1
$\begingroup$

I want to prove that every dual space is weak*-sequentially complete.

Let $X$ be a normed linear space and let $(f_n)$ be a weak* Cauchy sequence in $X^*$. Thus for all $x\in X$, $(f_n(x))$ is a Cauchy sequence in $\mathbb K$. Thus for all $x\in X$, $\lim\limits_{n\to \infty}f_n(x)$ exists. If I define $f(x)=\lim\limits_{n\to \infty}f_n(x)$ for all $x\in X$, then $f$ is linear. But how to show that it is bounded. Had $X$ been given a Banach space I could have done it by using Banach-Steinhauss theorem. But now how to proceed? Any hint is appreciated.

$\endgroup$
  • 5
    $\begingroup$ Do you mean $(f_n)$ is weak* Cauchy in $X^*$? I don't think your result is true. Let $X$ be the set of real sequences of finite support with the $\ell_1$ norm. Define $f_n\in X^*$ by $f_n(x)=x_1+2x_2+\cdots+n x_n$. Then for $x\in X$, $\lim_{n\rightarrow\infty} f_n(x) $ exists; so $(f_n)$ is weak* Cauchy. $(f_n)$ cannot be weak* convergent, though, since its "limit" must be $(1,2,3,\ldots)$. $\endgroup$ – David Mitra Mar 25 at 3:36
  • $\begingroup$ Yes I meant $(f_n)$ weak*-Cauchy in $X^*$. I have corrected it now. Your example clearly shows that my assertion is false. This will be true if $X$ is a Banach space. $\endgroup$ – Anupam Mar 25 at 8:11
1
$\begingroup$

Hint : You can use the following corollary of Banach-Alaoglu Theorem :

Let $ X $ be a normed space. Then every bounded subset of $ X^* $ is relatively weakly$^*$ compact.

$\endgroup$
  • $\begingroup$ It's not clear how this helps, because we would still need to show that the sequence $f_n$ is bounded. $\endgroup$ – Nate Eldredge Mar 25 at 3:35
  • $\begingroup$ And as David Mitra's example shows, in fact $f_n$ does not have to be bounded, and the desired statement is actually false. $\endgroup$ – Nate Eldredge Mar 25 at 3:38
  • $\begingroup$ @NateEldredge This result is true for Banach space $ X $ (In David Mitra's example $ X $ is not complete). $\endgroup$ – Math Fanatic Mar 25 at 5:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.