bounded linear functionals on ${L}^{\mathrm{\infty}}(\mu )$
For any measure space^{} $(X,\U0001d510,\mu )$ and $g\in {L}^{1}(\mu )$, the following linear map can be defined
${\mathrm{\Phi}}_{g}:{L}^{\mathrm{\infty}}(\mu )\to \mathbb{R},$ | ||
$f\mapsto {\mathrm{\Phi}}_{g}(f)\equiv {\displaystyle \int fg\mathit{d}\mu}.$ |
It is easily shown that ${\mathrm{\Phi}}_{g}$ is bounded^{} (http://planetmath.org/OperatorNorm), so is a member of the dual space^{} of ${\u0141}^{\mathrm{\infty}}(\mu )$. 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) $\nu :\U0001d510\to \mathbb{R}$ by
$$\nu (A)=\mathrm{\Phi}({1}_{A})$$ |
for every set $A\in \U0001d510$, 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}g\mathit{d}\mu $ for every $A\in \U0001d510$. Then, the equality
$$\mathrm{\Phi}(fh)=\int fg\mathit{d}\mu $$ |
is satisfied for $f={1}_{A}$ with any $A\in \U0001d510$ and the functional monotone class theorem extends this to any bounded and measurable $f:X\to \u2102$, 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 |