0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $L$ 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. $\endgroup$ – adfriedman Nov 30 '17 at 23:18
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.