Fix points of strict contractions over limits

Question

Considering a strict contraction $\phi$ and a corresponding sequence of iterated composition for a given initial value $x_0$, I.e. a sequence $(x_n)$ with $x_{n+1}=\phi(x_n)$, a Limit can be found for any such given initial value and contraction. I need to show that this limit is indeed the fix point of $\phi$.

I’m aware of the argument involving the continuity of $\phi$ but I’m forced to not use it, which is where my problem is.

Answer 1

If $A $ is a closed subset of $\Bbb R $,

$x_0\in A $ and $\phi (A)\subset A .$

then if $x_n $ converges to $L $,

$L $ will be in $A $ .

if $\phi $ is continuous at $L $ then

$\phi (x_n)$ will go to $\phi (L ) $.

ans since $x_{n+1}=\phi (x_n) $, we will get $$L=\phi (L).$$

Answer 2

What do you mean by not being able to use continuity of $\phi$? Because the fact of $\phi$ being a contraction implies it's (uniform) continuity, as is also the case for more general Lipschitz functions. Maybe you mean that you have to prove the existence of a limit before showing it is a fixed point?

The proof indeed has 'two parts'. I don't quite know your context, but in general, the theorem can be stated and proved in any complete metric space $(X,d)$ (for instance, a Banach space, a Hilbert space, any Euclidean space $\mathbb{R}^n$, etc). Then $\phi$ is a strict contraction iff there is some $L \in (0,1)$ such that $$d(\phi(x),\phi(x'))\le L d(x,x'),\quad\forall x,x' \in X.$$

First, to prove the existence of a limit of $x_n$ you have to prove that it is a Cauchy sequence, which implies it has a limit by definition of completeness.

You can see for instance that $$d(x_{k+1},x_1) = d\big(\phi(x_k),\phi(x_0)\big)\le L d(x_k,x_0)$$ or that $$d(x_{k+2},x_2) = d\big(\phi(x_{k+1}),\phi(x_1)\big)\le L d(x_{k+1},x_1)\le L^2 d(x_k,x_0);$$ by induction in $n\in \mathbb{N}$ is easily seen that for any $k\in \mathbb{N}$ $$d(x_{k+n},x_n) \le L^n d(x_k,x_0) \to 0 \quad \text{as}\quad n\to \infty.$$

This actually shows that $\{x_n\}$ is a Cauchy sequence and since $X$ is complete then $x_n$ is convergent in $X$ to some point $x^* \in X$.

The second part of the proof is the argument you mention, which is justified for the fact of $\phi$ being continuous (since it is absolutely continuous, since it is Lipschitz, since it is a contraction).