# $\lim_{n\to \infty}f(nx)=0$ implies $\lim_{x\to \infty}f(x)=0$

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.

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–  Byron Schmuland Jan 21 '12 at 20:26

Fix $\epsilon>0$ and put $F_n:=\left\{x\geq 0,\forall k\geq n, |f(kx)|\leq \varepsilon\right\}$. Then for all $n$ $F_n$ is closed since $f$ is continuous and $\bigcup_n F_n=[0,+\infty[$, so 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 $\geq n_0$ such that $\frac 1{n_1}<r$ and let $x\geq t_0$. Then we can write $x=n_xx_0+\beta$ where $n_x$ is an integer $\geq n_1$ and $0\leq \beta<x_0$. So $$|f(x)|=|f(n_xx_0+\beta)|=\left|f\left(n_x\left(x_0+\frac{\beta}{n_xx_0}\right)\right)\right|\leq \varepsilon$$ since $n_x\geq n_1$ and $x_0+\frac{\beta}{x_0}\in ]x_0-r,x_0+r[\subset F_{n_0}$ since $\left|\frac{\beta}{x_0n_x}\right|\leq \frac 1{n_x}\leq \frac 1{n_1}<r$.

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

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

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How is $n_x x_0 + \beta = n_x \Big( x_0 + \dfrac{\beta}{n_x x_0} \Big)$ ? –  user123733 Jan 2 at 14:18

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).

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