Proposition 3

a) Let $k\in \mathscr{N}^{p,q}(X\times Y)$. For $f\in \mathscr{L}^p(X)$ and $g\in \mathscr{L}^q(Y)$, the function

is $(\mu\otimes \nu )$-integrable and there exists a unique mapping $u_k$ from $L^p(X)$ into $L^{q'}(Y)$ such that

for all $f\in L^p(X)$ and $g\in L^q(Y)$. The mapping $u_k$ is linear and continuous, and its norm is $\leqslant \|k\|_{p,q}$;

b) Let $k\in \mathscr{N}^{p,q}(X\times Y)$. One has $u_k= 0$ if and only if $k$ is locally $(\mu\otimes \nu )$-negligible;

c) The mapping which associates $u_k$ with $k$ is a continuous linear mapping of the semi-normed space $\mathscr{N}^{p,q}(X\times Y)$ into the Banach space $\mathscr{L}(L^p(X); L^{q'}(Y))$, such that $\|u_k\|\leqslant \|k\|_{p,q}$ for every $k\in \mathscr{N}^{p,q}(X\times Y)$.

Let $k\in \mathscr{N}^{p,q}(X\times X)$. By definition, the function from $X\times Y$ into $\mathbf{C}$ defined by $(x, y)\mapsto k(x, y)f(x)g(y)$ is $(\mu\otimes \nu$)-integrable when $f\in \mathscr{L}^p(X)$ and $g\in \mathscr{L}^q$(Y), and the mapping

is a continuous bilinear mapping from $L^p(X)\times L^q(Y)$ into $\mathbf{C}$. For every $f\in L^p$(X), there therefore exists a unique $h\in L^{q'}(Y)$ such that $b_k(f, g) =$ $\langle g, h\rangle$ for every $g\in L^q(Y)$; let us write $u_k(f) =h$. By (2), one has $\|u_k(f)\|_{q'}\leqslant \|k\|_{p,q}\|f\|_p$; the mapping $u_k$ is linear and continuous, of norm $\leqslant \|k\|_{p,q}$. Moreover, the mapping $k\mapsto u_k$ is linear, and assertions a) and c) follow from this.

Let us prove b). Suppose that $k$ is locally $(\mu\otimes \nu$)-negligible. Let $f\in \mathscr{L}^p(X)$ and $g\in \mathscr{L}^q(Y)$; the function defined on $X\times Y$ by $(x, y)\mapsto k(x, y)f(x)g(y)$ is $(\mu\otimes \nu$)-moderate. Since it is locally $(\mu\otimes \nu$)-negligible, it is negligible (INT, V, p. 7, § 1, No.$^o2$, Corollary 1). Thus, one has $\langle u_k(f), g\rangle =b_k(f, g) = 0$. It follows that $u_k= 0$.

Conversely, let $k\in \mathscr{N}^{p,q}(X\times Y)$ be such that $u_k= 0$. The function $k$ is locally $(\mu\otimes \nu$)-integrable (Lemma 1). For all functions $f\in \mathscr{K}(X)$ and $g\in \mathscr{K}$ (Y), one has

therefore the measure $k\cdot (\mu\otimes \nu )$ on $X\times Y$ is zero (cf. INT, III, p. 82, § 4, No.$^o1$, Theorem 1). This means that $k$ is locally $(\mu\otimes \nu$)-negligible (INT, V, p. 46, §5, No.$^o3$, Corollary 2).