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Hermite functions form a graph core for position multiplication

Statement

Assume the Axiom of Countable Choice The Axiom of Countable Choice ( ${A C}_{ω}$ ). Let $H = L^{2} (R, d x; C)$ with its first-variable-linear integral pairing, and let $Q f (x) = x f (x)$ on $D (Q) = {f \in H : x f \in H}$ . Let $(g_{n})_{n \geq 0}$ be the Hermite functions of Hermite recurrence. Then $Q$ is self-adjoint on $D (Q)$ , and the finite linear span of the $g_{n}$ is a core for $Q$ : it is dense in $D (Q)$ for the graph norm $∥ f ∥_{Q} = (∥ f ∥_{2}^{2} + ∥ Q f ∥_{2}^{2})^{1 / 2}$ .

Facts & Assumptions

Given: Countable Choice; the complex Hilbert space $L^{2} (R)$ ; the real multiplication map $Q f = x f$ on its maximal domain; and the Hermite functions from Hermite recurrence.

[A1]

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

[F1]

$L^{2} (R)$ is complete for its integral pairing and satisfies Cauchy–Schwarz. $L^{2}$ with the integral pairing is a Hilbert space

[F2]

The oscillator functions have the form $g_{n} = H_{n} g_{0} / \sqrt{2^{n} n!}$ , where $g_{0} (x) = π^{- 1 / 4} e^{- x^{2} / 2}$ , each $H_{n}$ has degree $n$ with nonzero leading coefficient, and $\sqrt{2} x g_{n} = \sqrt{n} g_{n - 1} + \sqrt{n + 1} g_{n + 1}$ . Hermite recurrence

[F3]

The $L^{1}$ Fourier transform uses $\hat{h} (ξ) = \int h (x) e^{- 2 π i ξ x} d x$ ; under Countable Choice it is injective on $L^{1} (R)$ . Fourier transform on complex L1 classes Uniqueness of the L1 Fourier transform

[F4]

The complex exponential is given by its everywhere convergent power series, and dominated convergence applies to integrable majorants. The complex exponential by its power series The complex exponential series converges absolutely for every complex argument Dominated convergence

[F5]

Polynomial multiples of a Gaussian and shifted Gaussian moments are integrable. Euclidean Gaussian transform with the 2π normalization

[F6]

In a Hilbert space, a subspace with zero orthogonal complement is dense. Orthogonality and the orthogonal complement The double orthogonal complement of a subspace is its closure

Proof

1.1givenF1F4algebra

For $f \in H$ , put $f_{m} = f 1_{[- m, m]}$ . Then $f_{m} \in D (Q)$ because $∣ x ∣ \leq m$ on its support, and $f_{m} \to f$ in $L^{2}$ by dominated convergence applied to $∣ f_{m} - f ∣^{2} \leq ∣ f ∣^{2}$ ; hence $D (Q)$ is dense. For $f, g \in D (Q)$ , $x f \overline{g}$ is integrable by Cauchy–Schwarz, and real-valuedness of $x$ gives $⟨ Q f, g ⟩ = ⟨ f, Q g ⟩$ .

1.2A1F1F2F5construct

Equip $G = {f : \int_{R} (1 + x^{2}) ∣ f (x) ∣^{2} d x < \infty}$ with the explicit weighted inner product $⟨ f, g ⟩_{G} : = \int_{R} (1 + x^{2}) f (x) \overline{g (x)} d x$ . Multiplication by $\sqrt{1 + x^{2}}$ is an onto isometry from $G$ to $L^{2} (R)$ , with inverse multiplication by $(1 + x^{2})^{- 1 / 2}$ , so $G$ is Hilbert. Its norm equals $∥ f ∥_{Q}$ on $D (Q)$ . Suppose $f \in G$ is orthogonal to every $g_{n}$ in this weighted inner product. Since the $H_{n}$ have successive degrees and nonzero leading coefficients [F2], their finite span is all polynomials. Therefore $h (x) = (1 + x^{2}) f (x) g_{0} (x)$ lies in $L^{1}$ by Cauchy–Schwarz and $\int h (x) x^{j} d x = 0$ for every $j \geq 0$ .

2.1step 1.1F1algebra

Let $g \in D (Q^{*})$ and write $k = Q^{*} g$ . For each $m$ , the vector $f_{m} = (x g - k) 1_{[- m, m]}$ belongs to $D (Q)$ since $x$ is bounded there and $g, k \in L^{2}$ . The adjoint identity for $f_{m}$ yields $0 = ⟨ Q f_{m}, g ⟩ - ⟨ f_{m}, k ⟩ = \int_{- m}^{m} ∣ x g - k ∣^{2} d x$ . Thus $x g = k$ almost everywhere on every $[- m, m]$ , hence on $R$ ; since $k \in L^{2}$ , $g \in D (Q)$ and $Q^{*} g = Q g$ . Conversely, symmetry in step 1.1 gives $D (Q) \subseteq D (Q^{*})$ and $Q^{*} f = Q f$ . Therefore $Q = Q^{*}$ .

2.2step 1.2F1F3F4F5algebra

Fix $ξ \in R$ and let $S_{N} (x) = \sum_{j = 0}^{N} (- 2 π i ξ x)^{j} / j!$ . By [F4], $S_{N} (x) \to e^{- 2 π i ξ x}$ and $∣ S_{N} (x) ∣ \leq e^{2 π ∣ ξ ∣ ∣ x ∣}$ . The majorant $∣ h (x) ∣ e^{2 π ∣ ξ ∣ ∣ x ∣}$ is integrable: Cauchy–Schwarz bounds its integral by $∥ f ∥_{G} (\int (1 + x^{2}) g_{0} (x)^{2} e^{4 π ∣ ξ ∣ ∣ x ∣} d x)^{1 / 2}$ , and the latter integral is finite because completing the square gives $- x^{2} + 4 π ∣ ξ ∣ ∣ x ∣ = - (∣ x ∣ - 2 π ∣ ξ ∣)^{2} + (2 π ∣ ξ ∣)^{2}$ , so [F5] applies. Dominated convergence and the vanishing moments from step 1.2 imply $\hat{h} (ξ) = \lim_{N} \int h S_{N} = 0$ . Since $ξ$ was arbitrary, $\hat{h}$ vanishes identically.

3.1step 1.2step 2.2F2F3F6∎

Fourier uniqueness [F3] gives $h = 0$ almost everywhere. Since $1 + x^{2} > 0$ and $g_{0} (x) > 0$ for every $x$ , this implies $f = 0$ in $G$ . Thus the orthogonal complement of the Hermite span in $G$ is zero; [F6] makes that span dense in $G$ . Because the $G$ norm is the graph norm of $Q$ , the Hermite span is a graph core, proving the statement.

Hermite functions form a graph core for position multiplication

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources