0
$\begingroup$

If $f:(A,\rho)\to(A,\rho)$ is defined on all of $A$ and there is a constant $\alpha\in(0,1)$ such that $\rho(f(u),f(v))\leq\alpha\rho(u,v)$ for all $(u,v)\in A\times A$ so I have not understood the connection of the proof of the contraction Mapping theorem to this statement. The contraction mapping theorem states: If $f$ is a contraction of a complete metric space $(A,\rho)$, then the equation $f(u)=u$ has a unique solution. The proof makes use of the following axiom $\rho(u,v)=\rho(f(u),f(v))$, how can this be true? How can $\rho(u,v)=0$? And $u=v$? I have stuck on this for two days. Thanks for readingenter image description hereenter image description here

$\endgroup$
  • $\begingroup$ It is good idea to sketch the proof in your question. $\endgroup$ – abcd Mar 7 '17 at 15:49
  • 2
    $\begingroup$ What's the problem with saying that if $u=f(u)$ and $f(v)=v$, then $\rho(u,v)=\rho(f(u),f(v))$ ? $\endgroup$ – user228113 Mar 7 '17 at 15:59
  • $\begingroup$ If you have a function $f:(\Re,d)\to(\Re,d)$ such as the function$ f(x)=x^2$, it not true $f(x)=x$. However I am not certain $(\Re,d)$ is complete maybe that is the problem. Note: $\Re$ stands for the set of real numbers. $\endgroup$ – Pedro Gomes Mar 7 '17 at 16:04
  • $\begingroup$ @PedroGomes - standard notation for the reals is obtained by \mathbb R, or just \Bbb R: $\Bbb R$. $\endgroup$ – Paul Sinclair Mar 7 '17 at 21:27
1
$\begingroup$

The part of the proof you're asking about only shows the uniqueness of the fixed point. If there is no fixed point, the uniqueness is vacuously true and if there is only one we are done.

So, one needs to show that if $u,v\in A$ are two fixed points, then $u=v$. By definition this means $f(u)=u$ and $f(v)=v$. It follows that $\rho(f(u),f(v)) = \rho(u,v)$. The axioms of a metric space imply that if $\rho(u,v)=0$, then $u=v$. Suppose $\rho(u,v)\neq 0$. Since $\alpha<1$, this yields $$\rho(u,v)=\rho(f(u),f(v))\le \alpha \rho(u,v)<\rho(u,v),$$ which is a contradiction. Thus, $\rho(u,v)=0$ and $u=v$ as desired.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ First, thanks for your answer. The problem arises when intuition comes into play. I do not know if I am wrong. I have made this point in an earlier comment. As far as I know $\Bbb R$ is a complete set, I guess. If I have a function $f:(\Bbb R,d)\to(\Bbb R,d)$ as for example $f(x)=x^2$, it is not true $f(x)=x$. Is the metric space $(\Bbb R, d)$ in which d can be the Euclidian metric complete? At least there are some Cauchy sequences that converge. Thanks! $\endgroup$ – Pedro Gomes Mar 9 '17 at 14:26
  • 1
    $\begingroup$ @PedroGomes Well, you have $f(0)=0$ in this case, so there is a fixed point. And yes, in this case completeness holds. Anyhow, as I tried to indicate in my answer, the existence of fixed points is irrelevant in this part of the proof. What you want to prove is the uniqueness, i.e. if two fixed points exist, then they must be the same. $\endgroup$ – Vincent Boelens Mar 9 '17 at 14:58
  • $\begingroup$ I think I am starting to understand it now due to your last reply. Thank you, I was misunderstanding the theorem because I was thinking the condition would hold for the entire spaces. $\endgroup$ – Pedro Gomes Mar 9 '17 at 17:10

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.