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A closed L2 subspace with trivial orthogonal complement fills L2

Statement

Assume the Axiom of Choice. Let $(X, A, μ)$ be a measure space, let $K$ be $R$ or $C$ , let $L^{2} (μ)$ be the quotient space of $K$ -valued square-integrable functions modulo the almost-everywhere null functions, with the norm of The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ and the inner product $⟨ f, g ⟩ = \int f \overline{g} d μ$ (bilinear in the real case, linear in the first variable and conjugate-linear in the second in the complex case), and let $V \subseteq L^{2} (μ)$ be a closed linear subspace. If $V^{⊥} = {0}$ , where $V^{⊥} = {u \in L^{2} (μ) : ⟨ u, v ⟩ = 0 for all v \in V}$ , then $V = L^{2} (μ)$ .

Facts & Assumptions

Given: AC, a measure space $(X, A, μ)$ , a closed linear subspace $V \subseteq L^{2} (μ)$ with $V^{⊥} = {0}$ , and an element $x \in L^{2} (μ)$ .

[F1]

Hilbert structure and choice. AC supplies Countable Choice, under which the integral pairing gives a complete real or complex Hilbert space with the quotient norm. The Axiom of Choice AC supplies countable selections and prescribed serial paths $L^{2}$ with the integral pairing is a Hilbert space

[F2]

Parallelogram identity from the pairing. Expanding the pairing gives $∥ u + v ∥^{2} = ∥ u ∥^{2} + 2 Re ⟨ u, v ⟩ + ∥ v ∥^{2}$ and the analogous minus identity; their sum is $2 ∥ u ∥^{2} + 2 ∥ v ∥^{2}$ . This holds over both scalar fields. $L^{2}$ with the integral pairing is a Hilbert space

[F3]

Distance and continuity. Since $0 \in V$ , the set of distances from x to V is nonempty, bounded below by zero and has finite infimum d. The reverse triangle inequality implies continuity of the norm. Every nonempty set bounded below has an infimum The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$

[F4]

Minimizing sequence. For every integer n>=1, $\sqrt{d^{2} + 1 / n} > d$ , so the infimum property gives some $v \in V$ with $∥ x - v ∥^{2} < d^{2} + 1 / n$ . Countable Choice supplied by AC selects one such v_n for every n. Every nonempty set bounded below has an infimum The Axiom of Choice AC supplies countable selections and prescribed serial paths

[F5]

Linear structure. The given V is a linear subspace, hence closed under midpoints and real multiples, and under multiplication by i in the complex case. The quotient is a normed vector space. The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$

Proof

technique · direct

1.1

Choose a minimizing sequence $(v_{n})_{n \geq 1} \subseteq V$ with $∥ x - v_{n} ∥_{2}^{2} \leq d^{2} + 1 / n$ by [F4], where $d = \inf_{v \in V} ∥ x - v ∥_{2}$ .

F3F4

2.1

The sequence is Cauchy: applying the parallelogram law [F2] to $u = x - v_{n}$ and $v = x - v_{m}$ gives $∥ v_{n} - v_{m} ∥_{2}^{2} = 2 ∥ x - v_{n} ∥_{2}^{2} + 2 ∥ x - v_{m} ∥_{2}^{2} - 4 ∥ x - (v_{n} + v_{m}) / 2 ∥_{2}^{2}$ , and $(v_{n} + v_{m}) / 2 \in V$ by convexity, so $∥ x - (v_{n} + v_{m}) / 2 ∥_{2}^{2} \geq d^{2}$ ; hence $∥ v_{n} - v_{m} ∥_{2}^{2} \leq 2 (d^{2} + 1 / n) + 2 (d^{2} + 1 / m) - 4 d^{2} = 2 / n + 2 / m \to 0$ .

F2F5step 1.1

3.1

The limit lies in $V$ : by [F1] the complete space $L^{2} (μ)$ contains a limit $v$ of $(v_{n})$ ; since $V$ is closed, $v \in V$ , and by continuity of the norm $∥ x - v ∥_{2} = d$ .

F1step 2.1

4.1

Orthogonality by perturbation: for every $w \in V$ and every real $t$ , $v + t w \in V$ by [F5], so $∥ x - v ∥_{2}^{2} \leq ∥ x - v - t w ∥_{2}^{2} = ∥ x - v ∥_{2}^{2} - 2 t Re ⟨ x - v, w ⟩ + t^{2} ∥ w ∥_{2}^{2}$ ; the quadratic in $t$ is nonnegative with value $0$ at $t = 0$ only if its linear coefficient vanishes, so $Re ⟨ x - v, w ⟩ = 0$ . In the complex case apply the same argument with the real parameter $t$ to $i w$ (which lies in $V$ by [F5]) to get $Re ⟨ x - v, i w ⟩ = Im ⟨ x - v, w ⟩ = 0$ ; hence $⟨ x - v, w ⟩ = 0$ in both cases.

F1F2F5step 3.1

5.1

Conclusion: step 4.1 shows $x - v \in V^{⊥} = {0}$ , so $x = v \in V$ ; since $x \in L^{2} (μ)$ was arbitrary, $L^{2} (μ) \subseteq V$ , and $V \subseteq L^{2} (μ)$ by definition, so $V = L^{2} (μ)$ .

step 3.1step 4.1given

6.1

If x belongs to V, the constant sequence v_n=x is minimizing. If V={0}, its orthogonal complement is the whole Hilbert space, so the hypothesis forces the Hilbert space to be zero and the conclusion follows. This does not force the underlying measure to vanish: on a singleton of measure infinity, the only square-integrable function is zero although the measure is nonzero. The argument needs neither separability nor an orthonormal basis nor a projection theorem. AC supplies both the Countable Choice inherited in completeness and the selection in [F4]; no choice of projections for a family of x is made. The complex sign in step 4.1 follows from conjugate-linearity in the second variable.

F1F4step 4.1step 5.1∎

Source notes

Van der Vaart, Theorem 6.6, uses this closed-subspace fact as the first step of the Brownian martingale representation theorem: if the range of the terminal Ito integral has trivial orthogonal complement, then it fills the mean-zero $L^{2}$ space. The proof above is the standard nearest-point argument through the parallelogram law and the perturbation characterization of orthogonality.

Depends on

Used by

Brownian-filtration martingale representation Theorem

Dependency tree · two levels

31 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

Aad van der Vaart, Martingales, Diffusions and Financial Mathematics (preliminary notes), Theorem 6.6 closed-range argument (standard reference, not scraped)