$W$ 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 $W$?
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.