Detail in the algebraic proof of the polar decomposition

Most of the time one finds the proof for bounded operators on a Hilbert space, but Sakai in his book "C*-algebras and W*-agebras" gives a purely algbraic one, (Thm 1.12.1 p.27-28, partially at his link):

• Instead of taking directly $\sqrt{a^*a}$ as the positive part he considers the sequence $h(n):= (a^*a+\frac{1}{n}1)^{1/2} $. (I guess the point is that this is invertible because it is strictly positive) Anyway, he also defines $a(n):=a h(n)^{-1}$ where $a$ is the element we want to "polar" decompose.
• Of course $h(n)\rightarrow \sqrt{a^*a}$ uniformly, but we can reformulate this as $\sqrt{a^*a} \in h(n)+ \epsilon S $ for $n \geq n_0$ where $S$ is the unit "sphere" (understand ball) and $\epsilon>0$. Then multiplying by $a(n)$ gives $$\forall\ n\geq n_0,\quad a(n) \sqrt{a^*a}\in a+\epsilon S $$ because ($a(n)h(n)=a$ by construction) and one prooves that $\lVert a(n)\rVert\leq 1$
• Applying a theorem saying that the unit ball in a von Neumann algebra is weakly compact, one obtains that $a(n) \sqrt{a^*a}$ has an accumulation point $b$ ( or very very explicitly $\frac{a(n) \sqrt{a^*a}-a}{\epsilon}\in S$). Because this is true for all $\epsilon$ we actually have the equality $$a = b \sqrt{a^*a}$$ which is the polar decomposition.
• Now we want to check some claims that $b$ (with some adjustment...) is a partial isometry and that $b^*b$ and $bb^*$ are the support of $\sqrt{a^*a}$ and $\sqrt{aa^*}$, denoted $p$ and $q$. (Recall, the support $s$ of a self-adjoint element $c$ is the smallest projection such that $sc=c$ which is equivalent in that case to $cs=c$)

Here is the detail I don't get, he easily obtains $\sqrt{a^*a}(p-pb^*qbp)\sqrt{a^*a}=0 $ and then says

"Since $\lVert b\rVert\leq 1 $, we conclude that $p=pb^*qbp$."????

Of course with the interpretation that $p$ is the projection on the image of $\sqrt{a^*a}$ it works, but I don't get his argument. And also directly from the decomposition $a = b \sqrt{a^*a}$ and the property of a Banach norm we can already say that $\lVert b\rVert= 1$

Solution 1:

The point is that $r=p-pb^*qbp$ is positive: $$ pb^*qbp\leq pb^*bp\leq p\|b\|^2p\leq p^2=p. $$ So you have $$ 0=\sqrt{a^*a}\,r\,\sqrt{a^*a}=(r^{1/2}\sqrt{a^*a})\,r^{1/2}\sqrt{a^*a}. $$ Then $r^{1/2}\sqrt{a^*a}=0$. Multiplying by $r^{1/2}$, we get $r\sqrt{a^*a}=0$. Now $p$ is the range projection of $\sqrt{a^*a}$, so we have $rp=0$. As $rp=r$, we get $r=0$.

Edit: here is a version that seems to avoid using that $b$ is contractive, and also seems to use material available up to that state in Sakai's book.

I will write $x=(a^*a)^{1/2}$, $r=p-pb^*qbp$. We know that $pr=rp=r$, and that $xrx=0$. If you look at definition 1.10.3 in the book, where $a$ there is $x$ here, you get that $xr$ is in $\mathscr L$, so $xr=xr(1-p)$, implying $xrp=0$. Then $xr=0$. Now take adjoint and repeat the process to get $r=0$.