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The question was this: Suppose $f$ is continuous on $(0,\infty)$ and for every $t>0$ $$\lim_{n\to\infty}f(t/n)=0.$$Does it follow that $f(t)\to0$ as $t\to0$ from above?

(Evidently it doesn't go without saying: The question is whether the limit exists - of course it equals $0$ if so.)

I couldn't believe I didn't know the answer. Of course if $f$ is not continuous one could concoct a counterexample, but.

I believe the answer is yes. So the remaining questions are vague: Is this obvious to someone for some reason I don't see? Is it something everybody knows? Edit: Maybe it's not trivial; assuming just $f(t/n)\to0$ for almost every $t$ is not enough.

Here's a proof, I think. Suppose not. Then since $f$ is continuous there exist $\epsilon>0$, $t_k\to0$ and $\delta_k>0$ such that $$|f(t)|\ge\epsilon\quad(|t-t_k|<\delta_k).$$

Let $I_k=(t_k-\delta_k,t_k+\delta_k)$ and $$B_k=\bigcup_{n=1}^\infty nI_k,$$where $nS=\{ns:s\in S\}$.

Note that if $0<a<b$ and $t_k<b-a$ then $$[a,b]\cap B_k\ne\emptyset.$$

This allows you to find a nested sequence of intervals $[a_n,b_n]$ such that every element of $[a_n,b_n]$ lies in $B_k$ for at least $n$ values of $k$. So by compactness there exists $t$ such that $t$ lies in infinitely many $B_k$; hence $f(t/k)$ does not tend to $0$.

Chuckle: An example showing that assuming $f(t/n)\to0$ for almost every $t$ is not enough:

Say $t_k\to0$. Choose $\delta_k>0$ so that the $I_k$ are pairwise disjoint and also $$\sum_km((0,A)\cap B_k)<\infty$$for every $A>0$. Let $f$ be a continuous function on $(0,\infty)$ such that $f=0$ everywhere except on $I_k$, where $f$ has a spike of height $1$.

Almost every $t$ lies in only finitely many $B_k$, and for every such $t$ we have $f(t/n)\to0$.

David C. Ullrich
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  • I'm not exactly an expert, but I think this is not obvious, and I think your proof is optimal, since continuity at $0$ is not assumed. Of course, if you did have continuity at $0$, then I think it would be obvious. – A. Thomas Yerger Jan 06 '18 at 03:40
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    Similar problems have been asked before (equivalent formulation), there are proofs using nested intervals or the Baire theorem https://math.stackexchange.com/questions/63870/a-classical-problem-about-limit-of-continuous-function-at-infinity-and-its-conne – clark Jan 06 '18 at 04:03
  • Far from obvious (to me, at least). – copper.hat Jan 06 '18 at 05:25
  • This is not obvious at all if by "obvious" you mean "the proof can be supplied in mind without writing anything with almost little effort". On the face of it it may seem intuitively obvious and many beginners won't doubt it. – Paramanand Singh Jan 06 '18 at 09:42
  • @ParamanandSingh Heh-heh: If it seems intuitively obvious to you it follows that your intuition is either much better than mine or much worse. The only thing that seemed clear to me at the start was that it's false if we don't assume $f$ is continuous - it was clear that a continuous counterexample would be much harder but not at all clear that it was impossible. First actual progress was the "almost every $t$" counterexample at the bottom - finally saw why it was impossible to improve that to "every $t$"... – David C. Ullrich Jan 06 '18 at 14:17
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    I reiterate: it may seem intuitively obvious. I must say analysis is one of those fields where most of the results are counter-intuitive. Perhaps our human intuition was designed to deal with finite/discrete stuff. – Paramanand Singh Jan 06 '18 at 14:59
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    As far as I know this result is known as Croft's lemma. [H.T. Croft (1957): A question of limits. Eureka 20, 11-13] – r9m Jan 07 '18 at 08:53

1 Answers1

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This is a standard application of Baire Category Theorem. Considering $f(\frac 1 t)$ we can state this as follows: f continuous on $(0,\infty)$ and $f(nt) \to 0$ as $n \to \infty$ for all t implies $f(t) \to 0$ as $t \to \infty$. Write $(0,\infty)$ as union of the sets $\{t:|f(nt)| \leq \epsilon$ for all $n \geq m\}$ where $\epsilon >0$ is fixed. Since $(0,\infty)$ is of second category there exists m and $a<b$ such that $(a,b)$ is contained in $\{t:|f(nt)| \leq \epsilon$ for all $n \geq m\}$. Let $x>bm$ and $x(\frac 1a -\frac 1 b)>1$ The interval $(\frac x b ,\frac xa)$ has length exceeding 1 so it contains an integer k. Since $k>\frac x b >m$ and $\frac x k \in (a,b)$ it follows that $|f(k \frac x k)| \leq \epsilon$. Thus $|f(x)| \leq \epsilon$ for $x>bm$ and $x(\frac 1a -\frac 1 b)>1$.

Kavi Rama Murthy
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