Let $T:X\rightarrow Y$ be a linear continuous map between Banach spaces $X$ and $Y$ and denote by $T':Y'\rightarrow X'$ the norm adjoint of $T$. Let $M\subseteq U'$ be a subset of the unit sphere $U'$ of $Y'$. Let $C\subseteq M$ be any weak*-closed convex subset. Is the image $T'(C)$ weak*-closed in $X'$? Or more generally, assuming that $C$ is convex, under which additional conditions on $C$ is the image $T'(C)$ weak*-closed in $X'$?
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$\begingroup$ I have forgotten to mention that we can also assume that $T'$ is injective on $M$! $\endgroup$– Andy TeichCommented Jun 13, 2012 at 8:56
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$\begingroup$ Do you really mean the sphere, or do you mean the ball? Convex subsets of spheres might not be very big... $\endgroup$– Yemon ChoiCommented Jun 13, 2012 at 9:03
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$\begingroup$ Yes, it I really mean the sphere. Think of $Y'$ as being an $L^1$-space... $\endgroup$– Andy TeichCommented Jun 13, 2012 at 9:06
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1$\begingroup$ This is very similar to a question here: math.stackexchange.com/questions/157069/… $\endgroup$– Matthew DawsCommented Jun 13, 2012 at 12:10
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2$\begingroup$ @Peter: No; but if you look on this site, you'll see that it's generally considered good to cross-link. You probably wouldn't know that as a new user, so I thought I'd highlight this for you. $\endgroup$– Matthew DawsCommented Jun 13, 2012 at 12:27
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1 Answer
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When you say that $C$ is weak* closed I'm not sure whether you mean as a subset of $Y'$ or in the relative weak* topology on $M$. If the latter, the answer is obviously "no": take $T$ to be the identity map and let $C = M$ be a convex subset of the unit sphere which is not weak* closed. (Examples are easy to find, even in finite dimensions.) If the former, the answer is "yes", for then $C$ is a weak* closed subset of the closed unit ball, hence it is weak* compact, hence its image under a weak* continuous map (which $T'$ is) must also be weak* compact, and hence weak* closed.
answered Jun 13, 2012 at 12:14
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$\begingroup$ If I understand you correctly, it also would suffice to assume that $M$ itself is weak*-closed...? $\endgroup$ Commented Jun 13, 2012 at 12:45
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$\begingroup$ Weak* compact (which is the same as weak* closed, if it's bounded). $\endgroup$ Commented Jun 13, 2012 at 12:51
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$\begingroup$ So it is sufficient to assume that $C$ is weak*-closed and $M$ is weak*-closed. Is it also necessary to have that $C$ is weak*-closed?Can we give conditions on $C$, $M$ or $T'$ to have that for any $C\subseteq M$ the image $T'(C)$ is weak*-closed in $X'$? $\endgroup$ Commented Jun 13, 2012 at 13:26
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3$\begingroup$ Since the question is changing, maybe you'd better edit your original post to clarify what it is you really want. $\endgroup$ Commented Jun 13, 2012 at 13:37