Exercise 17, § 4

Let E, F be two Banach spaces, $u$ a strict morphism from E onto F. Then there exists a number $m > 0$ such that for every $\varepsilon \in ]0, m[$ and for every $y \in F$, there exists $z \in E$ such that $u(z) = y$ and $\|z\| \leq (m - \varepsilon)^{-1}\ \|y\|$. a) Let B be a closed ball $\|x - a\| \leq r$ in E, and let $\omega$ be a mapping from B into F satisfying the following conditions : $1^o$ $\sup_{x \in B} \|w(x)\| = M < + \infty$; $2^o$ there exists a number $k > 0$ such that for all $x, x'$ in B, we have $\|w(x) - w(x')\| \leq k \|x - x'\|$ ("Lipschitz condition"). Show that if $k < m$ and $M < r(m - k)$, then the image of B under the mapping $x \mapsto u(x) + w(x)$ contains a ball with center $b = u(a)$. (Show that for every $y \in F$ close enough to $b$, we can define a sequence $(x_n)_{n \geq 0}$ of points of B such that $u(x_0) = y$ and $u(x_n) = y - w(x_{n-1})$ for $n \geq 1$, and such that $(x_n)$ converges to a point of B.) b) Suppose that $w$ is a mapping from all E into F, such that $\|w(x) - w(x')\| \leq k \|x - x'\|$ for every $x, x'$ in E. Show that if $k < m$, $u + w$ is a surjective and open mapping from E onto F. c) In particular, if $w$ is a continuous linear mapping from E into F such that $\|w\| < m$, then $v = u + w$ is a strict morphism from E onto F. If $\varepsilon \in ]0, m[$, then for every $x \in v^{-1}(0)$ with $\|x\| = 1$, there exists $x_0 \in u^{-1}(0)$ such that $\|x - x_0\| \leq \|w\|/(m - \varepsilon)$; if in addition $\|w\| < m - \varepsilon$, then for every $x_0 \in u^{-1}(0)$ with $\|x_0\| = 1$, there exists $x \in v^{-1}(0)$ such that

Deduce that if there exists a closed vector subspace G of E which is the complement of $u^{-1}(0)$, then G is also the complement of $v^{-1}(0)$ whenever $\|w\|$ is small enough (argue by reductio ad absurdum : show that the projection of $v^{-1}(0)$ onto $u^{-1}(0)$ cannot be contained in a closed hyperplane of $u^{-1}(0)$; on the other hand, note that there exists $a > 0$ such that every point $x \in G$ such that $\|x\| = 1$ is at a distance $\geq a$ from $u^{-1}(0)$).