Proposition 12

Let $E$ and $F$ be two Fréchet spaces and $u$ a continuous linear mapping of $E$ on $F$. Then there exists a section of $u$ that is continuous though not necessarily linear.

By prop. 11 there exists a distance $d$ in $E$, invariant under translations, defining the topology of $E$ and for which open balls are convex. Given $y$ and $y'$ in $F$, let $\delta(y, y')$ be the distance apart of the closed sets $u^{-1}(y)$ and $u^{-1}(y')$ in $E$. As $u$ is a strict morphism (I, p. 17, th. 1) the remark of GT, IX, § 3.1 shows that $\delta$ is a distance on $F$ defining the topology of $F$. We shall construct, inductively, a sequence of continuous mappings $(s_n)_{n \in \mathbf{N}}$ of $F$ in $E$ satisfying the following inequalities for all $y \in F$:

Suppose then that either $n = 0$, or $n \geq 1$ and that $s_{n-1}$ has been constructed. Let $y_0 \in F$; as $u$ is surjective, the set $u^{-1}(y_0)$ is non-empty, and for $n \geq 1$, we have $d(u^{-1}(y_0), s_{n-1}(y_0)) < 2^{-n+1}$ by the induction hypothesis. Therefore there exists a point $x_0$ of $E$ such that $u(x_0) = y_0$ and for $n \geq 1$, $d(x_0, s_{n-1}(y_0)) < 2^{-n+1}$. As the mapping $s_{n-1}$ is continuous, the set of points $y$ of $F$ which satisfy the inequalities $\delta(y, y_0) < 2^{-n}$ and $d(x_0, s_{n-1}(y)) < 2^{-n+1}$ is an open neighbourhood of $y_0$. Hence there exist an open covering $(V_i)_{i \in I}$ of $F$ and constant mappings $s_{n,i}$ of $F$ in $E$ which satisfy the inequalities (2) and (3) in $V_i$ where one replaces $s_n$ by $s_{n,i}$. As the space $F$ is metrisable, there exists a continuous partition of unity $(f_i)_{i \in I}$, that is locally finite and subordinate to the covering $(V_i)_{i \in I}$ (GT, IX, § 4.5, th. 4 and § 4.4, cor. 1). For every $y \in F$, put $s_n(y) = \sum_{i \in I} f_i(y) \cdot s_{n,i}(y)$. The mapping $s_n$ of $F$ in $E$ is continuous; as the open balls are convex in $E$ and in $F$, the mapping $s_n$ satisfies the inequalities (2) and (3) for all $y \in F$.

From inequality (3) the mappings $s_n : F \to E$ form a Cauchy sequence, for uniform convergence. As $E$ is complete, the sequence $(s_n)_{n \in \mathbf{N}}$ converges uniformly to a continuous mapping $s : F \to E$ (GT, X, § 1.6); formula (2) shows that $u \circ s$ is the identity mapping of $F$, thus $s$ is a continuous section of $u$.