An inequality about operator norm

Question

Let $H$ be a Hilbert space and $T\in B(H)$, with $T_{i}\rightarrow T$ in strong operator topology.

Then can we prove that $\liminf_{i\rightarrow \infty}||T_{i}||\geq ||T||$ ?

Answer 1

Fix $x\in H$ of norm $1$ and $\varepsilon>0$. Since $T_ix\to Tx$, we can find an integer $i_0$ such that if $i\geqslant i_0$, then $\lVert Tx\rVert\leqslant \lVert T_ix\rVert+\varepsilon$. We thus have that for $j\geqslant i_0$, $$\lVert Tx\rVert\leqslant \inf_{i\geqslant j}\lVert T_ix\rVert+\varepsilon$$ and taking the limit as $j$ goes to infinity, we get $$\lVert Tx\rVert\leqslant \liminf_{i\to\infty}\lVert T_ix\rVert+\varepsilon.$$ Since the norm of $x$ is $1$, we have $\lVert T_ix\rVert\leqslant \lVert T_i\rVert$, hence for each $x$ of norm $1$, $$\lVert Tx\rVert\leqslant \liminf_{i\to\infty}\lVert T_i\rVert+\varepsilon.$$ To conclude, take the supremum over these $x$.