I am trying to prove that if $f : \mathbb{R}^n \mapsto \mathbb{R_+}$ is a strictly convex function then $g(x)=\sqrt{f(x)}$ will not be necessarily convex.
I have started with using the definition that $f(x)$ is strictly convex, so it must hold that
$f(\theta x+ (1-\theta)y) < \theta f(x) + (1-\theta)f(y) \qquad$ for all $x,y \in \mathbb{R^n}$ and $\theta \in (0,1)$
Then since $f : \mathbb{R}^n \mapsto \mathbb{R_+}$, I proceed with
$\sqrt{f(\theta x+ (1-\theta)y)} < \sqrt{\theta f(x) + (1-\theta)f(y)} \qquad$
$\sqrt{f(\theta x+ (1-\theta)y)} < \sqrt{\theta f(x)} + \sqrt{(1-\theta)f(y)} \qquad$
$\ \ \ \ {g(\theta x+ (1-\theta)y)} < \sqrt{\theta} g(x) + \sqrt{(1-\theta)}g(y) \qquad$
But this does not prove anything..
I have figured out that what happens is that g(x) can be concave, but I cannot figure out how to reverse the inequality to show that..
Any hints?
Thanks in advance!
Best Answer
You cannot really expect to write a "proof" that something fails sometimes.
In this case, since $x^r$ is strictly convex when $r>1$ and not convex when $r<1$, you can take $g(x)=|x|^{3/2}$, and then $\sqrt{g(x)}=|x|^{3/4}$ is not convex.