Proposition 7

Two norms $p, q$ on a vector space $E$ are equivalent if and only if there exist two numbers $a > 0, b > 0$ such that

(I)

for all $x \in E$.

These inequalities are sufficient, for it follows from the relation $a . p(x) \leq q(x)$ that, for each $r > 0$, the closed ball with centre o and radius $ar$ (relative to the norm $q$) is contained in the closed ball with centre o and radius $r$ (relative to the norm $p$); hence the topology defined by $q$ is finer than the topology defined by $p$. Similarly the inequality $q(x) \leq b . p(x)$ shows that the topology defined by $p$ is finer than that defined by $q$, and hence $p$ and $q$ are equivalent.

Let us now show that the inequalities (I) are necessary. If the topology defined by $q$ is finer than the topology defined by $p$, then the unit ball with respect to $p$ contains a closed ball with centre o and radius $\alpha > 0$ with respect to $q$; i.e., the relation $q(x) \leq \alpha$ implies $p(x) \leq 1$. If $t_0 \in K$ is such that $0 < |t_0| < 1$, then for each $x \neq o$ in $E$ there is a unique rational integer $k$ such that $\alpha |t_0| < q(t_0^k x) \leq \alpha$; therefore $p(t_0^k x) \leq 1$, so that

putting $a = \alpha |t_0|$ we have therefore $a . p(x) \leq q(x)$ for all $x \neq o$, and this inequality is also valid when $x = o$. Similarly we show that if the topology defined by $p$ is finer than that defined by $q$, there exists $b > 0$ such that $q(x) \leq b . p(x)$.