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Problem. Find all the $f: \mathbb{R} \to \mathbb{R}$ (not supposed continuous) such that for every real sequence $(a_n)$ we have : $$\sum a_k \; \text{is convergent} \Longrightarrow \sum f(a_k) \; \text{is convergent}$$

I'm trying to prove that the only functions are linear in a neighborhood of $0$. It is clear that those functions work but for the reciprocal it is much harder since $f$ is not supposed of any regularity. I have proven that :

  1. $f(0)=0$

  2. $f$ is continuous in $0$ : if it's not the case we can set $ \varepsilon >0$ and a sequence $(a_n)$ such that $|a_n| \leqslant 2^{-n}$ and $|f(a_n)| \geqslant \varepsilon$ which is absurd.

Any ideas to show that $f(x+y)=f(x)+f(y)$ near $0$ ? (which would be sufficient to prove the result)

Thanks !

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    $\begingroup$ The title should refer to convergent series, not sequences... $\endgroup$ – David C. Ullrich Nov 1 '15 at 15:40
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    $\begingroup$ @Nate River: Take $a_n$ such that $a_{3k}=a_{3k+1}=\frac{1}{b_k}$ and $a_{3k+2}=-\frac{2}{b_k}$ with $b_k$ equal to the cubic root of $k$ $\endgroup$ – Kelenner Nov 1 '15 at 16:01
  • $\begingroup$ This is definitely a duplicate. Let me try to find the earlier post. $\endgroup$ – PhoemueX Nov 1 '15 at 16:08
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    $\begingroup$ @NateRiver Yes it does. The partial sums $s_{3n}$ all vanish; since $a_n\to0$ it follows that the sum is $0$. $\endgroup$ – David C. Ullrich Nov 1 '15 at 16:09
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    $\begingroup$ Yes: If $S_n$ is the sum of the $n$ first terms, gouping the terms $3$ by $3$ with the possible exception of the term $a_n$, or the terms $a_n$ and $a_{n-1}$, give $S_n\to 0$. $\endgroup$ – Kelenner Nov 1 '15 at 16:10
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EDIT: Turns out I found Smith's proof of Wildeberg's theorem. Knew it must be an old question, thought the nice little proof might be interesting. Oh well...

As you point out, it's enough to show that $f(x+y)=f(x)+f(y)$ for all sufficiently small $x$ and $y$. Suppose this is false.

Then there exist sequences $x_k\to0$ and $y_k\to0$ such that $$f(x_k+y_k)-f(x_k)-f(y_k)\ne0.$$

Consider the series that consists of the three terms $(x_1+y_1) - x_1 - y_1$ repeated $N_1$ times, followed by the three terms $(x_2+y_2) - x_2 - y_2$ repeated $N_2$ times, etc. Since $x_k\to0$ and $y_k\to0$ it's easy to see that this sum converges.

But the sum of the first $3N_1$ terms of the series with $f$ applied is $$N_1(f(x_1+y_1)-f(x_1)-f(y_1)),$$which we can make as large as we like by taking $N_1$ large enough. Similarly we can make the sum of the next $3N_2$ terms of the modified series as large as we like. Taking an appropriate sequence $N_k$ we see that the new series diverges.

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