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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!

nntienvn
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    You have some continuity for $F$, I suppose. And bounded? What is $X$? – martini May 30 '13 at 08:10
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    Welcome. If you've mentioned the details of your attempt, it'd be helpful for both you and the people who are gonna help you. – f.nasim May 30 '13 at 08:30
  • $X$ is a Banach space. $\Omega$ is a subset of a Banach space $Y$, $\Omega\neq\emptyset$. – nntienvn May 30 '13 at 12:54
  • I read Multivalued Diffential Equations (Klaus Deimlin), in p.6. On the 6th line from the bottom, the write "If $F$ has bounded values then they are uniformly bounded, since $\Omega$ is compact." – nntienvn May 31 '13 at 12:28

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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.)

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