Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an increasing function, i.e., $x<y \Rightarrow f(x) < f(y)$ for all $x,y$.
Assume that $\lim_{|x| \rightarrow \infty} |f(x)| = \infty$
Define $F(x) = \int_{0}^{x} f(y) dy$.
Prove or disprove that $F$ has compact sublevel sets.
Trial: The statement is false if the assumption $\lim_{|x| \rightarrow \infty} f(x) = \infty$ does not hold. In fact, $F'=f$ can fail to be strongly convex. Instead, I am wondering if the integral function $F$ is such that $\lim_{|x| \rightarrow \infty} F(x) = \infty$.
There exists $b$ such that $f(x)\ge 1$ for $x\ge b.$ So for $x\ge b,$ $F(x)\ge\int_0^b f + (x-b),$ which tends to $\infty$ as $x\to \infty.$
Similarly, there exists $a$ such that $f(x)\le -1$ for $x\le a.$ So for $x\le a,$
$$F(x) = \int_0^a f + \int_a^x f = \int_0^a f + \int_x^a (-f) \ge \int_0^a f + (a-x),$$
which tends to $\infty$ as $x\to -\infty.$ This answers your "I am wondering …" question and should lead to a solution.