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When I google "baire category theorem", I get a link to Ben Green's website. And at the end of the paper, he mentioned such a classic problem:

Suppose that $f:\mathbb{R}^+\to\mathbb{R}^+$ is a continuous function with following property: for all $x\in\mathbb{R}^+$, the sequence $f(x), f(2x), f(3x),\dots$ tends to $0$. Prove that $\lim_{t\to\infty}f(t)=0$.

I find the problem 1.17 on P.27 of the book "Selected problems of real analysis", and on P.169 gives the answer by prove the following lemma:

If $G$ is a unbounded open set of $\mathbb{R}^+$, then for any closed interval $[p,q]\ (0<p<q)$, there exist a $x_0\in [p,q]$ such that $G$ contains infinitely many points of the form $nx_0\ (n\in\mathbb{N})$.

But, on the above book, it also says:

If $\lim_{n\to\infty}f(nx)$ exists only for points $x$ in a nonempty closed set without isolated points, then $\lim_{x\to\infty}f(x)$ also exist.

I didn't find a proof for this result, I want to know whether for any nonempty closed set with no isolated points, the above lemma is true? and could someone tell me why Ben Green mentioned this problem on his paper (see the hint of his paper)?

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  • $\begingroup$ Do you know how to prove that it's true for a closed interval? $\endgroup$
    – joriki
    Sep 12 '11 at 12:20
  • $\begingroup$ It might help if you give a proof or a reference for the special case you can show. $\endgroup$
    – Srivatsan
    Sep 12 '11 at 13:41
  • $\begingroup$ @gylns: When you say "if $\lim_{n\to\infty}f(nx)$...", is your $f$ continuous? $\endgroup$
    – leo
    Sep 14 '11 at 15:37
  • $\begingroup$ @leo:yes, $f$ is always continuous. $\endgroup$
    – gylns
    Sep 15 '11 at 5:09
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Since I see you asked this a while ago, I'll answer one of your questions now; I'll give you a solution to your first problem that relates it to the Baire Category Theorem. I have some ideas about your second question, and if no one else pops in to answer it in the meantime, I'll try to come up with something later. (Added: See below.)


If $f:\mathbb R^+\to\mathbb R^+$ is continuous and $\lim_{n\to\infty}{f(nx)}=0$ for every $x\in\mathbb R^+$, we want to prove that $\lim_{x\to\infty}{f(x)}=0$.

Fix some $\epsilon>0$. The sets $E_N=\{x: n \geq N \implies f(nx)\leq \epsilon\}$ are closed (write $E_N$ as the intersection over $n\geq N$ of the sets $\{x:f(nx)\leq\epsilon\}$, which are closed by the continuity of $x\mapsto f(nx)$). On the other hand, the assumption $f(nx)\to0$ that is made for every $x>0$ ensures that you can write $\mathbb R^{+}$ as the union of the $E_N$. The Baire Category Theorem says that at least one of them, say $E_{N}$, contains an open segment $(a,b)$. Thus if $n\geq N$ and $t\in(na,nb)$, then $f(t)\leq \epsilon$. But for some positive integer $M\geq N$ (any $M$ bigger than $a/(b-a)$ will suffice), \begin{equation*} (Ma,\infty)=\bigcup_{n\geq M}(na,nb). \end{equation*} So if $t>Ma$, then $f(t)\leq\epsilon$, which proves that $f(t)\to0$ as $t\to\infty$.


Here are some ideas for your second question, for now—like I said I'll update later if no one else finishes it off. Let $$ P=\left\{x>0:\lim_{n\to\infty}{f(nx)}\text{ exists}\right\}. $$ If $P=\mathbb R^+$, the above argument can be tinkered with to ensure that $\lim_{x\to\infty}{f(x)}$ exists, or you can use the argument in the book you linked to. You want to assume only that $P$ is closed with no isolated point, i.e., that $P$ is perfect. In that case, I believe I can modify the BCT argument to show that $\lim_{x\to\infty\atop x\in P}{f(x)}$ exists. (Edit: After thinking about this a bit I don't believe it's as simple as it initially looked to show that $\lim_{x\to\infty\atop x\in P}{f(x)}$ exists; in fact, I'm not even completely certain it's necessarily true.)

Here are a couple of easy observations right off the bat (though I'm not sure they lead in the right direction...):

  1. If $x\in P$, then $nx\in P$ for $n=1,2,3,\dots$

  2. If $P$ contains an interval, then $P=\mathbb R^+$. Hence if $P$ is anywhere dense, then $P=\mathbb R^+$. If $P$ contains points arbitrarily close to zero, then $P$ is dense by 1., so $P=\mathbb R^+$. So we can assume that $P\subset[\epsilon,\infty)$ for some $\epsilon>0$, and that $P$ is nowhere dense.


Added: After speaking to several more experienced people about your second problem (as I've phrased it above), it seems that it is likely false, but that constructing a counterexample may be difficult. Here's the gist of the difficulty. If you take a totally disconnected perfect subset $E$ of $[1,2]$ and let $P=\bigcup nE$, then $P$ is a totally disconnected perfect subset of $\mathbb R^+$ satisfying 1. above. You can take $g$ to be a function which is constant on $P$, and extend $g$ to a continuous function $f$ on $[0,\infty)$ such that $\lim_{x\to\infty}{f(x)}$ does not exist. If $G$ is the complement of $P$, the difficulty lies in answering the question of whether $\lim_{n\to\infty}{f(nx)}$ necessarily exists for some $x\in G$. I'm not exactly sure how to do that, but if the answer is no, then we have a counterexample.

Finally, I'm not quite certain the original problem was asking for all of this. The wording in the book you cite is as follows:

Prove that if $f\in C([0,+\infty))$ and the limit $\lim_{n\to\infty}{f(nx)}$ exists for any $x\geq0$, then the limit $\lim_{x\to\infty}{f(x)}$ exists. Prove this if the limit $\lim_{n\to\infty}{f(nx)}$ exists only for points $x$ in some nonempty closed set without isolated points.

To take this quite literally, note that $\lim_{n\to\infty}{f(n\cdot0)}$ certainly exists, so the set on which $\lim_{n\to\infty}{f(nx)}$ exists is a perfect set $E$ containing zero with the property 1. above, i.e., $x\in E\implies nx\in E$ for all positive integers $n$. From this it follows that $E=[0,\infty)$, and the problem reduces to the initial question. It was the word "nonempty" in the hypotheses that initially led me to believe that this was not the authors' intended interpretation, but the more I think about it, the more likely it seems that it was (especially considering the fact that the book is translated).

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  • $\begingroup$ Well done. This is a nice example where the Baire Category Theorem is very useful. $\endgroup$
    – JT_NL
    Sep 22 '11 at 11:31
  • $\begingroup$ very nice. I only know the argument of using nested closed interval and never try to use the BCT. $\endgroup$
    – gylns
    Sep 22 '11 at 12:26
  • $\begingroup$ You said "the sets $\{x:f(nx)\leq\epsilon\}$, which are closed by the continuity of $x\mapsto f(nx)$)" (In the eighth line from your answer).Why the sets $\{x:f(nx)\leq\epsilon\}$ are closed? From the continuity of $x\mapsto f(nx)$ and $x\in (0,\infty)$,we only conclued they are relatively closed in $(0,+\infty)$,not closed in $R$. Look forward to your reply. $\endgroup$ Apr 25 '17 at 0:38
  • $\begingroup$ @K.Kwai That's true but it's not relevant for the application of the Baire Category Theorem. $\endgroup$ Apr 26 '17 at 4:45
  • $\begingroup$ Absolutely,it does not affect the result at all .Please forgive me,I just wanted to point out a subtle bug appearing in your nice answer. $\endgroup$ Apr 26 '17 at 13:15

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