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?

  • $\begingroup$ $X$ 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

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, $d(F(x), x) = 0$.


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