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

If $\mu$ is a positive measure on a set $X$, $1\le p\le \mathrm{\infty }$, and $g\in L^q(\mu )$, where $q$ is the Hölder conjugate of $p$, then Hölder’s inequality implies that the map $f\mapsto {\int }_{X}fg𝑑\mu$ is a bounded linear functional on $L^p(\mu )$. It is therefore natural to ask whether or not all such functionals on $L^p(\mu )$ are of this form for some $g\in L^q(\mu )$. Under fairly mild hypotheses, and excepting the case $p=\mathrm{\infty }$, the Radon-Nikodym Theorem answers this question affirmatively.

Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{M}\mathrm{,}\mu \mathrm{)}$ be a $\sigma$-finite measure space, , and $q$ the Hölder conjugate of $p$. If $\mathrm{\Phi }$ is a bounded linear functional on $L^p\mathit{}\mathrm{(}\mu \mathrm{)}$, then there exists a unique $g\mathrm{\in }{L}^q\mathit{}\mathrm{(}\mu \mathrm{)}$ such that

$$\mathrm{\Phi }(f)={\int }_{X}fg𝑑\mu$$ (1)

for all $f\mathrm{\in }{L}^p\mathit{}\mathrm{(}\mu \mathrm{)}$. Furthermore, $\mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\mathrm{=}{\mathrm{\parallel }g\mathrm{\parallel }}_q$. Thus, under the stated hypotheses, $L^q\mathit{}\mathrm{(}\mu \mathrm{)}$ is isometrically isomorphic to the dual space of $L^p\mathit{}\mathrm{(}\mu \mathrm{)}$.

If , then the assertion of the theorem remains valid without the assumption that $\mu$ is $\sigma$-finite; however, even with this hypothesis, the result can fail in the case that $p=\mathrm{\infty }$. In particular, the bounded linear functionals on $L^{\mathrm{\infty }}(m)$, where $m$ is Lebesgue measure on $[0,1]$, are not all obtained in the above manner via members of $L^1(m)$. An explicit example illustrating this is constructed as follows: the assignment $f\mapsto f(0)$ defines a bounded linear functional on $C([0,1])$, which, by the Hahn-Banach Theorem, may be extended to a bounded linear functional $\mathrm{\Phi }$ on $L^{\mathrm{\infty }}(m)$. Assume for the sake of contradiction that there exists $g\in L^1(m)$ such that $\mathrm{\Phi }(f)={\int }_{[0,1]}fg𝑑m$ for every $f\in L^{\mathrm{\infty }}(m)$, and for $n\in {\mathbb{Z}}^{+}$, define $f_n:[0,1]\to ℂ$ by $f_n(x)=\max \{1-nx,0\}$. As each $f_n$ is continuous, we have $\mathrm{\Phi }(f_n)=\phi (f_n)=1$ for all $n$; however, because $f_n\to 0$ almost everywhere and $|f_n|\le 1$, the Dominated Convergence Theorem, together with our hypothesis on $g$, gives

$$1=\underset{n\to \mathrm{\infty }}{lim}\mathrm{\Phi }(f_n)=\underset{n\to \mathrm{\infty }}{lim}{\int }_{[0,1]}{f}_{n}g𝑑m=0\text{,}$$

a contradiction. It follows that no such $g$ can exist.

Title	bounded linear functionals on $L^p(\mu )$
Canonical name	BoundedLinearFunctionalsOnLpmu
Date of creation	2013-03-22 18:32:57
Last modified on	2013-03-22 18:32:57
Owner	azdbacks4234 (14155)
Last modified by	azdbacks4234 (14155)
Numerical id	15
Author	azdbacks4234 (14155)
Entry type	Theorem
Classification	msc 28B15
Related topic	LpSpace
Related topic	HolderInequality
Related topic	ContinuousLinearMapping
Related topic	BanachSpace
Related topic	DualSpace
Related topic	ConjugateIndex
Related topic	RadonNikodymTheorem
Related topic	BoundedLinearFunctionalsOnLinftymu
Related topic	LpNormIsDualToLq