How do I prove that $T(\overline\Omega)\subset\overline\Omega$, where $$T:\overline\Omega\rightarrow X$$ is compact (X Banach, $\Omega\subset X$ convex and bounded) and $$T(\partial\Omega)\subset\Omega$$
Is this even true for any convex subset of X?
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Let $T:\mathbb{R}\to \mathbb{R}$, with $Tx=2x$ and $\Omega = (-1,1)$, then $T(\overline{\Omega})=[-2,2]\not \subset [-1,1]$? – Lorenzo Q Feb 24 '18 at 19:29
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1Your example doens't satisfy the second condition of T :($T(\partial(-1,1))\subset(-1,1)$). – Danut Preda Feb 24 '18 at 19:38
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I'm sorry, I thought that was another claim (yeah, doesn't make much sense). – Lorenzo Q Feb 24 '18 at 19:39
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First $\overline{\Omega}$ is a convex set, and you can show pretty easily that $T\big( \overline{\Omega} \big)$ is convex. Furthermore $\overline{\Omega}$ is bounded and therefore by compactness of $T$ we know therefore that $T \big( \overline{\Omega} \big)$ is compact (can be seen by sequential compactness). By the Krein-Milman theorem we know that:
$ \overline{conv} \Big( Ext \big( T\big( \overline{\Omega} \big) \big) \Big)= T\big( \overline{\Omega} \big) $
Then it just remains to show that $Ext \big( T\big( \overline{\Omega} \big) \big)\subseteq \overline{\Omega}$. But since for a given set $A$, we know that $Ext(A)\cap int(A)=\emptyset$, and $\partial A = \overline{A}\setminus int(A)$ we know then that $Ext(A)\subseteq \partial A$. Applied to the problem hand it specifically gives us that:
$ Ext\Big( T\big( \overline{\Omega} \big) \Big) \subseteq T\big( \partial \Omega \big) $
And finally by the assumption $T\Big( \partial \Omega \Big) \subseteq \Omega$, and therefore $Ext \big( T\big( \overline{\Omega} \big) \big)\subseteq \Omega$ as needed.
Keen-ameteur
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1$Ext(T(\overline\Omega))\subset T(\partial\Omega)?$ Isn't it $Ext(T(\overline\Omega))\subset \partial T(\Omega)?$ – Danut Preda Feb 24 '18 at 21:15
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I also want to say that $Ext\big( T( \overline{\Omega}) \big)= T \big( Ext(\overline{\Omega} ) \big)$, which shows the inclusion I wrote. – Keen-ameteur Feb 24 '18 at 21:54
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But $Ext(T(\overline{\Omega}))$ is not $T(Ext(\overline{\Omega}))$ in general. Take $X=\mathbb{R}^2$, $\overline{\Omega}=D^2$ the unit disc and $T(x,y)=(x/2,0)$. In this case, we have $Ext(T(\overline{\Omega}))={(-1/2,0),(1/2,0)}$ and $T(Ext(\overline{\Omega}))=[-1/2,1/2]\times{0}$. – BlueDiamond Feb 27 '18 at 17:04