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Given a normed separable vector space $X$, the Banach-Alaoglu theorem states that the closed unit ball in the dual space $X^*$ is compact in the weak-* topology. Since $X$ is separable, that topology is metrizable.

Now, the Alaoglu theorem states that for any normed vector space, the closed unit ball is compact iff the space is finite dimensional. Applying this to $X^*$ allegedly yields that $X^*$ is finite dimensional (obviously false).

Where is my mistake?

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  • $\begingroup$ If $X$ is not finite-dimensional then the weak$^*$ topology on $X^*$ is strictly weaker than the norm-induced topology of $X^*. $ $\endgroup$ – DanielWainfleet Apr 17 '17 at 7:50
  • $\begingroup$ the closed unit ball is compact iff the space is finite dimensional: Riesz theorem not Banach-Alaoglu $\endgroup$ – Zbigniew Aug 22 '19 at 10:15
  • $\begingroup$ Finite dimensional is the trivial cases. You do not see anything. That is the space is trivially separable and the weak-* topology coincide with the norm topology and then the space is separable, the unite ball is compact. $\endgroup$ – Zbigniew Aug 22 '19 at 10:20
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"Compact" in the second theorem refers to the norm topology, not the weak-* topology. The weak-* topology may be metrizable, but that metric does not come from a norm.

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  • $\begingroup$ You mean a norm topology? Or am I missing why should there be a unique norm on that space (up to equivalence)? Also, isn't the metric induced by the norm $\|x\|=d(x,0)$? $\endgroup$ – BOS Apr 16 '17 at 23:18
  • $\begingroup$ My point is, it still seems to me that the weak-* topology is induced by a norm, and therefore the Alaoglu theorem applies to the compactness w.r.t that norm $\endgroup$ – BOS Apr 16 '17 at 23:23
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    $\begingroup$ No, the weak-* topology is not induced by a norm. $\endgroup$ – Robert Israel Apr 16 '17 at 23:38
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    $\begingroup$ If $d$ is a metric corresponding to the weak-* topology, $d(x,0)$ is not a norm. One reason it can't be a norm: every weak-* neighbourhood of $0$ contains infinite rays from the origin. $\endgroup$ – Robert Israel Apr 17 '17 at 2:27
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    $\begingroup$ No, it does not induce the weak * topology. $\{\psi: \|\psi\| < 1\}$ is not a weak-* neighbourhood of $0$. $\endgroup$ – Robert Israel Apr 19 '17 at 6:57

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