Lipschitz Continuity of Optimal Value Let $m$ be a probability measure on $W \subseteq \mathbb{R}^p$, so that $m(W) = 1$.
Consider a function $f : X \times Y \times W \rightarrow [0,1]$, where $X \subset \mathbb{R}^n$, and $Y \subset \mathbb{R}^m$ are compact sets.
Assume that for all $w \in W$ the mapping $(x,y) \mapsto f(x,y,w)$ is continuous, and for all $(x,y) \in X \times Y$ the mapping $v \mapsto f(x,y,w)$ is measurable.
It can be shown - mainly invoking the Dominated Convergence Theorem - that the mapping $F: X \rightarrow [0,1]$, defined as
$$ F(x) := \sup_{ y \in Y } \int_W f(x,y,w) m(dw), $$
is continuous.
Question.
If for each $w \in W$, the map $(x,y) \mapsto f(x,y,w)$ is Lipschitz continuous (on $X \times Y$), is $F$ also Lipschitz continuous (on $X$)? If not, please provide a counterexample.
Comment.
I think it is interesting to show more regularity of $F$, given more regularity of $f$. Seems it may be true, but I am not able to prove it.

For any $y\in Y$ you have
$$
\left|\int_W f(x',y,w)m(\mathrm dw) - \int_W f(x'',y,w)m(\mathrm dw)\right| = \left|\int_W f(x',y,w) - f(x'',y,w) m(\mathrm dw)\right|
$$
$$
  \leq \int_W|f(x',y,w) - f(x'',y,w)|m(\mathrm dw)\leq\int_W\alpha\|x'-x''\|m(\mathrm dw)=\alpha\|x'-x''\|.
$$
On the other hand, as it follows from this trick, it holds that $|F(x') - F(x'')|\leq \alpha\|x' - x''\|$.
