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The Nyquist no-aliasing condition

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $h > 0$ and let $E \subseteq R$ be Lebesgue measurable which, up to a Lebesgue null set, is contained in an interval of length $1 / h$ (for instance $λ_{1} (E ∖ [- 1 / (2 h), 1 / (2 h)]) = 0$ ). Then the reciprocal translates $E + m / h$ ( $m \in Z$ ) are pairwise disjoint up to null sets: $λ_{1} ((E + m / h) \cap (E + n / h)) = 0$ for $m \neq n$ . Consequently, if $f \in L^{2} (R)$ has Plancherel transform $\hat{f}$ vanishing almost everywhere off $E$ , then for almost every $ξ$ at most one term of $Q (ξ) : = \sum_{m \in Z} \hat{f} (ξ - m / h)$ is nonzero, and $Q (ξ) = \hat{f} (ξ)$ for almost every $ξ \in E$ . These assertions are independent of the measurable representative of $\hat{f}$ . For Schwartz $f$ , Sampling at a lattice produces periodisation of the spectrum over the dual lattice identifies $h^{- 1} Q$ with the Fourier transform of the sampled distribution; the corollary does not extend that lemma's distributional identity to arbitrary $L^{2}$ inputs. If $E$ is essentially contained in the centered band $[- 1 / (2 h), 1 / (2 h)]$ , Shannon sampling for band-limited $L^{2}$ functions also gives the stated reconstruction. A general translated interval of length $1 / h$ has the same no-overlap property, but is not itself that centered-band hypothesis.

Facts & Assumptions

Given: Countable Choice, $h > 0$ , a Lebesgue measurable $E \subseteq R$ with $E \subseteq I \cup N$ for an interval $I$ of length $1 / h$ and a null set $N$ , and an $L^{2}$ class $f$ whose Plancherel transform $\hat{f}$ vanishes almost everywhere off $E$ (The space $L^{p} (μ)$ as the quotient by null functions, Measure-null sets and almost-everywhere statements relative to a measure).

[F1]

Translation invariance: $λ_{1} (F + t) = λ_{1} (F)$ for every Lebesgue measurable $F$ and $t \in R$ , and measurability is preserved by translation (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F2]

Sampling periodisation: for Schwartz $f$ and $Λ = h Z$ , $F (\sum_{k} f (h k) δ_{h k}) = h^{- 1} \sum_{m \in Z} \hat{f} (\cdot - m / h)$ , the periodisation of the spectrum over the dual lattice $h^{- 1} Z$ (Sampling at a lattice produces periodisation of the spectrum over the dual lattice).

[F3]

Shannon sampling holds for $L^{2}$ functions whose transform vanishes almost everywhere off the band $[- 1 / (2 h), 1 / (2 h)]$ , with the convergence modes stated there (Shannon sampling for band-limited $L^{2}$ functions).

[F4]

A countable union of measurable null sets is null, by the countable-subadditivity inequality of Finite and countable subadditivity of measures. Singletons have measure zero by the degenerate-box case of A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included.

Proof

technique · direct

1.1F1F4givenalgebra

Let $m \neq n$ be integers. Then $(E + m / h) \cap (E + n / h)$ is contained in $((I + m / h) \cap (I + n / h)) \cup (N + m / h) \cup (N + n / h)$ . The interval translates have length $1 / h$ and their positions differ by $∣ m - n ∣ / h \geq 1 / h$ , so their intersection contains at most one point. This intersection and both translates of $N$ are null by [F1] and [F4]; hence the measurable intersection of the translates of $E$ is null.

2.1step 1.1F1F4givenalgebra

Fix a measurable representative $g$ of $\hat{f}$ and a null set $Z$ outside which $g$ vanishes off $E$ . The set $T : = ⋃_{m \neq n} ((E + m / h) \cap (E + n / h)) \cup ⋃_{m} (Z + m / h)$ is null by step 1.1, [F1] and [F4]; both unions are countable. For $ξ \notin T$ , at most one index $m$ satisfies $ξ - m / h \in E$ , and all other terms $g (ξ - m / h)$ vanish. For $ξ \in E ∖ T$ , the only possible index is $m = 0$ , giving $Q (ξ) = g (ξ)$ . Changing $g$ on a null set affects the translated terms only on its countable union of reciprocal translates, again null by [F1] and [F4], so the conclusions are representative-independent.

3.1step 1.1step 2.1F2F3given∎

Thus $h^{- 1} Q = h^{- 1} \hat{f}$ almost everywhere on $E$ , with no contribution from a nonzero reciprocal shift. For Schwartz inputs, [F2] identifies $h^{- 1} Q$ as the transform of the sampled distribution. For a centered containing band, [F3] gives Shannon reconstruction; its cutoff is $1 / (2 h)$ and its total band length is $1 / h$ . The closed band's endpoints can differ by $1 / h$ , so the disjointness assertion remains an almost-everywhere assertion. Countable Choice is inherited from the stated measure and Fourier suppliers.

Depends on

Used by

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Dependency tree · two levels

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Sources

Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)
Lior Silberman, Fourier series and the Poisson summation formula (Math 604/613 notes, UBC) (standard reference, not scraped)