Question about polar decomposition. Is the book wrong? An introduction to the classification of amenable C*-algebra.
page 140 Lemma 3.5.1 Let $x\in A$ with the polar decomposition $x=u|x|$ in $A''$ and $B=\overline{x^*Ax}$. Then $ub\in B$ for every $b\in B$.
$A''$ refers to the enveloping C*-algebra of $A$, the weak closure of $A$ in $B(H)$ where $A$ is universally represented. But these are not important.
Let $A$ be $B(l^2(\mathbb N))$ and let $x$ be the shift operator such that $x(e_j)=e_{j-1}$. Then $\overline {x^*Ax}=B(0\oplus l^2(\mathbb N^+))$ (regarding $B(0\oplus l^2(\mathbb N^+))$ as a subalgebra of $B(l^2(\mathbb N))$). Since $x$ is a partial isometry itself so $x=x|x|$ and $|x|$ is the projection onto $0\oplus l^2(\mathbb N^+)$.
However, $x=x|x|\not\in \overline{x^*Ax}$ since no element in $\overline{x^*Ax}=\overline{|x|A|x|}$ has range larger than $0\oplus l^2(\mathbb N^+)$.
Am I wrong, or is the book wrong?
 A: It is my impression that Lin wanted his Lemma to say:
Lemma 3.5.1. Let $x\in A$ with the polar decomposition $x=u|x|$ in $A''$.  Also let   $B_1=\overline{x^*Ax}$ and $B_2=\overline{xAx^*}$.
Then $uB_1u^*\subseteq B_2$.
Here is a proof of this result broken into Lemmas, each of which might have some interest in itself.
Lemma 1.  $x = \lim_n x(x^*x)^{1/n}$.
Proof.  Compute $\Vert x  - x(x^*x)^{1/n}\Vert ^2$ using the C*-identity $\Vert y\Vert ^2 = \Vert y^*y\Vert $.
Lemma 2.  For every $\alpha >0$ one has that $u(x^*x)^\alpha = (xx^*)^\alpha u\in A$.
Proof.  Since both sides vanish on the kernel of $x$ (seen in any  given faithful Hilbert space representation), it is enough to
show that they agree on
$$
  \text{Ker}(x)^\perp = \overline{\text{Ran}(x^*)} = \overline{ \text{Ran}(|x|)}.
  $$
We have
$$
  u(x^*x)^\alpha |x| =
  u|x| (x^*x)^\alpha =
  x (x^*x)^\alpha =
  (xx^*)^\alpha  x =
  (xx^*)^\alpha u|x|.
  $$
This proves the identity in the statement, so let us now prove that $u(x^*x)^\alpha \in A$.
Approximate the function $f(t)=t^{2\alpha}$ on the spectrum of $|x|$ by  a
polynomial  $p$ without constant term and hence we may write $p(t) = t q(t)$ for some other polynomial $q$.  Then
$$
  u(x^*x)^\alpha =
  u |x|^{2\alpha} \sim
  u p(|x|) =
  u |x| q(|x|) =
  x q(|x|) \in  A.
  $$
QED.
Lemma 3.  $ux^*\in  \overline{xA}$.
Proof.
$$
  ux^* =
  \lim_n u(x^*x)^{2/n}x^* =
  \lim_n (xx^*)^{1/n}u(x^*x)^{1/n}x^* \in 
  \overline{xA}.
  $$
QED
Therefore  the  result we need,  namely
$$
  ux^*Axu^* \subseteq  \overline{xAx^*}
  $$
follows easily.
A: Yes, the book is wrong. It was supposed to be $ub \in A$ for all $b \in B$.
See also Farah's book Corollary 1.6.13.
