Proposition 2

Let E and F be locally convex spaces, let $u\in \mathscr{L}(E; F)$ and $v\in \mathscr{L}(F; E)$. Suppose that $u=u\circ v\circ u$. Then $u$ is a direct morphism. Moreover, $v\circ u$ is a continuous projector in E with kernel Ker($u$), and $u\circ v$ is a continuous projector in F with image Im($u$).

Put $p=v\circ u$ and $q=u\circ v$. One has

therefore $p$ and $q$ are projectors. They are continuous. Let $E_1$ and $E_2$ be the image and the kernel of $p$, and let $F_1$ and $F_2$ be the image and the kernel of $q$. Since $u\circ v\circ u=u$, we have

Ker($u$)$\subset$ Ker($v\circ u$)$\subset$ Ker($u\circ v\circ u$) $=$ Ker($u$),

therefore Ker($u$) $=$ Ker($v\circ u$) $=$ Ker($p$) $= E_2$. Similarly, it follows from the inclusions

Im($u$)$\supset$ Im($u\circ v$)$\supset$ Im($u\circ v\circ u$) $=$ Im($u$)

that the image of $u$ is $F_1$. The spaces $E_2$ and $F_1$ are direct vector subspaces of E and F respectively.

The vector space E is the topological direct sum of $E_1$ and $E_2$, and $u$ defines by restriction a bijective linear mapping $u_1$ of $E_1$ onto $F_1$. For every $x\in E_1$, we have $v(u(x)) =p(x) =x$, and the mapping $u^{-1}_1$ of $F_1$ onto $E_1$ coincides with $v$ on $F_1$. Consequently, $u_1$ is an isomorphism of $E_1$ onto $F_1$ and $u$ is a strict morphism of E into F.

Let E and F be locally convex spaces and let $u\in \mathscr{L}(E; F)$. In order that $u$ be a direct injective morphism (resp. surjective), it is necessary and sufficient that there exist $v$ in $\mathscr{L}(F; E)$ such that $v\circ u= 1_E$ (resp. $u\circ v= 1_F$) (cf. TG, III, p. 47 and 48).