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We can compare topologies on $B(H)$. For instance, Sot topology is stronger than wot topology or $\sigma-$ weak topology is equivalent to weak* topology. I would like to compare wot topology and weak topology. Please help me. Thanks so much.

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- Is $H^2\cap H_0^1$ equipped with the norm $\|f'\|_{L^2}$ complete?
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- distribution with point support

By the *weak topology* do you mean the Banach-space weak topology? If so then it is stronger than the weak operator topology.

Denote by $(\cdot, \cdot)$ the inner product in $H$ and by $\langle \cdot, \cdot \rangle$ the duality bracket between a Banach space and its dual. The weak operator topology is generated by the seminorms

$T\mapsto |( y, Tx )| = |\langle T, x\otimes y\rangle|$ where $x,y\in H$.

Here by $x\otimes y$ I understand the rank-one nuclear operator given by $(x\otimes y)z = (x,z)y$. Recall that the space of nuclear operators is the predual of $B(H)$, whence in particular it is a subspace of $B(H)^*$.

The weak topology of $B(H)$ comes from duality with the humongous dual space $B(H)^*$. It is generated by the seminorms

$T\mapsto |\langle \Phi, T\rangle|$ where $\Phi \in B(H)^*$.

Note that the seminorms defining the weak operator topology form a proper subset of the family of norms defining the weak topology, so the weak topology is not coarser than the weak operator topology. Let’s see that we have strict inclusion. Indeed, the unit ball of $B(H)$ is compact in the weak operator topology. It is however not compact in the weak topology as $B(H)$ is not reflexive.

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