ON THE SET OF ALL CONTINUOUS FUNCTIONS WITH UNIFORMLY CONVERGENT FOURIER SERIES

Transcription

1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 24, Number, November 996 ON THE SET OF ALL CONTINUOUS FUNCTIONS WITH UNIFORMLY CONVERGENT FOURIER SERIES HASEO KI Communicated by Andreas R. Blass) Abstract. In this article we calculate the exact location in the Borel hierarchy of UCF, the set of all continuous functions on the unit circle with uniformly convergent Fourier series. It turns out to be complete F σδ. Also we prove that any G δσ set that includes UCF must contain a continuous function with divergent Fourier series. Introduction There are many criteria for uniform convergence of a Fourier series on the unit circle. One can find those tests in [Zy]. In the present paper, we study UCF from the point of view of descriptive set theory. In [Ke], it was a conjecture that UCF is complete F σδ i.e., F σδ but not G δσ ). A lot of natural complete F σδ sets have been found. For example, the collection of reals that are normal or simply normal to base n [KL]; C T), the class of infinitely differentiable functions viewed as a2π-periodic function on R); and UC X, the class of convergent sequences in a separable Banach space X, are complete F σδ [Ke]. It turns out that UCF is complete F σδ. We give two different proofs for it. Ajtai and Kechris [AK] have shown that ECF, the set of all continuous functions with everywhere convergent Fourier series, is complete CA, i.e., coanalytic non-borel. We show that there is no G δσ set A such that UCF A ECF. Hence any G δσ set that includes UCF must contain a continuous function with divergent Fourier series. From this point of view, although there are many natural complete F σδ sets, we can claim that UCF is a very interesting set in analysis. Definitions and background Let N = {, 2, 3, }be the set of positive integers and N N the Polish space with the usual product topology and N discrete. Let X be a Polish space. A subset A of X is CA if there is a Borel funtion from N N to X such that f N N) = X A. ACA F σδ ) subset A of X is called complete CA F σδ ) if for any CA F σδ ) subset B of N N, there is a Borel continuous) function f from N N to X such that the preimage of A of f is B, i.e., B = f A). From the definition, it is easy to see that no Received by the editors May 26, 994 and, in revised form, May 2, Mathematics Subject Classification. Primary 04A5, 26A2; Secondary 42A20. Key words and phrases. Descriptive set theory, Fourier series, complete F σδ, uniformly convergent Fourier series. The author was partially supported by GARC-KOSEF c 996 American Mathematical Society

2 3508 HASEO KI complete CA F σδ ) set is Borel G δσ ). In paticular, if an F σδ subset A of a Polish space is complete F σδ and the continuous preimage of an F σδ subset B of a Polish space, then B is also complete F σδ. Let R be the set of real numbers. Let T denote the unit circle and I the unit interval. Let E be T or I. We denote by CE) the Polish space of continuous functions on E with the uniform metric df,g) =sup{ fx) gx) : x E}. CT) can also be considered as the space of all continuous 2π-periodic functions on R, viewing T as R/2πZ. Let UC denote the set of all sequences of continuous functions on I that are uniformly convergent, i.e., UC = {f n ) CI) N :f n ) converges uniformly }. To each f CT), we associate its Fourier series S[f] where ˆfn) = 2π 2π 0 ft)e int dt. Let S n f,t) = n= n k= n ˆfn)e inx, ˆfk)e ikt be the nth partial sum of the Fourier series of f. We say the Fourier series of f converges at a point t T if the sequence S n f,t)) converges. Similarly, we define the uniform convergence of the Fourier series of f. Let ECF denote the set of all continuous functions with everywhere convergent Fourier series. According to a standard theorem [Ka], the Fourier series of f at t converges to ft) ifitconverges. Hence we have ) ECF ={f CT) : t [0, 2π] S n f,t)) converges } ) ={f CT) : t [0, 2π] ft) = lim S nf,t) }. n We denote by NCF the complement of ECF. Let UCF denote the set all continuous functions with uniformly convergent Fourier series, i.e., UCF = {f CT) : the Fourier series of f converges uniformly }. Results Theorem [AK]). ECF is complete CA. See [AK].) Proposition. UCF and UC are F σδ.

3 FUNCTIONS WITH UNIFORMLY CONVERGENT FOURIER SERIES 3509 Proof. Let Q be the set of all rational numbers. We consider T as [0, 2π] with0 and 2π identified. By the definition of UCF, f UCF S N f ) converges uniformly a N b N c, d N e Q S b+c f,e) S b+d f,e) ) a f V a, b, c, d, e), a N b N c,d N e Q [0,2π] where V a, b, c, d, e) is the collection of f CT) such that S b+c f,e) S b+d f,e) /a, which is closed, since the function f ˆfn) is continuous. Hence UFC is F σδ. Similarly, so is UC. We are done. Lemma 2. The set C 3 = {α N N : lim n αn) = } is complete F σδ. See [Ke, p. 80].) This set will be used to prove our main theorem. Proposition 3. UC is complete F σδ. Proof. We define the function F from N N to CI) N as follows: for each β N N, ) F β) =. βn) Then it is easy to see that β C 3 F β ) converges to 0 F β ) converges uniformly, since F β) is a sequence of constant functions. Clearly, F is continuous. Hence UC is the continuous preimage of C 3. By Proposition and Lemma 2, UC is complete F σδ. Theorem 4. There is a continuous function H from N N to CT) such that for all A with UCF A ECF, and β C 3 H β ) A, β/ C 3 H β ) NCF. In particular, UCF is complete F σδ. By this theorem, we have the following corollary. Corollary 5. There is no G δσ set A such that UCF A ECF, i.e., any G δσ set that includes UCF must contain a continuous function with divergent Fourier series. Proof. Suppose a G δσ set A satisfies UCF A ECF. Then by Theorem 4, we obtain H A) =C 3.Since A is G δσ, so is C 3. By Lemma 2, this contradicts our assumption.

4 350 HASEO KI It is a basic fact of descriptive set theory [Ke] that any Borel set is coanalytic. Since ECF is complete CA by Theorem [AK], it is a very natural guess that the complement of C 3 can be reducible to ECF UCF. In fact, we have the following theorem. Theorem 6. There is a continuous function H from N N to CT) such that and In particular, UCF is complete F σδ. β C 3 H β ) UCF, β/ C 3 H β ) ECF UCF. In order to prove Theorem 4 and Theorem 6, we need the following criterion due to Dini and Lipschitz [Zy, p. 63]. Let f be defined in a closed interval J, and let ωδ) =ωδ;f)=sup{ fx) fy) : x, y J and x y δ}. The function ωδ) is called the modulus of continuity of f. The Dini-Lipschitz test. If f is continuous and its modulus of continuity ωδ) satisfies the condition ωδ)logδ 0, then the Fourier series of f converges uniformly. We introduce the Féjer polynomials, for given 0 <n<n Nand x R, n sin kx Qx, N, n) =2sinNx, k k= n sin kx Rx, N, n) =2cosNx. k These two polynomials were used in [Zy] to prove that there exists a continuous function whose Fourier series diverges at a point. k= Lemma 7. There are positive numbers C,C 2 >0 such that Q <C and R <C 2, i.e., these polynomials are uniformly bounded in x, N, n. Since n k= sin kx k is uniformly bounded in n and x, Lemma 7 follows. By Lemma 7, we immediately have the following. Proposition 8. Let N k ) and n k ) be any two sequences of positive integers, with n k <N k,andlet )be a sequence of real numbers such that α + α 2 + α 3 + <. Then the series Qx, N k,n k ), αk Rx, N k,n k ) converge to continuous functions.

5 FUNCTIONS WITH UNIFORMLY CONVERGENT FOURIER SERIES 35 Proof of Theorem 4. We fix A with UCF A ECF. Let = 2 k, n k = N k /2=2 2k k=,2,3, ). We define H from N N to CT) as follows: for all β N N, Hβ) = βk) Qx, N k,n k ). Claim. H is continuous and well-defined. Proof. By Proposition 8, H is well-defined. By Lemma 7, it is easy to see that H is continuous. We divide the rest of proof into two parts so that we have more intuition. Case. lim n βk). We want to show that Hβ) NCF. For each k N the inequality ) ) SNk +n k Hβ), 0 SNk Hβ), 0 = Ĥβ)l) Ĥβ)l) l N k +n k l N k = + 2 ) ) βk) + + nk > βk) log n k =2 k log 22k βk) = βk) log 2 holds. Since lim n βk), there exists a p N such that for infinitely many k s, βk) = p. Hence the Fourier series of Hβ) does not converge, since in ) we have /p log 2 for infinitely many k s. So we derive Hβ) NCF. Case 2. lim n βk) =. We show that H β) UCF. We will demonstrate that ωδ; Hβ)) log δ 0as δ 0.Then by the Dini-Lipschitz test, this shows that the Fourier series of Hβ) converges uniformly. We take any 0 <δ /2 and define ν = νδ) as the largest integer k satifying 2 2k /δ By Lemma 7, we have the following inequality: 2) k=ν+ 2C βk) Qx + δ, N k,n k ) k=ν+ k=ν+ 2C sup{ βk) βk) : k>ν} βk) Qx, N k,n k ) k=ν+ =4Csup{ βk) : k>ν}2 ν 4Csup{ log 2 : k>ν} βk) log δ. Now we calculate the rest of Hβ). We clearly have Q x, N, n) =NRx, N, n)+2sinnx n cos kx, Q NC +2n=nC, k=

6 352 HASEO KI for N =2nand C =2C+2. By the mean value theorem, we have the following inequality: βk) Qx + δ, N k,n k ) βk) Qx, N k,n k ) C δ β) + +2 ν 2 ) 3) 2ν βν) C 2 2 ν 2 k 2 2k βk) C ν log δ k k βk). By 2) and 3), we have the following: 4) ωδ; Hβ)) log δ max{4c sup{ βk) : k>ν}log 2,C 2 2 ν 2 2k k βk) }. Now if δ 0, then ν.so it suffices to show that the right part of 4) goes to0asν.since βν) as ν,sup{/βk) :k>ν}goes to 0. We need to show that the rest goes to zero as ν diverges to infinity. This requires the following small claim. Claim 2. 22k k 2 2ν ν+. Proof. Use induction on ν. For ν =,2 2 = =2 2.Suppose it is true for ν. By the induction assumption, k 22k +2 2ν+ ν+) 2 2ν ν+ +2 2ν+ ν+). It is enough to show that 2 2ν ν+ +2 2ν+ ν+) 2 2ν+ +. Letting θ =2 2ν,one can verify this inequality. Fix ɛ. Take N 0 such that /βk) <ɛfor all k N 0. For this N 0, we choose N>N 0 so that 2 2ν +ν k N 0 2 2k k <ɛfor all ν N. Then for all ν N, by claim 2, the following inequality is valid: 2 2k k 2C 2 2ν 2 ν < 2C 2 2ν +ν < 2C ɛ +2 2ν +ν ɛ 2 2 k k +ν βk) +2 2ν k N 0 N 0< 2 2k k ) 2 2k k βk) N 0< ) ) ν+4 < 2ɛC + 22ν 2 2ν ν =34ɛC. Hence the right side of 4) converges to zero as ν goes to the infinity, i.e., as δ 0. So we derive Hβ) UCF. By case and case 2, we obtain β/ C 3 Hβ) NCF and β C 3 Hβ) UCF, respectively. Since NCF and A are disjoint, we have the following: β/ C 3 H β ) NCF and β C 3 H β ) A.

7 FUNCTIONS WITH UNIFORMLY CONVERGENT FOURIER SERIES 353 We have shown the first assertion of the theorem. In particular, C 3 is the preimage of UCF. Hence by Lemma 2, the second assertion follows. We have completed the proofoftheorem4. Proof of Theorem 6. Instead of Q, we use R. With N k,n k, and as in the proof of Theorem 4, we define H from N N to CT) as follows: for each β N N, Hβ) = βk) Rx, N k,n k ). The same proof as before will demonstrate that this function is well-defined and continuous and that if lim n βn) =,then the Fourier series of Hβ) converges uniformly. So it suffices to show that if lim n βn), then Hβ) ECF UCF. Suppose lim n βn). The representation of Hβ) as Fourier series is av sin vx. We see that a v sin vx converges uniformly for δ x πfor any δ>0,since the partial sums of Rx, N k,n k ) are uniformly bounded in k and x, for δ x π. The series a v sin vx contains sines only, and hence it converges for x =0,and so everywhere. Now we will show that a v sin vx does not converge uniformly. It is easy to see that 3n k 2n k a v sin vx a v sin vx = v= v= So if we let x = π/4n k, then we have 3n k v=2n k + 3n k 2n k a v sin vx a v sin vx =2 k βk) v= v= since, for all v in the interval v n k, a v sin vx =2 k βk) n k v= sin2n k + v)x v 3 4 π π 2n k + v) π 4n k 2. So finally, 3n k 2n k a v sin vx a v sin vx 2 k βk) sin π 4 5) v= v= = log 2 2 βk). n k v= n k v= sin2n k + v)x. v 2 k βk) sin π 4 n k v= v 2 k βk) log n k sin π 4 Hence a v sin vx does not converge uniformly, since in 5) the same value appears for infinitely many k s. Hence, as in the proof of Theorem 4, we finish the proof of Theorem 6. v, Acknowledgment I wish to thank Professor A. Kechris for his constant guidance. The author is indebted to the referee and Professor A. R. Blass for many valuable comments.

8 354 HASEO KI References [AK] M. Ajtai and A. S. Kechris, The set of continuous functions with everywhere convergent Fourier series, Trans. Amer. Math. Soc ), MR 89b:04005 [Ka] Y. Katznelson, An introduction to harmonic analysis, Dover, New York, 976. MR 54:0976 [Ke] Alexander S. Kechris, Classical Descriptive Set Theory, Springer Verlag, 995. CMP 95:09 [KL] Haseo Ki and Tom Linton, Normal numbers and subsets of N with given densities, Fundamenta Mathematicae 2) ), MR 95e:04005 [Zy] A. Zygmund, Trigonometric series, 2nd ed., Cambridge Univ. Press, 959. MR 2:6498 Department of Mathematics, California Institute of Technology, Pasadena, California 925 Current address: GARC, Department of Mathematics, Seoul National University, Seoul 5-742, Korea address: haseo@math.snu.ac.kr