Consider the strong topology and $\sigma$-strong topology. By definition, strong topology is the topology induced by the family of seminorms $p_\xi$, $\xi \in H$ where $p_\xi(T)=||T\xi||$ and the $\sigma$– strong topology is the topology induced by the family of seminorms $p_{(\xi_n)}$, $(\xi_n)\subseteq H$ is such that $\sum_n||\xi_n||^2<\infty$ and $p_{(\xi_n)}(T)=(\sum_n||T\xi_n||^2)^{\frac{1}{2}}$.
It is clear that $\sigma$-strong topology is finer than strong topology. I want to prove that on the unit ball in $B(H)$, these two topologies are the same.
In the proof it is written that the seminorms defining $\sigma$-strong topology is the uniform limit of strongly continuous seminorms. I am not understanding the terminologies here mentioned uniform limit and strong continuity and how does this imply the conclusion?
Best Answer
Fix $\bar\xi = (\xi_n)_n$ one of your sequences in in $\ell^2[H]$. By the convergence of the sum $\sum_n \| \xi_n \|_2^2$, we have that, for every $\epsilon > 0$ we can pick an $N$ such that $ \sum_{n > N} \| \xi_n \|_2^2 < \epsilon. $ But that implies that the seminorm $p_{\bar\xi}(T)$ is bounded by $$ \begin{eqnarray} p_{\bar\xi}(T) & = & \Big( \sum_{n = 1}^N \| T \xi_n \|_2^2 + \sum_{n = N + 1}^\infty \| T \xi_n \|_2^2 \Big)^\frac12\\ & \leq & N^\frac12 \max_{n \leq N} \{ \, p_{\xi_n}(T) \, \} + \Big( \sup_{n \geq N + 1} \| T_n \|\Big) \epsilon^\frac12. \end{eqnarray} $$ From that it is easy to conclude, if $T_\beta \to T$ strongly and $T_\beta$, $T$ are in the unit ball, then for every $\epsilon$ you can take $N$ as above. Now use that the first term is a finite combinations of seminorms and therefore goes to $0$ to get that $$ \lim_\beta p_{\bar\xi}(T_\beta - T) \leq \Big( \sup_{\beta} \| T_\beta -T \|\Big) \epsilon^\frac12 \leq \sqrt{2} \epsilon^\frac12, $$ but since $\epsilon$ is arbitrary we can conclude.