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I am trying to show that if $f:[0,\infty)\rightarrow [0,\infty)$ is continuous and for any $x$ in $[0,\infty)$ the sequence $f(x),f(2x),f(3x),...$ tends to zero then $\lim_{x\rightarrow\infty}f(x)=0$.

I know that every set $\{x|f(nx)\le \varepsilon\}$ for fixed $\varepsilon>0$ is closed due to the continuity of $f$, and I was thinking I could perhaps use this property.

Could I use Baire category theorem here, or how should I proceed?

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  • $\begingroup$ Do you mean $\lim_{x\to\infty}$? $\endgroup$
    – Kenta S
    Jul 16 at 1:36
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    $\begingroup$ Would A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem address your question? Including the assertion about the BCT? $\endgroup$ Jul 16 at 1:40
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    $\begingroup$ That's good to know. You now have both the advanced and the non-advanced proof available. I believe the statement can be made stronger, the content of the so-called Croft-Kingman lemma. Indeed, the CK lemma is very strong : all you need (with $f$ continuous) is that for all $x$, $\lim_{x\to \infty} f(nx) \to L_x$ for some $L_x<\infty$ depending on $x$. The theorem states that $L_x$ are the same for all $x$ and $L = \lim_{x \to \infty} f(x)$. $\endgroup$ Jul 16 at 1:45
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    $\begingroup$ In fact , once you learn topology, the theorem is astonishingly enough true for $f: [0,\infty) \to X$ where $X$ is any space that admits a metric(also called a metrizable space). So $X$ could be $C[0,1]$(on $\sup$ distance) or $L^p(\mu)$ (with the $L^p$ distance, $p\geq 1$) , and this lemma looks even more amazing in such settings. The strength of the lemma is that it is used in MC theory, for example, to promote Ergodic arguments : beginning from any time point, we want to iterate this time period again and again, so proving something for each time point provides a result for the process. $\endgroup$ Jul 16 at 1:51
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    $\begingroup$ Oh yeah, and $f$ is uniformly continuous on $[0,\infty)$ as well if this occurs. This simple hypothesis has got so much running for it. $\endgroup$ Jul 16 at 1:58