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Function $f$ is continuous on $\left[0,+\infty\right)$,
does $\displaystyle\lim_{n\rightarrow\infty}f\left(nx\right)=0,\quad\forall x>0$

imply that $\displaystyle\lim_{x\rightarrow+\infty}f\left(x\right)=0$?
If $f$ is not continuous, it is false. I don't know how to make use of the continuity.

Antimonius
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1 Answers1

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Yes, the property holds for continuous functions (see also A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem)

For $\epsilon>0$, let $A_N=\bigcap_{n\geq N}\{x>0: f(nx)\leq \epsilon\}$. Then $A_N$ is closed by the continuity of $f$ and $\lim_{n\rightarrow\infty}f\left(nx\right)=0$ implies that $$\bigcup_{N\geq 0} A_N=(0,+\infty).$$ Hence, by the Baire Category Theorem, there exists some $N_0$ such that $A_{N_0}$ contains an open interval $(a,b)$ which means that for $n≥N_0$ and $t\in (na,nb)$ we have that $f(t)\leq \epsilon$. Finally note that for $n>a/(b-a)$, then $nb>(n+1)a$ and therefore
$$t\in\bigcup_{n>\max(a/(b-a),N_0)}(na,nb)=(\delta,+\infty)\implies 0\leq f(t)\leq \epsilon$$ That is $\lim_{x\rightarrow+\infty}f\left(x\right)=0$.

P.S. For a proof without Baire Category Theorem see Gowers's Weblog

Robert Z
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  • The formulation of Baire Category Theorem that you're using is: If a non-empty complete metric space is the countable union of closed sets, then at least one of these sets must have non-empty interior, but here $(0, +\infty)$ is not complete since $(\frac{1}{n})$ is a Cauchy sequence but converges to $0$ which does not belong to $(0, +\infty).$ So what am I missing here? – Aryaman Jal Dec 23 '17 at 10:40
  • @Aryaman Jal Every locally compact Hausdorff space, like $(0,+\infty)$ is a Baire space. See https://en.wikipedia.org/wiki/Baire_category_theorem – Robert Z Dec 23 '17 at 10:47
  • Understood, thanks for the clarification – Aryaman Jal Dec 23 '17 at 10:50