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

If $(X,𝔐,\mu )$ is a $\sigma$-finite measure-space and $p,q$ are Hölder conjugates (http://planetmath.org/ConjugateIndex) with , 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

$${\mathrm{\Phi }}_g:L^p\to ℂ,f\mapsto {\mathrm{\Phi }}_g(f)=\int fg𝑑\mu .$$

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

So, suppose that $\mathrm{\Phi }:L^p\to ℂ$ is a bounded linear map. It needs to be shown that there is a $g\in L^q$ with $\mathrm{\Phi }={\mathrm{\Phi }}_g$. As any $\sigma$-finite measure is equivalent to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), there is a bounded $h>0$ such that $\int h𝑑\mu =1$. Let $\tilde{\mathrm{\Phi }}:L^{\mathrm{\infty }}\to ℂ$ be the bounded linear map given by $\tilde{\mathrm{\Phi }}(f)=\mathrm{\Phi }(hf)$. Then, there is a $g_0\in L^1$ such that

$$\mathrm{\Phi }(hf)=\tilde{\mathrm{\Phi }}(f)=\int f{g}_0𝑑\mu$$

for every $f\in L^{\mathrm{\infty }}$ (see bounded linear functionals on $L^{\mathrm{\infty }}$ (http://planetmath.org/BoundedLinearFunctionalsOnLinftymu)). Set $g=h^{-1}{g}_0$ and, for any $f\in L^p$, let $f_n$ be the sequence

As $h^{-1}{f}_n\in L^{\mathrm{\infty }}$,

$${\parallel f_{n}g\parallel }_1={\parallel h^{-1}{f}_n{g}_0\parallel }_1=\mathrm{\Phi }(\mathrm{sign}(f{g}_0)f_n)\le \parallel \mathrm{\Phi }\parallel {\parallel f_n\parallel }_p.$$

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

$${\parallel fg\parallel }_1\le \underset{n\to \mathrm{\infty }}{lim\; inf}{\parallel f_{n}g\parallel }_1\le \parallel \mathrm{\Phi }\parallel {\parallel f\parallel }_p.$$

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

$$\int fg𝑑\mu =\underset{n\to \mathrm{\infty }}{lim}\int f_{n}g𝑑\mu =\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Phi }(f_n)=\mathrm{\Phi }(f)$$

so ${\mathrm{\Phi }}_g=\mathrm{\Phi }$ as required.

Title	proof of bounded linear functionals on $L^p(\mu )$
Canonical name	ProofOfBoundedLinearFunctionalsOnLpmu
Date of creation	2013-03-22 18:38:19
Last modified on	2013-03-22 18:38:19
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 46E30
Classification	msc 28A25