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Unitary Fourier transform for a two-component lattice sequence

Statement

Assume Countable Choice. Let $T = R / (2 π Z)$ and let $m_{T}$ be its normalized Borel probability measure, so in the angle coordinate its integral is $\int_{T} F d m_{T} = \int_{0}^{2 π} F (k) d k / (2 π)$ , obtained from the normalized one-dimensional torus by $k = 2 π x$ . On $C^{2}$ use $⟨ u, v ⟩_{C^{2}} = u_{1} \overline{v_{1}} + u_{2} \overline{v_{2}}$ and $∥ u ∥_{C^{2}}^{2} = ∣ u_{1} ∣^{2} + ∣ u_{2} ∣^{2}$ , linear in the first argument. Define $ℓ^{2} (Z; C^{2}) : = {x = (x_{j})_{j \in Z} : x_{j} \in C^{2}, \sum_{j \in Z} ∥ x_{j} ∥_{C^{2}}^{2} < \infty}, ⟨ x, y ⟩_{ℓ^{2}} : = \sum_{j \in Z} ⟨ x_{j}, y_{j} ⟩_{C^{2}} .$ Define $L^{2} (T; C^{2})$ to be the vector space of pairs of Borel-measurable functions $f = (f_{1}, f_{2}) : T \to C^{2}$ with $\int_{T} (∣ f_{1} ∣^{2} + ∣ f_{2} ∣^{2}) d m_{T} < \infty$ , modulo equality $m_{T}$ -almost everywhere, with inner product $⟨ f, g ⟩_{L^{2}} : = \int_{T} (f_{1} \overline{g_{1}} + f_{2} \overline{g_{2}}) d m_{T} .$ These pairings are linear in the first argument. The square-summable-family supplier gives the scalar $ℓ^{2}$ norm and pairing used coordinatewise. The $L^{2}$ space is the pair of scalar $L^{2}$ quotient spaces with the displayed sum inner product and completeness convention. Let $c_{00} (Z; C^{2}) : = {x \in ℓ^{2} (Z; C^{2}) : {j : x_{j} \neq 0} is finite}$ . Define the linear map $F_{0} x (k) = \sum_{j \in Z} x_{j} e^{i j k}, k \in T,$ componentwise, from $c_{00} (Z; C^{2})$ to $L^{2} (T; C^{2})$ . Its values are continuous finite sums and therefore belong to this $L^{2}$ space. Then $F_{0}$ preserves inner products on finitely supported sequences and has a unique extension to a surjective linear isometry (a unitary map) $F : ℓ^{2} (Z; C^{2}) \to L^{2} (T; C^{2})$ .

Statement references

The one-dimensional torus and its normalized Haar integral

$L^{2}$ with the integral pairing is a Hilbert space

Facts & Assumptions

Given: Countable Choice, the normalized one-dimensional torus, and two-component square-summable sequences.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume Countable Choice, with its exact family-of-nonempty-sets formulation.

[F1]

The one-dimensional torus and its normalized Haar integral defines the quotient torus $R / Z$ , its normalized Borel measure, and its integral; under $k = 2 π x$ these give $m_{T}$ and $d k / (2 π)$ . Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ defines the scalar $ℓ^{2}$ norm and pairing used for the two coordinates, while the vector-valued space and pairing are defined locally in the Statement.

[F2]

The Fourier basis and Parseval's identity on the finite torus: under Countable Choice the one-dimensional coordinate characters form an orthonormal basis of $L^{2} (T)$ , and the Fourier-coefficient map is a surjective linear isometry onto $ℓ^{2} (Z; C)$ ; its inverse is the finite-subset-net Fourier series.

[F3]

$L^{2}$ with the integral pairing is a Hilbert space: scalar $L^{2}$ functions are equivalence classes modulo equality almost everywhere, their pairing is $\int f \overline{g} d m_{T}$ , and this quotient space is complete.

Proof

1.1givenF1algebra

For finitely supported $x, y$ , expanding the finite sums and using $\int_{T} e^{i (j - ℓ) k} d k / (2 π) = δ_{j ℓ}$ gives $⟨ F_{0} x, F_{0} y ⟩_{L^{2}} = \sum_{j, ℓ} ⟨ x_{j}, y_{ℓ} ⟩_{C^{2}} \int_{T} e^{i (j - ℓ) k} \frac{d k}{2 π} = \sum_{j} ⟨ x_{j}, y_{j} ⟩_{C^{2}} = ⟨ x, y ⟩_{ℓ^{2}} .$ Thus $F_{0}$ is an isometry on its finite-support domain.

2.1givenA1F1F2F3step 1.1algebra∎

By [F2] under the explicit Countable Choice hypothesis [A1], the scalar Fourier-coefficient map is a surjective linear isometry from the complete space $L^{2} (T)$ of [F3] onto scalar $ℓ^{2} (Z; C)$ , with inverse given by the finite-subset-net Fourier series. This proves scalar $ℓ^{2}$ completeness: a Cauchy sequence pulls back by that inverse to a Cauchy sequence in $L^{2}$ , whose limit maps to its $ℓ^{2}$ limit. Applying this to each of the two coordinates proves completeness of $ℓ^{2} (Z; C^{2})$ with [F1]'s norm. The inverse Fourier map applied coordinatewise is a unitary and agrees with $F_{0}$ on finitely supported sequences by its finite polynomial formula. Those sequences are dense: for any $ε > 0$ , the finite-subset tail property in [F1] gives a finite $F \subset Z$ with $\sum_{j \notin F} ∥ x_{j} ∥^{2} < ε^{2}$ ; choosing $M$ with $F \subseteq [- M, M]$ makes the squared truncation error at most that tail, hence below $ε^{2}$ . Thus $c_{00}$ is dense and the continuous extension is unique.

Unitary Fourier transform for a two-component lattice sequence

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources