Is $\sqrt{f(x)}$ convex if $f(x)$ is convex?

Question

Assume $f(x)\geq 0$. Is $\sqrt{f(x)}$ convex if $f$ is convex?? I was guessing it to be indeed.

Answer 1

Let me assume that $f$ is defined on $\mathbb{R}$ and twice differentiable. Note that

$$\begin{align} \frac{d}{dx} \sqrt{f(x)} &= \frac{f'(x)}{2\sqrt{f(x)}} \\ \frac{d^2}{dx^2} \sqrt{f(x)} &= \frac{2\sqrt{f(x)}f''(x)-(f'(x))^2/\sqrt{f(x)}}{4f(x)} \end{align}$$ For the second derivative to be nonnegative, we need the numerator to be nonnegative in addition to $f(x)>0$: $$\begin{align} 2\sqrt{f(x)}f''(x)&\geq(f'(x))^2/\sqrt{f(x)} \\ f''(x)&\geq\frac{1}{2}\frac{(f'(x))^2}{f(x)} \end{align}$$ This is obviously a stronger condition than $f''(x) \geq 0$. It requires the function to be curved stronger than what convexity requires. Functions like $x^2$ and $\exp(x)$ satisfy this criterion, while $|x|$ does not.