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Sampling at a lattice produces periodisation of the spectrum over the dual lattice

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $f \in S (R^{n})$ and let $Λ$ be a full-rank lattice with covolume $c$ and dual $Λ^{*}$ . The product $f \cdot {comb}_{Λ} \in S^{'} (R^{n})$ equals the sampled distribution $\sum_{λ \in Λ} f (λ) δ_{λ}$ , and $F (f \cdot {comb}_{Λ}) = \frac{1}{c} \sum_{λ^{*} \in Λ^{*}} \hat{f} (\cdot - λ^{*}) = \frac{1}{c} ({comb}_{Λ^{*}} * \hat{f}),$ the periodisation of the spectrum over the dual lattice with scale $c^{- 1}$ . In particular for $Λ = h Z^{n}$ ( $h > 0$ ) sampling at spacing $h$ gives $F (\sum_{k \in Z^{n}} f (h k) δ_{h k}) = h^{- n} \sum_{m \in Z^{n}} \hat{f} (\cdot - m / h) .$

Facts & Assumptions

Given: Countable Choice, $f \in S (R^{n})$ , the full-rank lattice $Λ = A Z^{n}$ with covolume $c$ and dual $Λ^{*}$ (Full-rank lattices, covolume, and the dual lattice), the Dirac combs and deltas of Dirac comb and Dirac delta and its derivatives, and the distributional convolution $(u * φ) (x) = ⟨ u_{y}, φ (x - y) ⟩$ of Convolution of a tempered distribution with a schwartz function.

[F1]

${comb}_{Λ} \in S^{'} (R^{n})$ with $⟨ {comb}_{Λ}, ψ ⟩ = \sum_{λ} ψ (λ)$ absolutely convergent for Schwartz $ψ$ , and $F {comb}_{Λ} = c^{- 1} {comb}_{Λ^{*}}$ (The Dirac comb of a full-rank lattice transforms to the dual comb).

[F2]

Multiplication by a Schwartz function is transposition, $⟨ f \cdot u, ψ ⟩ = ⟨ u, f ψ ⟩$ , preserves $S^{'}$ , and $f ψ \in S$ for $ψ \in S$ (Smooth polynomially bounded multipliers on schwartz space).

[F3]

Product-to-convolution: for $u \in S^{'}$ and $φ \in S$ , $F (φ u) = F φ * F u$ under the distribution-first convention (Fourier transform converts allowed tempered convolutions to products); the transform of a tempered distribution is defined by $⟨ F u, ψ ⟩ = ⟨ u, F ψ ⟩$ (Fourier transform of a tempered distribution).

[F4]

$\hat{f} \in S (R^{n})$ (Fourier transform acts continuously on Schwartz space), and for the $L^{1}$ transform $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ the two notions agree on Schwartz functions (Fourier transform on complex L1 classes).

[F5]

$covol (h Z^{n}) = h^{n}$ and $(h Z^{n})^{*} = h^{- 1} Z^{n}$ : the diagonal matrix $h I$ has determinant $h^{n}$ , and $(h I)^{- T} Z^{n} = h^{- 1} Z^{n}$ (Full-rank lattices, covolume, and the dual lattice).

Proof

technique · direct

1.1F1F2givenalgebra

For $ψ \in S$ , [F2] gives $⟨ f \cdot {comb}_{Λ}, ψ ⟩ = ⟨ {comb}_{Λ}, f ψ ⟩ = \sum_{λ \in Λ} f (λ) ψ (λ)$ by [F1], since $f ψ \in S$ . The right-hand side is absolutely convergent by the shell estimate of [F1], and equals $⟨ \sum_{λ} f (λ) δ_{λ}, ψ ⟩$ because for compactly supported $ψ$ only finitely many terms remain and the general case is the absolutely convergent limit of the partial sums [F1]. Hence the product is the sampled distribution.

2.1step 1.1F1F3F4givenalgebra

By the product-to-convolution law [F3] applied with $φ = f$ and $u = {comb}_{Λ}$ , and the comb duality [F1], $F (f \cdot {comb}_{Λ}) = \hat{f} * F {comb}_{Λ} = c^{- 1} (\hat{f} * {comb}_{Λ^{*}})$ ; the convolution definition [F3] and [F4] give $(\hat{f} * {comb}_{Λ^{*}}) (x) = \sum_{λ^{*} \in Λ^{*}} \hat{f} (x - λ^{*})$ , so $F (f \cdot {comb}_{Λ}) = c^{- 1} \sum_{λ^{*}} \hat{f} (\cdot - λ^{*})$ in $S^{'}$ .

3.1step 2.1F5given∎

For $Λ = h Z^{n}$ one has $c = h^{n}$ and $Λ^{*} = h^{- 1} Z^{n}$ by [F5], so the identity reads $F (\sum_{k} f (h k) δ_{h k}) = h^{- n} \sum_{m} \hat{f} (\cdot - m / h)$ , the sampling periodisation claimed. Countable Choice is inherited from the comb duality and Schwartz Fourier suppliers above.

Depends on

Used by

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Sources

Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)
Andrew Sutherland, MIT 18.785 Lecture 16: The functional equation (course PDF) (standard reference, not scraped)