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Parseval equivalences for an orthonormal family

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(e_{i})_{i \in I}$ be an orthonormal family in a real or complex Hilbert space $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and for $x \in H$ and finite $F \subseteq I$ put

$P_{F} x : = \sum_{i \in F} ⟨ x, e_{i} ⟩ e_{i} .$

Then the following four assertions are equivalent:

(completeness) the closed linear span of $(e_{i})_{i \in I}$ is $H$ ;
(zero complement) the only vector orthogonal to every $e_{i}$ is $0$ ;
(Parseval) $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} = ∥ x ∥^{2}$ for every $x \in H$ , the sum being the supremum of the finite subsums (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ );
(net convergence) the finite-subset net $(P_{F} x)_{F \in Fin (I)}$ converges to $x$ for every $x \in H$ .

Assertion 2 is the statement $span {e_{i} : i \in I}^{⊥} = {0}$ (Orthogonality and the orthogonal complement).

Facts & Assumptions

[A1]

If $(e_{i})_{i \in I}$ is orthonormal and $a = (a_{i})_{i \in I} \in ℓ^{2} (I, F)$ , then the finite-subset net $\sum_{i \in F} a_{i} e_{i}$ converges to a limit $s$ with $⟨ s, e_{j} ⟩ = a_{j}$ for every $j$ and $∥ s ∥^{2} = \sum_{i \in I} ∣ a_{i} ∣^{2}$ (Square-summable orthogonal families have norm-convergent finite sums).

[A2]

The Bessel inequality holds: $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} \leq ∥ x ∥^{2}$ , so the coefficient family lies in $ℓ^{2} (I, F)$ (The Bessel inequality for an arbitrary orthonormal family).

[A3]

For every finite $F$ , $∥ x - P_{F} x ∥^{2} = ∥ x ∥^{2} - \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ (The finite Bessel inequality and best approximation by a finite orthonormal family).

[A4]

If $z \in H$ satisfies $⟨ z, e_{i} ⟩ = 0$ for every $i$ , then $⟨ z, v ⟩ = 0$ for every $v$ in the closed linear span of the family: conjugate-linearity, and in particular additivity, in the second argument passes the vanishing to the algebraic span, while $∣ ⟨ z, w ⟩ ∣ \leq ∥ z ∥ ∥ w ∥$ passes it to norm limits (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Orthogonality and the orthogonal complement).

[A5]

$\overline{A} = {x : d (x, A) = 0}$ for nonempty $A$ , so $x \in \overline{A}$ exactly when vectors of $A$ come arbitrarily close to $x$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[A6]

The partial sums $t_{F} : = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ are nondecreasing under inclusion with supremum $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ , and a nondecreasing net of reals converges to its supremum. In particular if every real $ε > 0$ satisfies $S - ε < t_{F} \leq S$ for some finite $F$ then the net converges to $S$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Epsilon characterisation of the supremum).

[A7]

For a linear subspace $M$ of the Hilbert space $H$ , $M^{⊥ ⊥} = \overline{M}$ (The double orthogonal complement of a subspace is its closure), and ${0}^{⊥} = H$ (Orthogonality and the orthogonal complement).

[A8]

The span of ${e_{i} : i \in I}$ is a linear subspace and its closure is the closed linear span of the family (Linear subspace of a vector space, Orthonormal families, complete orthonormal systems and Hilbert bases).

[A9]

The norm is continuous along convergent nets and the square of a convergent scalar net converges to the square of the limit (The reverse triangle inequality in a normed space).

Proof

technique · direct

Given: Countable Choice, an orthonormal family $(e_{i})_{i \in I}$ in the Hilbert space $H$ , and the partial sums $P_{F} x = \sum_{i \in F} ⟨ x, e_{i} ⟩ e_{i}$ .

1.1A1A2A4A8

Completeness implies net convergence. Assume the closed linear span of the family is $H$ and let $x \in H$ . The coefficient family lies in $ℓ^{2} (I, F)$ by Bessel, so by [A1] the net $(P_{F} x)$ converges to a limit $s$ with $⟨ s, e_{j} ⟩ = ⟨ x, e_{j} ⟩$ for every $j$ ; then $z : = x - s$ satisfies $⟨ z, e_{j} ⟩ = 0$ for every $j$ , so $⟨ z, v ⟩ = 0$ for every $v$ in the closed linear span by [A4] and hence $⟨ z, x ⟩ = 0$ ; therefore $∥ z ∥^{2} = ⟨ z, z ⟩ = ⟨ z, x ⟩ - ⟨ z, s ⟩ = 0$ and $z = 0$ , that is $s = x$ .

1.2A5A8

Net convergence implies completeness. Assume $(P_{F} x)$ converges to $x$ for every $x \in H$ and fix $x$ . Every $P_{F} x$ lies in the span of the family, so for every real $ε > 0$ some vector of that span is within distance $ε$ of $x$ ; hence $d (x, span {e_{i}}) = 0$ and $x \in \overline{span {e_{i}}}$ , the closed linear span, by [A5] and [A8]. As $x$ was arbitrary, the closed linear span is $H$ .

1.3A3A6A9

Net convergence implies Parseval. Assume net convergence and fix $x$ . Every finite $F$ satisfies $∥ x - P_{F} x ∥^{2} = ∥ x ∥^{2} - t_{F}$ with $t_{F} = \sum_{i \in F} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ by [A3]; since $P_{F} x \to x$ , continuity of the norm gives $∥ x - P_{F} x ∥^{2} \to 0$ , so $t_{F} \to ∥ x ∥^{2}$ as a net of reals. But $t_{F}$ is nondecreasing under inclusion with supremum $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2}$ , so it converges to that supremum by [A6] and the limit is unique; hence $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} = ∥ x ∥^{2}$ .

1.4A6A9

Parseval implies zero complement. Assume Parseval for every $x$ and let $y$ satisfy $⟨ y, e_{i} ⟩ = 0$ for every $i$ . Then all finite subsums $\sum_{i \in F} ∣ ⟨ y, e_{i} ⟩ ∣^{2}$ vanish, so the supremum $\sum_{i \in I} ∣ ⟨ y, e_{i} ⟩ ∣^{2}$ is $0$ ; Parseval applied to $y$ gives $∥ y ∥^{2} = 0$ , hence $y = 0$ by definiteness of the inner product.

1.5A7A8

Zero complement implies completeness. Assume no nonzero vector is orthogonal to every $e_{i}$ , and let $M : = span {e_{i} : i \in I}$ , a linear subspace with $M^{⊥} = {0}$ . Then $M^{⊥ ⊥} = \overline{M}$ by [A7], while $M^{⊥ ⊥} = (M^{⊥})^{⊥} = {0}^{⊥} = H$ by [A7]; hence the closed linear span $\overline{M}$ of the family is $H$ .

2.1step 1.1step 1.2step 1.3step 1.4step 1.5∎

The five implications of steps 1.1 to 1.5 form the cycle (completeness) $\Rightarrow$ (net convergence) $\Rightarrow$ (completeness), (net convergence) $\Rightarrow$ (Parseval) $\Rightarrow$ (zero complement) $\Rightarrow$ (completeness), so any one of the four assertions implies all the others and the four are equivalent.

Parseval equivalences for an orthonormal family

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources