Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

1 Answer

up vote 1 down vote accepted

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 $n\to\infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.