Polar decomposition

Question

Every $x\in B(H)$ has a representation such as $x= u|x|$ (by polar decomposition) where $u$ is a partial isometry. In a paper, the author claims that$$u = {\rm strong} - \lim_{\epsilon\to 0} x(|x|+\epsilon)^{-1},$$ but I do not know how show it. Please help me. Thanks

Answer 1

I'll use $X$, $U$ instead of $x$, $u$ because I want to use vectors, and it just looks wrong otherwise. I'm assuming you have some sort of functional calculus in order to form $|X|=(X^{\star}X)^{1/2}$. And that means you can verify that $|X|(|X|+\epsilon I)^{-1}$ is uniformly bounded in norm by $1$ for $\epsilon > 0$. I'll show you that $$ s-\lim_{\epsilon\downarrow 0}|X|(|X|+\epsilon I)^{-1} = P, $$ where $P$ is the orthogonal projection onto the closure of the range of $|X|$. That's enough to give you what you want because $VP=V$ and $$ X(|X|+\epsilon I)^{-1}=V|X|(|X|+\epsilon I)^{-1} $$ So that's the plan.

Because of the uniform bound on the operator norm of $|X|_{\epsilon}=|X|(|X|+\epsilon I)^{-1}$, it is enough to show that $|X|_{\epsilon}$ converges strongly as $\epsilon \downarrow 0$ on a dense set subspace of $H$ to $P$. The dense subspace of interest here is $$ \mathcal{M}=\mathcal{R}(|X|)\oplus\mathcal{N}(|X|). $$ This is dense because $\mathcal{R}(|X|)^{\perp}=\mathcal{N}(|X|)$. Notice that $|X|_{\epsilon}$ is $0$ on $\mathcal{N}(|X|)$; so the only challenge is the strong limit on $\mathcal{R}(|X|)$. However, if $x=|X|z$ for some $z$, $$ \begin{align} |X|(|X|+\epsilon I)^{-1}x & =x-\epsilon(|X|+\epsilon I)^{-1}x \\ & = x -\epsilon|X|(|X|+\epsilon I)^{-1}z. \end{align} $$ The right side converges to $x$ as $\epsilon \downarrow 0$ because of the uniform operator norm bound on $|X|(|X|+\epsilon I)^{-1}$. So that's enough to show that $$ s-\lim_{\epsilon\downarrow 0}|X|(|X|+\epsilon I)^{-1} = P \\ \implies s-\lim_{\epsilon\downarrow 0}X(|X|+\epsilon I)^{-1} = VP=V. $$

Answer 2

By Borel functional calculus, $\phi: B_\infty(\sigma(X))\to B(H)$, for every $\epsilon>0$ correspondence $z_\epsilon =z(|z|+\epsilon)^{-1} \in B_\infty$, we have $\phi(z_\epsilon) = X(|X|+\epsilon)^{-1}$. Easily we can show that net $\{z_\epsilon\}$ is point-wise Cauchy in $B_\infty(\sigma(X))$, so it's p.w convergent to $u$. Also $$U = \phi(u) = s-\lim \phi(z_\epsilon) = s-\lim N(|N|+\epsilon)^{-1}$$ For $\xi,\eta\in H$, $$\langle U|N|\xi , \eta\rangle =\lim \langle N(|N|+\epsilon)^{-1}|N|\xi,\eta \rangle = \langle N\xi,\eta\rangle$$ Also for $\xi \in \ker N$, and $\eta\in H$, $$\langle U\xi,\eta\rangle = \lim \langle N(|N|+\epsilon)^{-1}\xi,\eta\rangle = \lim\langle (|N|+\epsilon)^{-1}N\xi,\eta\rangle =0$$ Thus $U$ is a partial isometry and $\ker U=\ker N$. Uniqueness of polra decomposition implies that this is desired result.