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.