ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Projection onto the constants is the mean

Example

Assume the Axiom of Countable Choice. Let $(X, A, μ)$ be a measure space with $0 < μ (X) < \infty$ , let $K$ be $R$ or $C$ , and let $M \subseteq L^{2} (μ; K)$ be the one-dimensional subspace of classes of constant functions, spanned by $1$ with $⟨ [f], [1] ⟩ = \int_{X} f d μ$ . Then $M$ is closed and the Hilbert projection of $[f]$ onto $M$ is

$P_{M} [f] = (\frac{1}{μ (X)} \int_{X} f d μ) [1],$

the mean of $f$ ; in particular $P_{M} [f]$ is the unique constant $c$ with $\int_{X} (f - c) d μ = 0$ .

Facts & Assumptions

[A1]

$L^{2} (μ; K)$ with the pairing $⟨ [f], [g] ⟩ = \int f \overline{g}$ is a Hilbert space under countable choice, and the constants form a finite-dimensional, hence closed, subspace (The standard inner products make K n, ell two and quotient L two Hilbert spaces, A finite-dimensional normed subspace is closed).

[A2]

Cauchy–Schwarz bounds $∣ \int f \overline{g} ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ ; in particular $\int_{X} f d μ = ⟨ [f], [1] ⟩$ is finite when $μ (X) < \infty$ (Cauchy-Schwarz inequality for $L^{2}$ , The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

[A3]

The Hilbert projection is characterised by $P_{M} [f] \in M$ and $[f] - P_{M} [f] \in M^{⊥}$ (The Hilbert orthogonal projection onto a closed subspace).

[A4]

Countable Choice is the standing choice hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )), while existence and uniqueness of the projection onto a closed subspace are supplied by the Hilbert-projection interface (The Hilbert orthogonal projection onto a closed subspace).

Verification

technique · direct

Given: Countable Choice, a measure space with $0 < μ (X) < \infty$ , a class $[f] \in L^{2} (μ; K)$ and the constants $M = span {[1]}$ .

1.1

The integral $\int_{X} f d μ$ is finite by [A2], the constant $μ (X)$ is finite and positive, and $M$ is a closed one-dimensional subspace by [A1].

A1A2A4

2.1

Put $c_{0} = μ (X)^{- 1} \int_{X} f d μ$ and $m = c_{0} [1]$ ; then $m \in M$ and $⟨ [f] - m, [1] ⟩ = \int_{X} (f - c_{0}) d μ = \int_{X} f d μ - c_{0} μ (X) = 0$ , so $[f] - m \in M^{⊥}$ .

step 1.1A3algebra

3.1

By the characterisation of the Hilbert projection, $P_{M} [f] = m = (μ (X)^{- 1} \int_{X} f d μ) [1]$ ; conversely, any constant $c$ with $\int_{X} (f - c) d μ = 0$ has $[f] - c [1] \in M^{⊥}$ , so $c = c_{0}$ by uniqueness of the projection, and for a probability measure the constant is the mean of $f$ .

step 2.1A3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §1.3.3 and §5.3.1 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Lectures 15–16 (standard reference, not scraped)