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Polar decomposition of a closed densely defined operator

Statement

Here a partial isometry, including when either Hilbert space is zero, is a bounded linear map $U$ that is isometric on $(\ker U)^{⊥}$ and vanishes on $\ker U$ . The initial space is $(\ker U)^{⊥}$ and the final space is $ran U$ . On nonzero Hilbert spaces this is the earlier definition.

Assume the Axiom of Choice. Let $H_{1}, H_{2}$ be complex Hilbert spaces and let $A : D (A) \subseteq H_{1} \to H_{2}$ be a linear operator with dense domain and closed graph. Let $A^{*}$ be the unbounded adjoint defined earlier. For a densely defined linear operator $R : D (R) \subseteq K \to K$ on a complex Hilbert space $K$ , self-adjoint means $D (R) = D (R^{*})$ and $R x = R^{*} x$ on this domain; positive means self-adjoint and $⟨ R x, x ⟩ \in [0, + \infty)$ for every $x \in D (R)$ . A unitary is a surjective linear isometry. Then $T : = A^{*} A$ , on $D (T) = {x \in D (A) : A x \in D (A^{*})}$ , is positive and self-adjoint. Its positive square root $P : = T^{1 / 2}$ has domain $D (P) = D (A)$ and satisfies $∥ P x ∥ = ∥ A x ∥$ for every $x \in D (A)$ . There is a unique partial isometry $U : H_{1} \to H_{2}$ such that $A = U P, \ker U = \ker A .$ Its initial space is $\overline{ran P} = (\ker A)^{⊥}$ and its final space is $\overline{ran A}$ . If $A$ is injective and has dense range, then $U$ is unitary.

Statement references

Conjugate Hilbert space and adjoint of an antilinear operator

The Axiom of Choice

Spectral theorem for unbounded self-adjoint operators (PVM form)

Isometry coisometry and partial isometry

Facts & Assumptions

Given: Choice, complex Hilbert spaces $H_{1}, H_{2}$ , and a densely defined linear $A : D (A) \subseteq H_{1} \to H_{2}$ whose graph is closed.

[F1]

The graph is a closed linear subspace of the Hilbert direct sum $H_{1} \oplus H_{2}$ . Every closed subspace has an orthogonal projection under Countable Choice (Conjugate Hilbert space and adjoint of an antilinear operator, Orthogonal decomposition by a closed subspace, The Hilbert orthogonal projection onto a closed subspace).

[F2]

For densely defined $A$ , $y \in D (A^{*})$ exactly when $x \mapsto ⟨ A x, y ⟩$ is bounded in the $H_{1}$ norm; then $⟨ A x, y ⟩ = ⟨ x, A^{*} y ⟩$ (Conjugate Hilbert space and adjoint of an antilinear operator).

[F3]

The unbounded spectral theorem for a positive self-adjoint $T$ supplies its spectral measure, the domain calculus for $T^{1 / 2}$ , and bounded spectral cutoffs (Spectral theorem for unbounded self-adjoint operators (PVM form)).

[F4]

For a positive spectral measure $μ$ , nonnegative measurable functions $g_{n} ↑ g$ satisfy $\int g_{n} d μ ↑ \int g d μ$ (Monotone convergence for the integral).

[F5]

For a bounded linear operator $R$ between Hilbert spaces, $\overline{ran R} = (\ker R^{*})^{⊥}$ (Kernel–range orthogonality for Hilbert adjoints). This is used for the bounded resolvent in step 2.1. The corresponding range-kernel identity for the unbounded self-adjoint operator $P$ is proved directly in step 5.1.

[F6]

A partial isometry is an isometry on its initial space and zero on its orthogonal complement; its final space is the closure of its range (Isometry coisometry and partial isometry).

[A1]

Choice implies Countable Choice, the hypothesis for the graph projection and Hilbert orthogonal decompositions (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: [A1], $H_{1}, H_{2}$ , and $A$ as in the Statement.

1.1A1F1F2construct

If $H_{1} = {0}$ , density forces $D (A) = {0}$ and $A = 0$ . Put $T = P = U = 0$ with their zero-space domains. Then $T$ is positive and self-adjoint, $P = T^{1 / 2}$ , $\ker U = \ker A = {0}$ , both asserted range closures are ${0}$ , and $U$ is a partial isometry by the zero-space convention in the Statement. If $A$ also has dense range, then $H_{2} = {0}$ and $U$ is unitary. Thus the conclusion holds; assume henceforth $H_{1} \neq {0}$ . Project $(y, 0)$ orthogonally onto the closed graph $Γ (A)$ to obtain $(x, A x)$ . Orthogonality against every $(z, A z)$ with $z \in D (A)$ gives $⟨ A z, A x ⟩ = ⟨ z, y - x ⟩$ . By [F2], $A x \in D (A^{*})$ and $A^{*} A x = y - x$ , so $(I + T) x = y$ . Thus $I + T$ is onto; it is injective because $⟨ (I + T) x, x ⟩ = ∥ x ∥^{2} + ∥ A x ∥^{2}$ , which vanishes only for $x = 0$ .

2.1A1F2F5step 1.1algebra

Let $R = (I + T)^{- 1}$ . If $x = R y$ , the preceding identity and Cauchy–Schwarz give $∥ R y ∥ \leq ∥ y ∥$ , so $R$ is bounded. For $x = R y$ and $w = R z$ , the graph-projection identity applied with test vectors $w$ and $x$ gives $⟨ y - x, w ⟩ = ⟨ A x, A w ⟩$ and $⟨ z - w, x ⟩ = ⟨ A w, A x ⟩$ . Conjugating the second equality and comparing shows $⟨ R y, z ⟩ = ⟨ y, R z ⟩$ ; hence $R$ is self-adjoint. It is injective and its range is exactly $D (T)$ , so [F5] gives $\overline{D (T)} = \overline{ran R} = H_{1}$ . The operator $T$ is symmetric, because for $x, y \in D (T)$ , $⟨ T x, y ⟩ = ⟨ A x, A y ⟩ = ⟨ x, T y ⟩$ , and positive, because $⟨ T x, x ⟩ = ∥ A x ∥^{2}$ . Finally let $y \in D (T^{*})$ and choose $x \in D (T)$ with $(I + T) x = (I + T^{*}) y$ , possible by step 1.1. Then $(I + T^{*}) (y - x) = 0$ . If $v$ lies in this kernel, then $⟨ (I + T) z, v ⟩ = 0$ for every $z \in D (T)$ ; surjectivity of $I + T$ forces $v = 0$ . Thus $y = x \in D (T)$ and $T^{*} y = T y$ . Therefore $T$ is self-adjoint.

3.1F2F3F4step 2.1algebra

Let $E$ be the spectral measure of the positive self-adjoint $T$ . It is supported on $[0, + \infty)$ : if $E ([- N, - 1 / N]) \neq 0$ for some positive integer $N$ , a nonzero vector in its range lies in $D (T)$ and has $⟨ T z, z ⟩ \leq - N^{- 1} ∥ z ∥^{2} < 0$ , contradicting positivity; these intervals cover $(- \infty, 0)$ . Set $P = T^{1 / 2}$ . For $x \in D (T)$ , the spectral calculus and the adjoint identity give $∥ P x ∥^{2} = ⟨ T x, x ⟩ = ∥ A x ∥^{2}$ . Now take any $x \in D (A)$ and, for each integer $n \geq 1$ , define $f_{n} (λ) = n / (n + λ)$ on $[0, + \infty)$ and $x_{n} = f_{n} (T) x = n (n + T)^{- 1} x$ . Since $f_{n}$ and $λ f_{n}$ are bounded, $x_{n} \in D (T)$ and $T x_{n} = n (x - x_{n})$ . For $z \in D (A)$ , write $d = z - x_{n}$ . The adjoint identity gives $⟨ A d, A x_{n} ⟩ = ⟨ d, T x_{n} ⟩ = - n ⟨ d, x_{n} - x ⟩$ . Expanding squared norms therefore yields $∥ A z ∥^{2} + n ∥ z - x ∥^{2} = ∥ A x_{n} ∥^{2} + n ∥ x_{n} - x ∥^{2} + ∥ A d ∥^{2} + n ∥ d ∥^{2} .$ Taking $z = x$ shows $∥ A x_{n} ∥^{2} \leq ∥ A x ∥^{2}$ . If $μ_{x} (B) = ⟨ E (B) x, x ⟩$ , then $∥ P x_{n} ∥^{2} = ∥ A x_{n} ∥^{2} = \int_{[0, \infty)} λ f_{n} (λ)^{2} d μ_{x} (λ)$ . The integrands increase pointwise to $λ$ , so [F4] gives $\int λ d μ_{x} \leq ∥ A x ∥^{2}$ . By the domain formula for $P = T^{1 / 2}$ , $x \in D (P)$ and $∥ P x ∥ \leq ∥ A x ∥$ .

4.1F1F3step 3.1

Conversely, if $x \in D (P)$ , let $x_{n} = E ([0, n]) x$ . The spectral theorem gives $x_{n} \in D (T)$ and $x_{n} \to x$ ; also $P x_{n} \to P x$ , since the tails of the finite measure $μ_{x}$ and of the integrable function $λ$ with respect to $μ_{x}$ tend to zero. Step 3.1 applied to $x_{n} - x_{m} \in D (T)$ shows that $A x_{n}$ is Cauchy and $∥ A x_{n} - A x_{m} ∥ = ∥ P (x_{n} - x_{m}) ∥$ . Closedness of $A$ then gives $x \in D (A)$ and $A x = \lim_{n} A x_{n}$ ; taking norms yields $∥ A x ∥ = ∥ P x ∥$ . Together with step 3.1 this proves $D (P) = D (A)$ and the asserted norm identity.

5.1F1F2F5F6step 4.1algebra

First, $\ker P$ is closed because $P$ is closed. For $y \in H_{1}$ , $y ⊥ ran P$ exactly when $⟨ P x, y ⟩ = 0$ for every $x \in D (P)$ . The adjoint definition then gives $y \in D (P^{*})$ and $P^{*} y = 0$ ; since $P = P^{*}$ , this is equivalent to $y \in \ker P$ . Conversely, if $y \in \ker P$ , self-adjointness gives $⟨ P x, y ⟩ = ⟨ x, P y ⟩ = 0$ for all $x \in D (P)$ . Hence $(ran P)^{⊥} = \ker P$ . For $C = \overline{ran P}$ , we have $C^{⊥} = \ker P$ ; the orthogonal decomposition theorem applied to $C$ and $\ker P$ therefore gives $C = (\ker P)^{⊥}$ . The norm identity from step 4.1 also gives $\ker P = \ker A$ . On $ran P$ , define $V (P x) = A x$ . If $P x = P y$ , then $∥ A (x - y) ∥ = ∥ P (x - y) ∥ = 0$ , so $V$ is well defined; the norm identity makes it an isometry. Extend it continuously to $C$ and set it to zero on $\ker P$ . For $x = c + k$ with $c \in C$ and $k \in \ker P$ , $∥ U x ∥ = ∥ c ∥ \leq ∥ x ∥$ ; thus the resulting linear map $U$ is bounded. It has kernel $\ker P$ , is isometric on $C = (\ker A)^{⊥}$ , and $A = U P$ . Its range is $\overline{ran A}$ because the extension of $V$ has range the closure of $V (ran P) = ran A$ . If $A$ is injective with dense range, its initial and final spaces are all of $H_{1}$ and $H_{2}$ , respectively, so $U$ is unitary. Any partial isometry with the stated factorization and kernel agrees with $U$ on $ran P$ and vanishes on $\ker P$ ; continuity gives uniqueness.

6.1step 1.1step 2.1step 3.1step 4.1step 5.1∎

Steps 1.1–5.1 prove positivity and self-adjointness of $A^{*} A$ , the domain and norm identity for its positive square root, the polar factorization, and the asserted initial/final spaces and uniqueness.

Polar decomposition of a closed densely defined operator

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources