Proposition 6

Let E and F be normed spaces and $u\in \mathscr{L}(E; F)$. Let B denote the set of elements of E of norm $<1$. Put

$P =u(E)-u(B),Q =u(E)$ - $(\overline{u(B)}\cap u(E))$.

The conorm of $u$ is equal to the distance from 0 to P in F. If the normed space E is complete or if $u$ is a strict morphism, the conorm of $u$ is equal to the distance from 0 to Q in F.

Let N be the kernel of $u$; let $p: E\rightarrow E/N$ be the canonical surjection and let $v: E/N\rightarrow F$ be the mapping deduced from $u$ by passing to the quotient. The set $p(B)$ is the set of elements of $E/N$ of norm $<1$. We have

since the mapping $v$ is injective. In other words, P is made up of the elements of F of the form $v(y)$ with $y\in E/N$ and $\|y\|\geqslant 1$. Let $d_P$ denote the distance from 0 to P in F. We have

$d_P=$ inf$_{y\|\in yE\|\geqslant/N1}\|v(y)\|=$ inf$_{y\|\in yE\|=1/N}\|v(y)\|= ((u))$ by (6).

Suppose that $u$ is a strict morphism. Let $\varepsilon >0$. The set $\varepsilon u(B)$ is a neighbourhood of 0 in $u(E)$. The closure of $u(B)$ in $u(E)$ is equal to $\overline{u(B)}\cap u(E)$. It is contained in the set $u(B) +\varepsilon u(B)$, which is equal to $(1 +\varepsilon )u(B)$ since $u(B)$ is convex. Hence we have $(1 +\varepsilon )P\subset Q\subset P$, and the distance $d_Q$ from 0 to Q in F satisfies the inequalities $d_P\leqslant d_Q\leqslant (1 +\varepsilon )d_P$. Since this holds for every $\varepsilon >0$, we have $d_Q=d_P= ((u))$.

Suppose that $u$ is not a strict morphism, but that the normed space E is complete. Then $((u)) = 0$ (remark above). The closure of $u(B)$ in $u$(E), which is equal to $\overline{u(B)}\cap u$(E), is not a neighbourhood of 0 in $u(E)$ (EVT, I, p. 17, th. 1). There then exist points of Q arbitrarily near 0, whence $d_Q= 0 = ((u))$.