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Suppose $f$ is continuous, $f:[0,\infty)\rightarrow \mathbb{R}$ with the property that $f(nt)\rightarrow 0$ as $n \rightarrow \infty$ for all $t >0$. We want to show that $f(x) \rightarrow 0$ as $x \rightarrow \infty$.

My attempt:

Fix $\varepsilon > 0$. We consider sets $A_n=\{t>0| m\geq n \implies |f(mt)|< \varepsilon\}$. Since $\bigcup_{n \in \mathbb{N}}A_n=(0,\infty)$ is nonmeager we have by Baire that $\overline{A_n}$ has a nonempty intertior for some $n$. Thus WLOG we can take $[a,b] \in A_n$ with $n(b-a)\geq a$

Why is this allowed? I understand we get the interval, but why should I be guaranteed that inequality? Is this where I'm meant to use continuity?

Now I'm supposed to show $|f(x)| < \varepsilon$ if $x \geq na$. We have that if $x \in [ka,kb]_{k\geq n}$, $|f(x)|\leq \varepsilon$ by the definition of $A_n$. I claim for any intervals $[ka,kb]_{k \geq n}$ we have $(k+1)a \leq kb$. If this is true then the intervals $[ka,kb]_{k\geq n}$ will cover $[na,\infty)$ And I will be done. Clearly the statement is true for $k=n$ since $n(b-a)\geq a$. Suppose its true for $n+1,n+2...k-1$. Then:

$$(k+1)a=ak+a \leq (k-1)b + a = kb - (b-a)<kb$$

This completes the induction(does it?) and I'm done? I'm skeptical because I never (at least explicitly) use the continuity of $f$? Thanks in advance.

Muselive
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1 Answers1

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The reason you can guarantee $n(b-a)\ge a$ is because the sets get larger as $n\to\infty$. If you choose $n$ and $[a,b]\subseteq A_n$ where $n(b-a)\not\ge a$, you could just choose a large $N>n$ so that $N(b-a)>a$ does hold, and you will still have $[a,b]\subset A_{N}$.

The rest of the proof is correct. You did use the continuity of $f$ when you said that $\bigcup_{n=1}^\infty A_n= (0,\infty)$. The reason that every $t$ is a part of some set $A_n$ is because the sequence $f(t),f(2t),f(3t)\dots$ converges to $0$, so there exists an $m$ such that $n\ge m$ implies $|f(nt)|<\varepsilon$.

Mike Earnest
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