Let $K$ be a convex,balanced and compact subset of a Banach space $X$. We let $X_{K}:=span\{K\}$. Define the norm $\|x\|_{K}:=\inf\{\alpha>0: x\in \alpha K\}, x\in X_{K}$. Then $(X_{K},\|\|_{K})$ is a Banach space and the unit ball of $(X_{K},\|\|_{K})$ is $K$. My question is : Is the space $(X_{K},\|\|_{K})$ reflexive? The answer maybe has already been known, but I do not know.
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asked Jul 15, 2014 at 16:39
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$\begingroup$ When people take the trouble to respond to a question on MO, the OP should be respond in timely fashion. You are being discourteous by not responding. $\endgroup$– Bill JohnsonCommented Jul 20, 2014 at 14:10
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$\begingroup$ @BillJohnson: Oh! I am sorry for not responding in timely fashion. I'll respond timely later. $\endgroup$– Dongyang ChenCommented Jul 20, 2014 at 16:29
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2 Answers
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No it is not reflexive in general. Your assumptions basically say that $X_K$ is compactly embedded in $X$ because $K$ is more or less the unit ball of $X_K$ (it is equal to the closed unit ball I think).
Now there are obviously a ton of examples of non-reflexive Banach spaces that are compactly embedded into other (completely arbitrary!) Banach spaces like the Sobolev space $W^{1,\infty}(0,1)$ (=the space of Lipschitz continuous functions) which compactly embeddeds into $C^0[0,1]$.
answered Jul 15, 2014 at 17:07
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Even if $K$ is just weakly compact, $X_K$ is isometrically isomorphic to a dual space with the weak$^*$ topology agreeing with the weak topology from $X$ on $K$. You can get a weakly compact convex symmetric set $J\supset K$ s.t. $X_J$ is reflexive. See my paper with Davis, Figiel, and Pelczynski, ``Factoring weakly compact operators".
answered Jul 15, 2014 at 18:28
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$\begingroup$ What exactly do you mean when you say "weak topology from X and K" ? I would interpret the terminology "weak topology" as "coarsest topology on $X_K$ that makes $X_K\to X$ continuous" but that would be the subspace topology and additionaly does not depend on $K$. Or do you mean the subspace topology of $X_K\subseteq X$ where $X$ is endowed with its weak topology? That still doesn't depend on $K$ but it sounds more like the right topology for your statement. $\endgroup$ Commented Jul 15, 2014 at 22:33
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$\begingroup$ I corrected a typo, Johannes; "and" should have been "on". By "weak topology from $X$" I mean the weak topology from $X^*$ which of course does not depend on $K$. $\endgroup$ Commented Jul 16, 2014 at 7:09