Two locally convex topologies on $B(X)$.

Question

Let $X$ be a non-reflexive Banach space. It is supposed to compare two locally convex topologies on $B(X)$:

Let $w$ be the topology on $B(X)$ implemented by all seminorms given by $$B(X)\to [0,\infty) : T\to |\langle T^*x^*,x\rangle|$$ where $x\in X$ and $x^*\in X^*$.

We also denote $w^*$ by the topology implemented by all seminorms given by $$B(X)\to [0,\infty) : T\to |\langle T^*x^*,x^{**}\rangle|$$ where $x^{**}\in X^{**}$ and $x^*\in X^*$.

Question) It seems even when $X$ is separable these two topologies $w$ and $w^*$ are not the same, does not it?

Answer 1

Fix $x \in X$ and let $(f_\alpha)$ be a net in $X^*$. For each $\alpha$ let $T_\alpha$ be the rank-one operator $y \mapsto f_\alpha(y) x$. Then you want to compare the seminorms $$ |\langle T_\alpha^*(x^*), x \rangle| = |\langle x^*, x\rangle| |\langle f_\alpha, x\rangle| $$ against $$ |\langle T_\alpha^*(x^*), x^{**} \rangle| = |\langle x^*, x\rangle| |\langle x^{**}, f_\alpha \rangle|. $$ As $X$ is non-reflexive, we can find a bounded net $(f_\alpha)$ which is, say, weak$^*$-null, but for some $x_0^{**} \in X^{**}$ we have $\langle x^{**}_0, f_\alpha\rangle=1$ for all $\alpha$. So the topologies differ.