Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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The finite Bessel inequality and best approximation by a finite orthonormal family

Statement

Let (ei)iI be an orthonormal family in a real or complex inner-product space H (Orthonormal families, complete orthonormal systems and Hilbert bases), let FI be finite, and for xH put

PFx:=iFx,eiei.Then: 1. PFx lies in the span of {ei:iF}, and PFx2=iFx,ei2; 2. the residual xPFx is orthogonal to every ej with jF, hence to every vector of the span of {ei:iF}; 3. xPFx2=x2iFx,ei2, and therefore the finite Bessel inequality holds:iFx,ei2x2;4. PFx is the unique best approximation to x from the span of {ei:iF}: for every y in that span,xy2=xPFx2+PFxy2xPFx2, with equality if and only if y=PFx.

At F= the sum defining PFx is empty, so PFx=0 and the identities read x2=x2.

Facts & Assumptions

[A1]

The inner product is linear in the first argument and conjugate-linear in the second, and v2=v,v (Real and complex inner-product spaces and their induced length).

[A2]

ei,ej=δij, so in particular ei=1 and ei0 (Orthonormal families, complete orthonormal systems and Hilbert bases).

[A3]

Orthogonality is symmetry-compatible and means u,v=0; every vector is orthogonal to 0 (Orthogonality and the orthogonal complement).

[A4]

Pairwise orthogonal finite sums satisfy Pythagoras: jzj2=jzj2 when the zj are pairwise orthogonal, the empty sum being 0 (Pythagoras and finite orthogonal sums).

[A5]

The span of a set of vectors consists of its finite linear combinations and is a linear subspace (Linear subspace of a vector space).

Proof

technique · direct

Given: An orthonormal family (ei)iI in H, a finite FI, a vector xH, and PFx=iFx,eiei.

1.1

The vector PFx is the finite linear combination of the vectors ei, iF, with coefficients x,ei, so it lies in the span of {ei:iF} by [A5]; and F= gives PFx=0 with empty sums on both sides of the two norm identities.

A5A1
1.2

For every jF, linearity in the first argument and [A2] give PFx,ej=iFx,eiei,ej=x,ej, hence xPFx,ej=x,ejPFx,ej=0: the residual is orthogonal to every ej with jF.

A1A2A3
1.3

Expanding the pairing of PFx with itself, PFx,PFx=i,jFx,eix,ejei,ej=iFx,ei2, so PFx2=iFx,ei2 by [A1].

A1A2algebra
2.1

If y lies in the span of {ei:iF}, then y=iFλiei for suitable scalars, so conjugate-linearity in the second argument together with step 1.2 gives xPFx,y=iFλixPFx,ei=0; thus the residual is orthogonal to the whole span.

step 1.2A1A5
2.2

The decomposition x=PFx+(xPFx) has orthogonal summands by step 1.2, so Pythagoras and step 1.3 give x2=PFx2+xPFx2=iFx,ei2+xPFx2; since xPFx20 this yields the difference identity and the finite Bessel inequality.

step 1.2step 1.3A4algebra
3.1

For y in the span, the vector PFxy also lies in the span, so it is orthogonal to xPFx by step 2.1; the difference identity of step 2.2 applied to the orthogonal decomposition xy=(xPFx)+(PFxy) gives xy2=xPFx2+PFxy2, which is at least xPFx2 and is equal to it exactly when PFxy=0, that is exactly when y=PFx by definiteness of the norm.

step 2.1step 2.2A4A5A1
4.1

Steps 1.1 and 1.3 give the first claim, steps 1.2 and 2.1 the second, step 2.2 the third, and step 3.1 the fourth, so all four assertions hold for every finite F and every x.

step 1.1step 1.3step 2.1step 2.2step 3.1

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