The map $T\longmapsto \|T\|$ is not continuous in the strong operator topology of $\mathscr B(H)$

Question

In the context of Strong and Weak operator topologies on $\mathscr B(H)$ there is an statement that says: the map on $\mathscr B(H)$ that $T\longmapsto \|T\|$ is not continuous in the strong operator topology.

Is this statement trivial? Or need some proof?

so thanks

Answer 1

Without loss of generality, you can assume that $H$ is an infinite-dimensional, separable Hilbert space.

Let $(e_n)_{n=1}^\infty$ be an orthonormal basis for $H$. Let $$P_n = e_1\otimes e_1 + \ldots e_n\otimes e_n.$$ Then $(P_n)_{n=1}^\infty$ is a sequence of finite-rank projections converging pointwise to the identity. Consequently, $(I-P_n)_{n=1}^\infty$ converges pointwise to 0, but $\|I-P_n\| = 1$ for each $n$. In particular, $$\|I-P_n\|\not\to \|0\|=0\text{ as }n\to \infty.$$