Algebrability of the set of non-convergent Fourier series

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1 STUDIA MATHEMATICA 75 () (2006) Algebrability of the set of non-convergent Fourier series by Richar M. Aron (Kent, OH), Davi Pérez-García (Mari) an Juan B. Seoane-Sepúlvea (Kent, OH) Abstract. We show that, given a set E T of measure zero, the set of continuous functions whose Fourier series expansion is ivergent at any point t E is ense-algebrable, i.e. there exists an infinite-imensional, infinitely generate ense subalgebra of C(T) every non-zero element of which has a Fourier series expansion ivergent in E.. Introuction an preliminaries. Many examples of functions with some special or pathological properties have been foun in analysis. Examples such as continuous nowhere ifferentiable functions, everywhere surjective functions, or ifferentiable nowhere monotone functions have been constructe in the past. Given such a special property, we say that the subset M of functions which satisfy it is spaceable if M {0} contains a close infinite-imensional subspace. The set M will be calle (ense) lineable if M {0} contains an infinite-imensional (ense) vector space. At times, we will be more specific, referring to the set M as µ-lineable if it contains a vector space of imension µ. Also, we let λ(m) be the maximum carinality (if it exists) of such a vector space. We believe that the terms lineable an spaceable were first introuce in [5] an, later, in [2] an [7] (see also []). Some of these special properties are not isolate phenomena. In [2] it was shown that the set of everywhere surjective functions is 2 car(r) -lineable an that the set of ifferentiable functions on R which are nowhere monotone is lineable in C(R). Fonf, Gurariy an Kaets showe ([6]) that the set of nowhere ifferentiable functions on [0, ] is spaceable in C[0, ]. Some of these pathological behaviors occur in really interesting ways. For instance, in [] Roríguez-Piazza prove that every separable Banach space is isometric to a space of continuous nowhere ifferentiable functions, an in [8], Hencl showe that any separable Banach space is isometrically isomorphic 2000 Mathematics Subject Classification: Primary 42A20, 46E25; Seconary 42A6. Key wors an phrases: Fourier series, ivergent series, lineability, spaceability, algebrability. Partially supporte by MTM [83]

2 84 R. M. Aron et al. to a subspace of C[0, ] whose non-zero elements are nowhere approximately ifferentiable an nowhere Höler. This paper continues with the search for sets of functions enjoying special properties, in particular, continuous functions whose Fourier series expansion iverges on a set of Lebesgue measure zero. The convergence of Fourier series has been eeply stuie in the past. It came as a consierable surprise when Du Bois-Reymon prouce an example of a continuous function f : T C whose Fourier series is ivergent at one point (see [0, pp ] for a moern reference). This result can be improve by means of an example of a continuous function whose Fourier series expansion iverges on a set of measure zero ([9, p. 58]). This last result is the best possible, since the Fourier expansion of every continuous function converges almost everywhere (Carleson, see e.g. [0, p. 75]). Moreover, by means of Baire s theorem, the pathological behavior can be shown to be generic: there exists a G δ ense subset E T such that the set of continuous functions whose Fourier expansion iverges on this set is a G δ ense subset of C(T) ([2, p. 02]). While writing this paper, the authors became aware of recent work by F. Bayart ([4]) where using ifferent techniques it is shown that, if F E C(T) is the set of continuous functions whose Fourier series expansion iverges on a set of measure zero, E, then F E is ense-lineable. This result is a consequence of our main result, Theorem 2.. Let us recall the following efinition, introuce in [3]: Definition. Given a Banach algebra A an a subset B A, we say that: () B is algebrable if there is a subalgebra C of A so that C B {0} an the carinality of any system of generators of C is infinite; (2) B is ense-algebrable if, in aition, C can be taken ense in A. Recently ([3]) it was shown that the set of complex-value everywhere surjective functions is algebrable. The aim of this paper is to show that, given any set E T of measure zero, the set of continuous functions whose Fourier series expansion is ivergent at any point t E is ense-algebrable. Here, an from now on, T enotes R/2πZ, f(n) or f (n) will enote the nth Fourier coefficient of f C(T), Z(f) enotes the set {n Z : f(n) 0}, an S n (f, t) will enote the nth partial Fourier sum n u= n f(u)e iut. 2. The main result Theorem 2.. Let E T be a set of measure zero. Let F E C(T) be the set of continuous functions whose Fourier series expansion iverges at every point t E. Then F E is ense-algebrable.

3 Ù Non-convergent Fourier series 85 In orer to prove this result, we nee several ifferent tools, amongst them some technical lemmas an the construction of a ouble sequence. 2.. Preliminary elements. By [9, pp ], given a set E T of measure zero we can fin a sequence ( H k ) k of trigonometric polynomials, a sequence (n k ) k of positive integers, an a sequence (E k ) k of measurable subsets of T such that every t E belongs to infinitely many E k s an:. H k (s) for all s T, 2. S nk ( H k, t) 2 k2 for every t E k. Without loss of generality we may suppose that H k (t) = ã k r=0 H k (r)e irt, that is, Z( H k ) [0, ã k ], with (ã k ) increasing, ã k > n k for every k N. Let us efine H k (t) = + e i(ã k+)t Hk (t) an let us call a k = 2ã k + (to have Z(H k ) [0, a k ]). Let (Q j ) j be a sequence of trigonometric polynomials that is ense in C(T) an such that Z(Q j ) [ q j, q j ]. We may clearly suppose that the sequence (q j ) j is increasing, an we let b j = max{a j, q j }. Next, consier the following linear orer on N N: { j + k < j (j, k) (j, k + k or ) j + k = j + k an k < k This orer can be represente by the following iagram, in which each arrow connects a pair with its immeiate successor: (, ) (, 2) (, 3) (, 4) (2, ) (2, 2) (2, 3) (3, ) (3, 2) É (4, )

4 86 R. M. Aron et al. With this orer we efine by recursion the following ouble sequence (p j k ) j,k: p = 3b +, p j k = 3kb k + 2max{k, k }p j k +, where (j, k ) is the immeiate preecessor of (j, k). Of course, p j k < pj k if an only if (j, k) (j, k ). Moreover (p j k ) j,k is increasing in both inices an has, trivially, the following properties: (P) If (j, k) (j, k ) then b k < p j k. (P2) p i > 3b for every i,. (P3) If (j, k) (i, ), then p i > 2pj k + 3b. (P4) If (i, ) (j, k), then p j k > 2pi The main construction. We now use the above elements to state an prove a technical lemma that will be necessary for the proof of the main result: Lemma 2.2. Suppose that () p i + u = p j + + p j s + v an: (i) s n ; (ii) p i p i 2 p i n (that is, i i n ); (iii) p j p j 2 k2 p j s ; (iv) u b ; (v) b v b + + b + b. Then n = s, i r = j r, k r = for every r an u = v. Proof. If p i > p j, then p i (P3) > 2p j + 3b (i) 2sp j + 3b (P)+(iii) > p j + + p j s + b + + b + 3b an, by (iv) an (v), we cannot have equation (). If p j > p i, then p j (P4) > 2p i (P2) > p i + 3b (i) np i + 3b (ii) p i + 3b an, by (iv) an (v), we cannot have equation (). So i = j, k = an we can convert () into (2) p i 2 + u = p j 2 k2 + + p j s + v where b v b k2 + + b + ( + )b.

5 Non-convergent Fourier series 87 Reasoning in the same way we can obtain i 2 = j 2 an k 2 =, reucing equation (2) to p i 3 + u = p j 3 k3 + + p j s + v where b v b k3 + + b + ( + 2)b. Continuing with this proceure (recall that, by (i), s n) we arrive at p i s + u = p j s + v where b v b + ( + s )b. With the same arguments one can conclue that j s = i s an k s =. So it remains to prove that n = s. If not, then n s +, an from (), we obtain v = p i s+ + u with u b an b v ( + s)b 2b. Now v = p i s+ + u p i s+ + u (P2) > 3b b = 2b v, an we reach a contraiction. Next, for each j an m, let us efine the function m fj m (t) = 2 k e ipj k t H k (t). k= Than to the Weierstrass M-test we know that, for each j N, the sequence (fj m) m converges uniformly to a continuous function f j C(T) with f j 2. Define g j = (/j)f j + Q j an let A be the algebra generate by {g j } j. Since (Q j ) j is a ense sequence in C(T), it is clear that (g j ) j (an hence A) is also ense in C(T). This immeiately implies that A is infiniteimensional. We will see that every g A \ {0} has a Fourier series ivergent at any t E. So let us take a generic element of A: N g = α j g i j g i j sj j= where s s N (s j is the number of functions that we multiply in each summan of g) an i j ij s j for every j. The heart of the matter is the next technical lemma: Lemma 2.3. If max{s, i,...,in } then: (a) ĝ(p i + +p i s +u) = α (i i s ) 2 s Ĥ s (u) for 0 u s a. (b) ĝ(u) = 0 for u (p i ). (c) If l > i, then ĝ l(p i + u) = 0 for every 0 u s a an hence g l g.

6 88 R. M. Aron et al. Proof. First of all, notice that if we expan the sum N g = α j (Q i j + ) ( i j i j Q i j sj + f ) i j f i j sj, s j j= we obtain terms of the form f h f hr Q hr+ Q hs with s s an h j for every j s (we can suppose that h h r ). By the uniform convergence of the sequence (f m j ) m, the sequence (f m h f m h r Q hr+ Q hs ) (v) converges to (f h f hr Q hr+ Q hs ) (v) for every v. It is easy to see that this sequence is in fact constant from some m on. Now, for a fixe m, the function f m h f m h r Q hr+ Q hs is a trigonometric polynomial whose non-zero Fourier coefficients are containe in the set A m := {p h k + + p h r k r + v : q hr+ q hs v a + + a kr + q hr+ + + q hs, k,...,k r m}. Therefore, the non-zero Fourier coefficients of f h f hr Q hr+ Q hs are containe in A = m Am an hence in B := {p h k + + p h r k r + v : b v b + + b kr + b, k,..., k r }. Since B [ b, + ), (P2) gives (b). Now, Lemma 2.2 tells us that, if p i + u ( u b ) belongs to B, then r = s, h = i,..., h r = i s an k = = k r =. Thus, ĝ(p i + u) = α i (f i i f i s ) (p i + u), s for u b. Moreover, Lemma 2.2 also says that p i + u ( u b ) is ifferent from any other p i k s + v with b v b + + b kr + b. Thus, looking at the formula f m i = f m i s (t) m a,...,a k s k,...,k s = r,...,r s =0 one fins that, if 0 u s a, then (f m i 2 k k sĥ (r ) Ĥk s (r s )e i(pi f m i s ) (p i + u) = r + +r s =u 0 r,...,r s a which is exactly 2 s Ĥ s (u), an this gives (a). k + +p i s +r + +r s )t 2 s Ĥ (r ) Ĥ(r s ),

7 Non-convergent Fourier series 89 Finally, to obtain (c) we can reason in the same way. The non-zero Fourier coefficients of g l are in the set C = {p l k + v : 0 v b k}. Now, than to Lemma 2.2, if p l k +v = pi + u for some 0 v b k, 0 u s a b, we obtain i = l, which is impossible since l > i. This completes the proof of the lemma. Proof of Theorem 2.. Now, set β = α (i i s ) an take any max{s, i,...,in }. It follows that (3) S p i + +p i s +ã ++n (g, t) S p i + +p i s (g, t) +ã = (p i + +p i s +ã +) ĝ(u)e iut + p i + +p i s +ã ++n ĝ(u)e iut. (p i + +p i s +ã ++n ) p i + +p i s +ã + By Lemma 2.3(b), the first term of (3) is zero. Thus, by Lemma 2.3(a), (3) is equal to ã ++n β 2 s Ĥ s. (u)eiut u=ã + But, by the efinition of H, we have s H s (t) = k=0 ( s k ) e ik(ã +)t Hk (t) an, since Z( H k ) [0, ) for any k, we see that for u [ã +, ã ++n ], Ĥ s (u) = s H (u (ã + )) (recall that n ã ). Therefore (3) equals β s 2 s n H (u)e iut = β s 2 s S n ( H, t) β s 2 2 s u=0 for every t E. Since any t E belongs to infinitely many E s, we have lim n sup S n (g, t) = for every t E. To finish the proof of Theorem 2. it only remains to prove that A is infinitely generate. Now, if A were finitely generate, there woul exist an l N such that every g h, h N, can be expresse as (4) g h = N j= α j g i j g i j sj with s s N an l > i j ij s j for every j. Taking h = l in (4) this contraicts Lemma 2.3(c), which finishes the proof of Theorem 2..

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