Lipschitz functions that map $0$ to $0$ are convergence-preserving

Question

I know convergence-preserving functions have been discussed a fair amount in the past; however, I was a looking at another post, and I saw the following result: if $f$ is Lipschitz and $f(0)=0$, then $\sum |a_n| <\infty \Rightarrow \sum |f(a_n)|<\infty$. How exactly would I go about proving this statement? I'm also guessing from other posts that this is a sufficient but not necessary condition?

Edit: made it absolute convergence instead.

Marios Gretsas · Accepted Answer · 2019-11-25T16:16:04.337

0

HINT

$|f(a_n)-f(0)| \leq M|a_n|$ for some $M>0$

The converse is not true.I'm sure you can find a counterexample.

edited Nov 25 '19 at 16:16

answered Nov 25 '19 at 16:14

Marios Gretsas

21,906

Oh, I see, hahaha. I feel silly now. – Nov 25 '19 at 16:15

Robert Israel · Answer 2 · 2019-11-25T16:18:54.683

0

It's not true at all, unless you assume $a_n \ge 0$. The problem is that if $\sum_n a_n $ converges conditionally, $\sum_n f(a_n)$ can diverge even though $|f(a_n)| \le M |a_n|$.

edited Nov 25 '19 at 16:18

answered Nov 25 '19 at 16:16

Robert Israel

448,999

Oops, sorry, I meant absolute convergence! I'll make an edit. – Nov 25 '19 at 16:17
concrete example: $a_n=(-1)^n/n$, $f(x)=|x|$ – Hagen von Eitzen Nov 25 '19 at 16:57

Lipschitz functions that map $0$ to $0$ are convergence-preserving

2 Answers2