A function $h$ is said to be Clarke regular at $\bar{x} \in \mathbb{R}^m$ provided that if for all direction $y$, the directional derivative $h^{\prime}(\bar{x} ; y)=\lim _{t \downarrow 0} \frac{h(\bar{x}+t y)-h(\bar{x})}{t}$ exists, and $$ h^{\prime}(\bar{x} ; y)=h^{\circ}(\bar{x} ; y):=\limsup _{{x \rightarrow \bar{x}}, {t \downarrow 0}} \frac{h(x+t y)-h(x)}{t}. $$ It is well known that any convex or smooth function is Clarke regular.
It is also well known that if $h_i$ is Clarke regular, then $g(x)=\max_{i=1,2,\ldots,n}\{h_i(x)\}$ is Clarke regular.
Let $\mathbb{Z}$ be a compact set, and h(x,z) is Clarke regular over $\mathbb{Z}$. Then, I want to pose my question:
Under which conditions $g(x)=\max_{z\in\mathbb{Z}}\{h(x,z)\}$ is Clarke regular?
I have checked the case when $h$ is concave with respect to $z$. However, this condition is not enough.
Besides, I know that if $h$ is strongly concave with respect to $z$ for all $x$, then $g$ is smooth. In this case $g$ is of course Clarke regular. I want to check if there is any weaker condition than the strongly concavity.
Reference:
title={Optimization and nonsmooth analysis}, author={Clarke, Frank H}, year={1990}, publisher={SIAM}, Section 2