Let $a \in E$ and let $U$ be an open set of $E$ containing $a$. Let $f : U \rightarrow E$ be an application of class $\mathscr{C}^1$ on $U$ such that $df(a) = \operatorname{Id}_E$. We fix a real number $r > 0$ such that $\overline{B(a,r)} \subset U$ and $$\forall x_1, x_2 \in \overline{B(a,r)}, \quad \left\|f(x_1) - f(x_2)\right\| \geqslant \frac{1}{2} \left\|x_1 - x_2\right\|.$$ Let $y_0 \in E$ such that $\left\|y_0 - f(a)\right\| \leqslant \frac{r}{4}$, and let $x_0 \in B(a,r)$ be a point where the application $x \mapsto \left\|y_0 - f(x)\right\|^2$ attains its minimum on $\overline{B(a,r)}$. Show that $f(x_0) = y_0$.
Let $a \in E$ and let $U$ be an open set of $E$ containing $a$. Let $f : U \rightarrow E$ be an application of class $\mathscr{C}^1$ on $U$ such that $df(a) = \operatorname{Id}_E$. We fix a real number $r > 0$ such that $\overline{B(a,r)} \subset U$ and
$$\forall x_1, x_2 \in \overline{B(a,r)}, \quad \left\|f(x_1) - f(x_2)\right\| \geqslant \frac{1}{2} \left\|x_1 - x_2\right\|.$$
Let $y_0 \in E$ such that $\left\|y_0 - f(a)\right\| \leqslant \frac{r}{4}$, and let $x_0 \in B(a,r)$ be a point where the application $x \mapsto \left\|y_0 - f(x)\right\|^2$ attains its minimum on $\overline{B(a,r)}$.
Show that $f(x_0) = y_0$.