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Coercive polynomial form completion and injection

Statement

Assume Countable Choice. Use the finite data and notation of Finite-lattice polynomial Hamiltonian data: $N \geq 1$ , $H_{N} = L^{2} (R^{N}, d q; C)$ , $D_{0} = C_{c}^{\infty} (R^{N}; C)$ , and $W = V + C_{0} + 1 \geq 1 + c ∣ q ∣^{2}$ . For $u, v \in D_{0}$ define $a (u, v) : = \sum_{j = 1}^{N} a_{j} \int_{R^{N}} (\partial_{j} u) \overline{\partial_{j} v} d q + \int_{R^{N}} (W + 1) u \overline{v} d q,$ and $∥ u ∥_{Q} : = a (u, u)^{1 / 2}$ . Let $Q$ be the completion of this inner-product space, constructed as equivalence classes of $∥ \cdot ∥_{Q}$ -Cauchy sequences. Then $Q$ is a complex Hilbert space and the map sending a Cauchy class to its $L^{2}$ limit is a continuous injective linear map $Q ↪ H_{N}$ with dense image. Each $u \in Q$ has weak derivatives $\partial_{j} u \in L^{2}$ and a weighted function $\sqrt{W + 1} u \in L^{2}$ , and $∥ u ∥_{Q}^{2} = \sum_{j = 1}^{N} a_{j} ∥ \partial_{j} u ∥_{2}^{2} + ∥ \sqrt{W + 1} u ∥_{2}^{2} .$ Here $g_{j} = \partial_{j} u$ means explicitly that $\int u \overline{\partial_{j} φ} = - \int g_{j} \overline{φ}$ for every $φ \in D_{0}$ .

Facts & Assumptions

Given: Countable Choice, the data of Finite-lattice polynomial Hamiltonian data, and the $L^{2}$ and density results below.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function; only countably many representatives are selected in the completion argument.

[F1]

Finite-lattice polynomial Hamiltonian data: $H_{N}$ is complex $L^{2}$ with first-variable-linear integral pairing, $D_{0} = C_{c}^{\infty}$ , and $W = V + C_{0} + 1 \geq 1 + c ∣ q ∣^{2}$ .

[F2]

$L^{2}$ with the integral pairing is a Hilbert space: under Countable Choice, $H_{N}$ is complete for $∥ \cdot ∥_{2}$ and $⟨ u, v ⟩ = \int u \overline{v} d q$ is its inner product.

[F3]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ in $H_{N}$ .

[F4]

$C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ : under Countable Choice and for $p = 2$ , $D_{0}$ is dense in $H_{N}$ .

Proof

technique · direct

1.1F1F2algebra

Since $W + 1 \geq 2$ and every $a_{j} > 0$ , $a$ is a Hermitian positive-definite sesquilinear form on $D_{0}$ : in particular $a (u, u) = \sum_{j} a_{j} ∥ \partial_{j} u ∥_{2}^{2} + ∥ \sqrt{W + 1} u ∥_{2}^{2} \geq 2 ∥ u ∥_{2}^{2}$ , and $a (u, u) = 0$ forces $u = 0$ in $H_{N}$ . Thus its Cauchy-sequence completion with the stated equivalence relation is a well-defined inner-product space.

2.1A1step 1.1algebra

To see that the completion is complete, let $(X_{n})$ be a Cauchy sequence of its classes. By [A1], choose $u_{n} \in D_{0}$ with $∥ [u_{n}] - X_{n} ∥_{Q} < 2^{- n}$ for each $n$ . The triangle inequality makes $(u_{n})$ $Q$ -Cauchy; its class $X$ is defined by the Cauchy-sequence construction, and $∥ X_{n} - X ∥_{Q} \leq 2^{- n} + ∥ [u_{n}] - X ∥_{Q} \to 0$ . Hence $Q$ is complete and therefore Hilbert.

2.2F1F2step 1.1

For a $Q$ -Cauchy representative sequence $(u_{n})$ , the inequalities $∥ u_{n} - u_{m} ∥_{2} \leq ∥ u_{n} - u_{m} ∥_{Q}$ , $\sqrt{a_{j}} ∥ \partial_{j} u_{n} - \partial_{j} u_{m} ∥_{2} \leq ∥ u_{n} - u_{m} ∥_{Q}$ , and $∥ \sqrt{W + 1} (u_{n} - u_{m}) ∥_{2} \leq ∥ u_{n} - u_{m} ∥_{Q}$ show that $u_{n}$ , each $\partial_{j} u_{n}$ , and $\sqrt{W + 1} u_{n}$ converge in $L^{2}$ by [F2]; write their limits as $u$ , $g_{j}$ , and $h$ . The first limit defines a continuous map $Q \to H_{N}$ , since $∥ u ∥_{2} \leq ∥ [u_{n}] ∥_{Q}$ , and equivalent Cauchy sequences have the same $L^{2}$ limit.

3.1step 2.2F2F3algebra

For every $φ \in D_{0}$ , integration by parts gives $\int u_{n} \overline{\partial_{j} φ} = - \int (\partial_{j} u_{n}) \overline{φ}$ . Passing to the limit in both integrals by $L^{2}$ convergence and Cauchy–Schwarz yields $\int u \overline{\partial_{j} φ} = - \int g_{j} \overline{φ}$ , so $g_{j}$ is the stated weak derivative.

3.2step 2.2F1F2algebra

On each compact ball $B_{R}$ , the continuous polynomial $W + 1$ is bounded. Therefore $\sqrt{W + 1} u_{n} \to \sqrt{W + 1} u$ in $L^{2} (B_{R})$ , while the same sequence converges there to $h$ ; uniqueness of limits gives $h = \sqrt{W + 1} u$ almost everywhere on every integer-radius ball and hence almost everywhere on $R^{N}$ . Taking limits of the component norms in $a (u_{n}, u_{n})$ now gives the displayed formula for $∥ u ∥_{Q}^{2}$ .

4.1step 3.1step 3.2F3F4algebra∎

If $u = 0$ in $L^{2}$ , then step 3.1 says $⟨ g_{j}, φ ⟩ = 0$ for every $φ \in D_{0}$ ; this pairing is continuous by [F3] and $D_{0}$ is dense by [F4], so it vanishes on all of $H_{N}$ , and taking $φ = g_{j}$ gives $g_{j} = 0$ . Step 3.2 then gives $h = 0$ ; the norm formula yields $∥ [u_{n}] ∥_{Q} = 0$ , so the embedding is injective. Finally its image contains $D_{0}$ , which is dense in $H_{N}$ by [F4]; thus the continuous injection has dense image and all claims follow.

Coercive polynomial form completion and injection

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources