# Fixed point without the constant

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.

• Consider the map $g\colon x \mapsto d(x,Fx)$ on $X$. What do you know about that map? – Daniel Fischer Jan 20 '16 at 14:47
• HINT: in a compact metric space $X$, the sequence $F^n(x)$ has a convergent subsequence. – Crostul Jan 20 '16 at 14:47
• This is a more general result. For your problem, you just note a subset of Euclidean space is compact iff it's closed and bounded. – John Jan 20 '16 at 15:34

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.