Theorem 43.8.11 (Polar Decomposition (Unbounded)).label Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then there exists a unique partial isometry $U \in B(H)$ with $\ker U = \ker T$ such that $T = U|T|$, where $|T|$ is the absolute value of $T$.
Proof. By Definition 43.8.10, $|T|: D(T) \to H$ is a densely defined, self-adjoint, and positive operator with $\normn{T\xi}_{H} = \norm{|T|\xi}_{H}$ for all $\xi \in D(T)$ and $\ker(T) = \ker(|T|)$. Let
then $U$ extends into a well-defined isometry $\ol{|T|(D(T))}\to \ol{T(D(T))}$. Further extend $U$ to $H$ by setting $U|_{|T|(D(T))^\perp}= 0$, then $U: H \to \ol{T(D(T))}\subset H$ is a partial isometry with $T = U|T|$. As $|T|$ is self-adjoint,
Finally, let $V: H \to H$ be a partial isometry with $T = V|T|$ and $\ker(V) = \ker(T)$, then for each $\xi \in D(T)$, $V|T|\xi = T\xi = U|T|\xi$, so $U = V$ on $|T|(D(T))$. By continuity, $U = V$ on $\ol{|T|(D(T))}$. As
and $H = \ker(V) \oplus \ol{|T|(D(T))}$, $U = V$.$\square$
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