# A question regarding the fixed point of the Banach theorem

To my understanding the unique fixed-point $\xi$ that the Banach fixed-point theorem guarantees the existence of is essentially a contraction point since if $x\neq\xi \Rightarrow |f(x)-\xi|<|x-\xi|$. So basically $f(x)$ is closer to $f(\xi)=\xi$ than $x$ is to $\xi$. What I am unable to understand is why this point $f(\xi)$ actually has to equal $\xi$. Why is it that this contraction point necessarily has to equal the point it was mapped from? Intuitively it seems to me that we could construct a function that satisfies all the necessary conditions and instead makes it such that $x\neq\xi \Rightarrow |f(x)-f(\xi)|<|x-\xi|$ with $f(\xi)\neq\xi$. Why can this not be the case?

By assumption, $f$ is a function such that $x \neq \xi \implies |f(x) - f(\xi)| < | x - \xi |$ for all $x, \xi$ in your complete metric space $X$.

[In fact, $f$ is assumed to satisfy the stronger condition that there exists a $q \in [0,1)$ such that $|f(x) - f(\xi)| \leq q | x - \xi |$ for all $x, \xi \in X$.]

The claim that there exists a unique $\xi \in X$ such that $f(\xi) = \xi$ is the conclusion of the theorem. In other words, the existence of such a point $\xi$ (the thing that is not yet clear to you) is precisely the non-trivial thing that must be justified by the proof of the theorem.

A sketch of the justification is as follows:

• Pick any point $x$. (It doesn't matter which one.) Define a sequence $x_n$ recursively by setting $x_0 = x$ and $x_{n+1} = f(x_n)$. Use the fact that $f$ is a contraction to show that the sequence $x_n$ is Cauchy.
• Since $X$ is complete, the Cauchy sequence $x_n$ converges to a limit as $n \to \infty$. Call this limit $\xi$.
• Take $\lim_{n \to \infty}$ of both sides of the equation $x_{n+1} = f(x_n)$ to deduce that $f(\xi) = \xi$.

Thus we have succeeded in exhibiting a fixed point for the function $f$: the $\xi$ that we constructed as the limit of the sequence $x_n$ is the desired fixed point. All that remains now is that $f$ doesn't have any other fixed points besides $\xi$, and this is an easy consequence of the contraction property of $f$.