bounded linear functionals on $L^{\mathrm{\infty }}(\mu )$

For any measure space $(X,𝔐,\mu )$ and $g\in L^1(\mu )$, the following linear map can be defined

${\mathrm{\Phi }}_g:L^{\mathrm{\infty }}(\mu )\to ℝ,$

$f\mapsto {\mathrm{\Phi }}_g(f)\equiv {\displaystyle \int fg𝑑\mu }.$

It is easily shown that ${\mathrm{\Phi }}_g$ is bounded (http://planetmath.org/OperatorNorm), so is a member of the dual space of ${Ł}^{\mathrm{\infty }}(\mu )$. However, unless the measure space consists of a finite set of atoms, not every element of the dual of $L^{\mathrm{\infty }}(\mu )$ can be written like this. Instead, it is necessary to restrict to linear maps satisfying a bounded convergence property.

Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{M}\mathrm{,}\mu \mathrm{)}$ be a $\sigma$-finite (http://planetmath.org/SigmaFinite) measure space and $V$ be the space of bounded linear maps $\mathrm{\Phi }\mathrm{:}{L}^{\mathrm{\infty }}\mathit{}\mathrm{(}\mu \mathrm{)}\mathrm{\to }\mathrm{R}$ satisfying bounded convergence. That is, if $\mathrm{|}{f}_n\mathrm{|}\mathrm{\le }\mathrm{1}$ are in $L^{\mathrm{\infty }}\mathit{}\mathrm{(}\mu \mathrm{)}$ and $f_n\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\to }\mathrm{0}$ for almost every $x\mathrm{\in }X$, then $\mathrm{\Phi }\mathit{}\mathrm{(}{f}_n\mathrm{)}\mathrm{\to }\mathrm{0}$.

Then $g\mathrm{\mapsto }{\mathrm{\Phi }}_g$ gives an isometric isomorphism from $L^{\mathrm{1}}\mathit{}\mathrm{(}\mu \mathrm{)}$ to $V$.

Proof.

First, the operator norm $\parallel {\mathrm{\Phi }}_g\parallel$ is equal to the $L^1$-norm of $g$ (see $L^p$-norm is dual to $L^q$ (http://planetmath.org/LpNormIsDualToLq)), so the map $g\mapsto {\mathrm{\Phi }}_g$ gives an isometric embedding from $L^1$ into the dual of $L^{\mathrm{\infty }}$. Furthermore, dominated convergence implies that ${\mathrm{\Phi }}_g$ satisfies bounded convergence so ${\mathrm{\Phi }}_g\in V$. We just need to show that $g\mapsto {\mathrm{\Phi }}_g$ maps onto $V$.

So, suppose that $\mathrm{\Phi }\in V$. It needs to be shows that $\mathrm{\Phi }={\mathrm{\Phi }}_g$ for some $g\in L^1$. Defining an additive set function (http://planetmath.org/Additive) $\nu :𝔐\to ℝ$ by

$$\nu (A)=\mathrm{\Phi }(1_A)$$

for every set $A\in 𝔐$, the bounded convergence property for $\mathrm{\Phi }$ implies that $\nu$ is countably additive and is therefore a finite signed measure. So, the Radon-Nikodym theorem gives a $g\in L^1$ such that $\nu (A)={\int }_{A}g𝑑\mu$ for every $A\in 𝔐$. Then, the equality

$$\mathrm{\Phi }(fh)=\int fg𝑑\mu$$

is satisfied for $f=1_A$ with any $A\in 𝔐$ and the functional monotone class theorem extends this to any bounded and measurable $f:X\to ℂ$, giving ${\mathrm{\Phi }}_g=\mathrm{\Phi }$. ∎

Title	bounded linear functionals on $L^{\mathrm{\infty }}(\mu )$
Canonical name	BoundedLinearFunctionalsOnLinftymu
Date of creation	2013-03-22 18:38:08
Last modified on	2013-03-22 18:38:08
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Theorem
Classification	msc 28A25
Related topic	BoundedLinearFunctionalsOnLpmu
Related topic	RadonNikodymTheorem
Related topic	LpNormIsDualToLq