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Conjugate Hilbert space and adjoint of an antilinear operator

Definition

Assume Countable Choice. All inner products are linear in their first argument. If $H$ is a complex Hilbert space, its conjugate Hilbert space $\overline{H}$ consists of symbols $\overline{x}$ for $x \in H$ , with $\overline{x} + \overline{y} = \overline{x + y}$ , $a \overline{x} = \overline{\overline{a} x}$ for $a \in C$ , and $⟨ \overline{x}, \overline{y} ⟩_{\overline{H}} : = \overline{⟨ x, y ⟩_{H}} .$ The canonical map $c_{H} : H \to \overline{H}$ , $c_{H} x = \overline{x}$ , is a conjugate-linear surjective isometry.

For complex Hilbert spaces $H_{1}, H_{2}$ , a linear operator $A$ is a linear map $A : D (A) \to H_{2}$ on a linear domain $D (A) \subseteq H_{1}$ . Its graph is $Γ (A) = {(x, A x) : x \in D (A)} \subseteq H_{1} \oplus H_{2}$ , with the direct-sum Hilbert norm. It is densely defined when $D (A)$ is dense in $H_{1}$ , and closed when $Γ (A)$ is closed.

For a densely defined linear operator $A : D (A) \subseteq H_{1} \to H_{2}$ , define $D (A^{*})$ to consist of those $y \in H_{2}$ for which $x \mapsto ⟨ A x, y ⟩_{H_{2}}$ is bounded on $D (A)$ in the $H_{1}$ norm. The bounded functional extends uniquely to $H_{1}$ ; its Riesz vector $A^{*} y \in H_{1}$ is specified by $⟨ A x, y ⟩_{H_{2}} = ⟨ x, A^{*} y ⟩_{H_{1}} (x \in D (A)) .$ This defines the linear adjoint $A^{*} : D (A^{*}) \subseteq H_{2} \to H_{1}$ .

An operator $S : D (S) \subseteq H \to H$ is antilinear when $S (a x + b y) = \overline{a} S x + \overline{b} S y$ . Its associated linear operator $\tilde{S} : D (\tilde{S}) \subseteq \overline{H} \to H$ is $D (\tilde{S}) = c_{H} (D (S))$ and $\tilde{S} (\overline{x}) = S x$ . For densely defined antilinear $S$ , define $S^{*} : = c_{H}^{- 1} {\tilde{S}}^{*},$ on the domain $D (S^{*}) = D ({\tilde{S}}^{*})$ . Thus $S^{*}$ is antilinear and $⟨ S x, y ⟩_{H} = ⟨ S^{*} y, x ⟩_{H} (x \in D (S), y \in D (S^{*})) .$

Definition references

Hilbert space

Real and complex inner-product spaces and their induced length

Riesz representation for Hilbert spaces

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Remarks

The direct-sum inner product is $⟨ (x, u), (y, v) ⟩ = ⟨ x, y ⟩_{H_{1}} + ⟨ u, v ⟩_{H_{2}}$ . It is complete because each summand is complete. The scalar and inner-product rules above make $\overline{H}$ an inner-product space, and $c_{H}$ preserves norms and is onto; consequently $\overline{H}$ is complete as well.

For fixed $y \in H_{2}$ , the map $x \mapsto ⟨ A x, y ⟩_{H_{2}}$ is linear. Riesz representation on $H_{1}$ therefore gives the unique adjoint vector whenever this map is bounded. Linearity and uniqueness show that $D (A^{*})$ is a linear subspace and $A^{*}$ is linear. For the antilinear convention, write ${\tilde{S}}^{*} y = \overline{z}$ . Then $⟨ S x, y ⟩_{H} = ⟨ c_{H} x, {\tilde{S}}^{*} y ⟩_{\overline{H}} = \overline{⟨ x, z ⟩_{H}} = ⟨ z, x ⟩_{H},$ so the displayed formula gives $S^{*} y = z$ . The definition makes no claim that $S$ is closable or that either adjoint domain is dense.

Facts & Assumptions

Given: Countable Choice, complex Hilbert spaces, a densely defined linear operator $A$ , and a densely defined antilinear operator $S$ .

[F1]

The inner product is linear in its first argument and conjugate-linear in its second; conjugating both scalar coefficients defines the complex conjugate vector space (Real and complex inner-product spaces and their induced length).

[F2]

Each bounded linear functional on a Hilbert space has a unique Riesz representing vector (Riesz representation for Hilbert spaces).

[F3]

Countable Choice is the explicit standing choice assumption (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: [F1]–[F3], the spaces and operators in the Definition.

1.1F1algebra

The scalar rules for $\overline{H}$ follow from conjugation in the underlying scalar multiplication, and $⟨ c_{H} x, c_{H} y ⟩_{\overline{H}} = \overline{⟨ x, y ⟩_{H}}$ . Thus $c_{H}$ preserves norms and is onto. A Cauchy sequence in $\overline{H}$ pulls back under $c_{H}$ to a Cauchy sequence in $H$ , whose limit maps forward; so $\overline{H}$ is complete.

1.2F1F2F3given

For fixed $y \in H_{2}$ , the map $x \mapsto ⟨ A x, y ⟩$ is linear. If it is bounded on the dense domain, continuity extends it uniquely to $H_{1}$ , and [F2] gives a unique $A^{*} y$ with the displayed adjoint identity. Uniqueness and linearity of the form in $y$ show that $D (A^{*})$ is a linear subspace and $A^{*}$ is linear there.

2.1F1F2step 1.1algebra∎

The domain $c_{H} (D (S))$ is a complex linear subspace because $D (S)$ is linear and $c_{H}$ is conjugate-linear. Additivity of $S$ makes $\tilde{S}$ additive, and for $a \in C$ , $\tilde{S} (a \overline{x}) = \tilde{S} (\overline{\overline{a} x}) = S (\overline{a} x) = a S x$ , so $\tilde{S}$ is linear. Since $c_{H}^{- 1}$ is conjugate-linear and ${\tilde{S}}^{*}$ is linear, $S^{*}$ is antilinear on its declared domain. If $y \in D ({\tilde{S}}^{*})$ and ${\tilde{S}}^{*} y = c_{H} z$ , then $⟨ S x, y ⟩_{H} = ⟨ c_{H} x, {\tilde{S}}^{*} y ⟩_{\overline{H}} = \overline{⟨ x, z ⟩_{H}} = ⟨ z, x ⟩_{H} = ⟨ c_{H}^{- 1} {\tilde{S}}^{*} y, x ⟩_{H} .$ This proves the stated antilinear-adjoint identity with its exact domain.

Conjugate Hilbert space and adjoint of an antilinear operator

Definition

Definition references

Remarks

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources