Convergene of and the Lip α lass Christoph Aistleitner Abstrat By Carleson s theorem a trigonometri series k osπkx or k sin πkx is ae onvergent if < (1) Gaposhkin generalized this result to series of the form k () for funtions f satisfying f(x + 1) = f(x), 1 f(x) = 0 and belonging to the Lip α lass 0 for some α > 1/ In the ase α 1/ ondition (1) is in general no longer suffiient to guarantee the ae onvergene of () For 0 < α < 1/ Gaposhkin showed that () is ae onvergent if kk 1 α (log k) β < for some β > 1 + α (3) In this paper we show that ondition (3) an be signifiantly weakened for α [1/4, 1/) In fat, we show that in this ase the fator k 1 α (log k) β an be replaed by a fator whih is asymptotially smaller than any positive power of k 1 Introdution and statement of results In 1966 Carleson showed that trigonometri series of the form are ae onvergent if k os πkx or k sin πkx k < (4) Graz University of Tehnology, Institute of Mathematis A, Steyrergasse 30, 8010 Graz, Austria e-mail: aistleitner@mathtugrazat Researh supported by the Austrian Researh Foundation (FWF), Projet S9603-N3 This paper was written while the author was a partiipant of the Oberwolfah Leibniz Fellowship Programme (OWLF) of the Mathematial Researh Institute of Oberwolfah, Germany Mathematis Subjet Classifiation: Primary 4A61, 4A0 Keywords: Almost everywhere onvergene, Lipshitz lasses 1
(see [7] for Carleson s paper; f also the monographs of Mozzohi [15] and Arias de Reyna [4]) Gaposhkin [11] showed that (4) also implies the ae onvergene of series of the form, (5) where, here and throughout the paper, f is a measurable funtion satisfying f(x + 1) = f(x), 1 0 f(x) dx = 0, and belonging to the Lip α lass for some α > 1/ If the ondition that f Lip α for some α > 1/ is dropped the onvergene of the sum (4) will in general no longer be suffiient to guarantee the ae onvergene of (5) One possibility to meet this fat is to onsider series of the form k f(n k x) (6) instead of (5), where (n k ) k 1 is a fast growing sequene of positive integers A typial growth ondition in this ase is Hadamard s ondition, requesting that n k+1 n k > q, k 1, for some q > 1 Under this ondition (4) is still suffiient to have ae onvergene of (6) for f belonging to the Lip α lass for any α > 0 (Ka [13]; for reent results in the field f eg Fukuyama [9], Aistleitner and Berkes [], Aistleitner [1]) An other possibility to get ae onvergene results for series of the form (5) is to impose a stronger ondition than (4) on the sequene ( k ) k 1, typially depending on the modulus of ontinuity of f For example, for f Lip 1/ Gaposhkin [10] proved the ae onvergene of (5) under the ondition k (log k)β <, β > 3 (later, Berkes and Weber [6] showed that it is suffiient to assume β > ) On the other hand, Berkes [5] showed that there exist a funtion f Lip 1/ and a sequene ( k ) k 1 satisfying k < suh that (5) is ae divergent In the ase 0 < α < 1/, Gaposhkin [10] showed that (5) is ae onvergent, provided k k1 α (log k) β < for some β > 1 + α, and Berkes proved that there exists a funtion f Lip α and a sequene ( k ) k 1 suh that (5) is ae divergent, although k (log k)γ < for all γ < 1 α
Reapitulating these results, we see that there is a large gap between the neessary and suffiient onditions on ( k ) k 1 to guarantee the ae onvergene of (5), where f Lip α, 0 < α 1/ In the ase α = 1/ the neessary and suffiient ondition has to be somewhere between k < and k (log k)β <, β >, and in the ase 0 < α < 1/ between and k (log k)γ <, γ [0,1 α) k k1 α (log k) β <, β > 1 + α Conerning this problem Berkes and Weber [6] wrote: It is possible that in the ase 0 < α < 1/ the ondition k (log k)γ < for a suitable γ > 0 suffies for the ae onvergene of kf(n k x), but this remains open The purpose of this paper is to give a strong improvement of Gaposhkin s result for the ase α [1/4,1/) We show that in this ase for onvergene in () the fator k 1 α in (3) an be replaed by k ε for any ε > 0, or even by ( ) log k exp (7) log log k (Remark: here, and in the sequel, we write exp(x) for e x Also, to simplify notation, we understand log x as (1, log x)) Observe that the funtion in (7) is asymptotially smaller than any positive power of k More preisely, we will prove the following theorem: Theorem 1 Let f Lip α for some α [1/4,1/) Then onverges ae provided ( ) log k k exp < log log k We note that, despite our improvement, the exat best possible ondition for ( k ) k 1 to imply ae onvergene of () remains unknown In partiular, the funtion ( ) log k exp log log k 3
in our theorem grows faster than (log n) β for any β R Therefore, the question whether k (log k)β < is suffiient to have the ae onvergene of () remains open (f the aforementioned remark of Berkes and Weber) Auxiliary results Lemma 1 Let f Lip α for some α [1/4,1/), and write Then for any m 1 for some onstant f a 0 + a j os πjx + b j sin πjx m+1 j=1 j= m +1 ( a j + b j) mα This is formula (33) in Zygmund [16, p 41] Lemma Let J and N be positive integers and let (n k ) 1kN be a sequenes of distint non-zero integers Then the number of solutions to j 1 n k1 = j n k with is bounded by where is a onstant 1 j 1,j J, 1 k 1,k N ( ) 5log N JN exp, log log N This is a speial ase of Harman and Dyer [8, Theorem ] (also ontained in Harmans monograph [1] as Theorem 39) We have already used this result in an earlier paper in a related ontext (f Aistleitner, Mayer and Ziegler [3]) 3 Preparations Throughout the rest of the paper we will assume that f and ( k ) k 1 are fixed, and that f Lip α for some α [1/4,1/) Wlog we will assume that f is an even funtion, ie that the Fourier series of f is a pure osine-series (the proof in the general ase is exatly the same), and that f 1 and k 1, k 1 We write f(x) a j os πjx (8) j=1 4
for the Fourier series of f and define ( 1 1/ f(x) = f(x) dx) Furthermore, we define f(x) = 0 a j os πjx j=1 Let K be a set of positive integers Then by the orthogonality of the trigonometri system k f(kx) k f(kx) (9) We will write for appropriate positive onstants, not always the same First we will prove the following Lemma 3 Let (n k ) k 1 be a stritly inreasing sequene of positive integers, and let M < Nbe positive integers Let Then Using this we an show K = {k : n k [M + 1,N]} and K = #K ( ) 5log K f(n k x) K exp Lemma 4 Let M < N be positive integers Let L > 1 Then ( ) 5log N log L exp M<kN 4log log N M<kN This yields k L 1 Lemma 5 Let N > N 1 be positive integers Then ( ) N 1 <MN 3log N N + exp log log N N 1 <km To dedue Theorem 1 from Lemma 5 we will finally need the following k 1/ N 1 <kn k Lemma 6 Assume that for every given ε > 0 there exists an M 0 suh that M M>M 0 ε Then is ae onvergent k=m 0 +1 5 1/
Proof of Lemma 3: For s = 0,1,, we define r s (x) = s+1 j= s +1 a j os πjx (the a j s are defined in (8)) Then by Minkowski s inequality f(n k x) r s (n k x) s 0 (10) By Lemma 1 we have s+1 j= s +1 a j sα, (11) where is a onstant Let η [0,1] (we will hoose the exat value of η later) By Minkowski s inequality and the orthogonality of the trigonometri system r s (n k x) (1) a j os πjn k x + a j os πjn k x sα sα(1 η) <a j, sα sα(1 η) a j s <j s+1, s <j s+1 a j s+1 os πjn k x + sα sα(1 η) os πjn k x sα sα(1 η) <a j, j= s +1 s <j s+1 a j os πjn k x (13) j: sα sα(1 η) <a j + sα+sαη s+1 j= s +1 os πjn k x (14) By the orthogonality of the trigonometri system and (11) the term (13) is bounded by K sα sα(1 η) <a j, s <j s+1 a j s+1 sα+sα(1 η) K K sαη j= s +1 a j 6
By Lemma the term (14) is bounded by sα+sαη Therefore (1) is bounded by We hoose and see that (1) is bounded by s+1 k 1,k K j 1,j = s +1 1(j 1 n k1 = j n k ) sα+sαη s/ ( ) 5log K K exp ( ) 5log K K s(α 1/)+sαη exp K sαη + K s(α 1/)+sαη exp η = 1 1 4α, K sα+s/4 exp ( ) 5log K 1/ ( ) 5log K By (10) this implies f(n k x) ( ) 5log K K s(α 1/4) exp s 0 ( ) 5log K K exp, whih proves Lemma 3 Proof of Lemma 4 : Let M < N and L be given By (9) we have M<kN k L 1 M<kN k L 1 s: 1 s L s: 1 s L k f(kx) M<kN s s 1 k < s M<kN s 1 k < s k f(kx) f(kx) 7
Using Lemma 3 we get M<kN s 1 k < s f(kx) ( ) 5log K(s) K(s) exp, (s) where Trivially K(s) N M K(s) = # { M < k N : s 1 k < s} Using the Cauhy-Shwarz inequality we finally get, s ( ) 5log K(s) K(s) exp M<kN (s) s: 1 s L k L 1 ( ) 5log N log L exp s K(s) 4log log N s: 1 s L ( ) 5log N log L exp 4log log N M<kN k 1/ 1/ This proves Lemma 4 Proof of Lemma 5: Let N > N 1 be given We have N 1 MN N 1 <km N 1 MN k f(kx) + N 1 <km k N 3 N 1 MN N 1 <km k N 3 (15) The first term in (15) is bounded by N 1 MN N 1 <km k N 3 N 1 <kn k N 3 k N (16) To estimate the value of the seond term in (15) we use Lemma 4, where we hose L = N 3 8
Then we have N 1 <kn k N 3 ( ) 5log (log N ) 1/ N exp k 4log log N N 1 <kn 1/ (17) Imitating the proof of the Rademaher-Menshov inequality (see [14, p 13]), we an easily show that ( ) 5log N 1 MN k f(kx) (log N ) 3/ N exp 1/ k 4log log N N 1 <km N 1 <kn k N 3 ( ) 3log N exp 1/ k log log N N 1 <kn Combining this with (16) we have ( ) N 1 MN 3log N N + exp log log N N 1 <km N 1 <kn k 1/ This proves Lemma 5 Proof of Lemma 6: Assume that for every given ε > 0 there exists an M 0 suh that sup N ε (18) N>M 0 By Minkowski s inequality this implies sup Therefore and inf M N >N 1 >M 0 k=m 0 +1 N k=n 1 ε inf N sup M N >N 1 >M k=n 1 = 0 sup N >N 1 >M whih implies the ae onvergene of N k=n 1 = 0 ae, 9
4 Proof of Theorem 1 Assume that ( k ) k 1 satisfies ( ) log k k exp < (19) log log k As a onsequene of (19) we have for r 1 r+1 k= r +1 ( log( r ) k exp ) log log( r ) By the monotone onvergene theorem and Minkowski s inequality we have, for any m 1, sup M M> m k f(kx) = lim M sup w k= m +1 m <M m+w k f(kx) k= m +1 M lim w k f(kx) mrm+w 1 r <M r+1 k= r +1 Together with Lemma 5 this implies sup M M> m k f(kx) k= m +1 M r m r <M r+1 k f(kx) k= r +1 r m ( r ) + exp m + r m ( log( r ) ) exp log log( r ) ( 3log( r ) ) log log( r k ) r <k r+1 1/ (0) For any given ε > 0, we an hoose m so large that (0) is smaller than ε Therefore, by Lemma 6 the series is ae onvergent Referenes [1] C Aistleitner The law of the iterated logarithm for k f(n k x) In Dependene in Probability, Analysis and Number Theory A Volume in Memory of Walter Philipp, pages 1 34 Kendrik Press, 010 10
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