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

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

 $\displaystyle\Phi_{g}\colon L^{\infty}(\mu)\rightarrow\mathbb{R},$ $\displaystyle f\mapsto\Phi_{g}(f)\equiv\int fg\,d\mu.$

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

###### Theorem.

Let $(X,\mathfrak{M},\mu)$ be a $\sigma$-finite (http://planetmath.org/SigmaFinite) measure space and $V$ be the space of bounded linear maps $\Phi\colon L^{\infty}(\mu)\rightarrow\mathbb{R}$ satisfying bounded convergence. That is, if $|f_{n}|\leq 1$ are in $L^{\infty}(\mu)$ and $f_{n}(x)\rightarrow 0$ for almost every $x\in X$, then $\Phi(f_{n})\rightarrow 0$.

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

###### Proof.

First, the operator norm $\|\Phi_{g}\|$ 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 $g\mapsto\Phi_{g}$ gives an isometric embedding from $L^{1}$ into the dual of $L^{\infty}$. Furthermore, dominated convergence implies that $\Phi_{g}$ satisfies bounded convergence so $\Phi_{g}\in V$. We just need to show that $g\mapsto\Phi_{g}$ maps onto $V$.

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

 $\nu(A)=\Phi(1_{A})$

for every set $A\in\mathfrak{M}$, the bounded convergence property for $\Phi$ 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 $\nu(A)=\int_{A}g\,d\mu$ for every $A\in\mathfrak{M}$. Then, the equality

 $\Phi(fh)=\int fg\,d\mu$

is satisfied for $f=1_{A}$ with any $A\in\mathfrak{M}$ and the functional monotone class theorem extends this to any bounded and measurable $f\colon X\rightarrow\mathbb{C}$, giving $\Phi_{g}=\Phi$. ∎

Title bounded linear functionals  on $L^{\infty}(\mu)$ BoundedLinearFunctionalsOnLinftymu 2013-03-22 18:38:08 2013-03-22 18:38:08 gel (22282) gel (22282) 5 gel (22282) Theorem msc 28A25 BoundedLinearFunctionalsOnLpmu RadonNikodymTheorem LpNormIsDualToLq