# Von Neumann algebra. Question regarding a proof in Kehe Zhu.

Theorem 17.1 i Kehe Zhu's "An introduction to operator algebras" is the following statement:

Suppose $A$ is a von Neumann algebra acting on $H$. If {${T_{\alpha}}$} is an increasing net of self-adjoint operators in $A$ which is bounded above, then {${T_{\alpha}}$} is strong operator convergent to a self-adjoint operator $T$ in $A$. Furthermore, $T$ is the least upper bound of {${T_{\alpha}}$}.

I have two questings regarding the proof:

1) The proof starts out by claiming, that we can assume "without loss of generality" that there exists a constant c>0, such that $-cI \leq T_{\alpha} \leq cI$ for all $\alpha$. But why is this so?

2) The proof proceeds by first showing that {${T_{\alpha}}$} has a subnet which converges in the weak operator topology. Then the book claims, that (as net {${T_{\alpha}}$} is increasing), this implies that the whole net is convergent in the weak-operator topology. But why is this so?

• Do you mean that $\{T_\alpha\}$ is bounded $\textit{above}$, rather than below? Since $T_\alpha = \alpha I$ for $\alpha\in\mathbb{R}_{\geq 0}$ is bounded below but clearly not convergent to anything. – Owen Sizemore Mar 20 '17 at 18:29
• Yes, thank you. That was a typo. "Above" is correct. – Kimarokko Mar 20 '17 at 18:31

## 1 Answer

1. Because to test convergence you only care about "the tail" of the net. So you can start on a fixed $$T_{\alpha_0},$$ and now $$-\|T_{\alpha_0}\|\leq T_{\alpha_0}\leq T_\alpha\leq c$$ for all $$\alpha$$.

2. I'm not sure what this has to do with the proof, but assume a subnet $$\{T_{\alpha_\beta}\}_\beta$$ converges wot to $$T.$$ Fix $$x\in H$$. For $$\varepsilon>0$$, there exists $$\beta_0$$ such that $$\langle Tx,x\rangle-\langle T_{\alpha_{\beta_0}}x,x\rangle<\varepsilon.$$ Then, for any $$\alpha\geq {\alpha_{\beta_0}}$$, we have $$\langle T_{\alpha_{\beta_0}}x,x\rangle\leq\langle T_{\alpha}x,x\rangle$$, and so $$\langle Tx,x\rangle-\langle T_{\alpha }x,x\rangle\leq \langle Tx,x\rangle-\langle T_{\alpha_{\beta_0}}x,x\rangle<\varepsilon.$$ So (by polarization), $$T_\alpha\to T$$ wot. Now, using that $$T-T_\alpha\geq0$$, $$\|(T-T_\alpha)x\|^2=\langle (T-T_\alpha)^2x,x\rangle \leq\|T-T\alpha\|\,\langle (T-T_\alpha)x,x\rangle\to0.$$

The last inequality, if not obvious, is justified as follows: for a positive operator $$S$$, we have, by Cauchy-Schwarz, $$\langle S^2x,x\rangle=\langle S\,S^{1/2}x,S^{1/2}x\rangle\leq\|S\|\,\|S^{1/2}x\|\,\|S^{1/2}s\| =\|S\|\,\|S^{1/2}x\|^2=\|S\|\,\langle Sx,x\rangle.$$

• @Argerami. Thank's! I have edited the post a bit to make it more clear, how point 2 is used in the proof (as you can see, you explained a bit more, than I was asking for - but it was enlightening, so it's appreciated). – Kimarokko Mar 21 '17 at 7:41
• @Argerami. Based on your answer to point 1, I would complete the argument like this: We start out by considering only the truncated net, as you suggest. Then we let UB be an upper bound of the net. Then for all $\alpha$ we have: $-\|T_{\alpha_0}\| I \leq T_{\alpha_0}\leq T_\alpha\leq UB \leq \|UB\| I$. Hence, if we let c:=max{$\|T_{\alpha_0}\|$, $\|UB\|$}, we get the desired inequality. – Kimarokko Mar 21 '17 at 8:04