Proposition 7

Let $k\in \mathscr{L}^{p',q'}(X,Y)$. The linear mapping $u_k$ of $L^p(X)$ into $L^{q'}(Y)$ is compact.

Suppose first that $k$ is such that $k_{\circ}\in \mathscr{K}(Y;\mathscr{L}^{p'}$(X)), and denote by A the support of $k_{\circ}$. Let F be the subspace of $\mathscr{K}(Y,A;\mathscr{L}^{p'}(X, \mu))$ generated by the functions $y\mapsto f_2(y)f_1$, where $f_2\in \mathscr{K}(Y,A)$ and $f_1\in \mathscr{L}^{p'}(X)$. When $m\in \mathscr{K}(Y;\mathscr{L}^{p'}(X))$ satisfies $m_{\circ}\in F$, the linear mapping $u_m$ is of finite rank, therefore compact. For $m_1$ and $m_2$ in $\mathscr{K}(Y,A;\mathscr{L}^{p'}$(X)), one has moreover

$\int'1/q'$

$\leqslant \nu (A)^{1/q'}$ sup$_{y\in Y}N_{p'}(m_{1,y}-m_{2,y})$.

Since the space F is dense in $\mathscr{K}(Y,A; L^{p'}(X))$ for the topology of uniform convergence in A (INT, III, p. 41, § 1, n$^o1$, prop. 1 and INT, III, p. 46, § 1, n$^o2$, prop. 5), it follows that the linear mapping $u_k$ is the limit of a sequence of linear mappings of finite rank. It is therefore compact (cor. 1 of III, p. 4).

Consider the general case. Let $k\in \mathscr{L}^{p',q'}(X,Y)$. For every $\varepsilon >0$, there exists a mapping $k_{\varepsilon ,\circ}\in \mathscr{K}(Y;\mathscr{L}^{p'}(X))$ such that

The corresponding function $k_{\varepsilon}: X\times Y\rightarrow \mathbf{C}$ then satisfies

Since $u_{k_{\varepsilon}}$ is compact by what precedes, the mapping $u_k$ is likewise compact (prop. 2 of III, p. 4).

Put $p= 1$ and suppose $q >1$. There may exist mappings $k: X\times Y\rightarrow \mathbf{C}$ such that $k_y\in \mathscr{L}^{\infty}(X)$ for every $y\in Y$, and such that the integral

is finite (in particular, condition (2) of III, p. 27 is valid), but such that the linear mapping $u_k$ from $L^1(X)$ into $L^{q'}(X)$ is not compact (exercise 2 of III, p. 119).

In particular, one deduces from the proposition the following corollary: