# Unique fixed point mapping from an open ball

We are asked to prove:

Thorem: Let $r>0$ and $D=B(0,r) \subset \mathbb{R}^n$, and let $f:D \rightarrow \mathbb{R}^n$ be Lipschitz continuous with a Lipschitz constant $L<1$. If $\|f(0)\|<r(1-L)$, then $f$ has a unique fixed point.

I understand that I can't use the Banach fixed point thorem because we aren't mapping from set $X \rightarrow X$.

Is this a valid proof:

Proof of existence: We note that $r(1-L) < 1$, and we create the ball $B(0, r(1-L))$. By definition, $f(0) \in B(0,r(1-L))$ We further note that $\lim\limits_{L \rightarrow 1} r(1-L) =0$. Since, for any $L$, $\|f(0)\| \in B(0,r(1-L))$, the formula $\lim\limits_{L \rightarrow 1} r(1-L) =0$ requires $f(0)=0$, providing the existence of a fixed point.

Proof of uniqueness: Assume, to the contrary, that there exists more than one fixed point. Then, exists $x \text{ and } y \in D$ where $x\neq y$ but $f(x)=x$ and $f(y)=y$. By the definition of fixed points, we have $$(i): \|x-y\|=\|f(x)-f(y)\|$$, And by the definition of Lipschitz continuous, we have $$(ii): \|f(x)-f(y)\|\leqslant L\|x-y\|$$. Combining $(i)$ and $(ii)$, we obtain $\|x-y\| \leqslant L\|x-y\|$. Since we defined $L<1$, it follows that $\|x-y\| = L\|x-y\|$, and since $L>0$, it follows that $\|x-y\|=0$ which provides $x=y$, providing uniqueness.

• $L$ is fixed, so it is unclear why you are trying to let $L\to 1$ and there is no obvious reason (at least to me) that $f(0)=0$ necessarily, but I don't have time to come up with a counter example. – adfriedman Nov 30 '17 at 23:18

## 1 Answer

Actually $f$ does map the ball to itself: $$\|f(x)\| \leq \|f(x)-f(0)\| + \|f(0)\| < L\underbrace{\|x-0\|}_{<\,r} + r(1-L) < r$$ and $f$ is a contraction, so the Banach fixed point theorem may be applied.

• Thanks... that's elegant. – Izzy M. Nov 30 '17 at 23:20