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 $\lim_{x\rightarrow+\infty}f(x)=+\infty$?

2$\begingroup$ Hm, I find the question quite reasonable and not obvious. Usually we do not close questions of such level. $\endgroup$– Fedor PetrovCommented Dec 27, 2016 at 16:25

2$\begingroup$ @叶胥达 the tag "functionalanalysis" is not very appropriate for this question; "realanalysis" would fit better $\endgroup$– Pietro MajerCommented Dec 27, 2016 at 19:24

1$\begingroup$ and the tag 'counterexamples' is a posteriori misleading $\endgroup$– Fedor PetrovCommented Dec 28, 2016 at 6:50

$\begingroup$ Sorry about that...I am new here. $\endgroup$– Xuda YeCommented Dec 28, 2016 at 13:32
2 Answers
Yes, this is true and well known. One of the references I know is the problem book of B. Makarov, M. Goluzina, A. Lodkin and A. Podkorytov (Selected problems in real analysis, Translations of Mathematical Monographs 107, AMS 1992), problem II.1.25 (it is slightly different, but essentially the same).
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.

$\begingroup$ If we remove the continuity hypothesis, what's a counterexample? $\endgroup$ Commented Dec 28, 2016 at 13:50

1$\begingroup$ 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. $\endgroup$ 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 nonempty open interval $(a,b)\subset\mathbb{R}_+$ and any $n\in\mathbb{N}$, the set $\cup_{k> n}(ka,kb)$, that contains a rightunbounded 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.

$\begingroup$ 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 $\delta>0$. $\endgroup$ Commented Dec 27, 2016 at 17:06