Multivalued function: lower semicontinuity equivalent condition.

Question

I have to prove that: $$ F\text{ - l.s.c. in } x_0\Rightarrow\forall (x_n)_n\subseteq X\text{ s.t. }x_n\rightarrow x_0,\forall y_0\in F(x_0),\exists(y_n)_n\subseteq Y \text{ s.t. } y_n\rightarrow y_0 \text{ with } y_n\in F((x_n)_n) $$

where: $(X,d_X),\ (Y,d_Y)\text{ - metric spaces},\ x_0\in X,\ F\colon X\multimap Y\text{ - multivalued function.}$

Can you give me a hint? How to define the sequence $(y_n)_{n\in\mathbb N}$?

Basics: Multivalued function is relation $F\subseteq X\times Y$ s.t.: $$\forall\ x\in X,\exists\ y\in Y\text{ with } (x,y)\in F.$$

Multivalued function is lower semicontinuous (l.s.c) in $x_0$ iff: $$\forall\ V\subseteq Y\text{ s.t. } V\text{ - open }, F(x_0)\cap V\neq\emptyset,\exists\ U_{x_0}\text{ - neighbourhood of } x_0\text{ with } U_{x_0}\subseteq F^-(V)$$ where $$F^-(V)=\{x\in X\colon\ F(x)\cap V\neq\emptyset\}$$.

Answer 1

I though that after such a long time it would be nice to answer this question. Here is the solution:

Take any $\{x_n\} \to x_0,$ and $y_0 \in F(x_0).$ For $m \in \Bbb N,$ consider $$V_m= B\left(y_0,\frac{1}{m}\right)=\{y\in Y : d_Y(y,y_0)\leq \frac{1}{m}\}.$$ Since $y_0 \in V_m,$ we have $F(x_0)\cap V_m \neq \emptyset.$ Since $F$ is l.s.c at $x_0,$ there exists $\epsilon(m)>0$ such that

$$d_X(x,x_0)<\epsilon(m) \Longrightarrow F(x) \cap V_m \neq \emptyset. $$ Since $x_n \to x_0,$ there exists $N(m)$ such that $n\geq N(m)$ implies $d_X(x_n,x_0)<\epsilon(m).$

Without lost of generality , we may assume that $\{N(m)\}$ is strictly increasing, that is, $N(m)< N(m+1)$ for each $m\in \Bbb N,$ and that $$\lim_{m\to \infty}N(m)=+\infty.$$ This means that, for any $n$ large enough, (in fact, for $n\geq N(1)$) there exists a unique $m \in \Bbb N$ such that

$$N(m)\leq n< N(m+1).$$ Define the function $\gamma:\Bbb N\to \Bbb N$ as $\gamma(n)=m,$ where $m$ is the unique natural number that satisfies

$$N(m)\leq n< N(m+1).$$ Note that $\gamma$ is well defined. Finally, for each $n\in \Bbb N, n\geq N(1),$ choose

$$y_n \in F(x_n)\cap V_{\gamma(n)}.$$ Note that such $y_n$ always exists because $n\geq N(\gamma(n))$ implies $d_X(x_n,x_0)<\epsilon(\gamma(n)).$

By construction, we have $$d_Y(y_n,y_0)\leq \frac{1}{\gamma(n)}.$$ Since $$\lim_{n\to \infty} \gamma(n)=+\infty,$$ we will have

$$y_n\to y_0,$$ as desired.

Note: For $n

The converse is also true:

If $F$ is not l.s.c at $x_0,$ then, there must be an open set $V$ and a sequence $x_n \to x_0,$ such that $F(x_n)\cap V =\emptyset, \; F(x_0)\cap V\neq \emptyset.$ Now take $y_0 \in F(x_0)\cap V$ arbitrarily.

Then this is a contradiction with the existence of a sequence $\{y_n\} $ such that $y_n\in F(x_n)$ and $y_n \to y_0$ because eventually $y_n\in V$ and hence $$y_n \in F(x_n)\cap V.$$