3
$\begingroup$

I will copy the definition I am using just to make things clearer.

Def. Let $(X,d)$ be a metric space and let $F:A(\subset X)\rightarrow X$. We say F is a contraction if there exists $\lambda$ where $ 0\leq \lambda <1$ such that: $$ d(F(x),F(y))\leq \lambda d(x,y) $$ for all $x,y\in X$.

Now the contraction mappping theorem states that if $(X,d)$ is a complete metric space, then $F$ has a unique fixed point.

If we let $x_0$ be any point in $X$, we can define a sequence $(x_n)_{n=1}^{\infty}$ such that $x_1=F(x_0)$, $x_2=F(x_1)$ $\dots$

I understood why it follows that $x_n$ is Cauchy. Then since $(X,d)$ is complete, $(x_n)$ converges to a limit in $(X,d)$, say $x$. The second claim of the theorem is that $x$ is a fixed point. But it was not totally clear to me why this is the case. The proof I have seen says: $x$ is a fixed point iff $d(F(x),x)=0$ (so far so good). Then for any $n$: $$ d(x,F(x)) \leq d(x,x_n)+d(x_n,F(x)) \\= d(x,x_n) + d(F(x_{n-1}),F(x))\\ \leq d(x,x_n)+\lambda d(x_{n-1},x) \rightarrow 0 \text{ as } n\rightarrow \infty $$

But this only proves that $d(F(x),x)\rightarrow0$ and not $d(F(x),x)=0$. How can I be certain that $x$ is indeed a fixed point?

$\endgroup$
  • $\begingroup$ $X$ needs to be complete and non-empty. If it is empty, then it is vacuously complete and the only function on it is vacuously a contraction, but it has no fixed-point. $\endgroup$ – Ittay Weiss May 19 '14 at 3:50
1
$\begingroup$

It does prove $d(F(x), x) = 0$. Keep in mind that $d(F(x),x)$ is a constant. Thus, if we can bound it above by something that goes to $0$ in the limit, it is at most $0$ (think about why this must be true; if $c<\epsilon$ for every $\epsilon > 0$, prove $c\le 0$). But the distance function on a metric space is a nonnegative function. Therefore, $d(F(x), x) = 0$.

$\endgroup$

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.