# Proving that Uniform operator convergence implies strong operator convergence implies weak

Let $H$ be a Hilbert Space. Let $\{T_n\}$ be a sequence of bounded operators in $H$.

I'm trying to prove that Uniform Operator Convergence implies Strong Operator Convergence implies Weak Operator Convergence.

My proof:

Suppose Uniform Operator Convergence prevails between the sequence $\{T_n\}$ and $T$.

This means that $||T_n-T|| \rightarrow 0$ as $n \rightarrow \infty$

Therefore, for any $x \in H$,

$||T_nx-Tx|| \le ||T_n-T||\,||x|| \rightarrow0$ as $n \rightarrow \infty$

and the implication uniform $\Rightarrow$ Strong $\Rightarrow$ Weak is proved.

Am I missing anything? Will this be enough to show that strong implies weak?

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You proved that uniform $\implies$ strong. You also need to show that strong $\implies$ weak. – Jesse Madnick Mar 20 '13 at 5:12
I guess that's where I need the help. How can I show that? Let an epsilon > 0. There is an index N such that $||T_n-T||< e$ whenever n is greater than or equal to N. – User69127 Mar 20 '13 at 5:17
Isn't it evident that strong implies weak on the above statement? – User69127 Mar 20 '13 at 6:06

The weak operator topology means that $(T_nx,y)\to (Tx,y)$ for all $x,y\in H$, where $(\cdot,\cdot)$ is our inner product. This is certainly the case if we could show, for all $x,y\in H$ and $\varepsilon>0$, that there exists $N\in\mathbb N$ such that $$|(Tx-T_nx,y)|<\varepsilon$$ for all $n>N$. (If that doesn't make sense, you should verify it). Well, by Cauchy-Schwarz, $$|(T_nx-Tx,y)|\leq\|T_nx-Tx\|\cdot\|y\|.$$ Since $\|y\|=M$ is finite, we can choose $N\in\mathbb N$ such that $\|T_nx-Tx\|<\varepsilon/M$ for $n>N$, which in turn shows that $$(T_Nx-Tx,y)\leq |(T_Nx-Tx,y)|<\varepsilon$$ for all $n>N$. Because this works for all $\varepsilon>0$, we have $(T_nx,y)\to(Tx,y)$ as $n\to\infty$.