Function $f: (0, \infty) \to \mathbb{R}$ is continuous. For every positive $x$ we have $\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$. Prove that $\lim\limits_{x \to 0}f(x)=0$. I have tried to deduce something from definition of continuity, but with no effect.

  • $\begingroup$ What level course is this for? To me this is the kind of question that seems like a Baire Category Theorem thing. $\endgroup$ – Moya Jan 7 '19 at 22:41
  • $\begingroup$ $n$ is an integer? $\endgroup$ – leonbloy Jan 7 '19 at 22:46
  • $\begingroup$ But can you just not rewrite the limit such that $y=\frac{x}{n}$ and so you get what you need by letting $y$ going to 0, which does happen as $n$ goes to infinity and $x$ is any positive real and f is continuous of course so the limit exists? $\endgroup$ – Marco Bellocchi Jan 7 '19 at 22:46
  • $\begingroup$ @Moya introductory level $\endgroup$ – user4201961 Jan 7 '19 at 22:48
  • $\begingroup$ This I believe is just a bit more general than the following, you can proceed in a similar manner math.stackexchange.com/questions/3045776/… $\endgroup$ – Marco Bellocchi Jan 7 '19 at 22:59

The proof relies on the following lemma:

Let $I \subset (0,\infty)$ be a closed bounded interval. Let $U \subset (0,\infty)$ be an open subset accumulating at $0$. Then there exists some integer $N \geq 2$ and some closed interval $J \subset U$ such that $N \cdot J \subset I$.

Sketch of proof: take some $x \in U$ with $x$ lower than the length of $I$ and the minimum of $I$ . For some integer $N \geq 2$, $Nx \in \overset{\circ}{I}$, thus, there exists some compact interval $J \subset U$ such that $N \cdot J \subset I$.

Now, assume $f$ does not go to $0$ at $0$. Then, for some $\epsilon > 0$, $U=\{|f| > \epsilon\}$ is an open subset of $(0,\infty)$ accumulating at $0$.

By iterating the lemma, we can construct sequences of compact intervals $(I_n)$ and integers $(N_n)$ such that $N_{n+1}I_{n+1} \subset I_{n-1}$, $N_n \geq 2$.

Now, let $A_n=N_1 \ldots N_n$, then $A_n \rightarrow \infty$ and $K_n=A_n \cdot I_n$ is a non-increasing sequence of nonempty compact intervals in $I_0$. Thus, there exists $x$ that is in every $K_n$.

So, $x/A_n \in I_n \subset U$, for all $n$, thus $|f(x/A_n)| > \epsilon$ for all $n$, a contradiction.

  • $\begingroup$ (edited) This looks correct, but I think you can remove some of the intermediate claims without affecting the argument. In particular, you don't need any information about $\sup I_n$. Also, when you say the sequence $K_n$ is non-decreasing, I think you mean that it is decreasing, or in more familiar language, descending. And in the final sentence, "all $n$" should be "infinitely many $n$". $\endgroup$ – Chris Culter Jan 7 '19 at 23:41
  • $\begingroup$ Thanks for the remarks, I edited. However, in the last sentence, I definitely meant « for all $n$ » and not « for infinitely many $n$», however this does not affect the result. $\endgroup$ – Mindlack Jan 8 '19 at 0:02
  • $\begingroup$ Reminds me of the Axiom of Archimedes. $\endgroup$ – marty cohen Jan 8 '19 at 0:49

This is a standard application of Baire Category Theorem (BCT). Since $(0,\infty)=\cup_n \{x:|f(\frac x k)| \leq\epsilon \,\forall k \leq n\}$ there exists $n$ such that $\{x:|f(\frac x k)| \leq \epsilon \, \forall k \geq n\}$ contains some open interval $(a,b)$. [ Because $(0,\infty)$, being an open subset of $\mathbb R$ has an equivalent complete metric so BCT applies]. This gives UNIFORM convergence of $f(\frac x n)$ to $0$ on some open interval from which the result follows easily. [ Details: let $0<\delta <b-a$ and $\delta < \frac a n$. Then, for any $x <\delta $ the interval $(\frac a x,\frac b x)$ has length exceeding $1$, so it contains an interger $k$. Hence $kx \in (a,b)$. Also, $k >\frac a x > \frac a {\delta} > n$ so $|f(x)|=|f(\frac {kx} k)| <\epsilon$].

  • $\begingroup$ Shouldn’t it be $\bigcup_n\{x,\,\forall k\geq n,\, |f(x/k)| \leq \epsilon\}$? $\endgroup$ – Mindlack Jan 8 '19 at 0:54
  • $\begingroup$ @Mindlack You are right. That was a silly mistake I made. $\endgroup$ – Kavi Rama Murthy Jan 8 '19 at 5:01

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