# Two questions from Dixmier's book on Von Neumann algebras

It seems something is going wrong with the preview I linked in some of my previous questions, so I will just type out the question. I am having trouble with Dixmier's proof of Corollary 5 on p. 46. That one states "Let $A$ be a Von Neumann Algebra, and $m$ a two-sided ideal of A, and $\bar m$ its weak closure. For each $T \in (\bar m)^+$ there exists an increasing [net] $F \subset m^+$ such that $T$ is the supremum of $F$." He then proves it as follows:

"Let $(T_i)_{i \in I}$ be a maximal family of non-zero operators of $m^+$ such that $\sum_{i \in J} T_i \leq T$ for every finite subset $J$ of $I$. The operators $\sum_{i \in J} T_i \leq T$ form an increasing [net in $m^+$] whose supremum $S$ is an element of $(\bar m )^+$ majorized by $T$. Let $R=T-S \in (\bar m)^+$. As $A \in m$ converges weakly to the greatest projection $E$ of $\bar m$, $R^{1/2}AR^{1/2}$ converges weakly to $R^{1/2}ER^{1/2}=R$; hence if $R \neq 0$, we have $R^{1/2}AR^{1/2}\neq 0$ for some $A \in m$, hence for some $A \in m^+$, such that $0 \leq A \leq I$. But we then have $R^{1/2}AR^{1/2}\leq R$, $R^{1/2}AR^{1/2} \in m^+$ and this contradicts the maximality of the family $(T_i)_{i \in I}$. Hence $R=0$."

Everything here is fine in my book until he speaks at the very end as if the $R^{1/2}AR^{1/2}$ he furnishes is not redundant with any of the $T_i$. That is where I don't understand why this proof works.

Another question I have is on the first page of Chapter 4. Let $A$ be a *-algebra of operators on $H$. He says that all positive linear functionals on $A$ $\phi$ (not necessarily norm continuous) turn out to be norm continuous, with norm $\phi(I)$ He argues for this using the following string of inequalities. I disagree with the second inequality:

$|\phi(T)|^2 \leq \phi(I)\phi(T^*T) \leq \phi(I)^2 ||T^*T||=\phi(I)^2||T||^2$.

Can you explain why the second inequality holds?

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The second inequality follows from the fact that $\phi(T^{\ast} T)$ is a spectral value of $T^{\ast} T$ and from the fact that the spectral radius is at most the norm. – Qiaochu Yuan Jul 22 '12 at 1:33
The $\phi(I)$ is squared in the 3rd expression. I don't see why $\phi(T^*T)$ must belong to the spectrum of $T^*T$ – Jeff Jul 22 '12 at 1:39
Ah, sorry, I made some extra assumptions about $\phi$. – Qiaochu Yuan Jul 22 '12 at 2:10

As to your second question, it is true in any C*-algebra, containing a positive element $A$ (i.e., $A=S^*S$ for some element $S$), that $A\leq \|A\|I$. Indeed, the identity function $z\mapsto z$ is smaller than the constant function $z\mapsto \|A\|$ on the spectrum of $A$. Thus the continuous functional calculus implies that $A\leq \|A\|I$.
Thus, if $\phi$ is a positive linear functional, it follows from the above inequality that $\phi(A)\leq \phi(I)\|A\|$.