Is the q<1 power of a p-norm concave?

Question

I came across the following function:

$f(\mathbf{x})=\|\mathbf{x}\|_p^q$

Here, $p>1$ and $0<q<1$. Naturally, each entry of $\mathbf{x}$ is greater than $0$.

I guess $f$ is concave, but I'm uncertain about how to prove it. In my previous experience, I could demonstrate convexity (or concavity) using rules of convexity, such as the convexity of the sum of two functions or certain composition rules. Or, I could prove the positive semi-definiteness of the Hessian. However, I haven't been able to apply any of these techniques to analyze $f$.

Answer 1

Not true. Counter-example: $p=2,q=\frac 1 2, \|x\|_p=1, y=2x$. In this case the inequality goes the wrong way: $\|tx+(1-t)(2x)\|_p^{q}< t\|x\|_p^{q}+(1-t)\|(2x)\|_p^{q}$.