Proposition 6

a) The space $\mathscr{L}^{p'}(X\times Y)$ is contained in $\mathscr{L}^{p',p'}(X,Y)$, and for $k\in \mathscr{L}^{p'}(X\times Y)$, one has $N_{p', p'}(k) =\|k\|_{p'}$;

b) The space $\mathscr{L}^{p',q'}(X,Y)$ is contained in $\mathscr{N}^{p,q}(X\times Y)$ and for $k\in \mathscr{L}^{p',q'}(X,Y)$, one has $\|k\|_{p,q}\leqslant N_{p',q'}(k)$.

Let us prove assertion a). Let $k\in \mathscr{L}^{p'}(X\times Y)$. By INT, V, p. 96, § 8, n$^o4$, th. 1, a), the function $k_y$ belongs to $\mathscr{L}^{p'}(X)$ for $\nu$-almost all $y\in Y$.

Let $\varepsilon >0$. There exists $h\in \mathscr{K}(X\times Y)$ such that $\|k-h\|_{p'}< \varepsilon$. The function $h_{\circ}$ is a continuous function with compact support from Y into $L^{p'}(X)$ satisfying

by Lemma 2. It follows that $k_{\circ}$ belongs to $\mathscr{L}_{L^{p'}(X)}^{p'}(Y, \nu )$, that is to say that $k$ belongs to $\mathscr{L}^{p',p}(X,Y)$. Lemma 2 then proves that $N_{p', p'}(k) =\|k\|_{p'}$.

Let us prove assertion b). Let $k\in \mathscr{L}^{p',q'}(X,Y)$ and let $f\in \mathscr{L}^p(X)$. For every $y\in Y$, Hölder's inequality (INT, IV, p. 208, § 6, n$^o4$, th. 2) applied to $k_y$ implies

Let $g\in \mathscr{L}^{q'}(Y)$. The mappings $f$ and $g$ are moderate, since $p$ and $q'$ are finite. Consequently, the mapping $(x, y)\mapsto f(x)g(y)$ is $(\mu\otimes \nu )$-moderate (cf. INT, V, p. 92, § 8, n$^o3$, cor. 2), and therefore the mapping $(x, y)\mapsto k(x, y)f(x)g(y)$ is $(\mu\otimes \nu$)-moderate. By INT, V, p. 93, § 8, n$^o3$, prop. 7, a), it follows that

again using Hölder's inequality. This concludes the proof.