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The finite Bessel inequality and best approximation by a finite orthonormal family

Statement

Let $(e_{i})_{i \in I}$ be an orthonormal family in a real or complex inner-product space $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), let $F \subseteq I$ be finite, and for $x \in H$ put

$P_{F} x : = \sum_{i \in F} ⟨ x, e_{i} ⟩ e_{i} .$ Then: 1. $P_{F} x$ lies in the span of ${e_{i} : i \in F}$ , and $∥ P_{F} x ∥^{2} = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ ; 2. the residual $x - P_{F} x$ is orthogonal to every $e_{j}$ with $j \in F$ , hence to every vector of the span of ${e_{i} : i \in F}$ ; 3. $∥ x - P_{F} x ∥^{2} = ∥ x ∥^{2} - \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ , and therefore the finite Bessel inequality holds: $\sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2} \leq ∥ x ∥^{2};$ 4. $P_{F} x$ is the unique best approximation to $x$ from the span of ${e_{i} : i \in F}$ : for every $y$ in that span, $∥ x - y ∥^{2} = ∥ x - P_{F} x ∥^{2} + ∥ P_{F} x - y ∥^{2} \geq ∥ x - P_{F} x ∥^{2},$ with equality if and only if $y = P_{F} x$ .

At $F = \emptyset$ the sum defining $P_{F} x$ is empty, so $P_{F} x = 0$ and the identities read $∥ x ∥^{2} = ∥ x ∥^{2}$ .

Facts & Assumptions

[A1]

The inner product is linear in the first argument and conjugate-linear in the second, and $∥ v ∥^{2} = ⟨ v, v ⟩$ (Real and complex inner-product spaces and their induced length).

[A2]

$⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ , so in particular $∥ e_{i} ∥ = 1$ and $e_{i} \neq 0$ (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: $∥ \sum_{j} z_{j} ∥^{2} = \sum_{j} ∥ z_{j} ∥^{2}$ when the $z_{j}$ 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 $(e_{i})_{i \in I}$ in $H$ , a finite $F \subseteq I$ , a vector $x \in H$ , and $P_{F} x = \sum_{i \in F} ⟨ x, e_{i} ⟩ e_{i}$ .

1.1A5A1

The vector $P_{F} x$ is the finite linear combination of the vectors $e_{i}$ , $i \in F$ , with coefficients $⟨ x, e_{i} ⟩$ , so it lies in the span of ${e_{i} : i \in F}$ by [A5]; and $F = \emptyset$ gives $P_{F} x = 0$ with empty sums on both sides of the two norm identities.

1.2A1A2A3

For every $j \in F$ , linearity in the first argument and [A2] give $⟨ P_{F} x, e_{j} ⟩ = \sum_{i \in F} ⟨ x, e_{i} ⟩ ⟨ e_{i}, e_{j} ⟩ = ⟨ x, e_{j} ⟩$ , hence $⟨ x - P_{F} x, e_{j} ⟩ = ⟨ x, e_{j} ⟩ - ⟨ P_{F} x, e_{j} ⟩ = 0$ : the residual is orthogonal to every $e_{j}$ with $j \in F$ .

1.3A1A2algebra

Expanding the pairing of $P_{F} x$ with itself, $⟨ P_{F} x, P_{F} x ⟩ = \sum_{i, j \in F} ⟨ x, e_{i} ⟩ \overline{⟨ x, e_{j} ⟩} ⟨ e_{i}, e_{j} ⟩ = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ , so $∥ P_{F} x ∥^{2} = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ by [A1].

2.1step 1.2A1A5

If $y$ lies in the span of ${e_{i} : i \in F}$ , then $y = \sum_{i \in F} λ_{i} e_{i}$ for suitable scalars, so conjugate-linearity in the second argument together with step 1.2 gives $⟨ x - P_{F} x, y ⟩ = \sum_{i \in F} \overline{λ_{i}} ⟨ x - P_{F} x, e_{i} ⟩ = 0$ ; thus the residual is orthogonal to the whole span.

2.2step 1.2step 1.3A4algebra

The decomposition $x = P_{F} x + (x - P_{F} x)$ has orthogonal summands by step 1.2, so Pythagoras and step 1.3 give $∥ x ∥^{2} = ∥ P_{F} x ∥^{2} + ∥ x - P_{F} x ∥^{2} = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2} + ∥ x - P_{F} x ∥^{2}$ ; since $∥ x - P_{F} x ∥^{2} \geq 0$ this yields the difference identity and the finite Bessel inequality.

3.1step 2.1step 2.2A4A5A1

For $y$ in the span, the vector $P_{F} x - y$ also lies in the span, so it is orthogonal to $x - P_{F} x$ by step 2.1; the difference identity of step 2.2 applied to the orthogonal decomposition $x - y = (x - P_{F} x) + (P_{F} x - y)$ gives $∥ x - y ∥^{2} = ∥ x - P_{F} x ∥^{2} + ∥ P_{F} x - y ∥^{2}$ , which is at least $∥ x - P_{F} x ∥^{2}$ and is equal to it exactly when $∥ P_{F} x - y ∥ = 0$ , that is exactly when $y = P_{F} x$ by definiteness of the norm.

4.1step 1.1step 1.3step 2.1step 2.2step 3.1∎

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 $x$ .

The finite Bessel inequality and best approximation by a finite orthonormal family

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources