Exercise 12, § 6

Consider a real or complex complete normable space E and a closed vector subspace F of E such that $F^0$ admits no topological supplement in $E'$.†

(a) Let A (resp. B) be the complete normable space of continuous endomorphisms of E (resp. F). Let C be the set of $u \in A$ such that $u(F) \subset F$. Let $\alpha$ be the mapping $u \mapsto (u, u|F)$ of C into $A \times B$. Then $\alpha$ is an isomorphism of the complete normable space C onto a closed vector subspace of $A \times B$ and $\alpha(C)$ admits no topological supplement in $A \times B$. (Let $x \in E$ be such that $x \notin F$. Let

† Let $c_0$ (resp. $l^1, l^\infty$) be the Banach space of real or complex sequences $(x, x_2, \ldots)$ satisfying $\lim_{n \to \infty} x_n = 0$ (resp. $\sum_n |x_n| < +\infty$, resp. $\sup_n |x_n| < +\infty$) with the norm $\|x\| = \sup_n |x_n|$ (resp. $\sum_n |x_n|$, resp. $\sup_n |x_n|$). There exists a continuous morphism $\pi$ of $l^1$ onto $c_0$; let F be its kernel. Then $(c_0)' = l^1$ is identified with $F^0$. But a vector subspace of countable type of $l^\infty$ cannot be a topological direct factor in $l^\infty$ (cf. A. Grothendieck, Sur les applications linéaires faiblement compactes d’espaces du type $C(K)$, Can. J. Math., 5 (1953), p. 169).

$\xi \in F^0$ be such that $\langle x, \xi \rangle = 1$. If $\eta \in E'$, let $\tilde{\eta}$ denote the element $y \mapsto \langle y, \eta \rangle x$ of A. Suppose that there exists a projector $\pi$ of $A \times B$ onto $\alpha(C)$. We define $\tilde{\pi}: E' \to E'$ by $\tilde{\pi}(\eta) = t(\alpha^{-1}(\pi(\tilde{\eta}, 0)))(\xi)$. Then $\tilde{\pi}$ is a projector of $E'$ onto $F^0$, which is absurd.)

(b) A unital real or complex complete normable algebra M is constructed as follows: $M = \bigoplus_{i=0}^\infty M_i$ is graded by the $M_i$; $M_0$ is the set of scalar multiples of 1; $M_1$ is the set of scalar multiples of a non-zero element u; $M_2 = \{0\}$; $M_3 = F$; $M_4 = E$; $M_i = \{0\}$ for $i \geq 5$; $ux = x$ for all $x \in M_3$. Let N be the complete normable space of continuous endomorphisms of the complete normable space M. Let $N_1$ be the set of continuous derivations of M. Show, using (a), that $N_1$ has no topological supplement in N.

(c) Show that the group of bicontinuous automorphisms of M is a Lie quasi-subgroup of $\mathbf{GL}(M)$, but not a Lie subgroup of $\mathbf{GL}(M)$. (Use Corollary 2 to Proposition 18 and (b).)