# Convergent sequences transformed into convergent sequences [duplicate]

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 !

• The title should refer to convergent series, not sequences... – David C. Ullrich Nov 1 '15 at 15:40
• @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 $k$ – Kelenner Nov 1 '15 at 16:01
• This is definitely a duplicate. Let me try to find the earlier post. – PhoemueX Nov 1 '15 at 16:08
• @NateRiver Yes it does. The partial sums $s_{3n}$ all vanish; since $a_n\to0$ it follows that the sum is $0$. – David C. Ullrich Nov 1 '15 at 16:09
• 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 $S_n\to 0$. – Kelenner Nov 1 '15 at 16:10

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.