Strictly increasing and strictly convex function that does not go to negative infinity

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be continuous, strictly increasing, and strictly convex for all $x$. What are necessary and sufficient conditions for $f$ to be bounded below? Strong convexity is sufficient, but seems too strong.

Answer 1

Strong convexity is indeed too strong: there are no functions that are both strictly increasing on $\mathbb R$ and strongly convex.

Also, conditions for the second derivative are not suitable for the task since, e.g., $e^x$ and $x+e^x$ have the same second derivative, but only one of them is bounded from below.

The boundedness of $f(x)$ as $x\to-\infty$ is equivalent to $\int_{-\infty}^0 f'(t)\,dt$ being convergent. All we know about the function $f'$ is that it's monotone and nonnegative. So, the problem reduces to studying convergence of such improper integrals. E.g., $f'(x)=O(|x|^{-1-\epsilon})$, $x\to-\infty$, suffices for $f$ to be bounded below.