# Thread: Uniqueness in a contraction

1. ## Uniqueness in a contraction

Given a complete metric space X and a function f: X->X with a Lipschitz constant of r on (0,1) (i.e contracting function) show that there is a unique point f(x) = x
Use a sequence with arbitrary x0 and xn=f(xn-1)
to get x as limit of the sequence.

How do I get the limit and how do i use this in proving that f(x)=x?

2. Originally Posted by twerdster
Given a complete metric space X and a function f: X->X with a Lipschitz constant of r on (0,1) (i.e contracting function) show that there is a unique point f(x) = x
Use a sequence with arbitrary x0 and xn=f(xn-1)
to get x as limit of the sequence.

How do I get the limit and how do i use this in proving that f(x)=x?
If you define $x_n= f(x_{n-1})$, starting with chosen $x_0$, then $d(x_n, x_{n+1})= d(f(x_{n-1}, df(x_n))\le rd(x_{n-1}, x_n)$. Use induction to prove that $d(x_{n+1},x_n)\le r^nd(x_1, x_0)$.

That isn't quite enough to show that this is a Cauchy sequence. For a Cauchy sequence, you need to show that $d(x_n, x_m)$ goes to 0 as n and m both go to infinity independently. Use the fact that, for n> m, $d(x_n, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1}, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1},x_{n-2})+ d(x_{n-2},x_m)$, etc.

To prove uniqueness, suppose that f(x)= x and f(y)= y. Then $d(x, y)= d(f(x), f(y))\le rd(x, y)$

3. Originally Posted by HallsofIvy
If you define $x_n= f(x_{n-1})$, starting with chosen $x_0$, then $d(x_n, x_{n+1})= d(f(x_{n-1}, df(x_n))\le rd(x_{n-1}, x_n)$. Use induction to prove that $d(x_{n+1},x_n)\le r^nd(x_1, x_0)$.

That isn't quite enough to show that this is a Cauchy sequence. For a Cauchy sequence, you need to show that $d(x_n, x_m)$ goes to 0 as n and m both go to infinity independently. Use the fact that, for n> m, $d(x_n, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1}, x_m)\le d(x_n, x_{n-1})+ d(x_{n-1},x_{n-2})+ d(x_{n-2},x_m)$, etc.

To prove uniqueness, suppose that f(x)= x and f(y)= y. Then $d(x, y)= d(f(x), f(y))\le rd(x, y)$
Adding on, to prove that if $\xi=\lim_{n\to\infty}x_n$ then $f(\xi)=\xi$ we note that since the sequence coverges so does every subsequence to the same value and $f(\xi)=f\left(\lim_{n\to\infty}x_n\right)\overset{ {\color{red}(*)}}{=}\lim_{n\to\infty}f(x_n)=\lim_{ n\to\infty}x_{n+1}=\xi$ the red starrred part gotten by continuity (every contraction is continuous)

4. Why is every contraction necessarily continuous?

5. Originally Posted by twerdster
Why is every contraction necessarily continuous?
Because every Lipschitz function is uniformly continuous and thus continuous.

For a fixed $x_0\in X$ you can make $d(f(x_0),f(y))<\varepsilon$ by just taking $\delta=\varepsilon$. For then, $d(x_0,y)<\delta=\varepsilon\implies d(f(x_0),f(y))=rd(x,y)