Let $\mathcal{B} (H)$ denote the set of bounded (continuous) linear operators on a Hilbert space $H$. Consider the following two topologies on $\mathcal{B} (H)$ characterized by net convergence:
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Norm topology: $T_\alpha \overset{||\cdot||_{\mathcal{B} (H)}}{\longrightarrow} T $.
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Strong Operator Topology: $T_\alpha \overset{SOT}{\longrightarrow} T \iff T_\alpha x \overset{||\cdot||}{\longrightarrow} T x \, , \forall x \in H$.
I see why the first is stronger than the second, but I don't see why it is strictly stronger. Here is what I did:
$$T_\alpha \overset{||\cdot||_{\mathcal{B} (H)}}{\longrightarrow} T \iff ||T_\alpha-T||_{\mathcal{B} (H)} \to 0\\
\quad \quad \quad \quad \quad \quad \, \iff \sup_{||x|| \leq 1} ||(T_\alpha-T)x || \to 0\\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \iff ||(T_\alpha-T)x || \to 0\, , \, \forall x \in H \, , \, ||x|| \leq 1\\
\quad \quad \quad \quad \quad \quad \quad \quad \, \iff ||(T_\alpha-T)x || \to 0\, , \, \forall x \in H \\
\quad \quad \quad \quad \quad \quad \iff T_\alpha x \overset{||\cdot||}{\longrightarrow} T x \, , \, \forall x \in H \\
\quad \, \iff T_\alpha \overset{SOT}{\longrightarrow} T \, .$$
I guess there is something wrong. Which of the $\iff$ arrows is actually $\Rightarrow$?
Best Answer
$$\sup_{||x|| \leq 1} ||(T_\alpha-T)x || \to 0 \implies ||(T_\alpha-T)x || \to 0\, , \, \forall x \in H \, , \, ||x|| \leq 1.\\$$
Consider $e_k$ an orthonormal basis of $H$ and define $T_n(e_k)=e_k$ for $k\leq n$ and $T_n(e_k)=2e_k$ for $k >n $.
Then $$||(T_n-T)x || =||\sum_{k=n+1}^{\infty}\langle x,e_k\rangle e_k||\to 0\, , \, \forall x \in H \, , \, ||x|| \leq 1 $$ But $$\sup_{||x|| \leq 1} ||(T_n-T)x ||\geq ||(T_n-T)e_{n+1} ||=||e_{n+1}||=1\\$$