Is the "$p$-norm" with $0<p<1$ a concave function We know that for $p&gt;1$ by the triangle inequality 

$$\|tx + (1-t)y\|_p \leq t\|x\|_p + (1-t)\|y\|_p,$$

when $t\in[0,1]$. Now the "$p$-norm" for $0&lt;p&lt;1$ is not actually a norm but we can still define it to be

$$\|x\|_p=\left(\sum_{i=1}^n |x_i|^p\right)^{1/p},$$

we do not have the triangle inequality but can we at least say that this function is concave?
It is not concave, as
$$
\frac12\|x\|_p + \frac12 \|-x\|_p \ge \left\| \frac{x-x}2\right\|_p =\|0\|_p=0.
$$
In fact, every norm is a convex function. More generally, a positively homogeneous function $f$ (all norms are positively homogeneous) is convex if and only if $f(x+y)\leq f(x)+f(y)$, which is exactly the triangle inequality.