Proposition 3

— Let $\mu$ be a positive measure on X, F a real or complex Banach space, F' its strong dual, and $(z, z') \mapsto \langle z, z' \rangle$ the canonical bilinear form on $F \times F'$.

1° For every function $f \in \mathscr{L}_F^p$ ($1 \leq p \leq +\infty$),

as g runs over the set of functions in $\mathscr{L}_{F'}^q$, such that $N_q(g) \leq 1$.

2° For every function $g \in \mathscr{L}_{F'}^q$ ($1 \leq q \leq +\infty$),

as f runs over the set of functions in $\mathscr{L}_F^p$ such that $N_p(f) \leq 1$.

Let us first prove the relation (9); we distinguish between two cases.

(i) $1 \leq p < +\infty$. The relation (9) being trivial when $N_p(f) = 0$ (because f and $\langle f, g \rangle$ are then negligible), we can always suppose, on multiplying f by a scalar, that $N_p(f) = 1$. Suppose first that f is an integrable step function, $f = \sum_{k=1}^n a_k \varphi_{A_k}$, where the $A_k$ are pairwise disjoint (§ 4, No. 9, Lemma). Thus $\sum_{k=1}^n |a_k|^p \mu(A_k) = 1$ by hypothesis. For every $\varepsilon > 0$, there exists (for every index k) a vector $a'_k \in F'$ such that $|a'_k|^q = |a_k|^p$ if $p > 1$ (resp. $|a'_k| = 1$ if $p = 1$) and $\langle a_k, a'_k \rangle \geq (1 - \varepsilon)|a_k| \cdot |a'_k|$ (TVS, IV,

§1, No. 3, Prop. 8). Setting $g = \sum_{k=1}^n a_k' \varphi_{A_k}$, we have $\sum_{k=1}^n |a_k'|^q \mu(A_k) = 1$ if $p > 1$ (resp. $\sup_{1 \leq k \leq n} |a_k'| = 1$ if $p = 1$), thus $N_q(g) = 1$; on the other hand,

and, since $|a_k'| = |a_k|^{p/q} = |a_k|^{p-1}$ if $p > 1$ (resp. $|a_k'| = 1$ if $p = 1$), we have

which proves the relation (9) in this case.

Let us pass to the case that $f$ is any element of $\mathscr{L}_F^p$ such that $N_p(f) = 1$. For every $\varepsilon > 0$, there exists a step function $f_1 \in \mathscr{L}_F^p$ such that $N_p(f - f_1) \leq \varepsilon$ (§ 4, No. 10, Cor. 1 of Prop. 19). By what we have just seen, there exists a function $g \in \mathscr{L}_F^q$, such that $N_q(g) = 1$ and

Now,

and, by (6),

whence

which proves (9).

(ii) $p = +\infty$. We may again restrict ourselves to the case that $N_\infty(f) > 0$. Let $\alpha$ be any number such that $0 < \alpha < N_\infty(f)$; by hypothesis, the set of $x \in X$ such that $|f(x)| > \alpha$ is measurable and is not locally negligible, therefore it contains a compact set $K$ of measure $> 0$. Since $f$ is measurable, there exists a compact set $K_1 \subset K$ of measure $> 0$, such that the restriction of $f$ to $K_1$ is continuous. It follows that, for every $\varepsilon > 0$, there exists a partition of $K_1$ into a finite number of integrable sets, in each of which the oscillation of $f$ is $\leq \varepsilon$; at least one of these sets $A$ has measure > 0. Let $a$ be one of the values of $f$ in $A$; then $|a| > \alpha$ and $|f(x) - a| \leq \varepsilon$ for all $x \in A$. There exists a vector $a' \in F'$ such that $|a'| = 1$ and $|\langle a, a' \rangle| \geq |a| - \varepsilon$; the function $g = \varphi_A \cdot a'/\mu(A)$ is integrable and $N_1(g) = 1$; on the other hand,

Now, one can write

and since

we see that

since $\varepsilon$ is arbitrary and $\alpha$ is any number < $N_\infty(f)$, the relation (9) is also verified in this case.

One argues in exactly the same manner to prove the relation (10), on considering separately the case $1 \leq q < +\infty$ and the case $q = +\infty$, and using the fact that for every $z' \in F'$, $|z'| = \sup_{|z| \leq 1} |\langle z, z' \rangle|$ by the definition of the norm in $F'$.