Return to Revisions

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.