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

1.1

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 $s = x$ .

A1A2A4A8

1.2

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 $H$ .

A5A8

1.3

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 $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} = ∥ x ∥^{2}$ .

A3A6A9

1.4

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.

A6A9

1.5

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 $H$ .

A7A8

2.1

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.

step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

The unilateral shift obstructs a cyclic linear trace extension Counterexample
Integral operator trace under a valid diagonal hypothesis Example
Positive square root of a compact positive operator Lemma
Schatten p classes Remark
A Hilbert space with a given orthonormal basis is ℓ² of the index set Theorem
Cyclicity of the trace Theorem
Existence of a maximal orthonormal family, and maximality as completeness Theorem
Fourier expansion in a Hilbert space Theorem
Hilbert Schmidt operators form a two sided ideal Theorem
L two kernels give Hilbert–Schmidt operators Theorem
Singular value decomposition for compact operators Theorem
Spectral theorem for compact self adjoint operators Theorem
The Hilbert–Schmidt norm is basis independent Theorem
The Parseval identity for Fourier series Theorem
Trace class iff product of two Hilbert Schmidt operators Theorem
Trace is absolutely convergent and basis independent Theorem

Dependency tree · two levels

62 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §2.1, pp.50–51, Theorem 2.4 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — Exercise 2.64, p.87 (standard reference, not scraped)