Proof that Clarke generalized directional derivative is upper continuous

Question

Let $X$ be a reflexive Banach space, $z, x, v \in X$. If $f: X \times X \rightarrow \mathbb{R}$ is continuous regarding its first argument and locally Lipschitz regarding its second argument. $\{z_i\}, \{x_i\}$ and $\{v_i\}$ are arbitrary sequences converging to $z, x$ and $v$, respectively.

I wonder if generalized directional derivative $f^{0}(z, x; v)$ (in Clarke sense) satisfies the following inequality: $$ f^{0}(z, x ; v) \geqslant \limsup _{i \rightarrow \infty} f^{0}\left(z_{i}, x_{i} ; v_{i}\right) $$

And I also want to know how to prove it if the inequality holds.

Thanks for your help in advance.

Answer 1

$\newcommand\R{\mathbb R}$The answer is no.

Indeed, apparently, in your post the Clarke derivative is meant with respect to the second argument of $f$: $$f^0(z,x;v):=\limsup_{\substack{y\to x\\ t\downarrow0}} \frac{f(z,y+tv)-f(z,y)}t.$$

Let now $X:=\R$ and $$f(z,x):=xe^{-x/|z|}\,1(z\ne0,x>0)$$ for real $z,x$ -- that is, $f(z,x)=xe^{-x/|z|}$ if $z\ne0$ and $x>0$, and $f(z,x)=0$ otherwise. Then $f(z,x)$ is continuous in $z$ and $1$-Lipschitz in $x$.

However, $f^0(0,0;1)=1(z\ne0)$, so that $f^0(z,x;v)$ is not upper-semicontinuous at $(0,0,1)$.