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Let $f:(0,\infty)\to \Bbb R$ be continuous and such that for each $x>0$ the sequence $\{f(\frac{x}{n})\}\to 0$.

Does it imply that $\lim_{x\to 0^+} f(x)=0$?

My try:Fix $x>0$.Since $f$ is continuous then $\lim _{n\to \infty}f(\frac{x}{n})=f(\lim_{n\to \infty} \frac{x}{n})=f(0)\implies f(0)=0$

Since $f$ is continuous and $f(0)=0\implies \lim _{x\to 0^+} f(x)=0$

Is the solution correct?In the book it is given to use the Baire Category Theorem.

Did
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    your trick only works if $f(0)$ is defined, which it is not in this case, or if you already know that the limit exists. – Asinomás Aug 30 '16 at 04:57
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    This is a variation on a classic application of the Baire Category Theorem. The typical idea is to write your space as a union of closed sets, and use the theorem to conclude that one of these sets must contain an open interval. – Moya Aug 30 '16 at 05:31

2 Answers2

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Expanding on Moya's comment, here is a big hint.

Fix $\epsilon>0$. Let $$ K_n = \{x>0:|f({x}/m)|\le\epsilon \text{ for all } m\ge n\} $$ The idea is that for all $x$, the sequence $f(x/n)$ gets close 0 eventually, so we define $K_n$ to be the set of $x$ for which is this sequence is close to zero specifically at time $n$.

Next, prove that

  1. The sets $K_n$ are closed.
  2. The union of $K_n$ is $\mathbb R^+$.

Since (one form of) the Baire Category Theorem states that $\mathbb R^+$ cannot be written as a countable union of nowhere dense closed sets, you can conclude that one of the sets $K_n$ is somewhere dense, which for closed sets means it contains an interval $[x_1,x_2]$, for some $x_1<x_2$.

This is huge progress. Now, we have series of solid blocks of points $$[x_1/n,x_2/n],[x_1/(n+1),x_2/(n+1)],[x_1/(n+2),x_2/(n+2)]\dots,$$ where for any $x$ in these intervals, $|f(x)|<\epsilon$. Use this fact to find a $\delta>0$ such that $|f(x)|<\epsilon$ whenever $|x|<\delta$.

Mike Earnest
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I'll give you an outline:

Start with the definition of a limit.

$$\forall \epsilon \text{ } \exists \delta \text{ s.t. } |f(x)|\leq \epsilon \text{ whenever } x\in (0,\delta).$$

And the definition of $\{f(\frac{x}{n})\}\to0$

$$\forall \epsilon \text{ } \exists N \text{ s.t. } |f(\frac{x}{n})|\leq \epsilon \text{ whenever } n>N.$$

And continuity (*edit, accidentally wrote the definition for uniform continuity)

$$\forall \epsilon,x \text{ } \exists \delta \text{ s.t. } |f(x)-f(c)|\leq \epsilon \text{ whenever } |x-c|<\delta.$$

First force $|f(x)|\leq \frac{\epsilon}{2}$, then force $|f(x)-f(c)|\leq \frac{\epsilon}{2}$, that gives that $f(c)<\epsilon$. You'll have to be a bit more rigorous with the deltas, let me know if I should elaborate.

rikhavshah
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    Sooo... for example, $f:x\mapsto1/x$ would not be continuous on $(0,\infty)$? – Did Aug 30 '16 at 05:50
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    (I did not downvote) For me, it is hard to prove rigorously if a "sketch of proof" is flawed. However, my gut tells me that the problem is way harder than some gymnastics with definitions (namely, the book hints towards Baire's cathegory theorem). For instance, you said: "first force $\lvert f(x)\rvert<\frac\varepsilon2$". Not clear how you can (in a way that does not depend on some $N$). –  Aug 30 '16 at 05:51