Firstly (i) implies (ii), for let u be a strict surjective morphism and N be the kernel of u. Then u induces an isomorphism of E/N on F. But E is metrisable and complete, hence E/N is complete (GT, IX, § 3.1, prop. 4), therefore F is complete.

Next (ii) implies (iii). Let F be complete and u be surjective. The image of u is precisely F and therefore not meagre in F from Baire’s theorem (GT, IX, § 5.3).

The following lemma shows that (iii) implies (iv).

Let W be a balanced neighbourhood of 0 in E such that W + W ⊂ V (I, § 1.5, prop. 4). Let α be an element of K such that |α| > 1; then E is the union of the sets $\alpha^n W$ where n varies in $\mathbf{N}$; in fact, for all x ∈ E, there exists β ∈ K such that x ∈ βW (I, p. 7, prop. 4) and there exists an integer n ≥ 0 such that |β| < |α|^n, then x ∈ $\alpha^n W$ since W is balanced. Hence, u(E) is the union of the sequence of sets $u(\alpha^n W) = \alpha^n u(W)$, and as u(E) is not meagre in F, one at least of the sets $\alpha^n \overline{u(W)}$ possesses an interior point (GT, IX, § 5.3, def. 2) and therefore $\overline{u(W)}$ has an interior point.

Let $y_0$ be an interior point of $\overline{u(W)}$; since $-u(W) = u(W)$, and therefore $-\overline{u(W)} = \overline{u(W)}$ it follows that $0 = y_0 + (-y_0)$ is an interior point of $\overline{u(W)} + \overline{u(W)}$. As vector addition is a continuous mapping of $F \times F$ in $F$, the set $\overline{u(W)} + \overline{u(W)}$ is contained in the closure of the set

hence $\overline{u(V)}$ is a neighbourhood of 0 in $F$.

Before proving that (iv) implies (i) we prove the following lemma, where we make the convention that, in all metric spaces, $B_r(x)$ denotes the closed ball of centre $x$ and radius $r$.

In these conditions, for all $a > r$, the image $u(B_a(x))$ contains the ball $B_{\rho(r)}(u(x))$.

Let $(r_n)$ be an infinite sequence of numbers $> 0$ such that $r_1 = r$ and $a = \sum_{n=1}^{\infty} r_n$. For each index $n$ there exists a number $\rho_n > 0$ (with $\rho_1 = \rho(r)$) such that

for all $x \in E$; we can, and will, suppose that $\lim_{n \to \infty} \rho_n = 0$.

Let $x_0$ be a point of E, and $y$ be a point of $B_{\rho(r)}(u(x_0))$. We shall show that $y$ belongs to $u(B_a(x_0))$.

For this, a sequence $(x_n)_{n>0}$ of points of E is defined inductively such that, for all $n \geq 1$, we have $x_n \in B_{r_n}(x_{n-1})$ and $u(x_n) \in B_{\rho_{n+1}}(y)$. If the $x_i$ have been defined for $0 \leq i \leq n-1$ satisfying these relations, then we have $y \in B_{\rho_n}(u(x_{n-1}))$; since

there exists a point $x_n \in B_{r_n}(x_{n-1})$ whose image $u(x_n)$ belongs to the neighbourhood $B_{\rho_{n+1}}(y)$ of $y$, which establishes the existence of the sequence $(x_n)$.

Since the distance of $x_n$ from $x_{n+p}$ is less than $r_{n+1} + r_{n+2} + \cdots + r_{n+p}$, which is arbitrarily small when $n$ is large, the sequence $(x_n)$ is a Cauchy sequence in E. As E is complete, the sequence $(x_n)$ converges to a point $x$ of E. The distance of $x_0$ from $x$ is less than $\sum_{n=1}^{\infty} r_n = a$, thus $x \in B_a(x_0)$. But $u$ is continuous, thus the sequence $u(x_n)$ converges to $u(x)$; also $u(x_n) \in B_{\rho_{n+1}}(y)$, hence $y = u(x)$, and the lemma is proved.

We return to the theorem and show that (iv) implies (i). Suppose that $u$ satisfies condition (iv). For each of the spaces E and F, consider a distance that is invariant under translation and defines its topology (I, p. 16). By hypothesis, the set $\overline{u(B_r(0))}$ is a neighbourhood of 0 in F for every $r > 0$, and thus there exists a number $\rho(r) > 0$ such that $B_{\rho(r)}(0) \subset u(B_r(0))$. By translation we conclude that $B_{\rho(r)}(u(x)) \subset \overline{u(B_r(x))}$ for all $r > 0$ and all $x \in E$. From lemma 2, for every pair of real positive numbers $(a, r), a > r > 0$, we have $B_{\rho(r)}(0) \subset u(B_a(0))$; thus $u$ is a strict morphism of E on F. We have shown that (iv) implies (i) and the proof of the theorem is completed.

In particular, if E and F are two complete normed spaces, there exists a number $a > 0$ such that $\|u(x)\| \geq a.\|x\|$ for all $x \in E$.

The condition is necessary, because if $u$ is a strict morphism, the image $u(E)$, being isomorphic to the quotient $E/u^{-1}(0)$, is complete (I, p. 17) and therefore closed in F. The condition is sufficient, since, if $u(E)$ is closed in F, then $u(E)$ must be a complete metrisable vector space and thus by theorem 1 $u$ is a strict morphism of E on $u(E)$.

For $M \times N$ is a complete metrisable vector space and the mapping $(y, z) \mapsto y + z$ of $M \times N$ on E is continuous and bijective, therefore an isomorphism (cor. 1).

The condition is necessary since the graph of a continuous mapping into a Hausdorff space is closed (GT, I, § 8.1, cor. 2). To see that it is sufficient, note that it implies that the graph G of $u$, which is a closed vector subspace of the complete metrisable space $E \times F$, is itself metrisable and complete. The projection $z \mapsto \mathrm{pr}_1(z)$ of G on E is a bijective, continuous linear mapping, therefore an isomorphism (cor. 1); since its inverse mapping is $x \mapsto (x, u(x))$, it follows that $u$ is continuous in E.

We can express this corollary in the following form : $u$ is continuous if the following situation holds : if the sequence $(x_n)$ of points of E both converges to 0 and is such that the sequence $(u(x_n))$ converges to $y$, then it is necessarily the case that $y = 0$.

1) E is metrisable and complete. 2) There exists a sequence $(F_n)$ of complete metrisable vector spaces over K and, for each n, an injective continuous linear mapping $v_n$ of $F_n$ in F such that F is the union of the subspaces $v_n(F_n)$.

Then let u be a linear mapping of E in F. If the graph of u is closed in $E \times F$, then there exists an integer n and a continuous linear mapping $u_n$ of E in $F_n$ such that $u = v_n \circ u_n$ (which implies that u is continuous and that $u(E) \subset v_n(F_n)$).*

Let G be the graph of $u$ in $E \times F$. For all $n$, we consider the continuous linear mapping $w_n : (x, y) \mapsto (x, v_n(y))$ of $E \times F_n$ in $E \times F$; as G is closed, the set $w_n^{-1}(G) = G_n$ is a closed vector subspace of $E \times F_n$; if $p_n$ is the restriction to $G_n$ of the first projection $\mathrm{pr}_1$, we have $p_n(G_n) = u^{-1}(v_n(F_n))$. As $p_n$ is continuous and $G_n$ is complete (since $G_n$ is closed in the complete space $E \times F_n$), $p_n(G_n)$ is, by theorem 1, either meagre in E or it is the whole of E. But, by hypothesis, E is the union of the $p_n(G_n)$, and as E is complete, the $p_n(G_n)$ cannot all be meagre in E by Baire’s theorem (GT, IX, § 5.3, th. 1). Therefore there exists an integer n such that $p_n(G_n) = E$, or in other words $u(E) \subset v_n(F_n)$. Further, as $v_n$ is injective, $G_n$ is the graph of a linear mapping $u_n$ of E in $F_n$, and by the closed graph theorem (I, p. 19, cor. 5) $u_n$ is continuous; it follows then from the definitions that $u = v_n \circ u_n$.

Exercises