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 $\lVert b\rVert= 1$
 A: 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 $r=0$. 
A: More explicit answer: let $c\in\mathfrak{M}$ be an arbitrary element in a von Neumann algebra, and $s$ its right support, $cs=c$. We want to proove $\forall\ d\in\mathfrak{M}, dc=0 \Rightarrow ds=0 $


*

*It seems to be that it is very much related to the proposition 1.10.1 still in Sakai: there exists a unique projection $p$ such that a (ultraweakly?) closed left ideal can be written $I=\mathfrak{M}p $. 
Let's then consider the left ideal $K=\{x\in\mathfrak{M}, xc=0 \} $. By the proposition, there exists a projection such that $K=\mathfrak{M}p$

*$dc=0 \Rightarrow \exists f\in\mathfrak{M}, d=fp\ $ 
then $ds=fps$. 

*One actually has a sufficient condition for it to vanish: in the definition 1.10.3 of left and right support Sakai seems to assume that $p=1-s_l $ where $s_l$ is the left support but in our case since $\sqrt{a^*a}$ is selft adjoint, left and right support coincide, $p=1-s$.


So if there is something to clarify, it is the link between the left/right support and the projections generating the left/right ideal of annihilators: the statement is probably
"$s$ left support of $c$ iff $K=\mathfrak{M}(1-s)$ " 
and analogously for the right support and right annihilators.


*

*We have obviously $K\supseteq\mathfrak{M}(1-s)$ : for any $d$,  $\; d(1-s)c=d(1-s)sc=0$

*At the same time the projection $p$ is by definition such that $\mathfrak{M}pc=0=pc $
Hence $(1-p)c=c$, which is similar to the definition of left support. But the later is smallest that satisfies such a condition, ie. $s \leq (1-p) $ and then $p \leq (1-s) $

*But $p \leq (1-s) $, implies $p(1-s)=p $, cf. this, and then $K=\mathfrak{M}p=\mathfrak{M}p(1-s) \subseteq \mathfrak{M}(1-s)$

