Is $\limsup f(x) \le \liminf g(x)$ if $f(x) \le g(x)$ for all $x$

Question

I was wondering if the following is true. $\limsup_{x \to \infty} f(x) \le \liminf_{x \to \infty} g(x)$ if $f(x) \le g(x)$ for all $x$.

One more constrain we can impose is that $g(x)$ and $f(x)$ are monotone and non-negative.

Answer 1

The obvious counter-example to the original Question is to take $f(x)=g(x)$, so the latter inequality is always satisfied. Now take $f(x)$ to exhibit some oscillation, e.g. $f(x) = \cos x$, whence:

$$ \lim \sup_{x\to \infty} f(x) = 1 \gt -1 = \lim \inf_{x\to \infty} f(x) $$

If you impose the additional constraint that these are monontone and non-negative functions, then $\lim \sup_{x\to \infty} f(x) = \lim_{x\to \infty} f(x)$ and $\lim \inf_{x\to \infty} g(x) = \lim_{x\to \infty} g(x)$. Then $f(x) \le g(x)$ does imply that (on the extended real line) $\lim \sup_{x\to \infty} f(x) \le \lim \inf_{x\to \infty} g(x)$ because we are simply talking about their limits.

Answer 2

Define $f(n)=1$ if $n\in \mathbb{Z}$, otherwise $0$. Let $g=f$; then $\lim\sup f(x) = 1$ and $\lim\inf g(x) = 0$

If $f\leq g $ is monotone, then $\lim\inf f(x)=\lim\sup f(x) \leq \lim\inf g(x)=\lim\sup g(x)$