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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.

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    $\begingroup$ Consider the map $g\colon x \mapsto d(x,Fx)$ on $X$. What do you know about that map? $\endgroup$ – Daniel Fischer Jan 20 '16 at 14:47
  • $\begingroup$ HINT: in a compact metric space $X$, the sequence $F^n(x)$ has a convergent subsequence. $\endgroup$ – Crostul Jan 20 '16 at 14:47
  • $\begingroup$ 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. $\endgroup$ – John Jan 20 '16 at 15:34
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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).

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