Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 2F (Topics In Analysis)

Let $D$ be the closed unit disc and setThe displacement bound gives $\Vert f(x)\Vert \le 1$, so $f:D\to D$ is continuous. The Brouwer fixed-point theorem applies because $D$ is nonempty, compact and convex. At a fixed point $x_{\ast }$, $x_{\ast }=x_{\ast }-g(K{x}_{\ast })/K$. Therefore $\boxed{t=K{x}_{\ast },g(t)=0}$.

For arbitrary $y\in {\mathbb{R}}^2$, the continuous map $g_y(x)=g(x)-y$ satisfies $\Vert g_y(x)-x\Vert \le K+\Vert y\Vert$. Applying the preceding Brouwer fixed-point theorem argument with this positive bound gives $g_y(t)=0$. Hence $g(t)=y$, and $g$ is surjective.

Let $e=(1,0)$ and defineThen $\Vert g(x)-x\Vert \le 1$ everywhere, but $0$ is not in the image of a function. The map is discontinuous at $0$, giving a bounded-displacement map which is not surjective.