I am trying to prove that for a continuous function $f: (0, \infty) \to \mathbb{R}$ if $g(x) = f(x+1) - f(x)$ and $\limsup_{x\to\infty}{g(x)} > 0$ and $\liminf_{x\to\infty}{g(x)} > 0$ then $f(x) \to \infty$.
By now, I am stuck trying to prove the opposite is impossible: consider a sequence of closed intervals [m, m+1] as $m \to \infty$. If $f(x)$ is bounded, then there should be $l$ so that $\inf_{[l, l+1]}{f(x)} = f(l+1)$ $\Longrightarrow$ $f(l+1) - f(l) \leq 0$.
And here I am stuck, I do not know how correctly prove that there should be such a closed interval with $\inf_{[l, l+1]}{f(x)} = f(l+1)$.
Could you please help by giving me an advise or hint so that I can go further? Thank you all!
Direct proof is possible. Having $\lim\inf_{x\to\infty}g(x)>0$ does not only imply $g(x)>0$ for $x>A$, but also $g(x)>m$ for every $m<\lim\inf_{x\to\infty}g(x)$ and $x>A_m$.
Pick $m$ strictly positive, then reasoning with induction yields $f(A_m+n)>f(A_m)+nm$ with $n$ positive integer. But $f(A_m)+nm\xrightarrow[n\to\infty]{}+\infty$ Therefore, by lower bound, $f(A_m+n)\xrightarrow[n\to\infty]{}+\infty$ and $f$ is unbounded. Since $f$ is continuous, you get $f(x)\xrightarrow[x\to\infty]{}+\infty$
Note that you don't need continuity to prove that $f$ is unbounded.