Let $X$ be a reflexive Banach space, $z, x, v \in X$. $\{z\_i\}, \{x\_i\}$ and $\{v\_i\}$ are arbitrary sequences converging to $z, x$ and $v$, respectively.


**I would like to know** under which **conditions of the function $f: X \times X \rightarrow \mathbb{R}$**, generalized directional derivative $f^{0}(z, x; v)$ (in [Clarke sense](https://encyclopediaofmath.org/wiki/Clarke_generalized_derivative)) satisfy the following inequality:
$$
f^{0}(z, x ; v) \geqslant \limsup \_{i \rightarrow \infty} f^{0}\left(z\_{i}, x\_{i} ; v\_{i}\right)
$$


**Conditions of the function $f$ I can accept:** For this $f$, I need it is locally Lipschitz regarding its second argument at least. Regarding its first argument, I want the function to have properties like Lipschitz continuous or something similar to continuous.


**And I also want to know** how to prove it if there are these conditions.


Thanks for your help in advance.



The conditions used to prove upper semicontinuous of generalized directional derivative (in Clarke sense)
$\newcommand\R{\mathbb R}$The conditions that you can accept are not enough for the conclusion that you desire.


Indeed, as in the [previous answer](https://mathoverflow.net/a/417683/36721), let $X:=\R$ 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 $C$-Lipschitz in $z$ (with $C:=(2/e)^2$) 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 $(0,0,1)$.




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If we modify the above example a bit, as follows:
$$f(z,x):=xe^{-|x|/|z|}\,1(z\ne0)$$
for real $z$ and $x$, then we will see that $f(z,x)$ is, not only $C$-Lipschitz in $z$ and $1$-Lipschitz in $x$, but also continuously differentiable in $z$ and in $x$, and yet still with $f^0(0,0;1)=1(z\ne0)$, so that $f^0(z,x;v)$ is not upper-semicontinuous at $(0,0,1)$.




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On the other hand, the conclusion you desire will obviously hold if $f(z,x)$ has a partial derivative in $x$ which is continuous in $(z,x)$.


