András Máthé, Department of Analysis, Eötvös Loránd University, Pázmány Péter sétány 1/C, 1117 Budapest, Hungary (e-mail: amathe@cs.elte.hu)

A nowhere convergent series of functions converging somewhere after every non-trivial

change of signs

Abstract

We construct a sequence of continuous functions (h_{n}) on any given uncountable Polish space, such that ∑ h_{n} is divergent everywhere, but for any sign sequence (ε_{n}) ∈ {−1, +1}^{ℕ} which contains infinitely many −1 and +1 the series ∑ ε_{n}h_{n} is convergent at at least one point. We can even have h_{n} → 0, and if we take our given Polish space to be any uncountable closed subset of ℝ, we can require that every h_{n} be a polynomial. This strengthens a construction of Tamás Keleti and Tamás Mátrai.

1 Introduction

Let X be a topological space, f_{n} : X → ℝ, n ∈ ℕ be a sequence of continuous functions. One can ask about a condition on this sequence which guarantees that for a “typical” choice of signs ε_{n} = ±1 the series ∑ ε_{n}f_{n} diverges everywhere on X.

By “typical” choice of signs we mean that the set of the proper sign sequences is a residual (or dense G_{δ}) subset of S = {−1, +1}^{ℕ}. Here we consider S as a product of discrete topological spaces, which is clearly a Baire space. By ℕ we denote the set of the positive integers. By Polish space we mean complete separable metric space.

In [1, Theorem 4.1] for σ-compact X spaces a condition was given on the divergence of the partial sums of ∑ f_{n} implying that ∑ ε_{n}f_{n} diverges everywhere for a typical sign sequence (ε_{n}) ∈ S. Motivated by this result, S. Konyagin asked whether in case of compact metric spaces X, the pure fact that ∑ f_{n} diverges everywhere could imply that ∑ ε_{n}f_{n} diverges everywhere for a typical sign sequence. Tamás Keleti and Tamás Mátrai (see [2]) gave a negative answer to this question by showing an example of a sequence of continuous functions (f_{n}) on any uncountable Polish space, such that ∑ f_{n} is divergent everywhere, but for a typical sign sequence (ε_{n}) ∈ S, the series ∑ ε_{n}f_{n} is convergent at at least one point.

Mathematical Reviews subject classification: 40A30

This paper strengthens this construction by showing a sequence of continuous functions f_{n} such that ∑ f_{n} is divergent everywhere but for every sign sequence (ε_{n}) ∈ S_{0} = {(ε′_{n}) ∈ S | (ε′_{n}) contains infinitely many −1 and +1}, the series ∑ ε_{n}f_{n} is convergent at at least one point. Clearly S_{0} is the largest subset of S for which this could be true.

We will also construct an other series of continuous functions with the same properties which satisfies even that f_{n} → 0. Providing that the uncountable Polish space is ℝ (or a closed subset of ℝ) we can require every f_{n} to be a polynomial, see Remark 1.

2 The example

Theorem 1. ^{1} Let P be an uncountable Polish space. There exists a sequence of continuous functions h_{n} : P → ℝ such that ∑ h_{n} diverges everywhere on P, but for any (ε_{n}) ∈ {−1, +1}^{ℕ} sign sequence containing infinitely many −1 and +1 digits ∑ ε_{n}h_{n} converges at at least one point of P.

Proof. At first we define continuous functions f_{n} : S = {−1, +1}^{ℕ} → [−1, +1] such that ∑ f_{n} is divergent everywhere, but for any (ε_{n}) ∈ S_{0} = {(ε′_{n}) ∈ S | (ε′_{n}) contains infinitely many −1 and +1} the series ∑ ε_{n}f_{n} is convergent at at least one point, in fact, at (ε_{n}).

Consider a fix x ∈ S as the sequence of the −1 and +1 digits. Divide this sequence into blocks of type AAA … AB (where A and B stand for −1 and +1 in some order), with the property of containing at least one A and containing exactly one B at the end. We start the division in the beginning of the sequence. Occasionally we make one infinite block of type AAA . . .. Thus, the division is well defined.

For example,

−1 +1 −1 −1 −1 +1 +1 +1 −1 +1 −1 −1 +1 +1 +1 +1 …

Let n be a positive integer. We are going to define the real number f_{n}(x). Suppose that the n^{th} digit of x is in the k^{th} block of x and this digit is the i^{th} number in this block. Denote the size of the k^{th} block by l. (Thus 1 ≤ i ≤ l and l ≥ 2.)

If l is even, then let f_{n}(x) = \frac{(−1)^{i+1}}{k} if 1 ≤ i ≤ l − 1 and let f_{n}(x) = \frac{+1}{k} if i = l.

1Independently from the author, Gergely Zábrádi gave almost the same construction on ℝ at the same time.

If l is odd, then let f_{n}(x) = \frac{(−1)^{i+1}}{k} if 1 ≤ i ≤ l − 2, let f_{n}(x) = 0 if i = l − 1 and let f_{n}(x) = \frac{+1}{k} if i = l.

If l = ∞, then let f_{n}(x) = \frac{(−1)^{i+1}}{k}.

For example (writing f_{n}(x) below the n^{th} digit of x),

\begin{matrix}−1 +1 & −1 −1 −1 +1 & +1 +1 −1 & +1 −1 & −1 +1 & +1 +1 +1 … \\ \frac{+1}{1} \frac{+1}{1} & \frac{+1}{2} \frac{−1}{2} \frac{+1}{2} \frac{+1}{2} & \frac{+1}{3} 0 \frac{+1}{3} & \frac{+1}{4} \frac{+1}{4} & \frac{+1}{5} \frac{+1}{5} & \frac{+1}{6} \frac{−1}{6} \frac{+1}{6} … \end{matrix}

Claim 1. The function f_{n} is an S → [−1, +1] continuous function for every n ∈ ℕ.

Proof. It is easy to see that f_{n}(x) depends only on the first n + 1 digits of x. This implies continuity.

Claim 2. The series ∑ f_{n}(x) is divergent for every x ∈ S.

Proof. For a fixed x consider those positive integers n for which the n^{th} digits of x are in the fixed k^{th} block. For these n the sum of f_{n}(x) equals to 2/k if this block is finite. Hence ∑_{n∈ℕ} f_{n}(x) = ∞ if x has infinitely many blocks. Otherwise x has an infinite block so the terms of the series ∑_{n∈ℕ} f_{n}(x) are not converging to 0.

Claim 3. For every (ε_{n}) ∈ S_{0} there exists x ∈ S for which ∑ ε_{n}f_{n}(x) is convergent, namely x = (ε_{n}).

Proof. The sequence x = (ε_{n}) ∈ S_{0} has only blocks of finite size. Consider those positive integers n for which the n^{th} digits of x are in the same fixed block. For these n the sum of ε_{n}f_{n}(x) is exactly zero. The sequence of partial sums converges to 0, hence the series ∑ ε_{n}f_{n}(x) is convergent.

It is well known (see [3, Corollary 6.5]) that P contains a homeomorphic copy of the Cantor set, denote it by C. Clearly S is homeomorphic to the Cantor set, let φ be a C → S homeomorphism. Let g_{n} : P → [−1, +1] be a continuous extension of f_{n} ∘ φ : C → [−1, +1] for every n. On P let h_{n}(p) = g_{n}(p) + n · d(p, C), where d(p, C) denotes the distance of p from the closed set C. Clearly for p ∉ C the series ∑ h_{n}(p) diverges. On C we have h_{n} = f_{n} ∘ φ, hence by Claim 2 and Claim 3 we obtain that (h_{n}) satisfies all required properties.

Theorem 2. Requiring that h_{n} → 0, Theorem 1 remains true.

Proof. Just like in the proof of Theorem 1, at first we define functions f_{n} on S. Let x ∈ S be fixed. Consider the same blocks. Suppose that the k^{th} block is finite and contains the a^{th}, (a + 1)^{th}, …, b^{th} digits of x (a, b ∈ ℕ, b − a ≥ 2). Define f_{a}(x), f_{a+1}(x), …, f_{b}(x) to be respectively

\frac{+1}{k} \frac{−1}{2k} \frac{−1}{2k} \frac{+1}{3k} \frac{+1}{3k} \frac{+1}{3k} … \underbrace{\frac{+1}{(2m + 1)k} … \frac{+1}{(2m + 1)k}}_{2m+1} \underbrace{0 0 … 0}_{<4m+5} \frac{+1}{k}

where the number of zeros is less than 4m + 5 and maybe there are no zeros at all. This properly defines the value of m (m ∈ {0, 1, 2, …}). Note that ∑_{n=a}^{b} f_{n}(x) = \frac{2}{k} and if x = (ε_{n}) ∈ S_{0} then ∑_{n=a}^{b} ε_{n}f_{n}(x) = 0.

If the k^{th} block is infinite and contains the a^{th}, (a + 1)^{th}, … digits of x then define f_{a}(x), f_{a+1}(x), … to be respectively

\frac{+1}{k} \frac{−1}{2k} \frac{−1}{2k} \frac{+1}{3k} \frac{+1}{3k} \frac{+1}{3k} \frac{−1}{4k} \frac{−1}{4k} \frac{−1}{4k} \frac{−1}{4k} …

Note that ∑_{n=a}^{∞} f_{n}(x) diverges.

One can easily check that f_{n}(x) depends only on the first 2n + 2 digits of x, so these functions are continuous. It is clear that Claim 2 and Claim 3 also hold for this sequence of functions f_{n}, and −1 ≤ f_{n} ≤ +1 for every n ∈ ℕ. Define φ and g_{n} the same way as in the proof of Theorem 1. We modify the definition of function h_{n}, put

h_{n}(p) = (max(1 − d(p, C), 0))^{n} g_{n}(p) + \frac{d(p, C)}{n} .

If p ∉ C then h_{n}(p) ∼ \frac{1}{n}, hence ∑ h_{n}(p) diverges and h_{n}(p) → 0. For p ∈ C we have h_{n}(p) = f_{n} ∘ φ(p). Hence by Claim 2 and Claim 3 we obtain that (h_{n}) satisfies all required properties.

Remark 1. Let P be an uncountable closed subset of ℝ (hence P is a Polish space). There exists a sequence of polynomials p_{n} : P → ℝ such that p_{n} → 0 and ∑ p_{n} diverges everywhere on P, but for any sign sequence (ε_{n}) ∈ {−1, +1}^{ℕ} containing infinitely many −1 and +1, the series ∑ ε_{n}p_{n} converges at at least one point of P.

Proof. Consider the continuous functions h_{n} given by Theorem 2 for P. Let p_{n} be a polynomial on ℝ for which |p_{n}(x) − h_{n}(x)| ≤ \frac{1}{n^{2}} for every x ∈ P ⋂[−n, n]. Clearly p_{n}(x) → 0 for every x ∈ P. Since the series ∑ \frac{1}{n^{2}} converges, for every (ε_{n}) ∈ S the series ∑ ε_{n}p_{n} converges if and only if ∑ ε_{n}h_{n} converges. This completes the proof.

References

[1] F. Bayart, S. V. Konyagin, H. Queffélec, Convergence almost everywhere and divergence everywhere of Taylor and Dirichlet series, Real Analysis Exchange 29 (2003/04), no. 2, 557–586.

[2] T. Keleti, T. Mátrai, A nowhere convergent series of functions which is somewhere convergent after a typical change of signs, Real Analysis Exchange 29 (2003/04), no. 2, 891–894.

[3] A. S. Kechris, Classical Descriptive Set Theory, Springer-Verlag New York, 1995.