Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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The weighted orthogonal-polynomial Galerkin projection is well-defined

Statement

Under the hypotheses of Weighted orthogonal-polynomial Galerkin spaces and spectral scale, for every real measurable u such that ∫∣upj∣ dμ<∞ for 0≤j≤N, the formula

ΠNu=∑j=0N∫u(x)pj(x) dμ(x)hj pj

is finite and defines the unique element of VN(μ) whose residual is orthogonal to every pi, 0≤i≤N.

Facts & Assumptions

Given: A finite positive measure μ with the support condition in Weighted orthogonal-polynomial Galerkin spaces and spectral scale, an integer N≥0, and a real measurable u with ∫∣upj∣ dμ<∞ for 0≤j≤N.

[F1]

Weighted orthogonal-polynomial Galerkin spaces and spectral scale: the basis polynomials p0,…,pN are monic orthogonal polynomials and hj=∫pj2 dμ>0.

[F2]

The polynomial moment pairing is a real inner product: on each finite polynomial space allowed by the moment data, the moment pairing is a positive-definite real inner product.

[F3]

The given integrability hypotheses make each numerator ∫upj dμ a finite real number.

[F4]

The Lebesgue integral is linear on L1(μ): finite sums and scalar multiples of integrable functions are integrable and may be integrated term by term.

Proof

1.1givenF1F2F3algebra

Each numerator is finite by [F3], and each denominator hj is finite and strictly positive by [F1]–[F2]. Thus every coefficient in the displayed finite sum is a defined real number.

2.1F1F4step 1.1algebra

The integrability hypotheses and [F1] make every term in the finite sum integrable, and [F4] distributes its integral. Orthogonality of the pj then gives, for each i≤N, ∫(u−ΠNu)pi dμ=∫upi dμ−∑j=0N∫upj dμhj∫pjpi dμ=0, because all terms with j≠i vanish and the j=i term equals ∫upi dμ. Thus ΠNu∈VN(μ) has the required residual orthogonality.

3.1F1F2step 2.1algebra∎

If v∈VN(μ) has the same residual property, then w=v−ΠNu lies in VN(μ) and is orthogonal to every basis vector, hence to itself. Positive definiteness in [F2] gives ∫w2 dμ=0, so w=0 in L2(μ) and v=ΠNu. This proves uniqueness.

Depends on

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