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Let $f \, : \, \mathbb{R} \to \mathbb{R}$ be such that

  1. $f \in C^\infty(\mathbb{R})$
  2. $\lim_{n \to \infty}{f(nx)} = 0 \; \forall x \in \mathbb{R}$
  3. $0 \leq f(x) \leq 1 \, \forall x \in \mathbb{R}$
  4. $f(x) = 0 \, \forall x \in (-\infty,1]$

Prove that

$\lim_{x \to \infty}{f(x)} = 0$

I would like a proof that doesn't use Baire category theorem.

I thought that it would be enough to show that

$\lim_{\beta \to \infty}{\int_{0}^{\beta}{ f'(x) dx }} = 0$

But this doesn't seem to help.

Paul
  • 1,489
  • Just for the record: A proof using the Baire category theorem is here: https://math.stackexchange.com/questions/63870/a-classical-problem-about-limit-of-continuous-function-at-infinity-and-its-conne (with many other Q&As linked to it). – Martin R Oct 29 '22 at 18:23
  • Minor observation: let's call $n_{\epsilon} : \mathbb{R} \to \mathbb{N}$ the function such that $\forall n \ge n_{\epsilon}(x) $ we have $|f(nx)| < \epsilon$. If we suppose that $f(x)$ does not tendo to zero for $x \to \infty$, I can find a sequence $y_k \to 1 $ from below such that $\lim_{k \to \infty} n_{\epsilon}(y_k) \to \infty$ (for some $\epsilon$). The problem would be solved if $n_{\epsilon}$ is shown to be upper semicontinous (at least for monotone ascending sequences). – Andrea Marino Oct 29 '22 at 18:33

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