Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis

Statement

If (e0,,er1) is a finite orthonormal list and v is any vector, then

i<rv,ei2v2.

Equality holds exactly when vspan(e0,,er1). If the list is an orthonormal basis, then for all v,w,

v=i<rv,eiei,v,w=i<rv,eiw,ei,

and

v2=i<rv,ei2.

The empty-list case is included.

Facts & Assumptions

Given: A finite orthonormal list (ei)i<r and vectors v,w.

[L1]

Orthonormality gives ei,ej=0 for ij and ei,ei=1 (Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases).

[L2]
[L3]

span(S)=L(S), the set of finite linear combinations i<nλivi of elements of S (span(S) is exactly the set of linear combinations of finite lists of elements of S, and span()={0V}).

Proof

technique · direct
1.1

Put p=i<rv,eiei. For each j<r, [L1] gives vp,ej=0, so vp is orthogonal to p.

L1algebra
2.1

By [L2], v2=p2+vp2. A second use of orthonormality gives p2=i<rv,ei2, proving Bessel's inequality.

step 1.1L1L2
3.1

Equality holds in step 2.1 exactly when vp=0, hence exactly when v=p. By [L3], this is exactly v belonging to the listed span.

step 2.1L3
4.1

If the list is a basis, its span is V, so step 3.1 gives the coordinate expansion and the squared-length identity. Substitute the coordinate expansion of v into v,w and use conjugate symmetry to obtain the displayed inner-product formula.

step 3.1L1algebra
5.1

When r=0, [L4] makes every displayed sum zero; the list can be a basis only of the zero space, so all assertions remain valid.

L4

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources