(MS) For every semi-norm $p \in S(E)$, which is null on the kernel of u, there exists q in S(F) such that $p \leq q \circ u$.

Let N be the kernel and M the image of u; we introduce the canonical decomposition of u, let

The continuous semi-norms on E which are null on N, are the semi-norms $p_1 \circ \pi$ where $p_1$ ranges over S(E/N); similarly S(M) consists of the semi-norms $q_1$ for which there exists $q \in S(F)$ with $q_1 \leq q/F$. Finally, u is a strict morphism if and only if the bijective continuous linear mapping $\tilde{u}$ has a continuous inverse; this also means that every semi-norm in S(E/N) is of the form $q_1 \circ \tilde{u}$ with $q_1$ in S(M). Prop. 1 follows immediately from these remarks.

a) The image of $^t u$ is closed in $E'$ for $\sigma(E', E)$.

b) Every equicontinuous subset of $E'$, contained in the image of $^t u$ is the image under $^t u$ of an equicontinuous subset of $F'$.

If this is so, we have $\operatorname{Ker} ^t u = (\operatorname{Im} u)^{\circ}$ and $\operatorname{Im} ^t u = (\operatorname{Ker} u)^{\circ}$ and there exist canonical isomorphisms from Coker $^t u$ onto the dual of Ker u and from Ker $^t u$ onto the dual of Coker u.

Let N be the kernel and I the image of u. By cor. 2 of II, p. 47, the kernel of $^t u$ is the orthogonal of I, and the closure of the image of $^t u$ for $\sigma(E', E)$ is the orthogonal $N^{\circ}$ of N. The conjunction of a) and b) is then equivalent to the following condition:

b') Every equicontinuous subset of $E'$ contained in $N^{\circ}$ is the image under $^t u$ of an equicontinuous subset of $F'$.

Since $N^{\circ}$ can be identified with the dual of E/N, prop. 9, (i) of IV, p. 8, shows that the equicontinuous subsets of $E'$ contained in $N^{\circ}$ are the sets which are contained in a set of the form $H_p$, where p is a continuous semi-norm on E, vanishing on N. The condition b') then says that, for every semi-norm $p \in S(E)$ which is null on N, there exists $q \in S(F)$ such that $H_p \subset ^t u(H_q)$. By Hahn-Banach theorem (II, p. 23, cor. 1 and 2, p. 63, th. 1 and cor. 1), we have $^t u(H_q) = H_{q \circ u}$, and the relations $H_p \subset H_{p'}$ and $p \leq p'$ are equivalent for all semi-norms $p$ and $p'$ in S(E). Consequently, the relation $H_p \subset ^t u(H_q)$ is equivalent to the relation $p \leq q \circ u$. By prop. 1, property b') implies that u is a strict morphism.

Suppose that u is a strict morphism. We have already seen that the kernel of $^t u$ is the orthogonal of I and the image of $^t u$ is the orthogonal of N. The cokernel of $u$ is the space $F/I$ and its dual can be identified with $I^\circ = \mathrm{Ker}\ ^t u$. Similarly, the dual of $N = \mathrm{Ker}\ u$ can be identified with $E'/N^\circ$ (IV, p. 8), i.e. with the cokernel of $^t u$ since $N^\circ$ is the image of $^t u$.

The necessity follows from prop. 2.

Conversely, suppose that the image of $^t u$ is closed for $\sigma(E', E)$ and introduce the canonical decomposition of $u$ as in the proof of prop. 1. By the above remarks, the inverse mapping $\tilde{u}^{-1}$ of $\tilde{u}$ is continuous for the weakened topologies. But the subspace $M = u(E)$ of F is metrizable, hence bornological (III, p. 12, prop. 2); consequently (IV, p. 7, prop. 7, (ii)), $\tilde{u}^{-1}$ is continuous, hence $u$ is a strict morphism.

a) $u$ is a strict morphism. b) $u$ is a strict morphism for the weakened topologies. c) The image of $u$ is closed in F. d) $^t u$ is a strict morphism from $F'$ into $E'$ for the weak topologies. e) The image of $^t u$ is closed in $E'$ for the weak topology $\sigma(E', E)$. f) The image of $^t u$ is closed in $E'$ for the strong topology $\beta(E', E)$. g) $^t u$ is a strict morphism from $F'_c$ into $E'_c$ (the duals endowed with the topology of compact convergence).

The equivalence of a), b) and e) follows from prop. 3 of IV, p. 28 and the remark preceding it. That of a) and c) is precisely cor. 3 of I, p. 19. The remark of IV, p. 28, also shows that d) is equivalent to the fact that, the image of $u$ is closed for the weakened topology $\sigma(F, F')$ of F; the equivalence of c) and d) then follows from prop. 2 of IV, p. 4.

We now prove the equivalence of e) and f). It is enough to prove that f) implies e). Suppose that the image of $^t u$ is closed for $\beta(E', E)$ in $E'$. On account of the Banach-Dieudonné theorem (IV, p. 25, cor. 2), it is enough to prove that for every convex balanced neighbourhood U of 0 in E, the intersection $B = ^t u(F') \cap U$ is compact for $\sigma(E', E)$. The strong dual $E'_b$ of the Fréchet space E is complete (IV, p. 22, prop. 2), hence the closed subset B of $E'_b$ is complete, and so the normed space $E'_B$ is complete (III, p. 8, corollary). Let $(V_n)$ be a decreasing sequence forming a fundamental system of neighbourhoods of 0 in F. Then $F'$ is the union of sets $C = V_n^\circ$ which are compact for $\sigma(F', F)$, hence $E'_B = \bigcup_n B_n$, with $B_n = E'_B \cap ^t u(C_n)$. Since $E'_B$ is a Baire space, and each of the sets $B_n$ is convex balanced and closed, there exists a real number $r > 0$ and an integer $n$ such that $B \subset r.B_n$. Then we have $B = U^\circ \cap {}^t u(r.C_n)$; since the sets $U^\circ$ and $r.C_n$ are compact and ${}^t u$ is continuous for the weak topologies, $B$ is compact for $\sigma(E', E)$. This completes the proof of the equivalence of e) and f).

Finally the equivalence of g) and the preceding conditions follows from prop. 18 of GT, X, § 2, No. 10 and the following lemma.

By Mackey’s th. (IV, p. 2, th. 1) and the fact that on $E'$ (resp. F) the topology of compact convergence coincides with that of convex compact convergence (IV, p. 4), we can identity E (resp. F) with the dual $E'_c$ (resp. $F'_c$). Then $u$ is the transpose of ${}^t u$, and the equicontinuous subsets of E (resp. F) are the relatively compact sets. Lemma 1 then follows from prop. 2 (IV, p. 27), since $u(E)$ is closed in F if and only if it is closed for the weakened topology $\sigma(F, F')$ (IV, p. 4, prop. 2).

The implication (i) $\Rightarrow$ (ii) follows immediately from the equivalence of conditions a), d) and e) of th. 1 and from IV, p. 6, prop. 5. It is clear that (ii) implies (iii). Finally, we prove that (iii) implies (i) : if ${}^t u$ is surjective $u$ is a strict morphism by the equivalence of a) and e) in th. 1 ; that $u$ is injective follows from prop. 5 of IV, p. 6.

The equivalence of (i) and (ii) follows from Banach’s th. (I, p. 17, th. 1).

In view of the equivalence of a) and c) in th. 1, condition (ii) says that $u$ is a strict morphism and that its image is dense in F for $\sigma(F, F')$. The equivalence of (ii) and (iii) then follows from the equivalence of a) and d) in th. 1 and from prop. 5 of IV, p. 6.

If $u : E \to F$ is a strict morphism of Fréchet spaces, the transpose ${}^t u$ is not necessarily a strict morphism from $F'_b$ into $E'_b$ (IV, p. 62, exerc. 3). However, we have the following partial result :

(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).

The conditions (ii) and (iii) are in fact reformulations of conditions (ii) and (iv) of prop. 4 for Banach spaces.