Corollary 4

Let E be a complex Banach space, $\mathscr{L}(E)$ the Banach algebra of continuous endomorphisms of E and A a nonzero subalgebra of $\mathscr{L}(E)$ such that E is a simple A-pseudomodule (A, II, p. 176, Appendix).

a) Let $u$ be an endomorphism of E, not necessarily continuous, which commutes with A. Then $u$ is a homothety;

b) Let $u$ be an endomorphism of E, not necessarily continuous. For every integer $n\geqslant 1$ and every $(\xi_1, . . . , \xi_n)\in E^n$, there exists $v\in A$ such that

Let us prove a). Let $\widetilde{A}$ be the algebra obtained from A by adjunction of a unit element. Since E is a simple A-pseudomodule, it is a simple $\widetilde{A}$-module.

Let B be the commutant of $\widetilde{A}$ in the ring of endomorphisms of the $\mathbf{C}$-vector space E. The algebra B contains 1 and is the endomorphism algebra of the $\widetilde{A}$-module E. Since E is a simple $\widetilde{A}$-module, Schur’s Lemma (A, VIII, p. 43, corollary) shows that B is a field.

Let $\xi_0\in E$ be such that $A\xi_0\not=\{0\}$. One has therefore $A\xi_0= E$. For every $u\in B$, let $A_u$ be the set of $v\in A$ such that $v(\xi_0) =u(\xi_0)$. This set is nonempty, since $A\xi_0= E$. One then puts

$\|u\|_B=$ inf$_{v\in A_u}\|v\|$.

The mapping $u\mapsto \|u\|_B$ is a seminorm on B.

Let us prove that this mapping is a norm. Let $u$ be a nonzero element of B. For every $v\in A_u$, one has $\|v\|\geqslant \|v(\xi_0)\|/\|\xi_0\|=\|u(\xi_0)\|/\|\xi_0\|$, so that $\|u\|_B\geqslant \|u(\xi_0)\|/\|\xi_0\|$. It is therefore enough to prove that $u(\xi_0)\not= 0$. Let $\xi_1\in E$ be such that $u(\xi_1)\not= 0$. Since $A\xi_0= E$, there exists $w\in A$ such that $\xi_1=w(\xi_0)$. Then $wu(\xi_0) =uw(\xi_0) =u(\xi_1)\not= 0$, hence $u(\xi_0)\not= 0$.

On the other hand, let $u$ and $u'$ be elements of B. For every $\varepsilon >0$, there exist $v, v'\in A$ such that $v(\xi_0) =u(\xi_0),v'(\xi_0) =u'(\xi_0)$ and $\|v\|\leqslant \|u\|_B+\varepsilon$, $\|v'\|\leqslant \|u'\|_B+\varepsilon$. Then one has $vv'(\xi_0) =vu'(\xi_0) =u'v(\xi_0) =u'u(\xi_0)$, whence

and finally $\|u'u\|_B\leqslant \|u\|_B\|u'\|_B$. This shows that B, endowed with the norm $u\mapsto \|u\|_B$, is a normed algebra. Since it is a field, corollary 2 implies that $B =\mathbf{C}\cdot 1$, which is the desired conclusion.

Let us prove b). According to a), the commutant of A in End$_{\mathbf{C}}(E)$ is reduced to the homotheties of E. Its bicommutant is therefore End$_{\mathbf{C}}(E)$. Assertion b) therefore results from Jacobson's density theorem (theorem 1 of A, VIII, p. 434).