Let $f: E \rightarrow E$, $E$ a complete metric space. Assume that there exists $K$ such that $0 < K < 1$ and $d(f(x), f(y)) \le Kd(x,y)$ for all $x,y \in E$. Prove that there exists $a \in E$ such that $a=f(a)$.
This was a problem given to me by a friend, as I am self studying real analysis. I am unsure of how to proceed with this question. Could someone help me out? Thanks!
Try to prove that $x, f(x), f(f(x)), f(f(f(x))),\ldots$ is a Cauchy sequence. Let $a$ be its limit. Then see if you can show that for every $\varepsilon>0$, $d(a,f(a))<\varepsilon$.
A function $f$ for which there exists $K$ between $0$ and $1$ such that for all points $x,y$ one has $d(f(x),f(y))\le Kd(x,y)$ is called a contraction.