# proof of bounded linear functionals on $L^{p}(\mu)$

If $(X,\mathfrak{M},\mu)$ is a $\sigma$-finite measure-space and $p,q$ are Hölder conjugates (http://planetmath.org/ConjugateIndex) with $p<\infty$, then we show that $L^{q}$ is isometrically isomorphic to the dual space    of $L^{p}$.

For any $g\in L^{q}$, define the linear map

 $\Phi_{g}\colon L^{p}\rightarrow\mathbb{C},\ f\mapsto\Phi_{g}(f)=\int fg\,d\mu.$

This is a bounded linear map with operator norm $\|\Phi_{g}\|=\|g\|_{q}$ (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 $L^{q}$ to the dual space of $L^{p}$. It only remains to show that it is onto.

So, suppose that $\Phi\colon L^{p}\rightarrow\mathbb{C}$ is a bounded linear map. It needs to be shown that there is a $g\in L^{q}$ with $\Phi=\Phi_{g}$. As any $\sigma$-finite measure is equivalent      to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), there is a bounded    $h>0$ such that $\int h\,d\mu=1$. Let $\tilde{\Phi}\colon L^{\infty}\rightarrow\mathbb{C}$ be the bounded linear map given by $\tilde{\Phi}(f)=\Phi(hf)$. Then, there is a $g_{0}\in L^{1}$ such that

 $\Phi(hf)=\tilde{\Phi}(f)=\int fg_{0}\,d\mu$

for every $f\in L^{\infty}$ (see bounded linear functionals   on $L^{\infty}$ (http://planetmath.org/BoundedLinearFunctionalsOnLinftymu)). Set $g=h^{-1}g_{0}$ and, for any $f\in L^{p}$, let $f_{n}$ be the sequence

 $f_{n}=f1_{\{|h^{-1}f|

As $h^{-1}f_{n}\in L^{\infty}$,

 $\|f_{n}g\|_{1}=\|h^{-1}f_{n}g_{0}\|_{1}=\Phi(\operatorname{sign}(fg_{0})f_{n})% \leq\|\Phi\|\|f_{n}\|_{p}.$

Letting $n$ tend to infinity  , dominated convergence says that $f_{n}\rightarrow f$ in the $L^{p}$-norm, so Fatou’s lemma gives

 $\|fg\|_{1}\leq\liminf_{n\rightarrow\infty}\|f_{n}g\|_{1}\leq\|\Phi\|\|f\|_{p}.$

In particular, $\|g\|_{q}\leq\|\Phi\|$ (see $L^{p}$-norm is dual to $L^{q}$ (http://planetmath.org/LpNormIsDualToLq)), so $g\in L^{q}$. As $|f_{n}g|\leq|fg|$ are in $L^{1}$, dominated convergence finally gives

 $\int fg\,d\mu=\lim_{n\rightarrow\infty}\int f_{n}g\,d\mu=\lim_{n\rightarrow% \infty}\Phi(f_{n})=\Phi(f)$

so $\Phi_{g}=\Phi$ as required.

Title proof of bounded linear functionals on $L^{p}(\mu)$ ProofOfBoundedLinearFunctionalsOnLpmu 2013-03-22 18:38:19 2013-03-22 18:38:19 gel (22282) gel (22282) 4 gel (22282) Proof msc 46E30 msc 28A25