Here is a related question: Comparing weak and weak operator topology
I still have a question why the ultra-weak topology on $B(H)$ is weaker than the weak (Banach spaces) topology on $B(H)$. The strict inclusion is then follows by the same argument using compactness of the unit ball.
There is no need to explain again why the weak operator topology (WOT) is weaker than the ultra-weak...
Thank you!
Shirly, when you regard $B(H)$ merely as a Banach space, then it is not reflexive (unless $H$ is finite-dimensional) because it contains, for example, a subspace isomorphic to $\ell_\infty$. It is a dual space on its own, so it has the ultraweak (=weak* topology coming from the duality with nuclear operators $N(H)$). A Banach space $X$ is reflexive if and only if the weak and the weak* topologies in $X^*$ are equal.
Consequently, $N(H)$ is not reflexive, as $B(H)$ is not, and so the weak and ultraweak topologies in $B(H)$ are different.
