Let be a closed interval and be a function satisfying for all distinct . What we want to show is that converges to the unique fixed point of .
Proofs for the existence and uniqueness of the fixed point are standard. Additionally, since for distinct , also has a unique fixed point. We also need the “trivial” result that if converges, it converges to the unique fixed point, and likewise for ; what’s left is to show that this limit exists regardless of starting point.
For any , let . Also let be the fixed point of . Suppose to the contrary that does not converge. Consider the sequence , which is clearly positive and decreasing, and thus convergent. If it converges to 0, then , a contradiction. Therefore must be positive, and there exists some such that for all , .
If is eventually always positive or eventually always negative, then exists, which is a contradiction. If is eventually always negative, then is eventually always positive or eventually always negative, where as before we get a contradiction.
Consequently, there exist some such that
Then we have
but also, since ,
a contradiction. Therefore the initial supposition that is divergent is false, and the proposition is proven.
There’s probably a better way of proving this. This one relies on the structure of the real numbers even though it should hold true when generalised to arbitrary metric spaces. The caveats are that we can’t use the intermediate value theorem to show the existence of a fixed point and that it’s not necessarily true that an ordering exists at all, so splitting into the positive/negative/alternating/none-of-the-above cases is an invalid operation.