# $\forall \ x>0: \lim_{\ n\to\infty}f(a_nx)=0$, then $\lim_{\ x\to\infty}f(x)=0$

Prove or Disprove:

Let $(a_n)_{n=1}^{\infty}$ be an increasing monotonic sequence of positive real number such that:

$a_n\to\infty \$ and $\ \frac{a_{n+1}}{a_n}\to 1$.

If $f:\mathbb{R}\to\mathbb{R}$ is a continuous function such that:

$\forall \ x>0: \lim_{\ n\to\infty}f(a_nx)=0$, then $\lim_{\ x\to\infty}f(x)=0$.

• $\lim_{n\to\infty}f(x)=f(x)$ – user160738 Jan 27 '17 at 16:23
• @user160738 It was a typo – Don Fanucci Jan 27 '17 at 16:25
• Still wrong. The limit should be over $x$ – b00n heT Jan 27 '17 at 16:26
• @TrueTopologist You have still a bad typo...or a rather nonsensical question here. It seems to be that it should be " ...then $\;\lim\limits_{n\to\infty}f(\color{red}{a_n})=0\;$" . Check this. – DonAntonio Jan 27 '17 at 16:27
• The result is probably true. I know for sure that the result holds for $a_n=n$ (This special case can be proved using Baire Category theorem) – b00n heT Jan 27 '17 at 16:29

The claim is true, and the proof proceeds along similar lines to the classical $a_n = n$ case.

Fix $\epsilon > 0$, and define the sets

$$A_n = \{ x \in \mathbb R^{+} : |f(a_n x)| \leq \epsilon \}$$ $$B_n = \bigcap_{m \geq n} A_m$$

By continuity of $f$, both $A_n$ and $B_n$ are closed. Furthermore, the given condition implies that

$$\mathbb R^{+} = \bigcup_{n} B_n$$

By the Baire category theorem, it follows that one of the $B_n$, say $B_M$, contains a nonempty open interval, say $(a, b)$. Pick $\delta > 0$ such that $b(1 - \delta) > a$. By the convergence $a_{n+1} / a_n \to 1$, we can pick $N$ sufficiently large such that for all $n > N$, we have $a_n > (1 - \delta) a_{n+1}$. Thus, we have that

$$a_n b > a_{n+1} (1 - \delta) b > a_{n+1} a$$

and the intersection $(a_n a, a_n b) \cap (a_{n+1} a, a_{n+1} b)$ is nontrivial for $n > N$. The divergence $a_n \to \infty$ then implies that

$$S = \bigcup_{n > \max\{M, N\}} (a_n a, a_n b) = (C, \infty)$$

for some $C$. By definition of $B_M$ and the inclusion $(a, b) \subset B_M$, it follows that for all $x \in S$, we have that $|f(x)| \leq \epsilon$. Since $\epsilon > 0$ was arbitrary, it follows that $f(x) \to 0$ as $x \to \infty$.

• Can you explain a little bit more why for all x∈S, we have that |f(x)|≤ϵ ? Is it because $S$ is in the interval $(a,\infty)$? – Don Fanucci Jan 29 '17 at 7:28
• Oh got it, proved it by contradiction – Don Fanucci Jan 29 '17 at 11:59