# The set of functions which map convergent series to convergent series

Suppose $f$ is some real function with the above property, i.e.

if $\sum\limits_{n = 0}^\infty {x_n}$ converges, then $\sum\limits_{n = 0}^\infty {f(x_n)}$ also converges.

My question is: can anything interesting be said regarding the behavior of such a function close to $0$, other than the fact that $f(0)=0$?

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A simple example: If $\sum x_n$ converges absolutely, then any $f(x) = O(x)$ has the desired property. –  Antonio Vargas Mar 6 '12 at 5:19
All I can think is that , if the space you're in is complete, uniformly-continuous functions take Cauchy sequences to Cauchy sequences. This allows a unique continuous extension for functions defined in dense subsets.And continuous alone is not enough for this last, as f(x)=1/(x-2^{1/2}) is an example. –  AQP Mar 6 '12 at 5:21
Let me ask this: must $f$ be continuous at $0$? –  the_fox Mar 6 '12 at 5:25
I misread your question so deleted the comment I posted. –  Patrick Mar 6 '12 at 5:30
Are your spaces--initial and target--complete, or do you want a more general result for any spaces? From your f(0)=0, I assume your maps are from R to R? –  AQP Mar 6 '12 at 5:39

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 $0$.

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Ok, next question: must $f$ be differentiable at $0$? –  the_fox Mar 6 '12 at 5:54

Answer to the next question: no.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be defined by $$f(x)=\begin{cases} n\,x & \text{if } x=2^{-n}, n\in\mathbb{N},\\ x & \text{otherwise.} \end{cases}$$ Then $\lim_{x\to0}f(x)=f(0)=0$, $f$ transforms convergent series in convergent series, but $f(x)/x$ is not bounded in any open set containing $0$. In particular $f$ is not differentiable at $x=0$. This example can be modified to make $f$ continuous.

Proof.

Let $\sum_{k=1}^\infty x_k$ be a convergent series. Let $I=\{k\in\mathbb{N}:x_k=2^{-n}\text{ for some }n\in\mathbb{N}\}$. For each $k\in I$ let $n_k\in\mathbb{N}$ be such that $x_k=2^{-n_k}$. Then $$\sum_{k=1}^\infty f(x_k)=\sum_{k\in I} n_k\,2^{-n_k}+\sum_{n\not\in I} x_n.$$ The series $\sum_{k\in I} n_k\,2^{-n_k}$ is convergent. It is enough to show that also $\sum_{n\not\in I} x_n$ is convergent. This follows from the equality $$\sum_{n=1}^\infty x_n=\sum_{n\in I} x_n+\sum_{n\not\in I} x_n$$ and the fact that $\sum_{n=1}^\infty x_n$ is convergent and $\sum_{k\in I} x_n$ absolutely convergent.

The proof is wrong. $\sum_{k\in I} x_n$ may be divergent. Consider the series $$\frac12-\frac12+\frac14-\frac14+\frac14-\frac14+\frac18-\frac18+\frac18-\frac18+\frac18-\frac18+\frac18-\frac18+\dots$$ It is convergent, since its partial sums are $$\frac12,0,\frac14,0,\frac14,0,\frac18,0,\frac18,0,\frac18,0,\frac18,0,\dots$$ The transformed series is $$\frac12-\frac12+\frac24-\frac14+\frac24-\frac14+\frac38-\frac18+\frac38-\frac18+\frac38-\frac18+\frac38-\frac18+\dots$$ whose partial sums are $$\frac12,0,\frac12,\frac14,\frac34,\frac12,\frac78,\frac34,\frac98,1,\frac{11}8,\frac54,\dots$$ which grow without bound.

On the other hand, $f(x)=O(x)$, the condition in Antonio Vargas' comment, is not enough when one considers series of arbitrary sign. Let $$f(x)=\begin{cases} x\cos\dfrac{\pi}{x} & \text{if } x\ne0,\\ 0 & \text{if } x=0, \end{cases} \quad\text{so that }|f(x)|\le|x|.$$ Let $x_n=\dfrac{(-1)^n}{n}$. Then $\sum_{n=1}^\infty x_n$ converges, but $$\sum_{n=1}^\infty f(x_n)=\sum_{n=1}^\infty\frac1n$$ diverges.

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Can you please provide a short proof why $f$ sends convergent series to convergent series? –  the_fox Mar 6 '12 at 20:16
I have written the proof. –  Julián Aguirre Mar 6 '12 at 21:04
Not sure $\sum\limits_{k\in I}n_k/2^{n_k}$ always converges. Consider $(x_i)_{i\geqslant1}$ such that every $x_i$ is a negative power of $2$ and assume that $x_i=1/2^k$ roughly $2^k/k^2$ times. Then $\sum\limits_{i\geqslant1}x_i\approx\sum\limits_{k\geqslant1}1/k^2$ converges but $\sum\limits_{i\geqslant1}f(x_i)\approx\sum\limits_{k\geqslant1}1/k$ diverges. –  Did Mar 6 '12 at 21:35
Aside from Didier's point above, are you certain you can split the series? –  the_fox Mar 6 '12 at 21:42