# On weak*-sequentially completeness

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.

• 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 $(1,2,3,\ldots)$. – David Mitra Mar 25 at 3:36
• 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. – Anupam Mar 25 at 8:11

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