weak*-compactness of unit ball in equivalent norm

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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I love how two contradictory answers both get voted up for this question. –  Joseph Van Name Feb 25 at 0:35

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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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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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. –  pavel Feb 25 at 0:39