$f:(0,\infty)\rightarrow (0,\infty)$ is continuous, $\lim_{n\rightarrow\infty}f(nx)=0$ then $\lim_{x\rightarrow\infty}f(x)=0$

Question

Could any one tell me how to solve this one?

$f:(0,\infty)\rightarrow (0,\infty)$ is continuos, $\lim_{n\rightarrow\infty}f(nx)=0$ then $\lim_{x\rightarrow\infty}f(x)=0$

I have no clue how to solve this one. Thank you.

Answer 1

Lemma: If $U\subset(0,\infty)$ is an unbounded open set, then there exists an $x>0$ such that the sequence $(nx)_{n\geqslant0}$ falls into $U$ infinitely many times.

Using that lemma, you should be able to prove the contrapositive of what you want: if $f(x)$ does not tend to $0$ when $x\to\infty$, then there exists an $x>0$ such that $f(nx)$ does not tend to $0$ when $n\to\infty$.

Now it remains to prove the lemma. Consider the set $$ X = \{x>0 : \{ nx \}_{n\geqslant0}\cap U~\text{is infinite} \}. $$ Assuming that $U$ is an unbounded open set, we want to prove that $X$ is not empty. Let's rewrite $X$ as follows: \begin{align} X &= \{x>0 : \forall N>0,~\exists n>N,~nx\in U \} \\ &= \bigcap_{N>0}\bigcup_{n>N}\frac 1n U, \end{align} where $\frac 1n U$ is the set $\{u/n:u\in U\}$.

Now for any $N>0$, the open set $U_N=\bigcup_{n>N}\frac1nU$ is dense. To prove that, take some $y\in(0,\infty)$ and an $\varepsilon>0$, and let's find a $u\in U$ and an $n>N$ such that $|u/n-y|<\varepsilon$, or equivalently $|u-ny|<n\varepsilon$. The idea is to look at the intervals $I_n=(ny-n\varepsilon,ny+n\varepsilon)$, and to show that at some point one of them has to meet $U$. That's the case because if $n$ is large enough, two consecutive intervals $I_n$ and $I_{n+1}$ will start to overlap, so that $\bigcup I_n$ will cover a whole interval $[a,\infty)$.

More precisely, choose some $n_0>N$ such that $n_0>2y/\varepsilon$. Then $V=\bigcup_{n>n_0}(ny-2y,ny+2y)$ contains the interval $[n_0y,\infty)$, and since $U$ is unbounded, there is a point $u\in U\cap[n_0y,\infty)$. Now $u$ is in $V$, so there exists an $n>n_0$ such that $|ny-u|<2y$, which implies $$ \left|\frac{u}{n}-y\right| < \frac{2y}{n} < \frac{2y}{n_0} < \varepsilon. $$

Finally, Baire's theorem allows us to conclude that $X$ is dense, so in particular it's not empty.

Answer 2

That's a very hard problem. I got the proof from somebody and now, am just recalling. I'll write up the most significant part. Note that the condition of $f(x)\in(0,+\infty)$ is redundant. $\newcommand\abs[1]{\left|#1\right|}$

It suffices to prove that, $\forall\epsilon>0,\exists b>a>0,\exists N,\forall x\in[a,b],\forall n\ge N: \abs{f(nx)}\le\epsilon$.

Suppose $\exists\epsilon_0>0,\forall b>a>0,\forall N,\exists x\in[a,b],\exists n\ge N: \abs{f(nx)}>\epsilon_0$. Fix $\epsilon_0>0$, and since $\lim_{n\to\infty}f(nx)=0$, we can suppose $g(x)=\max\left\{\;n\in\mathbb Z^+\;\big|\;\abs{f(nx)}>\epsilon_0\;\right\}$ (if there's no such $n$, $g(x)=0$, that's negligible). The condition is equivalent to: $g$ is unbounded on every non-empty interval.

Let $a_0=1$ and $b_0=2$, we have $g(x)\ge0$ on interval $[a_0,b_0]$. We'll construct intervals $[a_n,b_n]$ inductively (logically, the axiom of dependent choice is needed, I think, because we'll do infinite many choices), such that

• $[a_0,b_0]\supset[a_1,b_1]\supset\cdots$
• $g(x)\ge n$ on $[a_n,b_n]$, for $n=0,1,\cdots$

It's done for $n=0$. Suppose the intervals for $n=0,\ldots,m$ are constructed, let's construct one for $n=m+1$.

Since $g$ is unbounded on $(a_m,b_m)$, $\exists x_0\in(a_m,b_m): g(x_0)\ge m+1$, hence $\exists n\ge m+1: \abs{f(nx_0)}>\epsilon_0$. Since $f$ is continuous, $\exists\delta>0,\forall x\in[x_0-\delta,x_0+\delta]\subset[a_m,b_m]: \abs{f(nx)}>\epsilon_0$, therefore $g(x)\ge n\ge m+1$ on $[x_0-\delta,x_0+\delta]$. Let $a_{m+1}=x_0-\delta$ and $b_{m+1}=x_0+\delta$, constructed.

By Cantor's intersection theorem, $\bigcap_{n=0}^\infty [a_n,b_n]\neq\emptyset$, let $\xi\in\bigcap_{n=0}^\infty [a_n,b_n]$, we have $g(\xi)\ge n$ for all $n\ge0$, contradiction!