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Hermite completeness

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $L^{2} = L^{2} (R, d y)$ be the complex Lebesgue space with inner product linear in the first argument, and let $g_{0}, P, (g_{n})_{n \in N}$ and $H_{n}$ be as in Hermite recurrence. Define $h : = f g_{0} \in L^{1} (R)$ for $f \in L^{2}$ and write $\hat{h} (ξ) = \int_{R} h (y) e^{- 2 π i ξ y} d y$ for its $L^{1}$ Fourier transform. Then:

(a) (moment criterion) if $\int_{R} f q g_{0} = 0$ for every polynomial $q \in C [y]$ , then $f = 0$ almost everywhere;

(b) the family $(g_{n})_{n \in N}$ is a complete orthonormal family, i.e. a Hilbert basis, of $L^{2}$ (Orthonormal families, complete orthonormal systems and Hilbert bases): the only $f \in L^{2}$ with $⟨ f, g_{n} ⟩ = 0$ for all $n$ is $f = 0$ , and consequently every $f \in L^{2}$ satisfies $f = \sum_{n} ⟨ f, g_{n} ⟩ g_{n}$ in $L^{2}$ with $∥ f ∥^{2} = \sum_{n} ∣ ⟨ f, g_{n} ⟩ ∣^{2}$ (Parseval equivalences for an orthonormal family);

Facts & Assumptions

Given: the orthonormal family $(g_{n})$ with $g_{n} = H_{n} g_{0} / \sqrt{2^{n} n!}$ and $H_{n}$ a real polynomial of degree $n$ with leading coefficient $2^{n}$ (Hermite recurrence(a),(f)); the measures and integrals are Lebesgue measure on $R$ .

[F1]

$(g_{n})$ is orthonormal in $L^{2}$ and $P$ is a linear subspace of $L^{2}$ containing every $g_{n}$ (Hermite orthogonality(a), Hermite recurrence(a)).

[F2]

For a complete orthonormal family the following are equivalent: the closed linear span is $H$ ; the only vector orthogonal to every member is $0$ ; Parseval holds; and the finite-subset net of partial sums converges to each vector (Parseval equivalences for an orthonormal family, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F3]

$∥ f g ∥_{1} \leq ∥ f ∥_{2} ∥ g ∥_{2}$ for $f, g \in L^{2}$ , so $h = f g_{0} \in L^{1}$ with $∥ h ∥_{1} \leq ∥ f ∥_{2}$ because $∥ g_{0} ∥ = 1$ (Cauchy-Schwarz inequality for $L^{2}$ , Hermite recurrence(a)).

[F4]

The $L^{1}$ Fourier transform is $\hat{h} (ξ) = \int h (y) e^{- 2 π i ξ y} d y$ (Fourier transform on complex L1 classes), and $\hat{h} = 0$ implies $h = 0$ a.e. (Uniqueness of the L1 Fourier transform); if $h_{N} \to h$ pointwise a.e. with $∣ h_{N} ∣ \leq G$ a.e. for a fixed $G \in L^{1}$ , then $\int h_{N} \to \int h$ (Dominated convergence).

[F5]

For real $w$ , $e^{w} = \sum_{j \geq 0} w^{j} / j!$ is the sum of an absolutely convergent series (The complex exponential by its power series, The complex exponential series converges absolutely for every complex argument), and for nonnegative terms every partial sum is at most the full sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[F6]

The Lebesgue Gaussian transform at $n = 1$ , $ξ = 0$ and parameter $1 / (2 π)$ gives $\int_{R} e^{- y^{2} / 2} d y = \sqrt{2 π} < \infty$ (Euclidean Gaussian transform with the 2π normalization).

Proof

1.1givenF1F2

[Reduction to the moment criterion] Since $(g_{n})$ is orthonormal by [F1], [F2] reduces completeness to the statement that no nonzero $f \in L^{2}$ is orthogonal to every $g_{n}$ ; the expansion and Parseval identities are part of the same equivalence. The moment criterion (a) is proved first, and then applied.

1.2givenF1F3algebra

[Vanishing moments] Let $f \in L^{2}$ satisfy the moment hypothesis of (a), and put $h : = f g_{0}$ . By [F3], $h \in L^{1}$ and $∥ h ∥_{1} \leq ∥ f ∥_{2}$ ; every polynomial moment is integrable by Cauchy–Schwarz because $q g_{0} \in L^{2}$ by [F1], and the hypothesis gives $\int h q = 0$ , in particular $\int h (y) y^{j} d y = 0$ for every $j \geq 0$ . If instead $f$ is orthogonal to every $g_{n}$ , then real-valuedness of $H_{n}$ and $g_{0}$ gives $0 = ⟨ f, g_{n} ⟩ = (2^{n} n!)^{- 1 / 2} \int f H_{n} g_{0}$ . The polynomials $H_{0}, \dots, H_{m}$ have respective degrees $0, \dots, m$ and nonzero leading coefficients, so successive subtraction of leading terms expresses every polynomial of degree at most $m$ as their linear combination. Thus orthogonality implies the moment hypothesis of (a). It remains to prove $h = 0$ under that hypothesis.

2.1step 1.2F3F4F5F6algebra

[The transform vanishes] Fix $ξ \in R$ , write $w : = 2 π ξ y$ and $S_{N} (y) : = \sum_{j = 0}^{N} (- 2 π i ξ y)^{j} / j!$ . Then $S_{N} (y) \to e^{- 2 π i ξ y}$ for every $y$ by [F5], and $∣ S_{N} (y) ∣ \leq \sum_{j = 0}^{N} ∣ 2 π ξ y ∣^{j} / j! \leq e^{2 π ∣ ξ ∣ ∣ y ∣} = : G (y)$ by [F5]. Moreover $h G \in L^{1}$ : by [F3], $\int ∣ h (y) ∣ G (y) d y \leq ∥ f ∥_{2} ∥ g_{0} G ∥_{2} = ∥ f ∥_{2} (\int G (y)^{2} g_{0} (y)^{2} d y)^{1 / 2}$ , and $G (y)^{2} g_{0} (y)^{2} = π^{- 1 / 2} e^{4 π ∣ ξ ∣ ∣ y ∣ - y^{2}} \leq π^{- 1 / 2} e^{8 π^{2} ξ^{2}} e^{- y^{2} / 2}$ because $4 π ∣ ξ ∣ ∣ y ∣ - y^{2} \leq 8 π^{2} ξ^{2} - ∣ y ∣^{2} / 2$ ; the last integral is finite by [F6]. Hence [F4] applies with $\hat{h} (ξ) = \lim_{N \to \infty} \int h S_{N} = \lim_{N \to \infty} \sum_{j = 0}^{N} \frac{(- 2 π i ξ)^{j}}{j!} \int h (y) y^{j} d y = 0$ , the first equality by dominated convergence, the interchange of the finite sum with the integral by linearity of the integral, and the final equality because all moments of $h$ vanish by step 1.2.

3.1step 1.1step 1.2step 2.1

[Conclusion of (a) and (b)] Since $\hat{h} = 0$ identically, [F4] gives $h = f g_{0} = 0$ a.e.; as $g_{0} (y) = π^{- 1 / 4} e^{- y^{2} / 2} > 0$ for every $y$ , $f = 0$ a.e. This proves (a), and with step 1.1 it proves that $(g_{n})$ is complete. By [F2] applied to the complete orthonormal family $(g_{n})$ , every $f \in L^{2}$ has the stated expansion and Parseval identity, which is (b).

4.1step 3.1F1∎

[Density of $P$ ] By [F2] and step 3.1 the closed linear span of $(g_{n})$ is $L^{2}$ . Since every $g_{n} \in P$ by [F1] and $P$ is a linear subspace, $P$ contains the span of the family; a subspace containing a dense subspace is dense, so $P$ is dense. This is (c).

Hermite completeness

Statement

Facts & Assumptions

Proof

Depends on

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