Let $f:[1,\infty) \longrightarrow \mathbb{R} $ continuous function such that $\forall x \geqslant 1$ we have $\lim_{n \longrightarrow \infty} f(nx)=0$. Prove that $\lim_{y \longrightarrow \infty}f(y)=0$

My professor gave me as a hint to use Baire's theorem.

This is what i have done so far:

Let $\epsilon>0$.

We define $A_m=\{x \geqslant 1: \forall n \geqslant m ,|f(nx)| \leqslant \epsilon \}$

Let $x_k \in A_m$ and $x_k \longrightarrow x$

Then $\forall n \geqslant m$ and from the continuity of $f$ we have $|f(nx)|= \lim_{k \longrightarrow \infty}|f(nx_k)| \leqslant \epsilon$

Thus $x \in A_m$ ,so each $A_m$ is closed.

Now let $x \geqslant 1$ .From our hypothesis $\lim_{n \longrightarrow \infty} f(nx)=0$ ,hence $\exists m_0 \in \mathbb{N}$such that $\forall n \geqslant m_0,|f(nx)| \leqslant \epsilon$,

thus $x \in A_{m_0}$.In other words $[1,\infty)=\bigcup_{m=1}^{\infty}A_m$

We also know that $[1,\infty)$ is a complete metric space with respect to the euclidean metric of the real line.

From Baire's theorem exists a set $A_m$ with non empty interior.

In other words exists $[a,b] \subseteq A_m$.

If $x \in A_m$ then $|f(nx)| \leqslant \epsilon$ $\forall n \geqslant m$.

Can someone help me to proceed form this point?

Thank you in advance!


You're almost there. Note that for any $k \ge \max(m, \lceil \frac{a}{b-a}\rceil)=M$ we have $bk\ge a(k+1)$ and therefore $$\bigcup_{k\ge M} [ak,bk] = [aM,\infty).$$ Thus for any $x\ge aM$ there is an integer $k\ge m$ such that $x \in [ak,bk]$ and so $\frac xk \in [a,b] \subset A_m$. Therefore $$|f(x)| = |f(k\cdot \frac xk)| \le \epsilon.$$

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