Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $f$ is a continuous and real function on $[0,\infty]$. How can we show that if $\lim_{n\rightarrow\infty}(f(na))=0$ for all $a>0$ then $\lim_{x\rightarrow+\infty} f(x)=0$?

share|cite|improve this question

$\newcommand{\orb}{\operatorname{orb}}$If $f(x)\not\to 0$ as $x\to\infty$, then there is an $\epsilon>0$ such that for every $m\in\Bbb N$ there is an $x_m\ge m$ such that $|f(x_m)|\ge\epsilon$. Since $f$ is continuous, for each $m\in\Bbb N$ there is a $\delta_m>0$ such that $|f(x)|>\frac{\epsilon}2$ for all $x\in(x_m-\delta_m,x_m+\delta_m)$. For $n\in\Bbb N$ let $$U_n=\bigcup_{k\ge n}(x_k-\delta_k,x_k+\delta_k)\;.$$

For $a\in(0,1)$ let $\orb(a)=\{na:n\in\Bbb Z^+\}$, and for $n\in\Bbb N$ let $$G_n=\{a\in(0,1):\orb(a)\cap U_n\ne\varnothing\}\;.$$ Suppose that $0<b<c<1$, and let $$V(b,c)=\bigcup_{n\in\Bbb Z^+}(nb,nc)=\bigcup_{x\in(b,c)}\orb(x)\;.$$ Let $m=\left\lfloor\frac{b}{c-b}\right\rfloor+1$; $(n+1)b<nc$ for each $n\ge m$, so $V(b,c)\supseteq(mb,\to)$. It follows that $V(b,c)\cap U_n\ne\varnothing$ for each $n\in\Bbb N$ and hence that $(b,c)\cap G_n\ne\varnothing$ for each $n\in\Bbb N$. Thus, each $G_n$ is a dense open subset of $(0,1)$, so by the Baire category theorem $G=\bigcap_{n\in\Bbb N}G_n$ is dense in $(0,1)$ and in particular, $G\ne\varnothing$.

Fix $a\in G$. Then $a\in G_n$ for each $n\in\Bbb N$, so $\orb(a)\cap U_n\ne\varnothing$ for each $n\in\Bbb N$. This clearly implies that $\left\{n\in\Bbb Z^+:|f(na)|>\frac{\epsilon}2\right\}$ is infinite, contradicting the hypothesis that $\lim_{n\to\infty}f(na)=0$, and we conclude that $\lim_{x\to\infty}f(x)=0$.

Added: Since you’re having trouble with the notion of proof by contradiction, let me note that I need not have phrased it that way: with a small change in wording this becomes a proof of the contrapositive of the desired statement. Since a statement and its contrapositive are logically equivalent, it proves the desired statement as well.

The desired statement has the form $A\land B\Rightarrow C$, where $A$ is the hypothesis that $f$ is continuous, $B$ is the hypothesis that $\lim_{n\to\infty}f(na)=0$ for each $a>0$, and $C$ is the desired conclusion, that $\lim_{x\to\infty}f(x)=0$. As I phrased my argument, it has the following form:

Assume $A,B$, and $\lnot C$, and infer $\lnot B$, thereby showing that $A\land B\land\lnot C\Rightarrow B\land\lnot B$. Since $B\land\lnot B$ is a contradiction, the hypthesis $A\land B\land\lnot C$ is false. But we’re given that $A$ and $B$ are true, so it must be $\lnot C$ that’s false, and therefore, given that $A$ and $B$ are true, $C$ must be true.

I could, however, have cast the argument in the following form with very minor changes in wording:

Assume $A$. Then $\lnot C\Rightarrow\lnot B$, which is logically equivalent to $B\Rightarrow C$, so $A\land B\Rightarrow C$.

Specifically, I could have written this for the last paragraph:

Fix $a\in G$. Then $a\in G_n$ for each $n\in\Bbb N$, so $\orb(a)\cap U_n\ne\varnothing$ for each $n\in\Bbb N$. This clearly implies that $\left\{n\in\Bbb Z^+:|f(na)|>\frac{\epsilon}2\right\}$ is infinite, and hence that $\lim_{n\to\infty}f(na)\ne 0$. That is, we’ve shown that if $f(x)\lnot\to 0$ as $x\to\infty$, then there is at least one $a>0$ such that $\lim_{n\to\infty}f(na)\ne 0$. This is logically equivalent to the assertion that if $\lim_{n\to\infty}f(na)=0$ for every $a>0$, then $f(x)\to 0$ as $x\to\infty$, which is what we wanted to prove.

share|cite|improve this answer
I'm not certain I understand the last bit. How can the proof for this question contradict $\lim_{n\to\infty}f(na)=0$ if that is part of the statement of the question? – Xuan Huang Oct 14 '12 at 21:27
@Danielle: Because it’s a proof by contradiction. My very first step was assuming that $\lim_{x\to\infty}f(x)$ was not $0$, and that led me to the conclusion that $\lim_{n\to\infty}f(na)\ne 0$. This contradicts the hypotheses, so my initial assumption must be false, and $\lim_{x\to\infty}f(x)$ is $0$ after all. – Brian M. Scott Oct 14 '12 at 22:30
That's what I thought, but you seem to suggest in the end that $\lim_{n\to\infty}f(na)=0$ is contradicted instead of $\lim_{n\to\infty}f(na)\ne0$ being contradicted. Is this a typo? – Xuan Huang Oct 15 '12 at 13:21
@Danielle: No typo: it is $\lim_{n\to\infty}f(na)=0$ (for a particular $a>0$) that’s being contradicted. To get a contradiction I assumed that $\lim_{x\to\infty}f(x)\ne 0$. From this assumption I was able to show that there is an $a>0$ such that $\left\{n\in\Bbb Z^+:|f(na)|>\frac{\epsilon}2\right\}$ is infinite. This contradicts the hypothesis that $\lim_{n\to\infty}f(na)=0$. Since I got a contradiction by assuming that $\lim_{x\to\infty}f(x)\ne 0$, this assumption must be false, and in fact $\lim_{x\to\infty}f(x)=0$, which is what I wanted to prove. – Brian M. Scott Oct 15 '12 at 13:37
The question asks, however, how we can show that if ($\lim_{n\rightarrow\infty}(f(na))=0$) for all $a>0$ then $\lim_{x\rightarrow+\infty} f(x)=0$. So it doesn't make sense at all to me that you are proving that by contradicting the statement of the proof itself. Perhaps I am misunderstanding something? We are asked to show that $\lim_{x\to\infty}f(x)=0 $ **when** $\lim_{n\to\infty}(f(na))=0.$ How can you prove $\lim_{x\to\infty}f(x)=0 $ **when** $\lim_{n\to\infty}(f(na))=0$ when $\lim_{n\to\infty}(f(na))\ne 0$? – Xuan Huang Oct 16 '12 at 2:46

This is a standard (moderately tough) exercise in applying the Baire category theorem. Tim Gowers did a presentation of this result on his blog under the title "What is deep mathematics?".

But as he writes:

If you haven’t seen this before and want to get the most out of this post then you should (of course) make a serious attempt to solve this beautiful problem before reading on.

share|cite|improve this answer
I am not sure I understand how to do the problem very well, as I have played with it a great deal, but to no avail. That is why I am asking. – Xuan Huang Oct 5 '12 at 13:26
In his post, Tim solves this with an elementary approache. Do you know where can I find a solution involving BCT? – Idan Oct 5 '12 at 17:52
@Idan: Read the comments. Matthew Folz gave an argument based on BCT. – kahen Oct 5 '12 at 18:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.