# The dual of a dual space with the topology of uniform convergence on compact subsets?

$W$ is a Banach space. The topology of $W^*$ is the uniform convergence on the compact subsets of $W$. That is generated by the family of seminorms $$p_K(f)=\sup_{x\in K}|f(x)|,$$ for all compact subset $K\subset W$.

The question is how to prove the dual of $W^*$ in this topology is $W$?

• Are you allowed to use that the dual of ($W^*$ endowed with the $w^*$ topology) is equal to $W$? Feb 25, 2014 at 9:43
• Yes. How to use that? Feb 26, 2014 at 1:56
• The dual (with compact-open topology) of a Banach space is a Smith space, and the dual of a Smith space is a Banach space. See Scholze's Lecture notes, Thm 3.8. Jan 26, 2021 at 12:46

## 1 Answer

I think I found the answer.It's a modification of the proof in "Introduction to Tensor Products of Banach spaces, R.A.Ryan" Prop4.6.

Suppose $x^{**}$ is an element of the dual of $W^*$ endowed with the given topology.

It's easy to see that there exsist a compact subset $K\subset X$, s.t. $$|x^{**}(f)|\leq \sup_{x\in K}|f(x)|,\forall f\in W^*.$$ Use that a compact set is a closed subset of the close convex hull of a sequence $(x_n)$ converge to $0$ in $W$. So $K$ can be replaced by the sequence. Now $\forall f\in W^*,(f(x_n))$ is an element in $c_0$. $x^{**}$ can be extended to a bounded functional on $c_0$, denote $(\phi_i)\in l_1$. The representation of $x^{**}$ in W is the weak limit of $\sum \phi_i x_i$.

Many thanks to the people read this question.