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Centring by translation and modulation preserves the variance product

Statement

Assume countable choice. Let f∈L2(Rn;C) be nonzero with finite second moments, spatial mean a, frequency mean b, and variances Vx(f),Vξ(f) as in Spatial and frequency centres and variances of an L2 function with finite second moments (Translation of a function on Rn). Define g(x):=e−2πi b⋅xf(x+a). Then g∈L2 is nonzero with finite second moments, its spatial mean is 0 and its frequency mean is 0, and for every j ∥xjg∥2=∥(xj−aj)f∥2,∥ξjg^∥2=∥(ξj−bj)f^∥2. Consequently Vx(g)=Vx(f), Vξ(g)=Vξ(f), and Vx(g)Vξ(g)=Vx(f)Vξ(f).

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice (ACω)), a nonzero f∈L2(Rn;C) with finite second moments and means and variances a,b,Vx(f),Vξ(f) as in Spatial and frequency centres and variances of an L2 function with finite second moments, and g(x)=e−2πib⋅xf(x+a).

[F1]

Countable choice is assumed; it is used by the change-of-variables interface and to select the L2-approximating sequence in step 2.2 (The Axiom of Countable Choice (ACω)).

[F2]

Complex L1 change of variables: for a C1 diffeomorphism T with absolute Jacobian determinant ∣det⁡DT∣ and h∈L1, ∫h(Tx)∣det⁡DT(x)∣dx=∫h(y)dy; the affine maps x↦x+a and ξ↦ξ+b have determinant 1 (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). Complex L2 carries the norm ∥⋅∥2 of Complex completeness, density, and inner product: the consumer interface.

[F3]

Translation and modulation: with τaf(x)=f(x−a) and Mbf(x)=e2πib⋅xf(x), for f∈S one has τaf^(ξ)=e−2πia⋅ξf^(ξ) and Mbf^(ξ)=f^(ξ−b) at every frequency (Translation, modulation, linear dilation and reflection laws, Translation of a function on Rn).

[F4]

Schwartz functions lie in L1∩L2; Schwartz space is dense in L2, the integral Fourier transform of any L1∩L2 function represents its Plancherel transform almost everywhere, and Plancherel is an isometry (Schwartz derivatives are integrable, Schwartz space is dense in L2, Agreement of the integral and L2 transforms, Plancherel theorem).

[F5]

Integrable functions that agree almost everywhere have equal integrals (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

Proof

technique · direct
1.1F2given

The centred function is admissible on the spatial side. Translation preserves null equivalence and the L2 norm by [F2], while modulation has unit modulus, so g∈L2 and ∥g∥2=∥f∥2>0. Substituting y=x+a gives ∫∣x∣2∣g(x)∣2dx=∫∣y−a∣2∣f(y)∣2dy≤2∫∣y∣2∣f(y)∣2dy+2∣a∣2∥f∥22<∞. Also ∫∣xj∣∣g(x)∣2dx<∞ by Cauchy--Schwarz from g,xjg∈L2. Thus the spatial mean and variance of g are defined; frequency-side finiteness is established in the frequency computation below.

2.1F2givenstep 1.1

Spatial side. Substituting y=x+a and using ∣g(x)∣2=∣f(x+a)∣2 [F2] gives, for each j, ∫xj∣g(x)∣2dx=∫yj∣f(y)∣2dy−aj∫∣f(y)∣2dy=∥f∥22aj−∥f∥22aj=0,∫xj2∣g(x)∣2dx=∫(yj−aj)2∣f(y)∣2dy. Hence the spatial mean of g is 0, ∥xjg∥2=∥(xj−aj)f∥2, and Vx(g)=∥g∥2−2∫∣x∣2∣g∣2=∥f∥2−2∫∣y−a∣2∣f(y)∣2dy=Vx(f).

2.2F1F2F3F4F5givenstep 1.1

Frequency side. Choose fk∈S with fk→f in L2, using [F4] and countable choice [F1], and set gk=M−b(τ−afk). By [F4], fk∈L1∩L2; [F2] shows translation and modulation preserve both spaces and their norms, so gk∈L1∩L2. Translation and modulation preserve L2 distances, hence gk→g in L2; Plancherel gives F2fk→F2f and F2gk→F2g. By [F3], for each k the integral transforms satisfy g^k(ξ)=e2πia⋅(ξ+b)f^k(ξ+b), and [F4] identifies these transforms with their Plancherel classes. Translation and multiplication by this unit-modulus phase are isometries on L2 by [F2], so passing to the norm limits proves the Plancherel-class identity g^(ξ)=e2πia⋅(ξ+b)f^(ξ+b) almost everywhere. First, the affine change of variables η=ξ+b gives ∫∣ξ∣2∣g^(ξ)∣2dξ=∫∣η−b∣2∣f^(η)∣2dη<∞, so the first moments are absolutely integrable by Cauchy--Schwarz. Using [F5] for representatives, the same substitution now yields ∫ξj∣g^(ξ)∣2dξ=∫(ηj−bj)∣f^(η)∣2dη=∥f∥22bj−∥f∥22bj=0,∫ξj2∣g^(ξ)∣2dξ=∫(ηj−bj)2∣f^(η)∣2dη. Thus g has finite frequency second moments and mean 0, ∥ξjg^∥2=∥(ξj−bj)f^∥2, and Vξ(g)=∥g^∥2−2∫∣ξ∣2∣g^∣2=∥f^∥2−2∫∣η−b∣2∣f^(η)∣2dη=Vξ(f); Plancherel and the unitary covariance give ∥g^∥2=∥g∥2=∥f∥2=∥f^∥2.

3.1step 2.1step 2.2∎

Conclusion. Steps 2.1 and 2.2 give Vx(g)=Vx(f), Vξ(g)=Vξ(f) and hence Vx(g)Vξ(g)=Vx(f)Vξ(f), and both centred means vanish.

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