I just get really nervous around lim sup and nets for some reason... Please help!
Problem I was given: Suppose $(x_i)_{i\in I} \subset B(\mathcal{H})$ converges to $x \in B(\mathcal{H})$ in the SOT (strong operator topology). Show that $$ ||x|| \leq \limsup_{i \rightarrow \infty} ||x_i||. $$
My attempt: For the operator norm, we have $$ ||x|| = \sup_{||\xi||=1} ||x\xi|| $$ and for a net $$ \limsup_{i} ||x_i|| = \lim_{i\in I} \sup_{j \geq i} ||x_j|| = \lim_{i\in I} \sup_{j \geq i} \sup_{||\xi||=1} ||x_j\xi|| $$ if that's what $i\rightarrow \infty$ means? And we know that $$ \lim_{i \rightarrow \infty} ||(x-x_i)\xi|| = 0 \quad \forall \xi \in \mathcal{H}. $$
Lim sup is taking the limit of a decreasing sequence, because taking sup over smaller and smaller sets can only get smaller. So if we want to show the limit is bounded below by $||x||$, it's enough to see that each sup is bounded by it. Suppose BWOC that for some $i \in I$, $$ \sup_{j \geq i} ||x_j|| < ||x||. $$ Then for all $j \geq i$ we had $||x_j|| < ||x||$ which means $$ 0 < ||x|| - ||x_j|| \leq ||x-x_j|| = \sup_{||\xi||=1} ||(x-x_j)\xi||, $$ via Reverse Triangle Inequality and the definition of the operator norm. What follows I'm not at all sure is valid: Because we have a strict inequality, there is some $N \in \mathbb{N}$ such that $ \sup_{||\xi||=1} ||(x-x_j)\xi|| \geq \frac{1}{N}$ for all $j \geq i$. But if we let $\xi'$ be the vector at which the sup is achieved, then for $\xi' \in \mathcal{H}$ we have that $$ \lim_{i \rightarrow \infty} ||(x-x_i)\xi'|| \neq 0. $$ This contradicts the SOT convergence of $(x_i)_{i\in I}$. Therefore we must have that $ \sup_{j \geq i} ||x_j|| \geq ||x||$ for each $i$ and thus $\limsup_{i} ||x_i|| \geq ||x||$.