Proposition 4

— Let E and F be two Fréchet spaces, and u a continuous linear mapping from E into F. The following conditions are equivalent :

(i) u is surjective.

(ii) For every semi-norm $p \in S(E)$, there exists $q \in S(F)$ such that we have $|f| \leq q$ for every linear form $f \in F'$ satisfying $|f \circ u| \leq p$.

(iii) For every semi-norm $p \in S(E)$, there exists $q \in S(F)$ satisfying the following property : if a linear form $f \in F'$ satisfies $|f \circ u| \leq p$, then $f$ is null on points where $q$ is null and for all $y \in F$, $r \in S(F)$, there exists $x \in E$ with $r(u(x) - y) = 0$.

(iv) For every semi-norm $p \in S(E)$, we have

We shall prove the proposition according to the following logical scheme

!Diagram showing the logical relationships between (i), (ii), (iii), and (iv)

If $u$ is surjective, it is a strict morphism (IV, p. 28, th. 1) then for every semi-norm $p \in S(E)$, there exists a semi-norm $q \in S(F)$ such that, for all $y \in F$ satisfying $q(y) \leq 1$, there exists $x \in E$ satisfying $p(x) \leq 1$ and $u(x) = y$. We deduce immediately that (i) implies (ii) and (iii). It is clear that (ii) implies (iv).

To prove that (iii) implies (iv), let $p$ and $q$ be as in (iii). Let $y \in F$, by (iii), there exists $x$ in $E$ such that $q(u(x) - y) = 0$. If $f \in F'$ satisfies $|f \circ u| \leq p$, then we have $f(u(x) - y) = 0$, hence

and the relation (1) is satisfied.

Finally we prove that (iv) implies (i). Let $p \in S(E)$ and let $q$ be the superior envelope of the functions $|f|$ for $f \in F'$ satisfying $|f \circ u| \leq p$. By (iv), $q$ is finite on $F$, and is evidently a lower semi-continuous semi-norm on $F$; since $F$ is barrelled (III, p. 25, corollary), we have $q \in S(F)$. Let $B_p$ (resp. $B_q$) denote the set of all $x \in E$ (resp. $y \in F$) such that $p(x) \leq 1$ (resp. $q(y) \leq 1$). We have $q \circ u \leq p$, and so $u(B_p) \subset B_q$. The polar of $u(B_p)$ in $F'$ consists of linear forms $f \in F'$ such that $|f \circ u| \leq p$, hence $|f| \leq q$; in other words, we get $u(B_p)^{\circ} \subset B_q^{\circ}$, and finally that $\overline{u(B_p)} = B_q$ follows from the theorem of bipolars (II, p. 45, cor. 3). If $U$ is a neighbourhood of 0 in $E$, there exists $p \in S(E)$ such that $B_p \subset U$, hence $\overline{u(U)}$ contains the neighbourhood $B_q$ of 0 in $F$. This implies that $u$ is surjective (I, p. 17, th. 1).