Proposition 9

Let E and F be normed spaces, $n\in \mathbf{N}$ and $u$ a continuous linear mapping from E into F whose kernel is of dimension $n$. The conorm of $u$ is equal to the distance $d$ from $u$ to the set of mappings $v\in \mathscr{L}(E; F)$ whose kernel is of dimension at least $n+ 1$.

When $u= 0$, one has $((u)) = +\infty$ and dim(E) $=n$, whence $d= +\infty$. Suppose henceforth that $u$ is non-zero, and therefore that $((u))<+\infty$. Let $v$ be an element of $\mathscr{L}(E; F)$ such that $\|u-v\|<((u))$ and let us prove that its kernel is of dimension $\leqslant n$. Let $x$ be an element of norm 1 of Ker($v$) and $y$ its image in $E/$ Ker($u$). One has (formula (5) of III, p. 61)

whence $\|y\|<1$. Now $\|y\|$ is the distance from $x$ to Ker($u$). Theorem 4 then implies that one has dim Ker($v$)$\leqslant$ dim Ker($u$) $=n$. This proves the inequality $((u))\leqslant d$. The converse inequality $d\leqslant ((u))$ results from the more precise lemma that follows.