Let $X$ be a Banach space with norm $\|\cdot\|$, and $\mathcal{B}(X)$ the space of bounded operators on $X$. If a sequence $(f_n)_n\subset \mathcal{B}(X)$ converges under the norm topology, $f_n \to f\in\mathcal{B}(X)$, then we also have pointwise-convergence: for all $x\in X$, $\|f_n(x)-f(x)\|\leq \|f_n-f\|_\text{OP}\cdot \|x\|\to 0$. So the sequence also converges under the strong operator topology. The same is true for nets.
If I'm not mistaken, the S.O.T is a coarser (weaker) topology than the norm topology. However, wouldn't the above results imply every norm-closed sets in $\mathcal{B}(X)$ is also S.O.T-closed? Wouldn't that then imply the S.O.T is finer than (at least as fine as) the norm topology? Can anyone tell me what I got wrong?
Best Answer
'wouldn't the above results imply every norm-closed sets in $\mathcal{B}(X)$ is also S.O.T-closed?'. No, it works the other way. If $M$ is S.O.T. closed, $(f_n) \in M$ and $f_n \to f$ in operator norm, then $f_n \to f$ S.O.T which implies $f \in M$. So S.O.T. closed implies norm closed.