The Uniqueness of a Fixed Point

Question

Let $D(0,r) = \{ x \in \mathbb{R}^n | \ \rVert x \rVert \le r \}$ and $f: D(0,r) \to \mathbb{R}^n$ be a map with $a) \rVert f(x) - f(y) \rVert \le 1/3 \rVert x - y \rVert, \ \ \ b) \rVert f(0) \rVert \le (2/3)r$.

I want to determine if there is a unique fixed point of $f$.

I think all I need to do is show that $f$ maps $\text{cl}(D(0,r))$ into itself. Then $f$ obviously satisfies the hypotheses of the contraction mapping principle, but I'm having difficulty showing $f$ maps $\text{cl}(D(0,r))$ into itself. Can someone give me a hint? Thanks!

EDIT: I just came up with a way of showing it. $$ \rVert f(x) \rVert \le \rVert f(x) - f(0) \rVert + \rVert f(0) \rVert \le \frac13 \rVert x \rVert + \frac23r \le r $$ So $f$ maps $D(0,r)$ to itself. So by the contraction mapping principle, $f$ has a unique fixed point. Can someone verify my proof?

Answer 1

First note that $\text{cl}(D(0,r)) = D(0, r)$ because according to your definition, $D(0,r) = \{ x \in \mathbb{R}^n \mid \; \rVert x \rVert \le r \}$ is already closed.

Now if $x \in D(0, r)$ then $\Vert x \Vert \le r$ and therefore $$ \Vert f(x) \Vert \le \Vert f(x) - f(0) \Vert + \Vert f(0) \Vert \\ \le \frac 1 3 \Vert x - 0 \Vert + \frac 23 r \le r $$ so that $f(x) \in D(0, r)$.