# Uniqueness in a contraction

• May 18th 2010, 04:51 AM
twerdster
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?
• May 18th 2010, 05:17 AM
HallsofIvy
Quote:

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 $d(x, y)= d(f(x), f(y))\le rd(x, y)$
• May 18th 2010, 08:50 AM
Drexel28
Quote:

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)
• May 18th 2010, 09:11 AM
twerdster
Why is every contraction necessarily continuous?
• May 18th 2010, 09:24 AM
Drexel28
Quote:

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)