ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Bessel's inequality is strict for a vector outside the span of a proper orthonormal set

Example

In standard $R^{3}$ , take the orthonormal list $(e_{0}, e_{1})$ and $v = (1, 2, 3)$ . Then Bessel's inequality is strict:

$∣ ⟨ v, e_{0} ⟩ ∣^{2} + ∣ ⟨ v, e_{1} ⟩ ∣^{2} = 5 < 14 = ∥ v ∥^{2} .$

The gap $9$ is the squared norm of the orthogonal remainder $3 e_{2}$ .

Facts & Assumptions

Given: The first two standard basis vectors and $v = (1, 2, 3)$ .

[L1]

Bessel's inequality becomes equality exactly when the vector lies in the span of the orthonormal list (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[L2]

Standard coordinate inner products are dot products (The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products).

Verification

technique · computation

1.1

By [L2], the two coefficients are $1$ and $2$ , while $∥ v ∥^{2} = 1^{2} + 2^{2} + 3^{2} = 14$ . Hence the Bessel sum is $5$ and the gap is $9$ .

L2algebra

2.1

The projection onto $span (e_{0}, e_{1})$ is $(1, 2, 0)$ , leaving $3 e_{2}$ of squared norm $9$ . Since $v$ is outside the span, strictness also follows from [L1].

step 1.1L1L2∎

Depends on

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