# Continuous functions and infinity

Suppose $f(x)$ is continuous on $\mathbb{R}$, for $\forall \delta>0, \forall x\in\mathbb{R}, \lim_{n\rightarrow\infty}f(x+n\delta)=+\infty$. Is it correct that $\lim_{x\rightarrow+\infty}f(x)=+\infty$?

• Hm, I find the question quite reasonable and not obvious. Usually we do not close questions of such level. Commented Dec 27, 2016 at 16:25
• @叶胥达 the tag "functional-analysis" is not very appropriate for this question; "real-analysis" would fit better Commented Dec 27, 2016 at 19:24
• and the tag 'counterexamples' is a posteriori misleading Commented Dec 28, 2016 at 6:50
• Sorry about that...I am new here. Commented Dec 28, 2016 at 13:32

An idea is that if, on the contrary, there are intervals $\Delta_k=[x_k,y_k]$ onto which $f$ is bounded, $x_k\to \infty$, then we may construct a nested family of closed intervals $T_i$, $T_1\supset T_2\supset T_3\dots$, and positive integers $n_1<n_2<\dots$, such that each interval of the form $n_iT_i$ is contained in some $\Delta_k$. Then for a common point $t=\cap T_i$ it appears that $f(nt)$ does not tend to infinity.
• take rationally independent numbers $x_1<x_2<\dots$ tending to infinity, define $f(x_i)=0$, $f(t)=t$ for $t\notin \{x_i\}$. Each arithmetic progression contains at most two $x$'s, so the limits along it equals infinity. Commented Dec 28, 2016 at 17:52
Yes, in fact, $$\inf_{\delta>0}\ \liminf_{n\to\infty}f(n\delta) =\liminf_{x\to+\infty}f(x).$$ Assuming w.l.o.g. $\liminf_{x\to+\infty}f(x)<\alpha<+\infty$, the open set $A=\{f<\alpha\}$ is unbounded. Therefore, for any non-empty open interval $(a,b)\subset\mathbb{R}_+$ and any $n\in\mathbb{N}$, the set $\cup_{k> n}(ka,kb)$, that contains a right-unbounded interval, meets $A$. Equivalently, for any $n\in\mathbb{N}$, the open set $B_n:=\cup_{k> n}{1\over k}A$ meets $(a,b)$, so that $B_n$ is dense in $\mathbb{R}_+$. By the Baire category theorem $\cap_{n\ge0}B_n$ is not empty, actually dense, meaning that there exist $\delta>0$ such that $n\delta\in A$ for infinitely many $n$, an this means $\liminf_{n\to\infty}f(n\delta)\le\alpha.$ Being $\alpha$ arbitrary, the claim follows.
• With a sligthly smaller, open unbounded set $A$ one shows that actually $\liminf_{n\to\infty}f(n\delta)=\liminf_{x\to\infty}f(x)$ holds for a dense set of $\delta>0$. Commented Dec 27, 2016 at 17:06