e- rlf(rei 8 )! 2 dr is a continuous func- 0 tion in 8, 0 ~ A ~2rc, whose indicator diagram is contained in the unit disc and whose Borel transform cp is not of class Ir(a \ D). REMARKS 1. By means of a conformal mapping argument the above counter- example can be shown to exist for domains bounded by convex curves which have a continuously turning tangent. The result of Levin con- cerning convex polygons quoted in the introduction shows, however, that the conjecture holds for convex polygons. 2. To find an entire function f with indicator diagram equal to the unit disc and violating the conjecture one simply adds to fan en- f z n~·/n2 (n!)! whose tire function of exponential type of the form L n=1 21 Borel transform has fwl = 1 as a natural boundary and is nevertheless in the class if(t \ D). If we integrate (2.14) with respect to 8 fran O to 2,c, then we 5. obtain the formula +co = J Icv( -i ~ 2 e 9 2 ( 1 + it) ) I de dt. -00 By means of the change of coordinates X + iy = ei& ( 1 + it), the latter integral can be written in the form 2: I s Icp(x iy) I2 cJx~2-+___,,,y2--_-,) + 2 X 00 2 J1 / 2r 2,c ,..Jr co Jo I I an r-n-1 e-i(nt1)912 d8 dr = - 1 n=O lan 12 r (n + ½) CO r f(z) d.x dy = +y~l co 1 = 2,c _, 2n+2 dr = --r-(-n+-l~).----, where Since r(n + -½)/n! - 1/Jn, we obtain that, 'if 22 I !anl /Jn 2 J0 e-2r!f{re18 )! 2 00 + 1 < 00, then n=O cp E If (c \ D) if and only if l • 1 dr E L {o,2n). Since 2 la 1 < oo, it follows now easily that n n=O J 00 2 e- r If(reie) I2 dr E L1 ( O, 2n:) need not imply that cp E If (C \ D). 0 00 deed, the function l n=1 zn/Jn • (n!) has the required properties. In- 23 REFERE~ES [1] R. P. Boas, Jr., ''Representations far Entire Functions of Exponential Type." Ann. of Math. (2) 39, (1938), 269-286. (2] R. P. Boas, Jr., Entire Functions, Academic Press, New York, 1954. (3) F. Carlson, "Quelques inegalites concernant les fonctions analytiques", Arkiv f8r (4] Ma.th., Astr. ochFysik. 29B, (11) (1943), 1-6. P. L. Duren, Theory of !!P Spaces, Academic Press, New York, 1970. [5] R. M. Gabriel, "Some Results Concerning the Integrals of Moduli of Regular Functions along Certain Curves", J. Lond. Ma.th. Soc. 2, (1927), 112-117. [ 6] V. E. Kacnel' son, "Generalization of the Paley-Wiener Theorem on the Representation of Entire Functions of Finite Degree" (Russian) Teor. Funkheit Funkcional Anal. i Prilozven. 1, (1964), 99-110. (7) B. J. Levin, Distributions of Zeros of Entire Functions, .American Mathematical Society, Providence, 1964. [8] M. K. Liht, "A Remark on the Theorem of Paley and Wiener on entire Functions of Finite Type." (Russian) Uspehi Mat. Nank. (no.1(115)), 19, 1964, 1 69-171 . [9] R. E. A. C. Paley and N. Wiener, Fourier Transforms in the Canplex Domain, American Math. Soc. Colloquium Publications, Vol.. 19, New York, 1934. 24 [ 10] M. Pla.ncherel and G. Polya, ''Fonctions Entiers et Integrales de Fourier Multiples," Comment. Ma.th. Helv. 9, (1937), 224-248, 10(19:38), 110-163. (11) G. Polya, "Untersuchen 'fiber L'ilcken und Singularitlf.ten von Potenzreihen", Ma.th. Zeitshi:rt 29(1929), 549-640.