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Schwartz density in the joint quadrature graph norm

Statement

Assume Countable Choice, let $r \geq 1$ and $ℏ > 0$ , and use the Schrödinger quadratures $Q_{j} f = \sqrt{ℏ} x_{j} f$ on $D (Q_{j}) = {f : x_{j} f \in L^{2}}$ and $P_{j} = F_{2}^{- 1} M_{2 π \sqrt{ℏ} ξ_{j}} F_{2}$ , where $F_{2}$ is the unitary $2 π$ -frequency Fourier transform on $L^{2} (R^{r})$ and the multiplier has its maximal $L^{2}$ domain. Put $D = {f \in L^{2} (R^{r}) : x_{j} f \in L^{2} (R^{r}), \partial_{j} f \in L^{2} (R^{r}) for every j} .$ For $∥ f ∥_{q u a d} : = ∥ f ∥_{2} + \sum_{j = 1}^{r} (∥ x_{j} f ∥_{2} + ∥ \partial_{j} f ∥_{2}),$ the Schwartz space $S (R^{r})$ is dense in $D$ . The weak derivatives in this definition are taken in the distributional sense and agree with the Fourier-defined momentum domains of the quadratures just defined.

Facts & Assumptions

Given: Countable Choice, finite $r \geq 1$ , $ℏ > 0$ , and $f \in D$ .

[A1]

Countable Choice is the hypothesis of the Fourier–Sobolev and compactly-supported smooth density interfaces (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Under Countable Choice the Fourier transform $F_{2}$ is unitary (Plancherel theorem). Schwartz space consists of the smooth functions with every seminorm $p_{α, β}$ finite (Schwartz space and its seminorms). The quadratures in the Statement are defined directly here; no AC-dependent representation uniqueness is used.

[F2]

The Fourier characterization of $W^{1, 2} (R^{r})$ identifies its weak-derivative norm with an equivalent norm requiring $f$ and every $ξ_{j} \hat{f}$ in $L^{2}$ ; hence $f \in D$ lies in $W^{1, 2}$ (Integer-order W^{k,2} and H^k agree with equivalent norms).

[F3]

For $0 < a < b$ a smooth bump exists that equals $1$ on ${\overline{B}}_{a} (0)$ and has support in $B_{b} (0)$ (A smooth bump between concentric Euclidean balls).

[F4]

Multiplication by a smooth function obeys the weak Leibniz rule, so $\partial_{j} (χ f) = χ \partial_{j} f + (\partial_{j} χ) f$ (Weak Leibniz rule with a smooth factor).

[F5]

Countable Choice implies that $C_{c}^{\infty} (R^{r})$ is dense in $W^{1, 2} (R^{r})$ (Compactly supported smooth functions are dense in W^{k,p}(R^n)).

[F6]

Dominated convergence applies to the integrable squared tails of $f$ , $x_{j} f$ , and $\partial_{j} f$ (Dominated convergence).

Proof

1.1A1F1F2F3F4F6givenalgebra

Choose $χ \in C_{c}^{\infty} (R^{r})$ with $0 \leq χ \leq 1$ , $χ = 1$ on ${\overline{B}}_{1} (0)$ , and support in $B_{2} (0)$ by [F3]. For $R \geq 1$ set $χ_{R} (x) = χ (x / R)$ and $f_{R} = χ_{R} f$ . Then $∣ \partial_{j} χ_{R} ∣ \leq C / R$ . By [F4], $f_{R} \in W^{1, 2}$ and $\partial_{j} (f_{R} - f) = (χ_{R} - 1) \partial_{j} f + (\partial_{j} χ_{R}) f .$ The first terms in $f_{R} - f$ , $x_{j} (f_{R} - f)$ , and $(χ_{R} - 1) \partial_{j} f$ tend to zero in $L^{2}$ by dominated convergence [F6], since $χ_{R} \to 1$ pointwise and $∣ χ_{R} - 1 ∣ \leq 1$ . The remaining derivative term has norm at most $C ∥ f ∥_{2} / R$ . Thus $∥ f_{R} - f ∥_{q u a d} \to 0$ .

2.1A1F3F4F5step 1.1givenalgebra∎

Fix $R$ so the preceding error is below $ε / 2$ . The function $f_{R}$ is supported in ${\overline{B}}_{2 R} (0)$ . By [F5], choose $ϕ \in C_{c}^{\infty} (R^{r})$ with $∥ ϕ - f_{R} ∥_{W^{1, 2}}$ as small as needed. Choose $θ \in C_{c}^{\infty}$ equal to $1$ on ${\overline{B}}_{2 R} (0)$ and supported in $B_{2 R + 2} (0)$ by [F3], and put $g = θ ϕ$ . Since $θ f_{R} = f_{R}$ , the weak product rule [F4] gives $g - f_{R} = θ (ϕ - f_{R}), \partial_{j} (g - f_{R}) = θ \partial_{j} (ϕ - f_{R}) + (\partial_{j} θ) (ϕ - f_{R}) .$ The bounded derivatives of $θ$ make the $W^{1, 2}$ norm of $g - f_{R}$ at most a fixed constant times $∥ ϕ - f_{R} ∥_{W^{1, 2}}$ . Both $g$ and $f_{R}$ are supported in ${\overline{B}}_{2 R + 2} (0)$ , so $∥ x_{j} (g - f_{R}) ∥_{2} \leq (2 R + 2) ∥ g - f_{R} ∥_{2} .$ Choose $ϕ$ close enough that $∥ g - f_{R} ∥_{q u a d} < ε / 2$ . Then $g \in C_{c}^{\infty} \subset S$ and $∥ g - f ∥_{q u a d} < ε$ by step 1.1. Since $f$ and $ε > 0$ were arbitrary, $S (R^{r})$ is dense in $D$ .

Schwartz density in the joint quadrature graph norm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources