Fixed point without the constant

Question

If $d(Fx,Fy)<d(x,y)$ for all $x,y$ in a closed bounded subset $X$ of Euclidean space and $F\colon X\rightarrow X$ then there is a unique fixed point $x_0$ and $\lim \limits _{n\to\infty} F^n(x)=x_0$. It looks similar to Banach FPT but because there is not constant for ALL the inequalities I can't think how I can approach. Please help.

Answer 1

Take the function $g(x)=d(x,F(x))$. Then $g$ is continuous (as $F$ is) on a compact set and hence it attains its minimum, which is nonnegative. Let $m$ be such minimum.

By contradiction, assume $m>0$ and let $g(x_0)=m$, so that $0<g(x_0) \leq g(x)$ for every $x \in X$. Now take $x=F(x_0)$. We have $$ g(x)=d(F(x),x)=d(F(F(x_0)),F(x_0)) < d(F(x_0),x_0)=g(x_0), $$ which contradicts the minimality of $g(x_0)$. Thus, $m=0$ and so $g(x_0)=0$ which implies $F(x_0)=x_0$, i.e. $x_0$ is a fixed point.

Once you know the fixed point exists, uniqueness and the limit condition can be proved in the same way as they are proved in the Banach fixed point theorem, only existence requires more work (also note compactness is necessary for it).