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$L^{1} [0, 1]$ fails the Radon--Nikodym property

Statement

Assume the Axiom of Choice. The real and complex Banach spaces $L^{1} ([0, 1], λ)$ for nonatomic Lebesgue measure do not have the Radon--Nikodym property.

Facts & Assumptions

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[L1]

AC supplies Countable Choice (AC supplies the countable and dependent choices used in Banach integration). Under Countable Choice, Lebesgue measure is a complete measure and an interval has its length (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, 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).

[L2]

Restriction to a measurable set is a measure (Restriction of a measure to a measurable set, The restriction of a measure to a measurable set is a measure).

[L3]

Real $L^{1}$ is the quotient by almost-everywhere equality, its integral formula is a norm, and it is complete under Countable Choice (The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ).

[L4]

The same quotient norm and completeness statements hold for complex $L^{1}$ (Complex Lp classes and Euclidean test-function conventions, Complex Holder, Minkowski, and the quotient norm, Complex Lp completeness and almost-everywhere subsequences).

[L5]

Under AC, a Banach space has RNP exactly when all its Lipschitz curves on $[0, 1]$ are norm differentiable almost everywhere (RNP and almost-everywhere differentiability of Lipschitz curves).

Proof

technique · counterexample

Given: AC and either the real or complex scalar field.

1.1

Fix the precise $L^{1} [0, 1]$ model. Let $λ_{[0, 1]} (E) = λ (E \cap [0, 1])$ on the Lebesgue sigma-algebra of $R$ . By [L1]--[L2] this is a finite measure. We use $L^{1} (λ_{[0, 1]})$ as the same-ambient realization of $L^{1} ([0, 1], λ)$ : values outside $[0, 1]$ have zero seminorm. The real space is Banach by [L3], and the complex space is Banach by [L4]; [A1] supplies the Countable Choice required by the completeness results.

givenA1L1L2L3L4

2.1

Construct the Lipschitz curve. For $0 \leq t \leq 1$ , put $F (t) = [1_{(0, t)}]$ . Every representative is measurable and integrable. If $0 \leq s < t \leq 1$ , then

L1L3L4step 1.1construct

$∥ F (t) - F (s) ∥_{1} = ∥ [1_{(s, t)}] ∥_{1} = λ ((s, t)) = t - s .$

Thus $F$ is an isometric, and in particular one-Lipschitz, curve in either the real or complex target.

3.1

Calculate two incompatible positive difference quotients. Fix $t \in (0, 1)$ and $0 < h < 1 - t$ . The positive difference quotient is

L1L3L4step 2.1algebra

$q_{h} : = \frac{F (t + h) - F (t)}{h} = [\frac{1}{h} 1_{(t, t + h)}] .$

On $(t, t + h / 2)$ the difference $q_{h} - q_{h / 2}$ has absolute value $1 / h$ , and on $(t + h / 2, t + h)$ it again has absolute value $1 / h$ . It vanishes elsewhere up to endpoints. Consequently

$∥ q_{h} - q_{h / 2} ∥_{1} = \frac{1}{h} \frac{h}{2} + \frac{1}{h} \frac{h}{2} = 1.$

4.1

Prove failure of differentiability at every interior point. Choose, for example, $h_{n} = (1 - t) / (n + 2)$ . If $F^{'} (t)$ existed in norm, both $q_{h_{n}}$ and $q_{h_{n} / 2}$ would converge to it, so their mutual distances would tend to zero. Step 3.1 says every one of those distances is one, a contradiction. Hence $F$ is norm nondifferentiable at every $t \in (0, 1)$ .

step 3.1

5.1

Apply the Lipschitz characterization and close the boundary cases. [A1, L5, step 1.1, step 2.1, step 4.1] If either $L^{1}$ target had RNP, [L5] would make the one-Lipschitz curve $F$ norm differentiable at almost every interior point, contrary to step 4.1. Thus both targets fail RNP. Endpoint values cause no issue: $F (0) = 0$ and $F (1) = [1_{(0, 1)}]$ , while differentiability is asserted only on the interior. Open, closed, or half-open versions of the intervals define the same $L^{1}$ classes because their endpoint differences are null. The complex proof uses the same real-valued representatives inside complex $L^{1}$ . AC is used through [L5] and to supply the Countable Choice in step 1.1.

A1L1L5step 4.1∎

Depends on

Used by

Sequence ell-one versus nonatomic L-one for the RNP Remark

Dependency tree · two levels

91 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

Jeff Cheeger and Bruce Kleiner, On the differentiability of Lipschitz maps from metric measure spaces to Banach spaces (standard reference, not scraped)
Gilles Pisier, Martingales in Banach Spaces (standard reference, not scraped)

L1[0,1]L1[0,1] fails the Radon--Nikodym property