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Let be $f:(0,1)\rightarrow \mathbb{R}$. For all $x\in (0,1)$ I have that $\lim_n{f(\frac{x}{n})}=0$.

It's true that $lim_{x\to0}f(x)=0$?

And if $f$ is continuous?

For the second point I just use the definition of continuity and limits: Let be $x_0\in(0,1)$ and let be the succession $\{f(\frac{x_0}{n})\}$, and I know that for all $\varepsilon_1>0$ $\exists N>0$ such that if $n>N$ then $|f(\frac{x_0}{n})|<\varepsilon_1$.

$f$ is continuos in $\frac{x_0}{n}$, so for all $\varepsilon_2>0$ $\exists \delta_2>0$ such that if $|x-\frac{x_0}{n}|<\delta_2$ then $|f(x)-f(\frac{x_0}{n})|<\varepsilon_2$.

So I have that $|x|\le|x-\frac{x_0}{n}|+|\frac{x_0}{n}|<\delta_2+|\frac{x_0}{n}|=\delta^*$, because $x_0$ is fixed.

On the other hand $|f(x)|\le|f(x)-f(\frac{x_0}{n})|+|f(\frac{x_0}{n})|<\varepsilon_1+\varepsilon_2$.

But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.

(Sorry for my bad English, I'm Italian)

Simmetrico
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2 Answers2

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This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/\sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.

More generally, define $f(x)=1$ for $x\in A,$ and $f(x)=0$ for $x\notin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $x\in A$ and $n\in\mathbb{N}$ with $n>1$ implies $x/n \notin A.$

Reiner Martin
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I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $\epsilon >0$ then $(0,1)=\cup_n A_n$ where $A_k=\{x:|f(\frac x n )| \leq \epsilon \forall k \geq n\}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(\frac x n )| <\epsilon$ whenever $a<x<b$ and $ n \geq n_0$ $\cdots (1)$. Now the length of the interval $(\frac a x, \frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $\frac a x >n_0$ or $x <\frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <\epsilon$ whenever $x$ is sufficiently small.

Kavi Rama Murthy
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