Set-valued maps: locally Lipschitz implies upper semicontinuity

Question

With $X$ and $Y$ metric spaces, let $F:X\rightrightarrows Y$ be a set-valued map satisfying the following

(P) For each compact set $K\subset \operatorname{dom} F$ there exists $L>0$ such that, for all $x_1,x_2\in K$, $\|F_1-F_2\|\le L\|x_1-x_2\|$ holds for all $F_1\in F(x_1)$ and $F_2\in F(x_2)$

Is it true that $F$ is upper semicontinuous?

My clumsy attempt: Pick $x\in X$ and let $V_\epsilon:= F(x)+ B_\epsilon$ (with $B_\epsilon$ the open ball on $Y$). With $\nu>0$, pick any $x_1,x_2\in x+B_\nu$ and let $K\subset X$ be any compact subset containing $x+B_\nu$. Then from (P) we have $\sup\{ \|f_1- f_2\|\,:\, f_1\in F(x_1),\, f_2 \in F(x_2)\} \le L\|x_1-x_2\|\le L\nu$ for some $L$ depending only on $K$. Thus, if $\nu\le \epsilon/(2L)$, then $F(x+B_\nu)\subset V_\epsilon$, and the result should follow.

I am on the right way? If that's correct I guess that this can be found in many books. However I could not find it. Do you have any reference?

-------------------------- edit

As @Michelle pointed out, (P) implies that $F$ is single valued, so the guess is true. Therefore I want to modify the question with the following assumption holding anstead of (P):

(P2) For any $x\in X$ and for any compact subset $K\subset X$ containing $x$, there exists $L>0$ such that $\forall x_1,x_2\in K$, $F(x_1)\subset F(x_2) + B_{L\|x_1-x_2\|}$

Thanks.

Answer 1

Version 1: incorrect form of Lipschitz continuity

for all $x_1,x_2\in K$, $\|F_1-F_2\|\le L\|x_1-x_2\|$ holds for all $F_1\in F(x_1)$ and $F_2\in F(x_2)$

This implies $\|F_1-F_2\| = 0$ for all $F_1,F_2\in F(x_1)$. So the map $F$ is actually single-valued. The set-valued upper semicontinuity then becomes regular continuity of $F$.

The latter still takes a bit of effort to prove since we only have the Lipschitz property on compact subsets. It goes like this: pick any convergent sequence $x_n\to x$. The set $K=\{x_n\}\cup \{x\}$ is compact, so $f$ is Lipschitz continuous on it. Hence $f(x_n)\to f(x)$. Since we are in a metric space, sequential continuity is equivalent to continuity.

Version 2: Lipschitz continuity with respect to Hausdorff distance

For any $x\in X$ and for any compact subset $K\subset X$ containing $x$, there exists $L>0$ such that $\forall x_1,x_2\in K$, $F(x_1)\subset F(x_2) + B_{L\|x_1-x_2\|}$

This does not imply upper semicontinuity. For example, let $F(x) = (x,x+1)\subset \mathbb R$ for $x\in \mathbb{R}$. The definition of upper semicontinuity says that for every open set $W\subset Y$, the set $\{x\in X: F(x)\subset W\}$ must be open in $X$. But this fails with $W=(0,1)$ for which the latter set is $\{0\}$.

Aside: The definition of upper semicontinuity makes more sense for maps such that $F(x)$ is compact for every $x$, or at least closed.

Version 3: Lipschitz continuity with respect to Hausdorff distance, compact-valued map

Suppose $F(x)$ is compact for every $x$, and is globally Lipschitz in the Hausdorff metric. (Not just Lipschitz on each compact set). Then upper semicontinuity holds. Indeed, take an open set $W\subset Y$ and a point $x_0$ such that $F(x_0)\subset W$. Let $\rho = \operatorname{dist}(F(x_0), Y\setminus W) $, which is a positive number. If $d(x,x_0)<\rho/L$, then the Lipschitz condition on $F$ implies $F(x)\subset W$, as required for upper semicontinuity.