Why is this composition of concave and convex functions concave?

Solution 1:

The convex function $j$ of a concave function $i$ is not necessarily concave. For example, if $j$ is strictly convex and $i$ is a constant function, then $j\circ i$ is strictly convex.

In your case, the $p$-"norm" is concave when $p<1$ because the Hessian matrix is negative semidefinite. More specifically, let $S=\sum z_i^p$. Then $$ \frac{\partial^2 S^{1/p}}{\partial z_i \partial z_j} =(1-p)S^{1/p-2} \left( z_i^{p-1}z_j^{p-1} - S z_i^{p-2} \delta_{ij} \right). $$ So the Hessian matrix is given by $H=(1-p)S^{1/p-2} D(uu^T-SI)D$, where $u=(z_1^{p/2},\ldots,z_n^{p/2})^T$ and $D=\operatorname{diag}(z_1^{p/2-1},\ldots,z_n^{p/2-1})$. As the eigenvalues of the matrix $uu^T-SI$ are $0$ (simple eigenvalue) and $-S$ (with multiplicity $n-1$), $H$ is negative semidefinite.