Theorem 4

— Let p and q be two conjugate exponents such that $1 \leq p < +\infty$. Every continuous linear form on $\mathscr{L}^p(T, \mu)$ is of the type $f \mapsto \int fg\,d\mu$, where g is a function in $\mathscr{L}^q(T, \mu)$ whose class in $L^q$ is well-defined.

Let $\theta$ be a continuous linear form on $\mathscr{L}^p$; thus, there exists a number $a \geq 0$ such that $|\theta(f)| \leq a \cdot N_p(f)$ for every function $f \in \mathscr{L}^p$. Consider the restriction of $\theta$ to the space $\mathscr{K}(T)$ of continuous functions with compact support: for every compact subset K of T and every function $f \in \mathscr{K}(T, K)$ (the space of continuous functions with compact support contained in K), one has $N_p(f) \leq (\mu(K))^{1/p} \| f \|$; therefore the topology induced on $\mathscr{K}(T, K)$ by that of $\mathscr{L}^p$ is coarser than the topology of uniform convergence, and the restriction of $\theta$ to each $\mathscr{K}(T, K)$ is consequently continuous for the latter topology. This means that the restriction of $\theta$ to $\mathscr{K}(T)$ is a real measure $\nu$ (Ch. III, §1, No. 3, Def. 2).

Let us show that $|\nu|(|f|) \leq a \cdot N_p(f)$ for every function $f$ in $\mathscr{K}(T)$. It suffices to prove this formula for $f \geq 0$. Now, for every function $\psi$ in $\mathscr{K}(T)$ such that $|\psi| \leq f$, we have

our assertion follows from the expression for the absolute value of a measure given in Ch. III, §1, No. 6, formula (12). The relation $|\nu|(|f|) \leq a (\mu(|f|^p))^{1/p}$ extends at once to the case that $f$ is the characteristic function of a compact set, by means of a passage to the lower envelope, and then implies that every $\mu$-negligible compact set is $\nu$-negligible, so that $\nu$ is a measure with base $\mu$ (No. 5, Th. 2).

Thus, there exists a locally $\mu$-integrable positive function $h_1$ such that $|\nu|(f) = \int f h_1 d\mu$ for every function $f \in \mathscr{K}(T)$. Let us show that $h_1$ is locally almost everywhere equal to a function in $\mathscr{L}^q$. If the function $f \geq 0$ in $\mathscr{K}(T)$ is such that $N_p(f) \leq 1$, then $\int f h_1 d\mu = |\nu|(f) \leq a$. For every continuous mapping $f_0$ of T into $[0,1]$ with compact support, we therefore have $\sup \int (f_0 h_1) f d\mu \leq a$ as $f$ runs over the set of functions $\geq 0$ in $\mathscr{K}(T)$ such that $N_p(f) \leq 1$. From this one deduces, by means of formula (11) of Chap. IV, §6, No. 4, that $N_q(f_0 h_1) \leq a$. It follows from this that $\sup_{K} N_q(\varphi_K h_1) \leq a$ as K runs over the set of compact subsets of T, and this proves our assertion (§1, Prop. 9).

Let $v$ be a universally measurable (real) function of absolute value 1 such that $\nu = v \cdot |\nu|$ (Cor. 3 of Th. 2) and let $g = v h_1$; then $\nu = g \cdot \mu$, and $g$ belongs to $\mathscr{L}^q$. For every function $f \in \mathscr{K}(T)$, we have $\theta(f) = \nu(f) = \int f g d\mu$. In other words, the continuous linear forms $\theta$ and $\theta_g$ coincide on $\mathscr{K}(T)$; they are therefore equal on $\mathscr{L}^p$, since $\mathscr{K}(T)$ is dense in $\mathscr{L}^p$, and this completes the proof.