15
$\begingroup$

Can anyone help me with this problem?

Let $f:[0,\infty)\longrightarrow \mathbb R$ be a continuous function such that for each $x>0$, we have $\lim_{n\to \infty}f(nx)=0$. Then prove that $\lim_{x\to \infty}f(x)=0$.

Our teacher told first to prove Baire's theorem, and then show that this is a consequence of that theorem. I proved Baire's theorem, and I spend a few hours thinking on how Baire's theorem is related to this problem, but I couldn't find anything. I'd really appreciate your help.

$\endgroup$

marked as duplicate by Nate Eldredge, saz, Claude Leibovici, user91500, Mankind Sep 13 '15 at 11:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

13
$\begingroup$

Fix $\epsilon>0$ and put $F_n:=\left\{x\geq 0,\forall k\geq n, |f(kx)|\leqslant \varepsilon\right\}$. Then for all $n$ $F_n$ is closed since $f$ is continuous and $\bigcup_n F_n=[0,+\infty[$. By Baire's theorem we can find $x_0\geq 0$, $r>0$ and $n_0$ such that $]x_0-r,x_0+r[\subset F_{n_0}$. Put $t_0:=n_1x_0$ where $n_1$ is an integer $\geqslant n_0$ and such that $\frac{x_0}{n_1}<r$. Take $x\geq t_0$. Then we can write $x=n_xx_0+\beta$ where $n_x$ is an integer $\geqslant n_1$ and $0\leqslant \beta<x_0$. So $$|f(x)|=\left|f(n_xx_0+\beta)\right|=\left|f\left(n_x\left(x_0+\frac{\beta}{n_x}\right)\right)\right|\leqslant \varepsilon$$ since $n_x\geqslant n_1$ and $x_0+\frac{\beta}{n_x}\in ]x_0-r,x_0+r[\subset F_{n_0}$ (this because $\left|\frac{\beta}{n_x}\right|\leqslant \frac{x_0}{n_x}\leqslant \frac{x_0}{n_1}<r$).

So we have shown that given a $\varepsilon>0$, we can find $t_0$ such that for $x\geqslant t_0$, $|f(x)|\leq\varepsilon$.


The result is more easy to establish when $f$ is supposed to be uniformly continuous on $[0,+\infty[$.

$\endgroup$
  • $\begingroup$ Nice answer! May I ask what this notation: $]a,b[$, means? $\endgroup$ – user2139 Apr 29 '18 at 6:16
  • 1
    $\begingroup$ @user2139 The open interval $(a,b)$. $\endgroup$ – Davide Giraudo Apr 29 '18 at 9:51
  • $\begingroup$ I encountered that notation in Europe. I believe it is to avoid confusion with the ordered pair $(a,b)$. $\endgroup$ – Justin Young Apr 30 '18 at 17:54
8
$\begingroup$

You might also be interested in this article by Timothy Gowers, where he describes a thought process of how one might arrive at an elementary proof of this statement (without using the Baire theorem).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.