The set of functions which map convergent series to convergent series

I'm quite late on this one, but I think the result is nice enough to be included here.

Definition A function $f : \mathbb R \to \mathbb R$ is said to be convergence-preserving (hereafter CP) if $\sum f(a_n)$ converges for every convergent series $\sum a_n$.

Theorem (Wildenberg): The CP functions are exactly the ones which are linear on some neighbourhood of $0$.

Proof (Smith): Clearly, whether $f$ is CP only depends on the restriction of $f$ on an arbitrary small neighbourhood of $0$. Since the linear functions are CP, the condition is clearly sufficient. Let's prove that it is also necessary.

We will prove two preliminary results.

Lemma 1: $f$ CP $\Rightarrow$ $f$ continuous at $0$.

Proof: Let's suppose that $f$ isn't continuous at 0. This implies that there exists a sequence $\epsilon_n \to 0$ and a positive real $\eta > 0$ such that $\forall n, |f(\epsilon_n)| \geq \eta$. But it is easy to extract a subsequence $\epsilon_{\phi(n)}$ such that $\sum \epsilon_{\phi(n)}$ converges (take $\phi$ such that $\epsilon_{\phi(n)} \leq 2^{-n}$, for instance). For such a subsequence, we still have that $|f(\epsilon_{\phi(n)})| \geq \eta$. This prevents $\sum f(\epsilon_{\phi(n)})$ to converge and, thus, $f$ to be CP, a contradiction.

Lemma 2: The function $(x, y) \mapsto f(x+y) + f(-x) + f(-y)$ vanishes on some neighbourhood of $0$.

Proof: If it didn't, one would be able to find sequences $x_n \to 0$ and $y_n \to 0$ s.t. $\forall n, f(x_n + y_n) + f(-x_n) + f(-y_n) \neq 0$. Up to some extraction, we can assume that $\delta_n = f(x_n + y_n) + f(-x_n) + f(-y_n)$ always has the same sign (let's say $\delta_n > 0$, for the sake of simplicity.)

Consider now the series $$\begin{array}{l@{}l} (x_0 + y_0) &+ (-x_0) + (-y_0) + \cdots + (x_0 + y_0) + (-x_0) + (-y_0)\\ & +(x_1 + y_1) + (-x_1) + (-y_1) + \cdots + (x_1 + y_1) + (-x_1) + (-y_1)\\ &+\cdots\\ &+(x_n + y_n) + (-x_n) + (-y_n) + \cdots + (x_n + y_n) + (-x_n) + (-y_n)+\cdots, \end{array}$$ where every triplet of termes $(x_i+y_i) + (-x_i) + (-y_i)$ is repeated $M_i > 0$ times, for some integer $M_i > 0$.

Because $x_n \to 0$ and $y_n \to 0$ and the three terms $x_i + y_i, -x_i, -y_i$ add to 0, it is easy to see that this series is convergent, regardless of the choice of the $M_i$'s.

On the other hand, if we choose $M_i \geq \delta_i^{-1}$, the image of our series by $f$ is $$\begin{array}{l@{}l} f(x_0 + y_0) &+ f(-x_0) + f(-y_0) + \cdots + f(x_0 + y_0) + f(-x_0) + f(-y_0)\\ & +f(x_1 + y_1) + f(-x_1) + f(-y_1) + \cdots + f(x_1 + y_1) + f(-x_1) + f(-y_1)\\ &+\cdots\\ &+f(x_n + y_n) + f(-x_n) + f(-y_n) + \cdots + f(x_n + y_n) + f(-x_n) + f(-y_n)+\cdots, \end{array}$$ which diverges, for every line adds to $M_i \delta_i > 1$. Again, this in direct contradiction with the CPness of $f$.

If we apply the result of lemma 2 with $y = 0$, we get that $f(-x) = -f(x)$. So we can rewrite lemma 2 in the following way: $\exists \eta > 0 : \forall x, y \in (-\eta, \eta), f(x+y) = f(x) + f(y)$.

This property and the continuity at 0 imply first the continuity on the whole of $(-\eta, \eta)$ and it is then not hard to adapt the classical proof to show that $f$ is linear on $(-\eta, \eta)$. Q.E.D.

If $f$ is not continuous at $0$, then we can find a sequence $x_n$ that converges to $0$ but $f(x_n)$ doesn't converge to $0$. First get a subsequence $y_n$ of $x_n$ with $|f( y_n)| > r$ for some $r>0$. Next choose some subsequence $z_n$ of $y_n$ so that $\sum z_n$ converges. However the series $\sum f(z_n)$ diverges and it follows that $f$ is continuous at $0$.