WHAT STRONG MONOTONICITY CONDITION ON FOURIER COEFFICIENTS CAN MAKE THE RATIO \\f-sn(f)\\/en(f) BOUNDED? S. P. ZHOU. (Communicated by J.

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1 proceedings of the american mathematical society Volume 121, Number 3, July 1994 WHAT STRONG MONOTONICITY CONDITION ON FOURIER COEFFICIENTS CAN MAKE THE RATIO \\f-sn(f)\\/en(f) BOUNDED? S. P. ZHOU (Communicated by J. Marshall Ash) Dedicated to Professor Tingfan Xiefor his sixtieth birthday Abstract. Let {</>n} JÍ, be a positive increasing sequence and <t>nf(n) decrease. We ask what exact conditions on {<j> ) make \\f Sn(f)\\/E (f) bounded? The present paper will give a complete answer to it. 1. Introduction Let C2n be the space of all real continuous functions of period 2n with norm = max_<x<00. For an even function / e C2n (we always assume that f(x) is an even function), denote its Fourier series by f(x)~yjf(n)cosnx> and the «th partial sum of the Fourier series by n S (f, x) = ^/(fc)cosa:x. k=0 In Fourier analysis, it is always an interesting question whether or not the approximation by partial sums of a continuous function can achieve the best approximation rate in uniform norm. It is well known that (cf. [Zyg]) (1) \\f-sn(f)\\ = 0(log(n + l)en(f)), where E (f) is the best approximation by trigonometric polynomials of degree n. We also know that (cf. [Zyg]), in general, the factor log(«+ 1) in (1) cannot be removed. However, people always hope that, for f(x) in some subclass of C2l[ (especially in those which can be easily characterized like continuous functions with restricted Fourier coefficients), one can obtain some better estimates, preferably, (2)_ \\f-sn(f)\\ = 0(En(f)). Received by the editors June 22, 1992 and, in revised form, October 10, Mathematics Subject Classification. Primary 42A10, 42A20, 42A32. Key words and phrases. Approximation, Fourier sums, coefficients, strong monotonicity. Supported in part by NSERC Canada American Mathematical Society /94 $ $.25 per page

2 780 S. P. ZHOU One result in this direction is due to [NeRi] showing that \\f-sn(f)\\<4een(f) holds for every function / e C2n with decreasing and logarithmically concave Fourier coefficients (that a sequence a is logarithmically concave means that it satisfies a2 >an+xan-x for every n). At the same time, we can easily see that for / e C2n with positive Fourier coefficients, consequently for / e C2n with monotone Fourier coefficients, \\f-sn(f)\\ = 0(cok(f,n-x)), k>l, obviously holds, where cok(f, t) is the modulus of smthness of order k of f(x), and the O in the above inequality depends upon k only (the same result also holds for odd continuous functions of period 2n by a little more complicated arguments). Probably from this point of view, [Xie] asked Problem XI. Does there exist a positive constant M such that, for every function / e C2n with positive Fourier coefficients and all n 1,2,..., \\f-sn(f)\\<men(f)1 Later, in a seminar, Xie asked further Problem X2. Does there exist a positive constant M such that, for every function / e C2n with decreasing Fourier coefficients and all «= 1,2,..., \\f-sn(f)\\<men(f)1 A little surprisingly, [Zho] constructed a counterexample showing that there exists a function / e C2n with strongly monotone Fourier coefficients (which means that nf(n) decreases, in symbol, nf(n) J.) such that Ums p!tftça>o. n- log ne (f) This gave a negative answer to Problem X2 as well as to Problem X1. Some natural questions now arise: What kinds of strong monotonicity conditions on Fourier coefficients can guarantee (2)? Is there any counterexample which satisfies nsf(n) j for sufficiently large s > 0 (the counterexample given in [Zho] does not satisfy this condition) but makes /- S (f)\\/e (f) unbounded? One can easily verify a trivial example that if q"f(n) j for some q > 1, then (2) holds. Therefore, a general statement of the problem must be as follows. Let O = {4>n} Li be a positive increasing sequence and 4> f(n) [. Then what are exact conditions on {</> } which make \\f - S (f)\\/en(f) bounded? The present paper will give a complete answer to this question. We always let C indicate some positive absolute constant which is not necessarily the same in every different occurrence. 2. Result and prf Given e > 0. Assume í> = {(f>n} xlx is a positive increasing sequence. Write L (0, e) to be the maximum length of string of Wk+x/MlZl not exceeding

3 1 + e, namely, BOUNDING \\f-s (f)\\/e (f ) 781 Ln(< >, e) = max {k-j+1: (f>j+x/(pj < 1 + e, X<j<k<n-X 4>j+2/4>j+x <l+e,..., (f>k+x/(pk < 1 + e}, in case there is no (pj+x/(pj < 1 + e for 1 < / < n- 1, we simply set Ln(<S>, e) 0. Since L +X(Q>, e) > Ln(<S>, e), we may define the limit L(<S>, e) = lim L (<D, e). n * Define J7+ = < / e C2n : f ~ ^2 f(n)cos nx ' 4nf(n) i > and &E = {fec2n: /-SB(/) We have the following result. = 0(En(f))}. Theorem 1. The necessary and sufficient condition for ^c5^e is L(&, e0) = 0(1) for some eo > 0. Lemma 1. L(0, e) = oc for all e > 0 implies that there are two increasing sequences of natural numbers {«/}/f] and {w/}g, with «/ + W/-I-1 < n +x, 1 = 1,2,..., as well as a decreasing sequence { /}/2i with lini/-^ S = 0 such that (3) (pn,+j+x I4>n,+j <1+S, j = 0,l,...,m, 1=1,2,... Prf. The argument is elementary and we omit it. D Lemma 2. Let L(0, e) = for all e > 0. Then there is a function f e J7q, such that «-» LnU ) Prf. Since <f>n is positive and increasing, there is a constant c > 0 such that (p > c for all n = 1, 2,.... Without loss of generality we therefore assume </) > 1 for all n = 1,2,...' From Lemma 1, we have three sequences {«/}, {m }, and {S } making (3) be satisfied. Then we may also assume o < 1/e2 and 3 < m < n for I = 1,2,.... We construct the required counterexample by induction. Let /(0) be any positive number and /( 1 ) = 1. Given f(k), we have the following two cases. Case 1. k ^ n, n + 1,..., n + m, I = 1,2,... In this case, we simply set (4) f(k+l) = f(k)/(2(f>k+l), in particular, f(n,) = f(nl-l)/(2<p l). Case 2. k = n for some I. Now we construct (5) /(*,+;)= hni) 2j<pni+xlo%(Mt + iy j =1,2,..., M := minl m, yjôfx \ <e, 'Otherwise write <j>$ = <j> /c. Then </>J > 1 and we use <t>'n instead.

4 782 S. P. ZHOU where [x] is the greatest integer not exceeding x, (6) f(ni + J+l)=iJ, 2<Pn,+j+X log(m/ + 1) In this way we have defined a function j = M, M, + l, m. f(x) = Y^f(n)cosnxn=0 Clearly, by condition (5), for / = 1, 2,..., n/+mi (7) g'/(fc)<^)<^l, while by conditions (4) and (6), for k ^ n + 1, n + 2,..., n + M - 1, (8) f(k+l)<\f(k). Altogether, n*lo f(n) < > which means / e C2n. We check that n +Mi f(0)-sn,(f,0)= Y f(k)+ Y k=n +X k=ni+m/+x f(k) > (pn +xlog(m, + l) logmi-o 2(j>ni+Ml+\ log(m + 1) (by (7), (8)) ^Crh <Pn,+X At the same time, En,(f) < By noting that M, M, Y Ant + ;') cos(n, + j)x - Y f(n + j) cos(n - j)x j=x ;=i + Y Ak). k=n,+m;+x M, M, Y f("i + J) cos(«, + j)x - Y f(ni + J) cos("/ - J)x j=x j=x M, f(n )sinn x ^ sinjx '(pm+xlog(mi + 1) ^ j=x j together with the well-known inequality (cf. [Zyg]) sup m>x Esmkx ~k~ k=x < 3^,

5 BOUNDING \\f-s (f)\\/en{f) 783 we have p (f^ 30t/(»/) ( f(n, + M,) \ E"7J ) ^ a 0 l+i.i«,itf.j.n log(m/ + 1) + ü \2(pni+Ul+x log(mi + 1)) = o{_m_v Therefore, I^SfMïC.o.M,. Since lim/^a// = by lim^^ôi = 0 and lim/^^an/ =, the above estimate yields _ ll/-fl»,(/)ll lim- /-co,(/) ' =. We now need to verify that / e ^>. If k n + 1, n + 2,...,n + M - 1, 1= 1,2,..., by (4), (6), When k = n + j, j = 1, 2,..., M 1, for some /,, - i., n - í/t\ /("/) (<t>n,+j+x <t>n,+j \ <pk+xf(k +1) - **/(*) = 2^+llog(M/ + 1) (-pr - J By Lemma 1, but for 1 <j <M -1, _ /(«/) ^h/+j (K+j+i_ l _ ]\ 2(j)ni+x l0g(m + 1) j + 1 V <Pn,+j j ) ' <f>ni+j+x/<t>n,+j < 1+àl, l/j>l/(ml-í)>x/jl, so that (pni+j+x/(/),+j - 1-1/7 < Si - y/ôi < 0. Altogether it follows that for k = n + j, j = 1, 2,..., M 1, Thus we are done. D <Pk+xf(k+l)-(t>kf(k)<0. Lemma 3. Assume L(0, n) = 0(1) for some en > 0. Let f e J7. Then for n>l, 00 Y f(k) = 0(f(n + l)). k=n+x Prf. Given n > 1. Since L(0, n) = 0(1), say, L(0, e0) < M, we have a sequence {«/} C {n + 1, n + 2,...} such that 0 < n - n - 1 < M, 0 < "i+x -n < M, and (9) </>,+1/</>«,> 1+eo.

6 784 S. P. ZHOU Write Ef(k)=( Y+E OO / «OO "/+! \ k=n+x \k=n+x l=x k=n +X/ Evidently, for k e {n + 2, n + 3,..., n +x}, /(*) f(k) = 4>J(k)± < 4>nl+xf(n, + l)j-k < f(n, + 1), and similarly for k = n+l,n + 2,...,n, f(k) < f(n + 1). Henee / \ Y f(k)<m f(n+l) + Yf(n, + l)\ k=n+x \ l=x ) < M (f(n + 1) + (f>ní+xf(nx + 1)Y ^-) = M[f(n + l) + f(nx + l)y <Pn + 1 /=1 <t>n,+x <Mf(n + l)ll+y(l+ o)-l\ (by (9)) = 0(f(n + l)). D Prf of Theorem 1. Necessity. By Lemma 2, we have the required result. Sufficiency. From Lemma 3, \\f-sn(f)\\= Y f(k) = 0(f(n + l)). k=n+x On the other hand, En(f)> (jon\f(x)-sn(f,x)\2dx) That is, or J7 ctt^e- O ( Y f2(k)\ k=n+x 7 \ '/2 \\f-sn(f)\\ = 0(En(f)), >/(«+ i). The following are two typical applications of Theorem 1. Corollary 1. Let s > 0 be any given number. Then there exists a function f e C2n with nsf(n) { such that /i-, <n\j! Corollary 2. If f e C2n satisfies q"f(n) Tor some q > 1, then \\f-sn(f)\\ = 0(En(f)). For sine series, we have a similar result (with a similar prf).

7 Theorem 2. Let g(x) ec2n, BOUNDING \\f-s (f)\\/e (f) 785 g(x)^y^(n)sinnxn=x Then the necessary and sufficient condition for J77 c <5* is L(<b, en) = 0(1) Tor some en > 0, where J7 = \ge C2n: g ^Yê(n)sinnx, (png(n) I >. Bibliography [NeRi] D. J. Newman and T. J. Rivlin, Approximation of monomials by lower degree polynomials, Aequationes Math. 14 (1976), [Xie] T. F. Xie, Problem no. 1, Approx. Theory Appl. 3 (1987), 144. [Zho] S. P. Zhou, A problem on approximation by Fourier sums with monotone coefficients, J. Approx. Theory 62 (1990), [Zyg] A. A. Zygmund, Trigonometric series, Cambridge Univ. Press, Cambridge, Dalhousie University, Department of Mathematics, Statistics and Computing Science, Halifax, Nova Scotia, Canada B3H 3J5 Current address: Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada T6G 2G1 address : zhouîlapprox. math. ualberta. ca