1

Can anyone give me a hint (just a hint will be enough) on the following problem?

Let $f$ be a continuous function on $(0,+\infty)$ such that $\lim_{n\to+\infty}f(nx_0)=0$ for all $x_0>0$. Show that $\lim_{x\to+\infty}f(x)=0$.

My problem is that here continuity looks not so useful, but uniform continuity does (actually if $f$ is uniformly continuous on $[a,+\infty)$ for some $a>0$ the problem will be easy). But it seems impossible to prove uniform continuity. Another possible way is to prove by contradiction. If the conclusion fails to hold, then we get some $\epsilon>0$ and a sequence $\{x_n\}\subseteq \mathbf R_{++}$ such that $x_n\to +\infty$ by $|f(x_n)|>\epsilon$ for all $n$. The problem here is how to construct an arithmetic sequence out of $\{x_n\}$

OnoL
  • 2,715

1 Answers1

2

HINT: Use the Baire category theorem to show that if $U\subseteq(0,\infty)$ is open and unbounded, then there is an $x>0$ such that $U\cap\{nx:n\in\Bbb N\}$ is infinite. This intermediate result was a question a few days ago; if you get stuck, you can find a proof here.

Community
  • 1
Brian M. Scott
  • 631,399