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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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The trigonometric characters are orthonormal in L2 of the torus

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). The family of characters (ek)kZ of Fourier coefficients and trigonometric polynomials on the torus is orthonormal in the complex Hilbert space L2(T;C) (L2 with the integral pairing is a Hilbert space, Orthonormal families, complete orthonormal systems and Hilbert bases):

ek,el=TekeldmT={1,k=l,0,kl.

Moreover the coefficient pairing of a trigonometric polynomial computes its coefficients: if p=kFckek and jF, then p^(j)=cj, and p^(j)=0 for jF.

Facts & Assumptions

[A1]

ekel=ekl and ek=1; in particular the function tem(q(t)) on R equals exp(2πimt)=cos(2πmt)+isin(2πmt), and it has period 1 (Fourier coefficients and trigonometric polynomials on the torus, exp(x+iy)=ex(cosy+isiny), exp(x+iy)=ex, and eiπ+1=0).

[A2]

For a continuous 1-periodic g:RC the torus integral equals the Riemann integral over [0,1]: TGdmT=01g(t)dt for G(q(t))=g(t), because the torus integral is represented on [0,1) and a bounded Riemann integrable function on [0,1] is Lebesgue measurable with the same integral, the endpoint being a null set (The one-dimensional torus and its normalized Haar integral, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[A4]

sin(mπ)=0 and cos(mπ)=(1)m for every integer m: the zero-set theorem gives the sine values, while the shift formula cos(x+π)=cosx and cos0=1 give the cosine values by integer induction (The zero sets of sine and cosine and the least positive common period 2 pi, Quarter-turn values and shifts by pi/2 and pi).

[A5]

The pairing on complex L2 is linear in the first variable and conjugate-linear in the second, with f,g=fg, and the Fourier coefficient is f^(k)=f,ek (L2 with the integral pairing is a Hilbert space, Fourier coefficients and trigonometric polynomials on the torus).

Proof

technique · direct

Given: Countable Choice and characters ek on T.

1.1

For k,lZ put m:=kl; then ekel=em, and the corresponding 1-periodic function on R is gm(t)=cos(2πmt)+isin(2πmt).

A1A5
2.1

The torus integral of ekel is the Riemann integral of gm over [0,1]: for m=0 the integrand is 1 and the integral is 1, while for m0 the real and imaginary parts have the primitives tsin(2πmt)/(2πm) and tcos(2πmt)/(2πm), whose values at t=0 and t=1 agree because sin(2πm)=sin0=0 and cos(2πm)=cos0=1, so each of the two definite integrals vanishes. Hence the integral is 1 when m=0 and 0 when m0.

step 1.1A2A3A4algebra
3.1

Therefore ek,el=1 for k=l and 0 for kl, so the characters are orthonormal.

step 2.1A5
4.1

For the coefficient claim, let p=kFckek and fix jZ; by linearity of the pairing and orthonormality, p^(j)=kFckek,ej equals cj if jF and 0 otherwise.

step 3.1A5
5.1

Steps 3.1 and 4.1 prove orthonormality of the characters and the coefficient formula for trigonometric polynomials.

step 3.1step 4.1

Depends on

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