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

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    $\begingroup$ $\lim_{n\to\infty}f(x)=f(x)$ $\endgroup$ – user160738 Jan 27 '17 at 16:23
  • $\begingroup$ @user160738 It was a typo $\endgroup$ – Don Fanucci Jan 27 '17 at 16:25
  • $\begingroup$ Still wrong. The limit should be over $x$ $\endgroup$ – b00n heT Jan 27 '17 at 16:26
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    $\begingroup$ @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. $\endgroup$ – DonAntonio Jan 27 '17 at 16:27
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    $\begingroup$ 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) $\endgroup$ – b00n heT Jan 27 '17 at 16:29
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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 $.

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  • $\begingroup$ 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 $(a,\infty)$? $\endgroup$ – Don Fanucci Jan 29 '17 at 7:28
  • $\begingroup$ Oh got it, proved it by contradiction $\endgroup$ – Don Fanucci Jan 29 '17 at 11:59

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