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Let $(X,\|\cdot\|)$ be a Banach space which is the dual of another Banach space. The Banach-Alaoglu theorem asserts that the closed unit ball in $X$ is compact in the weak*-topology. Assume that we have another norm $\|\cdot\|_2$ on $X$ which is equivalent to the given one, so that there is $C\geq1$ with

$\forall x\in X:\quad C^{-1}\|x\|\leq\|x\|_2\leq C\|x\|$.

Is it true that the closed unit ball in this second norm is also weak*-compact?

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  • $\begingroup$ The weak-*-topology is not normable! $\endgroup$
    – Zbigniew
    Commented Aug 22, 2019 at 10:37

3 Answers 3

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No. In fact, any non reflexive space can be equivalently renormed so that it is not isometric to a dual space. See

Davis, William J.; Johnson, William B. A renorming of nonreflexive Banach spaces. Proc. Amer. Math. Soc. 37 (1973), 486–488.

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  • $\begingroup$ Bill's comment below clarified the answer for me: the unit ball of a Banach space is compact in some weaker locally convex Hausdorff topology if and only if the Banach space is isometrically isomorphic to a dual space. $\endgroup$
    – pavel
    Commented Feb 25, 2013 at 0:39
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On the sequence space $l^1$, define the equivalent norm $$\Vert x \Vert =\sum |x_i|+2|\sum x_i|.$$ Let $e^n$ be the nth unit vector, and define $x^n=e^1-e^n$. Then $\Vert x^n\Vert=2$. But the weak-* limit of $x^n$ is $e^1$, and $\Vert e^1\Vert=3$.

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Of course, the topologies are the same.

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    $\begingroup$ The two weak* topologies are the same, but the two balls are different. $\endgroup$
    – Gerald Edgar
    Commented Feb 24, 2013 at 22:16
  • $\begingroup$ I guess I didn't look carefully at the last line. But it still true, isn't it? Alain $\endgroup$
    – Alain
    Commented Feb 24, 2013 at 22:26
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    $\begingroup$ Let me try to get myself into more trouble: What does "isometric" have to do with this? The second ball is a closed subset of a multiple of the first one, isn't it? And scalar multiplication is weak-* continuous. Alain $\endgroup$
    – Alain
    Commented Feb 24, 2013 at 22:49
  • $\begingroup$ Alain, the unit ball of a Banach space is compact in some weaker locally convex Hausdorff topology if and only if the Banach space is isometrically isomorphic to a dual space. $\endgroup$
    – Bill Johnson
    Commented Feb 25, 2013 at 0:29
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    $\begingroup$ Great, thanks. I stand corrected. That reminds me of a Groucho Marx recommendation: It is better to keep your mouth shut and seem stupid than opening it and leave no doubt about it. Alain $\endgroup$
    – Alain
    Commented Feb 25, 2013 at 1:08

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