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Integer exponentials are orthogonal for angular Lebesgue measure

Statement

Assume Countable Choice. For every integer k, with the unnormalized angular Lebesgue measure dλ on T from Spectral measures and the known-mean periodogram, ∫Teikλ dλ={2π,k=0,0,k≠0.

Facts & Assumptions

Given: Countable Choice, an integer k, and angular Lebesgue measure dλ on the circle, integrated over one period [−π,π).

[A1]

The Axiom of Countable Choice (ACω): the Countable Choice assumption required for the angular Lebesgue measure and its interval values.

[F1]

Spectral measures and the known-mean periodogram: dλ is ordinary Lebesgue measure on one angular period, with total mass 2π.

[F3]

The nonnegative Lebesgue integral, The integral of a nonnegative simple function, and Monotonicity and nonnegative homogeneity of the nonnegative integral: the Lebesgue integral is monotone on nonnegative functions, and the integral of a nonnegative step function is the sum of its values times the measures of its intervals.

[F5]

The Lebesgue integral is linear on L1(μ): the Lebesgue integral is linear on integrable real functions.

[F7]

The zero sets of sine and cosine and the least positive common period 2 pi and Parity and the Pythagorean identity for sine and cosine: for integer k≠0, sin⁡(kπ)=sin⁡(−kπ)=0 and cos⁡(kπ)=cos⁡(−kπ).

[F8]

exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0 and Integrable real and complex functions, and their integrals: eikλ=cos⁡(kλ)+isin⁡(kλ), and the integral of a complex function is the integral of its real part plus i times that of its imaginary part.

[F10]

A continuous map has Borel preimages of Borel sets: the preimage of every Borel set under a continuous map is Borel; in particular the continuous real function in step 1.1 is Borel measurable.

[F11]

For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε: for every δ>0 there is an N≥1 with 2π/N<δ.

Proof

Proof technique: compare the Lebesgue integral with Darboux sums, then use the fundamental theorem of calculus.

1.1A1F1F2F3F4F5F9F10F11given

Let g:[−π,π]→R be continuous and bounded by B≥0. For each N≥1, partition the interval into the fixed equal pieces Jj,N=[−π+2πj/N,−π+2π(j+1)/N), with the endpoint π added to the last piece. On each piece take the infimum and supremum of g+B on its closed hull and use them as the constant values of simple step functions ℓN and uN. They satisfy 0≤ℓN≤g+B≤uN. By [F3]–[F4], under [A1], their Lebesgue integrals with respect to λ1 are exactly the lower and upper Darboux sums of g+B for this partition. Given η>0, uniform continuity [F2] gives δ>0 such that the oscillation on any interval of length below δ is less than η. Choose N with 2π/N<δ by [F11]; then every oscillation is below η, and the gap between the sums is at most 2πη. Hence that gap tends to zero. Monotonicity of the Lebesgue integral places ∫[−π,π)(g+B) dλ1 between the same sums; the Darboux integral of g+B lies between them as well. Therefore the Lebesgue and Darboux integrals of g+B agree. By linearity [F5] and λ1([−π,π))=2π from [F1], the two integrals of g agree.

1.2A1F1given

If k=0, the integrand is the constant 1, so its integral is 2π by [F1] under [A1].

2.1A1F1F2F6F7F8step 1.1

If k≠0, apply step 1.1 to the continuous real functions g(λ)=cos⁡(kλ) and g(λ)=sin⁡(kλ). Their antiderivatives are respectively sin⁡(kλ)/k and −cos⁡(kλ)/k by [F6] and the chain rule. Their endpoint differences vanish by [F7], so both Lebesgue integrals on [−π,π) are zero. These functions are 2π-periodic; by the angular integral definition [F1], their integrals on T are those same integrals over [−π,π). Euler's identity and the complex-integral convention [F8] give ∫Teikλ dλ=0.

3.1givenstep 1.2step 2.1algebra∎

The cases k=0 and k≠0 exhaust the integers, and steps 1.2 and 2.1 give the stated formula.

Depends on

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