uniformly bounded on compact set

350 Views Asked by At

Let $F:\Omega\to 2^X\setminus\emptyset$, $\Omega$ is compact. If $F$ has bounded values then they are uniformly bounded. Please help me prove this! Thanks!

1

There are 1 best solutions below

0
On

A crucial assumption is missing from the question, as martini said. Quoting from the book:

... we have an $(\epsilon-\delta)$ usc $F:\Omega\to 2^X\setminus \varnothing$ with closed convex values.

Definition 1.2 on the same page explains the meaning of "$(\epsilon-\delta)$ usc": to every $\omega_0\in\Omega $ and $\epsilon>0$ there exists $\delta=\delta(\omega_0,\epsilon)>0$ such that $F(\omega)\subset F(\omega_0)+B_\epsilon(0)$ on $B_\delta(\omega_0)\cap \Omega$.

Now that the right assumptions are in place, the proof is easy. Let $\epsilon=1$. The sets $\{B_\delta(\omega_0): \delta=\delta(\omega_0,\epsilon), \omega_0\in\Omega\}$ form an open cover of $\Omega$. Pick a finite subcover, indexed by $\omega_1,\dots,\omega_N$. Let $A=\bigcup_{i=1}^N F(\omega_i)+B_1(0)$; this is a bounded set, and it contains $F(\omega)$ for every $\omega\in\Omega$.

(The part about having closed convex values does not come into play here.)