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?



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


