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On a finite-measure space, a bounded functional on $L^{p}$ defines a finite signed measure

Statement

Let $(X, A, μ)$ be a finite measure space, let $1 \leq p < \infty$ , and let $Λ : L^{p} (μ) \to R$ be a bounded linear functional. Define $ν (E) : = Λ ([1_{E}]) (E \in A) .$ Then $ν$ is a finite signed measure on $(X, A)$ .

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ , an exponent $1 \leq p < \infty$ , and a bounded linear functional $Λ : L^{p} (μ) \to R$ .

[L1]

A bounded linear functional on $L^{p} (μ)$ is linear and continuous with respect to the $L^{p}$ norm (A bounded linear functional on $L^{p} (μ)$ and its operator norm).

[L2]

A real-valued countably additive set function with finite values is a signed measure (A signed measure is countably additive and takes at most one infinite value).

[L3]

Dominated convergence applies to integrable majorants (Dominated convergence).

Proof

technique · Define $\nu(E)=\Lambda([\mathbf 1_E])$. Finite measure puts every indicator in $L^p$, and countable additivity comes from $L^p$ convergence of partial indicator sums plus continuity of $\Lambda$

1.1L1given

For every measurable $E \subseteq X$ , $\int ∣ 1_{E} ∣^{p} d μ = μ (E) \leq μ (X) < \infty,$ so $[1_{E}] \in L^{p} (μ)$ and $ν (E)$ is a finite scalar. Also $ν (\emptyset) = Λ (0) = 0$ .

1.2L3givenconstruct

Let $(E_{n})$ be pairwise disjoint measurable sets, and put $E : = ⋃_{n = 1}^{\infty} E_{n}, s_{N} : = \sum_{n = 1}^{N} 1_{E_{n}} = 1_{⋃_{n = 1}^{N} E_{n}} .$ Then $s_{N} \to 1_{E}$ pointwise and $∣ s_{N} - 1_{E} ∣^{p} = 1_{⋃_{n > N} E_{n}} \leq 1_{E} .$ Because $μ (X) < \infty$ , the majorant $1_{E}$ is integrable, so [L3] gives $∥ [s_{N}] - [1_{E}] ∥_{p}^{p} = \int ∣ s_{N} - 1_{E} ∣^{p} d μ ⟶ 0.$

2.1L1step 1.1algebra

If $A, B \in A$ are disjoint, then $1_{A \cup B} = 1_{A} + 1_{B}$ . By linearity of $Λ$ , $ν (A \cup B) = Λ ([1_{A} + 1_{B}]) = Λ ([1_{A}]) + Λ ([1_{B}]) = ν (A) + ν (B) .$ So $ν$ is finitely additive on disjoint measurable sets.

3.1L1L2step 1.1step 1.2step 2.1∎

By continuity of $Λ$ from [L1], finite additivity from step 2.1, and the convergence from step 1.2, $ν (E) = Λ ([1_{E}]) = \lim_{N \to \infty} Λ ([s_{N}]) = \lim_{N \to \infty} \sum_{n = 1}^{N} ν (E_{n}) .$ So $ν$ is countably additive. Together with step 1.1, [L2] shows that $ν$ is a finite signed measure.

On a finite-measure space, a bounded functional on $L^{p}$ defines a finite signed measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

On a finite-measure space, a bounded functional on LpLp defines a finite signed measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

On a finite-measure space, a bounded functional on $L^{p}$ defines a finite signed measure