Proposition 2

Let E be the numerical space of dimension $n$ and let A be a convex and compact subset of E of which 0 is an interior point. There exists a unique bijection $u$ of E onto itself satisfying the following three properties:

(i) For every $x\in E$ and every $t\in \mathbf{R}_+,u(tx) =tu(x)$;

(ii) For every $x\in E$, there exists $\lambda \in \mathbf{R}_+$ such that $u(x) =\lambda x$;

(iii) One has $u(A) =\mathbf{B}_n$.

The mapping $u$ is a homeomorphism and induces, by passing to subspaces, a homeomorphism of A onto $\mathbf{B}_n$, a homeomorphism of the interior of A onto the interior of $\mathbf{B}_n$ and a homeomorphism of the boundary of A in E onto the sphere $\mathbf{S}_{n-1}$.

Let $p_A$ be the gauge of the convex subset A. Let $x\in E$ and let $t\in \mathbf{R}_+$; in order that $x\in tA$, it is necessary and sufficient that $p_A(x)\leqslant t$. Since A is compact, there exists a real number M such that $\|x\|\leqslant M$ for every $x\in A$.

Let $u$ be a mapping satisfying the conditions of the proposition. We have $u(0) = 0$. Let $x\in E-\{0\}$ and let $\lambda \in \mathbf{R}_+$ be such that $u(x) =\lambda x$. For every $t\in \mathbf{R}_+$ such that $tx\in A$, we have $u(tx) =t\lambda x$. Since $u$ is injective, $\lambda = 0\not$ . Since $u(A)$ is contained in $\mathbf{B}_n$, we also have $\lambda \leqslant p_A(x)/\|x\|$. Put $z=x/\|x\|$; this is a point of $\mathbf{S}_{n-1}$. In order that $z$ should have a preimage in A under $u$, it is necessary and sufficient that the point $(\lambda \|x\|)^{-1}x$ should belong to A, that is, that $\lambda \|x\|\geqslant p_A(x)$, i.e. $\lambda \geqslant p_A(x)/\|x\|$. This implies the uniqueness of such a mapping $u$.

Let us now denote by $u$ the mapping of E into itself which maps 0 to 0 and $x$ to $(p_A(x)/\|x\|)x$, for every $x\in E-\{0\}$.

By Lemma $1,u$ is continuous at every point of E $-\{0\}$. We have $\|u(x)\|=p_A(x)$ for every $x\in E$ and $p_A(x)\rightarrow 0$ when $x\rightarrow 0$ (loc. cit.); consequently, $u$ is continuous at 0. Therefore $u$ is continuous.

The only preimage of 0 under $u$ is 0. Let $y\in E-\{0\}$. In order that an element $x$ of E should satisfy $u(x) =y$, it is necessary and sufficient that $x= (\|y\|/p_A(y))y$. It follows that $u$ is a continuous bijection of E onto itself. Its inverse is the mapping $v:y\mapsto (\|y\|/p_A(y))y$. Since $p_A$ is continuous and vanishes only at 0, the mapping $v$ is continuous at every point of E $-\{0\}$; the inequality $p_A(y)\geqslant \|y\|/M$ implies that $\|v(y)\|\leqslant M\|y\|$ for every $y\in E$. It follows that $v$ is continuous. Therefore the mapping $u$ is a homeomorphism of E onto itself. Since the relations $p_A(x)\leqslant 1$ and $x\in A$ are equivalent, we also have $u(X) =\mathbf{B}_n$. The proposition follows.