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$L^{2}$ with the integral pairing is a Hilbert space

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, A, μ)$ be a measure space and let $L^{2} (μ)$ be the quotient of $L^{2} (μ)$ by the almost-everywhere zero functions (The space $L^{p} (μ)$ as the quotient by null functions), with the quotient norm $∥ \cdot ∥_{2}$ (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

On real $L^{2} (μ)$ the formula $⟨ f, g ⟩ : = \int f g d μ$ is a well-defined inner product, linear in both variables, positive definite, with $⟨ f, f ⟩ = ∥ f ∥_{2}^{2}$ ; with it $L^{2} (μ)$ is a real Hilbert space (Hilbert space). 2. On complex $L^{2} (μ; C)$ the formula $⟨ f, g ⟩ : = \int f \overline{g} d μ$ is a well-defined inner product, linear in the first variable and conjugate-linear in the second, positive definite, with $⟨ f, f ⟩ = ∥ f ∥_{2}^{2}$ ; with it $L^{2} (μ; C)$ is a complex Hilbert space (Complex Lp classes and Euclidean test-function conventions).

In both cases the inner product induces exactly the established quotient $L^{2}$ norm.

Facts & Assumptions

[A1]

For $f, g \in L^{2} (μ)$ one has $\int ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2} < + \infty$ , so $f g \in L^{1} (μ)$ ; the integral is unchanged when a representative is replaced by an almost-everywhere equal one, and it is linear on $L^{1}$ (Cauchy-Schwarz inequality for $L^{2}$ , Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree, The Lebesgue integral is linear on $L^{1} (μ)$ ).

[A2]

$L^{2} (μ)$ is complete for $∥ \cdot ∥_{2}$ , and $∥ \cdot ∥_{2}$ is a norm on the quotient, so the quotient metrics are the ones in which completeness is asserted (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

[A3]

For a nonnegative measurable $h$ , $\int h d μ = 0$ if and only if $h = 0$ almost everywhere; consequently $⟨ f, f ⟩ = 0$ forces $f = 0$ in $L^{2} (μ)$ (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[A4]

On complex $L^{2}$ the pairing $\int f \overline{g}$ is representative-independent, linear in the first variable, conjugate-linear in the second, conjugate symmetric, positive definite, satisfies $⟨ f, f ⟩ = ∥ f ∥_{2}^{2}$ and Cauchy–Schwarz, and complex $L^{2}$ is complete (Complex completeness, density, and inner product: the consumer interface).

[A5]

An inner product is linear in the first variable, conjugate symmetric and positive definite, and induces the norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ (Real and complex inner-product spaces and their induced length).

Proof

technique · direct

Given: Countable Choice and a measure space $(X, A, μ)$ .

1.1A1

Real case: the pairing. For $f, g \in L^{2} (μ)$ the product $f g$ is integrable by [A1], so $\int f g d μ$ is defined and depends only on the classes of $f$ and $g$ by [A1]; the assignment is bilinear by linearity of the integral on $L^{1}$ and symmetric because multiplication of real functions is commutative.

1.2A4A5

Complex case. The published complex interface [A4] states that on complex $L^{2}$ the pairing $\int f \overline{g}$ is representative-independent, linear in the first variable, conjugate-linear in the second, conjugate symmetric and positive definite with $⟨ f, f ⟩ = ∥ f ∥_{2}^{2}$ , and that complex $L^{2}$ is complete for $∥ \cdot ∥_{2}$ ; hence complex $L^{2} (μ; C)$ is a complex Hilbert space for that pairing, with the established quotient norm.

2.1step 1.1A2A3A5

The real pairing is positive definite: $⟨ f, f ⟩ = \int f^{2} d μ \geq 0$ vanishes exactly when $f^{2} = 0$ almost everywhere, that is exactly when $f$ represents the zero class, by [A3]; moreover $⟨ f, f ⟩ = \int f^{2} d μ = ∥ f ∥_{2}^{2}$ because the $L^{2}$ norm is the square root of $\int ∣ f ∣^{2}$ and $f^{2} = ∣ f ∣^{2}$ for real $f$ . Hence the real pairing is an inner product inducing the quotient norm.

3.1step 2.1A2A5

The real quotient is complete for that norm by [A2], so with this inner product real $L^{2} (μ)$ is a real Hilbert space.

4.1step 1.2step 3.1∎

Steps 3.1 and 1.2 establish both claims for every measure space under Countable Choice, the norm in each case being the established quotient $L^{2}$ norm.

Depends on

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…and 65 more results.

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis — §§1.3.3 and 5.3.1, pp.38–41 and 235–237 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §2.5, pp.63–64 (standard reference, not scraped)

L2L2 with the integral pairing is a Hilbert space