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Given: X seperable Banach space. X' its dual. We have $M\subset X'$ a closed unit ball in X'. Choose a sequence $(x_{n}$) of nonzero elements in X which is dense in X. Define $d(x',y'):=\sum\limits_{n=1}^n \frac{1}{2^n} \mid{x'(x_n)-y'(y_n))}\mid $

Prove that $M\ni x'_m\rightharpoonup^*x'\in M$ iff $d(x_m',x')\to 0\:as\: m\to \infty $.

1.Do we need to use Banach-Alaoglu theorem?

2.Is it somehow related with the Schur's property or may be it gives us something for the proof? I was reading something which says that a Banach space has Schur's property if weak convergence of a sequence implies convergence in norm?

3.Or may be Rosenthal's $ l_1$ theorem can play a role. Statement :"Every bounded sequence in a Banach space contains a subsequence that is either weakly Cauchy or equivalent to the unit vector basis of $l_1$."

Any help is appreciated.

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  • $\begingroup$ no need for Banach-Alaoglu. Prove that the distance goes to $0$ iff every summand in the distance goes to $0$ (i.e. like $w^*$ convergence, but in place of all elements of $X$ you only use a dense subset) iff you have $w^*$ convergence. $\endgroup$ – user8268 May 25 '14 at 14:00
  • $\begingroup$ thank you for the hint user8268 :) $\endgroup$ – learner May 25 '14 at 14:03
  • $\begingroup$ @user8268, why don't you post thiss ans answer? $\endgroup$ – Norbert May 25 '14 at 14:17
  • $\begingroup$ I am not getting what "the summand in the distance" mean :( Again stack :( $\endgroup$ – learner May 25 '14 at 16:24
  • $\begingroup$ @learner I think it was supposed to be $d(x',y')=\sum_{n=1}^\infty\frac{1}{2^{n}}|x'(x_n)-y'(x_n)|$. $\endgroup$ – Luiz Cordeiro May 26 '14 at 1:13
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Let it be clear that items 2 and 3 have nothing to do with the simple (though important) fact stated in the first paragraph. You are to show that when $X$ is separable, the weak* topology of $X^*$ is metrizable on bounded subsets.

Aside: this fact does come up in the context of Banach-Alaoglu, allowing for a nice proof without Tychonov's theorem (see sequential Banach-Alaoglu). But you don't need any big theorems to establish the fact itself.

The proof consists of two parts.

Part 1: $d(x_m',x')\to 0$ if and only if $x_m'(x_n)-x'(x_n)\to 0$ for every $n$.

The direction $\Rightarrow$ is straightforward. For the direction $\Leftarrow$ split the series defining $d$ into "head" and "tail": the head is small by assumption, the tail is small because of $1/2^n$.

Part 2 $x_m'\to x'$ in weak* topology if and only if $x_m'(x_n)-x'(x_n)\to 0$ for every $n$.

Again, $\Rightarrow$ is straightforward. For the converse, fix $x\in X$. You can find $x_n$ close to $x$. Use the triangle inequality like this: $$\begin{split} |x_m'(x_n)-x'(x)| &\le |x_m'(x_n)-x'(x_n)| + |x_m'(x_n)-x_m'(x)| + |x'(x_n)-x'(x)| \\ &\le |x_m'(x_n)-x'(x_n)|+2\|x-x_n\| \end{split}$$

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