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Let $X$ be a normed $\mathbb K$-linear space . Then the weak$*$ topology is metrizable iff $X$ is finite dimensional.

We know that if X is finite dimensional then the weak topology on X is metrizable and also X is finite dimensional iff $X^*$ is finite dimensional.

From here can we approach for proof of the above theorem?

Please someone give hints.

thank you..

Mini_me
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2 Answers2

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I think you need $X$ to be a Banach space. Then first use the Banach-Steinhaus theorem to show that $(X^*,\sigma^*)$ is sequentially complete. If it is metrizable (hence a Frechet space) the open mapping theorem implies that the weak$^*$ topology coincides with the topology induced by the dual norm. For the unit ball $B$ of $X$ you thus get finitely many $x_1,\ldots,x_n\in X$ such that $\{x_1,\ldots,x_n\}^\circ \subset B^\circ$. Then apply the theorem of bipolars together with the fact that the absolutely convex hull $\Gamma(E)$ of a finite set is closed to get $B \subseteq \Gamma(\{x_1,\ldots,x_n\})$.

EDIT. As mentioned in the comment, for the space $X$ of scalar sequences with only finitely many non-zero terms endowed with the $\ell^2$-norm, the weak$^*$-topology on $X^*=\ell^2$ is the topology of coordinate-wise convergence which is metrizable.

Jochen
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  • No.. I also asked my professor whether the space is Banach or not, he replied no. – Mini_me Oct 24 '17 at 15:58
  • I do believe that for the countable dimensional space $X$ of real sequences with only finitely many non-zero terms endowed with the $\ell^2$- norm the dual (which is $\ell^2$) is metrizable in the weak$^*$ topology. – Jochen Oct 24 '17 at 16:46
  • Does your professor still say no? I would be interested in his or her argument. – Jochen Oct 25 '17 at 12:56
  • Any news from your professor? – Jochen Nov 02 '17 at 08:54
  • Sorry . I just forgot . He can give some hints but the question will remain same. – Mini_me Nov 02 '17 at 12:27
  • As you finally accepted the answer: Is your professor still convinced that the statement holds for normed spaces? – Jochen Dec 04 '17 at 09:10
  • @Jochen How to show that the topology of coordinate wise convergence in $X^*=(\ell^2,|.|_2)$ is metrizable? – Anupam May 19 '19 at 07:50
  • A possible metric is $d (x,y)=\sup{\min (|x_n-y_n|, 1/n): n \in \mathbb N} $. – Jochen May 19 '19 at 09:37
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Hint: in infinite dimensions weakly open neighborhood of origin always contain a (maximal) vector subspace.

Red shoes
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