ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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Two-step functions expose the $L^{2}$ conjugation convention

Example

Assume countable choice and use Lebesgue measure on $R$ . Set $f = 1_{[0, 1)} + i 1_{[1, 2)}, g = i 1_{[0, 1)} + 1_{[1, 2)} .$ Then $∥ f ∥_{2} = ∥ g ∥_{2} = \sqrt{2}$ , but $⟨ f, g ⟩ = 0$ while $\int f g = 2 i$ . Moreover $⟨ i f, f ⟩ = 2 i$ and $⟨ f, i f ⟩ = - 2 i$ . Thus the bilinear integral is not the $L^{2}$ inner product.

Facts & Assumptions

Given: Countable choice and the two disjoint unit intervals with the displayed complex coefficients.

[F1]

The $L^{2}$ pairing conjugates the second function, is linear in the first, and equals the squared norm on the diagonal (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

[F2]

Under countable choice each of the half-open intervals has Lebesgue measure its length, here one (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F3]

The nonnegative simple integral is the sum of values times measures (The integral of a nonnegative simple function).

[F4]

Complex integrals are linear (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Verification

technique · Integrate the two disjoint constant pieces and compute both scalar placements

1.1F2F3given

F2 gives measure one to both intervals; they are disjoint, and both functions vanish elsewhere. Their squared moduli are each $1_{[0, 1)} + 1_{[1, 2)}$ , so F3 gives $∥ f ∥_{2}^{2} = ∥ g ∥_{2}^{2} = 1 + 1 = 2$ . Thus both are integrable $L^{2}$ representatives.

2.1F1F2F4step 1.1

On the first interval $f \overline{g} = 1 \cdot (- i) = - i$ and on the second it is $i \cdot 1 = i$ . F1 and F4 therefore give $⟨ f, g ⟩ = (- i) \cdot 1 + i \cdot 1 = 0$ . The bilinear product instead has value $i$ on each interval, so $\int f g = i \cdot 1 + i \cdot 1 = 2 i$ .

3.1F1F2F4step 1.1step 2.1∎

The coefficients of $i f$ are $i, - 1$ and those of $\overline{f}$ are $1, - i$ , so $(i f) \overline{f}$ has values $i, i$ . Hence $⟨ i f, f ⟩ = 2 i$ . The coefficients of $\overline{i f}$ are $- i, - 1$ , so $f \overline{i f}$ has values $- i, - i$ and $⟨ f, i f ⟩ = - 2 i$ . These computations agree with first-variable linearity and second-variable conjugate-linearity and show concretely why the bilinear expression cannot replace the inner product.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)

Two-step functions expose the L2L2 conjugation convention