Proposition 11

Let E and F be Banach spaces. The set $\mathscr{Q}\mathscr{M}(E; F)$ is open in $\mathscr{L}(E; F)$. It is the interior of the set of mappings in $\mathscr{L}(E; F)$ whose kernel is of finite dimension.

Let A be the set of continuous linear mappings of E into F whose kernel is of finite dimension.

Let $u$ be an element of $\mathscr{Q}\mathscr{M}(E; F)$. Then $((u))>0$(III, p. 62, remark). Every element $v\in \mathscr{L}(E; F)$ whose distance from $u$ is $<((u))$ then belongs to A (III, p. 65, prop. 9), so that $u$ is an interior point of A.

Conversely, let $u$ be an element of A which is not a strict morphism. One has $((u)) = 0 ($III, p. 62, Remark). Let $v$ be an element of $\mathscr{L}(E; F)$ differing from $u$ by a linear mapping of finite rank; by Corollary 1 of III, p. 40, the morphism $v$ cannot be strict with closed image; since a strict morphism of E into F has a closed image in F (EVT, IV, p. 28, Theorem 1), this means that $v$ is not a strict morphism. Hence $((v)) = 0$. Let $\varepsilon$ be a real number $>0$. By the foregoing, Lemma 3 of III, p. 66 enables one to construct by induction a sequence $(u_m)_{m\in\mathbf{N}}$ of elements of $\mathscr{L}(E; F)$ satisfying the following conditions:

(i) One has $u_0=u$;

(ii) For every $m\geqslant 0,u_{m+1}-u_m$ is a continuous linear mapping of rank 1 and norm $\leqslant 2^{-m-1}\varepsilon$;

(iii) For every $m\geqslant 0$, the kernel of $u_m$ has dimension $n+m$ and is contained in that of $u_{m+1}$.

The sequence $(u_m)$ is a Cauchy sequence in the Banach space $\mathscr{L}(E; F)$. Let $u'$ be its limit. The kernel of $u'$ contains that of $u_m$ for every $m\geqslant 0$; it is infinite-dimensional. Since

the distance from $u$ to the complement of A is less than $\varepsilon$. Since this holds for every $\varepsilon >0$, one concludes that $u$ is not an interior point of A.