Lemma 2

Let E and F be two metric spaces, and suppose that E is also complete. Let u be a linear mapping of E in F having the following property: whatever the number $r > 0$, there exists a number $\rho(r) > 0$ such that, for all $x \in E$, we have

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.