a) For every $x\in E$, one has $\lambda x\in E$ and the mapping $u:x\mapsto \lambda x$ is a continuous endomorphism of E, whose spectrum in the Banach algebra $\mathscr{L}(E)$ is equal to the closure in $\mathbf{C}$ of the set of the $\lambda_i$, and whose norm is equal to sup$_i|\lambda_i|$;

b) The endomorphism $u$ is compact if and only if for every real number $\varepsilon >0$, the set of elements $i\in I$ such that $|\lambda_i|\geqslant \varepsilon$ is finite.

With the notation of the proposition, the endomorphism $x\mapsto \lambda x$ is called a diagonal endomorphism of $\ell^p_{\mathbf{C}}(I)$.

Let us prove the proposition. Let C = sup$_i|\lambda_i|$. Let $x\in E$. One has the inequalities

$\sum_{i\in I}|\lambda_ix_i|^p\leqslant C^p\|x\|^p$, if $p\not = +\infty$

sup$_{i\in I}|\lambda_ix_i|\leqslant C\|x\|$, if $p= +\infty$.

This proves that $\lambda x\in E$. The mapping $x\mapsto \lambda x$ is therefore an endomorphism of E, and these inequalities prove that it is of norm $\leqslant C$.

For $j\in I$, denote by $e_j$ the element $(x_i)_{i\in I}$ of $\ell^p_{\mathbf{C}}(I)$ such that $x_j= 1$ and $x_i= 0$ if $i\not =j$. We then have $u(e_j) =\lambda_je_j$ for all $j\in I$, which shows that $\lambda_j$ belongs to the spectrum of $u$. Since the spectrum of $u$ is closed, the closure in $\mathbf{C}$ of the set of the $\lambda_i$ is contained in the spectrum of $u$. Since the spectrum of $u$ is contained in the disk centered at 0 and of radius $\|u\|($I, p. 24, Theorem 1 and formula (3) of I, p. 21), we deduce the inequality $C\leqslant \|u\|$, whence $\|u\|= C$.

Conversely, if $\lambda \in \mathbf{C}$ is not adherent to the set of the $\lambda_i$, then the family $((\lambda_i-\lambda )^{-1})_{i\in I}$ is bounded. What precedes therefore shows that the linear mapping $x\mapsto ((\lambda -\lambda_i)^{-1}x_i)_{i\in I}$ is an endomorphism of E. It is inverse to $u-\lambda 1_E$, and hence $\lambda$ does not belong to the spectrum of $u$. This proves a).

Let us now prove b). Suppose that the endomorphism $u$ is compact. Let $\varepsilon >0$. Denote by J the set of the $i\in I$ such that $|\lambda_i|\geqslant \varepsilon$. The set of elements $e_i$ for $i\in J$ is bounded in E. Its image by $u$, formed by the elements $\lambda_ie_i$ for $i\in J$, is therefore relatively compact in E (Remark 1 of III, p. 2). Since $\|\lambda_ie_i-\lambda_je_j\|\geqslant \varepsilon$ for every pair of elements $i\not =j$ in J, this is possible only if the set J is finite.

Let us prove the converse assertion. Let J be a finite subset of I and denote by $z_J= (z_{J,i})_{i\in I}$, where $z_{J,i}=\lambda_i$ for $i\in J$ and $z_{J,i}= 0$ for $i\in I$- J ; the family $z_J$ defines an endomorphism of finite rank $u_J:x\mapsto z_Jx$ of E. According to what precedes, the norm of the linear mapping $u-u_J$ is bounded above by the upper bound in $\mathbf{R}_+$ of the family $(|\lambda_i|)_{i\in I-J}$. The assumption implies that for every $\varepsilon >0$, there exists a finite subset $J\subset I$ such that this upper bound is $\leqslant \varepsilon$. It follows that the mapping $u$ is adherent to $\mathscr{L}^f$(E), and hence that it is compact (III, p. 4, cor. of Prop. 2).

For every $y$ and $y'$ in Y, one has

By Ascoli's theorem (TG, X, p. 18, cor. 2), it follows that the image under $\widetilde{k}$ of the unit ball of $\mathscr{L}_{\mathbf{C}}^{\infty}(X;\mu)$ is relatively compact in $\mathscr{C}(Y)$.

Let us verify conditions (i) and (ii) for $k$. For every $y\in Y$, the mapping $k_y$ is continuous on the compact space X, hence $\mu$-integrable. Moreover, the mapping $y\mapsto k_y$ is continuous from Y into $\mathscr{C}(X)$ since X is compact (TG, X, p. 28, th. 3). Since $\|k_y-k_{y'}\|_1\leqslant \|\mu\| \|k_y-k_{y'}\|$ for every $(y, y')\in Y^2$, this mapping of Y into $\mathscr{L}_{\mathbf{C}}^1(X, \mu)$ is continuous. By prop. 2, the mapping $\widetilde{k}$ of $\mathscr{L}_{\mathbf{C}}^{\infty}(X, \mu)$ into $\mathscr{C}(Y)$ is compact. By composition with the canonical mapping $\mathscr{C}(X)\rightarrow \mathscr{L}_{\mathbf{C}}^{\infty}(X, \mu)$, one obtains a compact linear mapping of $\mathscr{C}(X)$ into $\mathscr{C}(Y)$.

The mapping $k$ is $(\mu\otimes \nu$)-measurable by hypothesis. Let A be a compact subset of $X\times Y$. There exist compact subsets $B\subset X$ and $C\subset Y$ such that $A\subset B\times C$. Applying (2) to the characteristic functions $\varphi_B$ of B and $\varphi_C$ of C, one obtains

whence the result (INT, V, p. 41, §5, n$^o1$, Proposition 1 and Definition 1).

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).

Let $Y'$ denote the set of $y\in Y$ such that the mapping $x\mapsto k(x, y)f(x)$ is not $\mu$-integrable. Let $y\in Y$ and let C be a compact neighbourhood of $y$, whose characteristic function is denoted by $\varphi$. By condition (2) applied to $f$ and $\varphi$, the function $(x, y)\mapsto k(x, y)f(x)\varphi (y)$ is integrable on $X\times Y$ with respect to the measure $\mu\otimes \nu$. By the Lebesgue-Fubini theorem (INT, V, p. 96, § 8, No.$^o4$, Theorem 1, a)), the function $x\mapsto k(x, y)f(x)\varphi (y)$ is $\mu$-integrable for $y$ outside a $\nu$-negligible set $Y_C$. Since $Y'\cap C\subset Y_C$, it follows that $Y'$ is locally $\nu$-negligible in Y (INT, IV, p. 172, § 5, No.$^o2$, Definition 3).

Let $h$ denote the mapping defined locally $\nu$-almost everywhere on Y by $y\mapsto \int k(x, y)f(x)d\mu$. It is locally $\nu$-integrable (INT, V, loc. cit.). Let $g\in \mathscr{K}(Y)$. We have

(INT, V, loc. cit.). Hence $u_k(f) =h$ locally $\nu$-almost everywhere (cf. INT, V, p. 46, § 5, No.$^o3$, Corollary 2).

Let $k\in \mathscr{L}^{p'}(X\times Y)$. The mapping $k$ is measurable (INT, IV, p. 84, § 5, No.$^o6$, Theorem 5). Let $f\in \mathscr{L}^p(X)$ and $g\in \mathscr{L}^p(Y)$. The function $h: (x, y)\mapsto f(x)g(y)$ belong to $\mathscr{L}^p(X\times Y)$ and $\|h\|_p=\|f\|_p\|g\|_p$ (INT, V, p. 95, § 8, No.$^o3$, Corollary 2). The function $hk$ is therefore integrable on $X\times Y$ and we have

by Hölder's inequality (INT, IV, p. 208, § 6, No.$^o4$, Theorem 2), which proves that $k\in \mathscr{N}^{p,p}(X\times Y)$ with $\|k\|_{p,p}\leqslant \|k\|_{p'}$.

Since $p'\not = +\infty$, every function $k\in \mathscr{L}^{p'}(X\times Y)$ is moderate (INT, V, p. 9, § 1, No.$^o3$, Corollary), and the last assertion then results from Proposition 3, b) and c).

For every function $k: X\times Y\rightarrow \mathbf{C}$ and every $y\in Y$, we denote by $k_y$ the mapping of X into $\mathbf{C}$ defined by $k_y(x) =k(x, y)$, and we denote by $k_{\circ}$ the mapping of Y into $\mathscr{F}(X;\mathbf{C})$ defined by $y\mapsto k_y$. Analogously, we shall denote by the same symbol the mapping of Y into the space of functions defined $\mu$-almost everywhere on X which associates to $y$ the class of $k_y$.

For every function $k$ of $X\times Y$ into $\mathbf{C}$, we put

It is an element of $[0,+\infty ]$.

Let $k\in \mathscr{L}^{p'}(X\times Y)$. By INT, V, p. 96, § 8, No.$^o4$, Theorem 1, a), the set $Y'$ of elements $y\in Y$ such that the function $|k_y|^{p'}$ is not $\mu$-integrable is $\nu$-negligible; moreover, the function defined by $y\mapsto \mu(|k_y|^{p'})$ on Y - $Y'$ and zero on $Y'$ is $\nu$-integrable and satisfies

We denote by $\mathscr{L}^{p',q'}(X,Y, \mu, \nu )$, or simply by $\mathscr{L}^{p',q'}(X,Y)$, the complex vector space of $(\mu\otimes \nu$)-measurable functions $k$ of $X\times Y$ into $\mathbf{C}$ such that, for $\nu$-almost every $y\in Y$, the mapping $k_y$ belongs to $\mathscr{L}^{p'}$(X), and such that the mapping $k_{\circ}$ belongs to $\mathscr{L}_{L^{p'}(X)}^{q'}(Y, \nu )$ (INT, IV, p. 129, § 3, No.$^o4$, Definition 2). We endow $\mathscr{L}^{p',q'}(X,Y)$ with the seminorm $k\mapsto N_{p',q'}(k)$; note that one then has $N_{p',q'}(k) =\|k_{\circ}\|_{q'}$, where $k_{\circ}$ is regarded as a mapping with values in the Banach space $L^{p'}(X)$.

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.

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: