Lemma 1

There exists an element $a$ of A such that the kernel of $ha-1_E$ is not reduced to 0.

There exists an element $x_0$ of E such that $h(x_0)\not = 0$. Let V be a closed neighbourhood of $h(x_0)$ not containing 0. Since the endomorphism $h$ is compact, one can choose a convex open neighbourhood U of $x_0$ such that $h(U)$ is a relatively compact subset of V. The closure C of $h(U)$ is therefore a compact convex subset of E; since it is contained in V, it does not contain 0.

Let $y$ be a point of C. Let $Ay$ denote the set of images of $y$ under the elements of A. It is a vector subspace of E stable under A; it is nonzero since it contains $y$. By assumption (i) of the theorem, the set $Ay$ is dense in E. There therefore exists an element $b$ of A such that $b(y)\in U$. Since the set C is compact, there exists a

$-1$

finite family $(b_1, . . . , b_n)$ of elements of A such that the sets $b_i(U)$ cover C. There exists a continuous partition of unity $(\varphi_1, . . . , \varphi_n)$ on C subordinate to this covering (TG, IX, p. 47, th. 3). Define a mapping $f: C\rightarrow E$ by putting

for every $x\in C$; it is a continuous mapping. Since U is convex and $b_i(x)$ belongs to U when $\varphi_i(x)$ is not zero, we have $f(C)\subset U$ and $h(f(C))\subset C$. By Leray-Schauder's theorem (Theorem 2 of III, p. 11), there exists an element $x\in C$ such that $h(f(x)) =x$. Put then

We have $h(a(x)) =h(f(x)) =x$ and $x$ is nonzero since 0 does not belong to C, hence the kernel of $ha-1_E$ is nonzero.

Let us complete the proof of Theorem 3. Let $a$ be an element of A such that the kernel T of $ha-1_E$ is nonzero (Lemma 1). Let $i$ denote the canonical injection of T into E. The linear mapping $i$ is equal to $h\circ a\circ i$, and is therefore compact. Hence the dimension of T is finite (Remark 3 of III, p. 2). Let $u$ be an element of $\mathscr{L}(E)$ which commutes with A. Since $u$ commutes with $h\circ a$, we have $u(T)\subset T$. Since T is of nonzero finite dimension over the algebraically closed field $\mathbf{C}$, the endomorphism of T induced by $u$ has an eigenvalue $\lambda$.

Put F = Ker($u-\lambda 1_E$). It is a closed nonzero vector subspace of E; it is stable under A, since $u$ commutes with the elements of A. By assumption (i), we have F = E, whence $u=\lambda 1_E$.