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Positive injective compact polynomial form resolvent

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and use $Q$ , $H_{N}$ , $a$ , and $W$ from Coercive polynomial form completion and injection. For each $f \in H_{N}$ , let $R f$ be the unique vector of $Q$ satisfying $a (v, R f) = ⟨ v, f ⟩_{H_{N}} (v \in Q),$ and view $R$ as an operator $H_{N} \to H_{N}$ by the inclusion of Cube mean estimate and compact form inclusion. Then $R$ is bounded, compact, self-adjoint, positive, injective, and has dense range; moreover $∥ R ∥ \leq \frac{1}{2}$ . This is the variational inverse of the shifted polynomial form. It establishes compactness and dense range but does not itself assert a differential-operator realization.

Facts & Assumptions

Given: Countable Choice, the coercive form Hilbert space $Q$ , the complex Hilbert space $H_{N}$ , and the compact inclusion $J : Q ↪ H_{N}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function.

[F1]

Coercive polynomial form completion and injection: $a$ is a Hilbert inner product on $Q$ , $∥ v ∥_{2} \leq ∥ v ∥_{Q}$ , and $a (v, v) \geq 2 ∥ v ∥_{2}^{2}$ .

[F2]

Cube mean estimate and compact form inclusion: $J : Q \to H_{N}$ is a compact linear inclusion.

[F3]

Riesz representation for Hilbert spaces: under Countable Choice, every bounded linear functional on a real or complex Hilbert space is uniquely represented by pairing in the first variable; the representing vector has the same norm as the functional.

[F4]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ in $H_{N}$ and in $Q$ .

[F5]

A bounded linear operator between normed spaces: a bounded linear operator is a linear map $T$ with $∥ T x ∥ \leq C ∥ x ∥$ for some $C$ .

[F6]

Compact linear operator: a linear operator is compact when every bounded set has image with compact closure.

[F7]

Self-adjoint, positive, unitary and normal operators: for a bounded operator on a Hilbert space, self-adjoint means $T^{*} = T$ and positive means $⟨ T x, x ⟩ \geq 0$ for all $x$ .

[F8]

The double orthogonal complement of a subspace is its closure: for a linear subspace $M$ of a Hilbert space, $M^{⊥ ⊥} = \overline{M}$ .

Proof

technique · direct

1.1A1F1F3F4given

For fixed $f \in H_{N}$ , the functional $v \mapsto ⟨ v, f ⟩_{H_{N}}$ is linear and satisfies $∣ ⟨ v, f ⟩ ∣ \leq ∥ v ∥_{2} ∥ f ∥_{2} \leq ∥ v ∥_{Q} ∥ f ∥_{2}$ by [F1] and Cauchy–Schwarz. Thus [F3] applied to the Hilbert space $(Q, a)$ gives a unique $R f \in Q$ with the stated variational identity. For $α, β \in C$ , conjugate-linearity in the second argument gives $a (v, α R f + β R g) = ⟨ v, α f + β g ⟩$ , so uniqueness makes $R$ linear.

2.1step 1.1F1F4F5algebra

Taking $v = R f$ in the identity gives $∥ R f ∥_{Q}^{2} = a (R f, R f) = ⟨ R f, f ⟩$ . Hence $∥ R f ∥_{Q}^{2} \leq ∥ R f ∥_{2} ∥ f ∥_{2} \leq ∥ R f ∥_{Q} ∥ f ∥_{2}$ , so $∥ R f ∥_{Q} \leq ∥ f ∥_{2}$ (with the zero case immediate). Consequently $R : H_{N} \to H_{N}$ is bounded by [F5].

2.2step 1.1F1F4algebra

The same identity and $a (R f, R f) \geq 2 ∥ R f ∥_{2}^{2}$ give $2 ∥ R f ∥_{2}^{2} \leq ∥ R f ∥_{2} ∥ f ∥_{2}$ ; division when $R f \neq 0$ yields $∥ R ∥ \leq \frac{1}{2}$ . If $R f = 0$ , the variational identity says $⟨ v, f ⟩ = 0$ for every $v \in Q$ ; since $Q$ is dense in $H_{N}$ by [F1], continuity of the inner product gives $⟨ f, f ⟩ = 0$ , hence $f = 0$ . Thus $R$ is injective.

2.3step 1.1F1F7algebra

For $f, g \in H_{N}$ , the variational identity and conjugate symmetry of $a$ yield $⟨ R f, g ⟩ = a (R f, R g) = ⟨ f, R g ⟩$ , so $R$ is self-adjoint by [F7]. Also $⟨ R f, f ⟩ = a (R f, R f) \geq 0$ , so it is positive.

3.1step 2.1F2F6

The map $f \mapsto R f$ is bounded from $H_{N}$ into $Q$ by step 2.1, and the inclusion $J : Q \to H_{N}$ is compact by [F2]. Therefore for every bounded $B \subseteq H_{N}$ , $R (B) = J (R (B))$ has compact closure, which is exactly compactness by [F6].

4.1step 2.2step 2.3F8∎

If $z ⊥ ran R$ , then $⟨ R f, z ⟩ = 0$ for every $f \in H_{N}$ . Self-adjointness from step 2.3 gives $⟨ f, R z ⟩ = 0$ for every $f$ , so $R z = 0$ and step 2.2 gives $z = 0$ . Thus $(ran R)^{⊥} = {0}$ , and [F8] gives $\overline{ran R} = (ran R)^{⊥ ⊥} = {0}^{⊥} = H_{N}$ . Therefore $R$ has dense range.

Positive injective compact polynomial form resolvent

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources