TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Reflexivity of Lp for one less p less infinity

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ . For every measure space $(S, A, μ)$ and every $1 < p < \infty$ , both $L^{p} (μ; R)$ and $L^{p} (μ; C)$ are reflexive.

Facts & Assumptions

Given: ${A C}_{ω}$ , an arbitrary measure space, $1 < p < \infty$ , and the conjugate exponent $q$ , so $1 < q < \infty$ and the conjugate exponent of $q$ is $p$ .

[F1]

A Banach space is reflexive exactly when its canonical evaluation map into the bidual is surjective (Reflexivity is surjectivity of the canonical map).

[F2]

Under Countable Choice, real $L^{r}$ duality over an arbitrary measure space identifies every member of $(L^{r})^{*}$ uniquely and isometrically with a bilinear integration density in $L^{r^{'}}$ , for $1 < r < \infty$ (For $1 < p < \infty$ , the same representation theorem holds on arbitrary measure spaces).

[F3]

Under Countable Choice, the same unique isometric bilinear-pairing identification holds for complex $L^{r}$ , $1 < r < \infty$ (Complex Lp duality from real Lp duality).

[F4]

Under Countable Choice, real $L^{r}$ is complete for every $1 \leq r \leq \infty$ (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ).

[F5]

Complex $L^{r}$ has a well-defined norm, and its real and imaginary parts have norm at most the complex norm while the complex norm is at most the sum of their norms (Complex Holder, Minkowski, and the quotient norm).

[F6]

Countable Choice is the assertion that every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · apply the $L^p$ representation theorem twice and identify the resulting map with canonical evaluation

1.1

First verify the Banach condition. Real $L^{p}$ is complete by [F4]. If $(f_{n})$ is Cauchy in complex $L^{p}$ , [F5] makes $(Re f_{n})$ and $(Im f_{n})$ Cauchy in real $L^{p}$ ; [F4] gives limits $u, v \in L^{p} (μ; R)$ . The upper component bound in [F5] gives $∥ f_{n} - (u + i v) ∥_{p} \leq ∥ Re f_{n} - u ∥_{p} + ∥ Im f_{n} - v ∥_{p} \to 0$ . Thus complex $L^{p}$ is complete as well, including the zero and empty measure spaces.

F4F5given

1.2

Fix either scalar field $K$ and write $E = L^{p} (μ; K)$ and $H = L^{q} (μ; K)$ . By [F2] in the real case and [F3] in the complex case, the map $T_{q} : H \to E^{*}$ defined by $(T_{q} h) (f) = \int f h d μ$ is a scalar-linear isometric bijection. The same theorem with $q$ in place of $p$ identifies $H^{*}$ isometrically with $L^{p} = E$ by the same bilinear formula.

F2F3given

2.1

Let $Φ \in E^{* *}$ . Since $T_{q}$ is a bounded linear map, $Φ \circ T_{q}$ lies in $H^{*}$ . The $q$ -duality assertion in step 1.2 therefore supplies $u \in E$ such that $Φ (T_{q} h) = \int h u d μ$ for every $h \in H$ . This includes $Φ = 0$ , for which uniqueness gives $u = 0$ .

step 1.2

3.1

Given any $ℓ \in E^{*}$ , surjectivity of $T_{q}$ supplies $h \in H$ with $ℓ = T_{q} h$ . Commutativity of scalar multiplication and the bilinear pairing then gives $Φ (ℓ) = Φ (T_{q} h) = \int h u = \int u h = (T_{q} h) (u) = ℓ (u) = (J_{E} u) (ℓ)$ . Hence $Φ = J_{E} u$ .

step 1.2step 2.1algebra

4.1

Every $Φ \in E^{* *}$ is therefore in the range of $J_{E}$ . Step 1.1 makes $E$ Banach, so [F1] proves reflexivity in both scalar fields. The proof uses Countable Choice only through the completeness and arbitrary-measure duality suppliers cited in steps 1.1–1.2; [F6] records that exact assumption. No Hahn–Banach or compactness principle is additionally invoked. The argument requires both $p$ and $q$ to lie strictly between one and infinity, so it makes no endpoint claim.

F1F6step 1.1step 1.2step 3.1∎

Remarks

Using the bilinear complex pairing is what makes the canonical-map calculation literal: the two scalar factors commute in $\int h u = \int u h$ . With a sesquilinear convention an explicit conjugation map would be required.

Depends on

Used by

Reflexivity of ℓᵖ and Lᵖ Example

Dependency tree · two levels

37 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 B. Folland, Real Analysis, 2nd ed. (standard reference, not scraped)
John K. Hunter, Measure Theory (standard reference, not scraped)