Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?

Question

Is the following statement true?

Let, $f:\mathbb{R} \rightarrow \mathbb{R}$ a function that satisfies that, for evety $a \in \mathbb{R}$, the sequence $\{ f(\frac{a}{n}) \}$ converges to $0$. Then $f$, has a limit in $0$.

I think this is true, but not sure how to prove it.

Thank you!

By "$f$, has a limit in $0$" do you mean, $\lim\limits_{x\to 0}f(x) = 0$? If so, this should be stated. — Jonas Meyer, Nov 30 '13 at 09:01
Somewhat related: http://math.stackexchange.com/questions/63870/a-classic-problem-about-limit-of-continuous-function-at-infinity-and-its-connect — Jonas Meyer, Nov 30 '13 at 09:16
@JonasMeyer: Based on your commented for my deleted answer, I suppose that the condition "$\lim_{n \to \infty}f(a_{n}) = 0$ whenver $a_{n} \to 0$" is much stronger than just considering sequences of type $a_{n} = a/n$. I mean all sequences $a_{n}$ tending to zero encompass any possible way of a variable $x$ tending to zero, but just the family of sequences $a/n$ for all numbers $a$ does not represent all ways of a variable $x$ tending to zero — Paramanand Singh, Nov 30 '13 at 09:26

Ewan Delanoy · Answer 1 · 2013-11-30T09:06:25.710

8

Hint Consider the function $f$ defined by

$$ f(x)=\left\lbrace \begin{array}{ll} x\sqrt{p}, & \text{if } x \text{ is a rational multiple of } \frac{1}{\sqrt{p}} \text{ for some prime} p \\ 0, & \text{otherwise} \end{array}\right. $$

edited Nov 30 '13 at 09:06

answered Nov 30 '13 at 08:58

Ewan Delanoy

61,600

@JonasMeyer Corrected, thanks. – Ewan Delanoy Nov 30 '13 at 09:07
But, that there is no $a \in \mathbb{R}$ such that, $\frac{a}{x}=x\sqrt{p}$? – user83081 Nov 30 '13 at 09:27
1

@user83081 I don’t understand your question. Note that the definition says “O otherwise”. – Ewan Delanoy Nov 30 '13 at 09:29
Maby it sounds weared, but somehow I miss something. The counterexample has to show a function that has no limit when $x \rightarrow 0$, but still, for every $a$, it $f(\frac{a}{n})$, converges to $0$. now, if you can find an $a$, such that, $x\sqrt{p}=\frac{a}{n}$, then where is the counter example? – user83081 Nov 30 '13 at 09:34
2

@user83081: In order to show that $f(a/n)$ converges to $0$ for each $a$, you must start with an arbitrary $a$, then consider the sequence $f(a), f(a/2), f(a/3),\ldots$. Ewan has given an example where you can give explicit reason why this sequence converges to $0$. You can break up your argument into $2$ cases: either $a$ is a rational multiple of $\dfrac{1}{\sqrt p}$ for some prime $p$, or it is not. The "$x$" appearing in the definition of the function is just for purposes of defining the function. Your question does not really make sense. – Jonas Meyer Nov 30 '13 at 09:38
Oh, sorry. I see now, during thinking of the problem I replaced the original condition of the question with, ${ f(\frac{a}{x}) }$.. Thanks!!! – user83081 Nov 30 '13 at 10:02

score 4 · Answer 2 · answered Nov 30 '13 at 09:08

4

Suppose that $f\left(\dfrac{1}{e^n}\right)=1$ for all $n\in \mathbb N$, and $f(x)=0$ if $x$ does not have the form $\dfrac{1}{e^n}$.

answered Nov 30 '13 at 09:08

Jonas Meyer

53,602

Can't we find an $a \in \mathbb{R}$ such that $\frac{a}{n}=\frac{1}{e^n}$? – user83081 Nov 30 '13 at 09:24
@user83081: I wonder if you might be overloading $n$ there. Taken literally, of course that would be the case when $a=n/e^n$. But then $f(a/k)=0$ when $k>n$, so there is no problem. More to the point, for every $a$, there is at most one $k$ such that there exists $n$ such that $a/n=1/e^k$, and therefore this $n$ is also unique, and thus $f(a/m)=0$ when $m>n$. This uses the fact that no power of $e$ is an integer. – Jonas Meyer Nov 30 '13 at 09:28
agian, sorry, I think I miss somethong here... The counterexample has to show a function that has no limit when $x \rightarrow 0$, but still, for every $a$, it $f(\frac{a}{n})$, converges to $0$. now, if you can find an $a$, such that, $\frac{1}{e^n}=\frac{a}{n}$, then where is the counter example? or, maby it can not hold for every n? But then, don't I have to prove it somehow? – user83081 Nov 30 '13 at 09:37
1

@user83081: (See also my comment on Ewan's answer.) To show that a given function $f$ satisfies the hypothesis means showing that if you start with arbitrary $a\in\mathbb R$, the sequence $f(a), f(a/2),f(a/3),\ldots$ converges to $0$. Pick any $a$ you want to, and it may or may not be the case that there exists some $n$ and some $k$ such that $1/e^k = a/n$ (again, there is no reason to assume $k=n$). Let your argument depend on whether or not such $n$ and $k$ exist. If they do, show that it happens only once, so all further terms in the sequence vanish. – Jonas Meyer Nov 30 '13 at 09:42
(continued) @user83081: And if they don't, then all the terms are $0$. – Jonas Meyer Nov 30 '13 at 09:43
Yes get it now, sorry, I missplaced the condition of the original statement with something else. Thank you!! – user83081 Nov 30 '13 at 10:03

score 1 · Answer 3 · answered Nov 30 '13 at 16:31

For any non-zero real number $a$, define $[a]$ to be the set of all numbers $a\frac{n}{m}$, where $n$ and $m$ are non-zero integers. Either $[a]=[b]$ or $[a]\cap[b]=\emptyset$; indeed, if $[a]$ and $[b]$ share a common element, then $a\frac{m}{n}=b\frac{p}{q}$ for non-zero integers $m,n,p,q$, which gives $a=b\frac{pn}{qm}$ and $b=a\frac{mq}{np}$.

Choose any irrational number $a_{1}\in(\frac{1}{2},1]$. Then choose an irrational number $a_{2}\in (\frac{1}{3},\frac{1}{2}]$ which is not in $[a_{1}]$. Next choose an irrational number $a_{3} \in (\frac{1}{4},\frac{1}{3}]$ which is not in $[a_{1}]\cup[a_{2}]$. Continue by induction to find a sequence of rational numbers $\{ a_{n}\}_{n=1}^{\infty}$ such that $a_{n} \in (\frac{1}{n+1},\frac{1}{n}]$ and so that none of the sets $[a_{n}]$ share a common element. Define a function $f$ so that $f(a_{n})=1$ and $f(x)=0$ otherwise. Then $\lim_{n}f(a_{n})\ne 0$, even though $\lim_{n}a_{n}=0$. So $\lim_{x\rightarrow 0}f(x)\ne 0$. If $b$ is an irrational number, then $\{ b,\frac{b}{2},\frac{b}{3},\ldots\}$ contains at most one $a_{n}$ and, hence, $\lim_{n\rightarrow\infty}f(\frac{b}{n})=0$. If $b$ is rational then $f(b)=0$ and, hence, $\lim_{n\rightarrow\infty}f(\frac{b}{n})=0$ because $b/n$ is also rational.

Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?

3 Answers3