The set $u(U)$ is a neighbourhood of 0 in I; it is closed in F, hence the subgroup I of F is locally closed at 0, and consequently it is closed (TG, III, p. 7, prop. 4).

Since Ker($u$) meets U only at 0, we have Ker($u$) $=\{0\}$ and the mapping $u$ is injective. Let V be a closed neighbourhood of 0 in E contained in U. Since $u(\mathring{V})$ is a neighbourhood of 0 in $u$(V), there exists a balanced neighbourhood W of 0 in F such that $u(V)\cap W\subset u(\mathring{V})$. The set $\overset{-1}{u}(W)$ is balanced, hence connected. It contains 0 and is contained in $\mathring{V}\cup (E$ - V). Since $\mathring{V}$ and E- V are disjoint open subsets of E, the set $\overset{-1}{u}(W)$ is contained in $\mathring{V}$, whence $W\cap I\subset u(\mathring{V})$. Consequently $u(\mathring{V})$ is a neighbourhood of 0 in I. This implies that $u$ induces a homeomorphism of E onto I.

(i) The mapping $u$ is a strict morphism, its kernel is of finite dimension and its image is closed in F ;

(ii) There exists a closed neighbourhood V of 0 in E such that the restriction of $u$ to V is proper (TG, I, p. 72).

(i) $=\Rightarrow$ (ii) : Suppose that $u$ satisfies condition (i). Since the kernel of $u$ is of finite dimension, it has a topological supplement $E_1$ in E (III, p. 55, prop. 1), and there exists a compact neighbourhood C of 0 in Ker($u$). Identify E with $E_1\times$ Ker($u$); the set $V = E_1\times C$ is then a closed neighbourhood of 0 in E. The restriction of $u$ to V is composed of the projection of $E_1\times C$ onto $E_1$ which is proper (TG, I, p. 77, cor. 5) and of the restriction $u_1$ of $u$ to $E_1$. Now $u_1$ is a homeomorphism of $E_1$ onto a closed subspace of F, hence is proper (TG, I, p. 72, prop. 2). The composite of two proper mappings is proper (TG, I, p. 73, prop. 5, a)), hence the restriction of $u$ to V is proper.

(ii) $=\Rightarrow$ (i) : Let V be a closed neighbourhood of 0 in E such that the restriction $v$ of $u$ to V is proper. The set $V\cap$ Ker($u$) $=\overset{-1}{v}(\{0\})$ is then compact (TG, I, p. 75, th. 1); consequently, the vector space Ker($u$) is locally compact, hence of finite dimension (EVT, I, p. 15, th. 3). Let $E_1$ be a topological supplement of Ker($u$) in E (prop. 1 of III, p. 55); put $V_1= E_1\cap V$. The set $V_1$ is closed in V. The mapping $u|V_1$ is proper (TG, I, p. 74, cor. 1) and injective, hence is a homeomorphism of $V_1$ onto a closed subset of F (TG, I, p. 72, prop. 2). By Lemma 1, the restriction of $u$ to $E_1$ is a homeomorphism of $E_1$ onto a closed subspace of F, hence $u$ is a strict morphism with closed image.

According to prop. 1 of III, p. 71, there exists a closed neighbourhood V of 0 in E such that the restriction of $u$ to V is proper. Since $h$ is compact, there exists a closed neighbourhood W of 0 contained in V such that the set $h(W)$ is relatively compact. Put $C =h(W)$. The restriction of $u$ to W is proper (TG, I, p. 74, cor. 1). The mapping $\alpha :x\mapsto (u(x), h(x))$ from W into $F\times C$ is proper because the mapping pr$_1\circ \alpha =u|W$ from W into F is proper (TG, I, p. 73, prop. 5, d)). The mapping $\beta : (y, z)\mapsto (y+z, z)$ from $F\times C$ into $F\times C$ is a homeomorphism, and the mapping pr$_1$ from $F\times C$ into F is proper (TG, I, p. 77, cor. 5). The composed mapping pr$_1\circ \beta \circ \alpha$ from W into F is therefore proper (TG, I, p. 73, prop. 5, a)). But this mapping is none other than the restriction of $u+h$ to W. According to the prop. 1 of III, p. 71$,u+h$ is a strict morphism, its kernel is of finite dimension and its image is closed.

(i) The cokernel of $u$ is of finite dimension ;

(ii) The mapping $^tu: F'_c\rightarrow E'_c$ is a strict morphism with closed image whose kernel is of finite dimension.

(i) $=\Rightarrow$ (ii) : Suppose that the cokernel of $u$ is of finite dimension. The mapping $u$ is then a strict morphism (lemma 6 of III, p. 52). According to prop. 2 of EVT, IV, p. 27, the image of $^tu$ is closed in $E'$ endowed with the weak topology, and fortiori in $E'_c$. The kernel of $^tu$ is the orthogonal of the image of $u($loc. cit.) ; it is therefore of finite dimension. Finally, $^tu$ is a strict morphism from $F'_c$ into $E'_c$ according to theorem 1 of EVT, IV, p. 28.

(ii) $=\Rightarrow$ (i) : Suppose that $^tu: F'_c\rightarrow E'_c$ is a strict morphism. According to EVT, IV, p. 28, th. 1, the image of $u$ is closed. The kernel of $^tu$ is the orthogonal of Im($u$) (EVT, IV, p. 27, prop. 2) ; if this kernel is of finite dimension, the image of $u$ is of finite codimension in F.

According to lemma 6 of III, p. 52, it suffices to show that the cokernel of $u+h$ is of finite dimension. Now, according to prop. 9 of III, p. 9, the mapping $^th$ from $F'_c$ into $E'_c$ is compact ; theorem 2 results from theorem 1 and lemma 2.

We shall prove the theorem in three steps.

A) Suppose that E and F are Banach spaces. Proposition 2 of III, p. 42 shows that the linear mapping $u$ is a strict morphism with closed image whose kernel and cokernel are of finite dimension. Since E and F are Banach spaces, Theorems 1 and 2 of III, p. 73 imply that, for every $t\in [0,1]$, the linear mapping $u_t=u+th$ has these same properties, and is therefore a Fredholm mapping (III, p. 42, Proposition 2). The mapping $t\mapsto u_t$ of $[0,1]$ into the set $\mathscr{F}(E; F)$ of Fredholm mappings of E into F is continuous. By Theorem 1 of III, p. 58, the mapping $t\mapsto$ ind($u_t$) is locally constant. Since the interval $[0,1]$ of $\mathbf{R}$ is connected, this mapping is constant. Hence ind($u$) $=$ ind($u+h$).

B) Suppose that E = F and $u= 1_E$. By Proposition 5 of III, p. 5, there exist a Banach space G, a continuous linear mapping $p: E\rightarrow G$ and a compact linear mapping $q: G\rightarrow E$ such that $h=q\circ p$. The endomorphism $p\circ q$ of G is compact. By A), $1_G+p\circ q$ is a Fredholm endomorphism of G of index zero. But then $1_E+h= 1_E+q\circ p$ is a Fredholm endomorphism of E of index zero (III, p. 49, Proposition 10, f )).

C) General case. Let $v$ be a quasi-inverse of $u$. The endomorphisms $u\circ v$ and $(u+h)\circ v$ of F differ from $1_F$ by compact linear mappings. By B), they are Fredholm endomorphisms of F of index zero. It follows that $u+h$ is a Fredholm endomorphism (III, p. 40, n$^o2$) of the same index as $u($III, p. 43, n$^o3$, formula (2)).

Since a continuous linear mapping of finite rank is compact, the stated condition is necessary.

Let $v$ be an element of $\mathscr{L}(F; E)$ such that the linear mappings $1_E-v\circ u$ and $1_F-u\circ v$ are compact. By Theorem 3$,v\circ u$ and $u\circ v$ are Fredholm mappings. Let $p$ and $q$ be quasi-inverses of $v\circ u$ and $u\circ v$ respectively. Put $w=p\circ v$ and $w'=v\circ q$. With the notations of No.$^o2$ of III, p. 40, one has $w\circ u\equiv 1_E$ and $u\circ w'\equiv 1_F$, whence $w\equiv w\circ u\circ w'\equiv w'$. It follows that $w$ is a quasi-inverse of $u$.

(i) The mapping $u$ is a Fredholm mapping of index zero;

(ii) There exists an isomorphism $v$ of E onto F such that $u-v$ is of finite rank ;

(iii) There exists an isomorphism $v$ of E onto F such that $u-v$ is compact.

(i) $=\Rightarrow$ (ii) : Suppose that $u$ is a Fredholm mapping of index zero. There exist decompositions into topological direct sums $E = E_1\oplus E_2$ and $F = F_1\oplus F_2$ with $E_2$ and $F_2$ finite-dimensional such that $u$ vanishes on $E_2$ and defines by restriction an isomorphism $u_1$ of $E_1$ onto $F_1($III, p. 42, prop. 2). If ind($u$) $= 0$, then the dimensions of $E_2$ and $F_2$ are equal and there exists an isomorphism $v$ of E onto F extending $u_1$. The kernel of $u-v$ contains $E_1$, hence $u-v$ is of finite rank.

We have (ii) $=\Rightarrow$ (iii) since every continuous linear mapping of finite rank from E into F is compact, and (iii) $=\Rightarrow$ (i) according to th. 3.

Let $r\geqslant 1$ be an integer such that $\|1_E-\pi (u)\|^r<\frac{1}{2}$. Let $P\in \mathbf{C}[X]$ be the polynomial $\frac{1-(1-X)^r}{X}$. Let $v= 1_E-(1_E-u)^r$. We thus have $v=uP(u)$ and $\|1_E-\pi (v)\|<\frac{1}{2}$. Since the endomorphisms $u$ and $P(u)$ of E commute, it suffices to prove that $v$ is a Riesz endomorphism of E (III, p. 49, prop. 9).

Since $\|1_E-\pi (v)\|<1/2$, there exists, by definition of the quotient norm in the space $\mathscr{C}$alk(E), a compact endomorphism $h$ of E and an endomorphism $w$ of E such that $v= 1_E+h+w$ and $\|w\|<\frac{1}{2}$. According to Corollary 1 of I, p. 22, the element $1_E+w$ is an automorphism of E. Since $h$ is compact, $v= (1_E+w) +h$ is a Fredholm endomorphism of E of index 0 (Cor. 2 of III, p. 74). For every integer $n\geqslant 0$, denote by $N_n$ the kernel of $v^n$. To prove that $v$ is a Riesz endomorphism of E, it suffices to prove that there exists an integer $n\geqslant 0$ such that $N_n= N_{n+1}($III, p. 46, def. 2 and III, p. 46, remark).

Reason by contradiction by supposing the sequence $(N_n)$ strictly increasing. For every $n\in \mathbf{N}$, let $p_n$ be the canonical mapping of E onto the normed space $E/N_n$. Let $c$ be a real number such that $2\|w\|< c <1$. Let $n\in \mathbf{N}$. Since $N_{n+1}$ is different from $N_n$, there exists an element $x_n\in N_{n+1}$ such that $\|p_n(x_n)\|=c$ and such that $\|x_n\|<1$ (indeed, there exists $y\in N_{n+1}/N_n$ of norm $c$ so, for every $\varepsilon >0$, there exists $x_n\in N_{n+1}$ such that $p_n(x_n) =y$ and $\|x_n\|\leqslant c+\varepsilon$ ).

Let $m,n$ be two integers such that $m > n\geqslant 0$. We have

since $\|x_m\|\leqslant 1$ and $\|x_n\|\leqslant 1$. Moreover, note that

But since $n < m$ and $v(N_{m+1})\subset N_m$, we have $v(x_m-x_n) +x_n\in N_m$, whence

The inequality (1) therefore gives the lower bound

The sequence $(h(x_m))_{m\in\mathbf{N}}$ therefore has no value of adherence in E, which contradicts the fact that the image by $h$ of the unit ball of E is relatively compact, since $h$ is compact (remark 1 of III, p. 2).

Since every continuous linear mapping of finite rank is compact, the condition is necessary (III, p. 47, prop. 6 d)). Let us prove that it is sufficient. Let $v$ be an element of $\mathscr{L}(E)$ which commutes with $u$ and such that the endomorphism $h= 1_E-u\circ v$ of E is compact. There exist a Banach space G, a continuous linear mapping $p: E\rightarrow G$ and a compact linear mapping $q: G\rightarrow E$ such that $h=q\circ p($III, p. 5, prop. 5). The endomorphism $p\circ q$ of G is compact. According to prop. 2, the linear mapping $1_G-p\circ q$ is a Riesz endomorphism of G. But then $u\circ v= 1_E-q\circ p$ is a Riesz endomorphism of E (III, p. 49, prop. 10, g)). Since $u$ and $v$ commute, $u$ is a Riesz endomorphism of E (III, p. 49, prop. 9).

Let $v$ be the canonical quasi-inverse of $u$. It commutes with $u$ and with $h($III, p. 47, prop. 6), hence with $u+h$. The endomorphism $1_E-(u+h)\circ v$ of E is the sum of the continuous linear mapping of finite rank $1_E-u\circ v$ and the compact linear mapping $-h\circ v$, hence is compact. It follows that $u+h$ is a Riesz endomorphism of E (corollary 1).

Since $\pi (A)$ is a unital subalgebra of $\mathscr{C}$alk(E), its closure B is a Banach algebra. If $u$ is a Riesz endomorphism of E, the element $\pi (u)$ possesses an inverse in $\pi (A)$ (cor. 1), hence in B. Conversely, suppose that $\pi (u)$ has an inverse $s$ in B. By definition of B, there exists an element $v$ of A such that

whence $\|1_E-\pi (u\circ v)\|<1$. According to prop. 2, the linear mapping $u\circ v$ is a Riesz endomorphism of E. The same is true of $u$ since $u$ and $v$ commute (III, p. 49, prop. 9).

Let $\mathscr{B}$ be a countable basis of the topology of X and let D be a discrete subset of X. For each element $x$ of D, there exists an element $B_x$ of $\mathscr{B}$ such that $B_x\cap D =\{x\}$. The mapping from D into $\mathscr{B}$ defined by $x\mapsto B_x$ is injective, whence the lemma.

The set Σ is also the set of $\lambda \in K$ such that $1_E-\lambda h$ is not injective and the set of $\lambda \in K$ such that $1_E-\lambda h$ is not surjective.

There exist a Banach space G, a continuous linear mapping $p: E\rightarrow G$ and a compact linear mapping $q: G\rightarrow E$ such that $h=q\circ p($III, p. 5, prop. 5). The endomorphism $h'=p\circ q$ of G is compact, and for $1_E-\lambda h$ to be an automorphism of E it is necessary and sufficient that $1_G-\lambda h'$ be an automorphism of G (prop. 10 of III, p. 49, e)). It is therefore enough for us to prove the theorem when E is a Banach space, which we henceforth suppose.

Let $\lambda \in K$. Since the index of a Riesz endomorphism is zero, it amounts to the same thing to say that the mapping $1_E-\lambda h$ is an automorphism of E, or that it is injective, or again that it is surjective.

The set Σ is closed in K since the set of automorphisms of E is open in $\mathscr{L}(E)$. Let $\lambda$ be an element of Σ. We have $\lambda \not = 0$. Denote by N the nilspace of $h$ and by I its conilspace, and denote by $h_I$ and $h_N$ the endomorphisms of I and N which $h$ defines by passing to the subspaces. Then $1_I-\lambda h_I$ is an automorphism of I, and there exists a neighbourhood U of $\lambda$ in K such that $1_I-\mu h_I$ is an automorphism of I for every $\mu\in U$. The endomorphism $1_N-\lambda h_N$ of N is nilpotent; for every $\mu\not =\lambda$, we have

therefore $1_N-\mu h_N$ is an automorphism of N. Consequently, the set $U\cap \Sigma$ is reduced to the single element $\lambda$ and the set Σ is discrete. It is countable by the lemma 3.

a) The spaces F and G have the same finite dimension $n\geqslant 0$;

b) For $y\in E$, there exists $x\in E$ such that $x-\lambda h(x) =y$ if and only if $\langle y, \ell \rangle = 0$ for all $\ell \in G$;

c) For $\ell \in E'$, there exists $f\in E'$ such that $f-\lambda^th(f) =\ell$ if and only if $\langle x, f\rangle = 0$ for all $x\in F$.

In particular, one and only one of the following conditions is valid:

(i) The space F is nonzero ;

(ii) For every $y\in E$, there exists $x\in E$ such that $x-\lambda h(x) =y$.

Replacing $h$ by $\lambda h$, we reduce to the case where $\lambda = 1$.

The endomorphism $^th$ is compact by corollary 1 of III, p. 9. The endomorphisms $w= 1_E-h$ of E and $w'= 1_{E'}-^th$ of $E'$ are therefore Riesz endomorphisms (cor. 2 of prop. 2 of III, p. 75). In particular, their kernels F and G have finite dimension. Since $w'=^tw$ and the index of $w$ is zero, the dimension of F is equal to that of G by formula (3) of III, p. 43.

For an element $y$ of E, the equation $x-h(x) =y$ has a solution $x\in E$ if and only if $x$ belongs to the image of $w$. Since the dual of the cokernel of $w$ identifies with the kernel G of Ker($^tw$) (EVT, IV, p. 27, prop. 2), this is the case if and only if $\langle y, \ell \rangle = 0$ for all $\ell \in G$.

For an element $\ell$ of $E'$, the equation $f-^th(f) =\ell$ has a solution $f\in E'$ if and only if $f$ belongs to the image of $w'$. Since the image of $^tw$ is the orthogonal of the kernel F of $w$ (EVT, IV, p. 27, prop. 2) this is the case if and only if $\langle x, f\rangle = 0$ for all $x\in F$.

If $g\in G_{\lambda}$ satisfies $g\cdot \mu= 0$, then formula (5) implies $g= 0$, hence the mapping $g\mapsto g\cdot \mu$ is injective.

Let $m\in E'$ be a measure belonging to the kernel of $1_{E'}-\lambda^th$. According to relation (6), one has

for every continuous function $f: X\rightarrow K$. The measure $m$ is therefore equal to the measure $g\cdot \mu$, where $g$ is the continuous function from X into K defined by

for every $y\in X$. Formula (8) then implies that the function $g$ belongs to $G_{\lambda}$.

Conversely, let $g: X\rightarrow K$ be a continuous function belonging to $G_{\lambda}$ and put $m=g\cdot \mu$. For every continuous function $f: X\rightarrow K$, we therefore have

so that $^th(m) =\lambda m$. This proves Lemma 4.

By applying Proposition 4 and Theorem 4 of III, p. 78. to the endomorphism $h$, one then obtains the following statements.

a) The sets $F_{\lambda}$ and $G_{\lambda}$ are finite-dimensional vector subspaces of $\mathscr{C}(X; K)$ and their dimensions are equal ;

b) In order that equation (3) have at least one solution $f\in \mathscr{C}(X,K)$, it is necessary and sufficient that one have $\int_Xa(x)g(x)d\mu(x) = 0$ for every function $g\in G_{\lambda}$. The set of solutions of (3) is then an affine subspace of $\mathscr{C}(X; K)$ with direction $F_{\lambda}$;

c) One of the following conditions, which exclude each other, is satisfied :

(i) For every function $a\in \mathscr{C}(X; K)$, there exists a unique

solution $f\in \mathscr{C}(X; K)$ of the integral equation (3) ;

(ii) The homogeneous equation (4) possesses a nonzero solution