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Given $(X,d)$ complete with $A \subset X$ closed, and $f: A \to A$ satisfying $$ d(f^{n}(x),f^{n}(y)) \leqslant a_{n}d(x,y) \hspace{3mm} \forall x,y \in A \hspace{3mm} n \in \mathbb{N}$$ where $a_{n} > 0$ and $\sum_{n=1}^{\infty} a_{n} < \infty$, show that $f$ has a unique fixed point.

Obviously, I want to show that $f$ has a unique fixed point. Since $A \subset X$ is closed in the complete space $(X,d)$, then $(A,d)$ is complete. Do I simply need to show that $f$ is a contraction? (Banach Fixed Point Theorem). I feel that I am missing something.

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    $\begingroup$ Hint: There exists some sufficiently large $n$ at which $a_n < 1$. Then you will apply the fixed pt theorem somehow. $\endgroup$ – Christopher A. Wong Feb 17 '16 at 0:58
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If some iterate of $f$ is a (strict)-contraction, then $f$ will have an unique fixed point, in the conditions of Banach's fixed point theorem (see generalizations).

Since $\sum_{n\geq 1} a_n < +\infty$, we have that $\lim_{n\to +\infty} a_n = 0$. Then there is $n_0 \geq 1$ such that $0 < a_n < 1$ for all $n\geq n_0$. So it follows that, say, $f^{n_0}$ is a contraction.

Since $A\subseteq X$ is closed and $X$ is complete, we have $A$ complete. Then by the Banach fixed point theorem, $f$ has a unique fixed point in $A$.

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    $\begingroup$ To the OP: Your main task is to the prove this generalization of Banach's theorem that Ivo has mentioned. $\endgroup$ – Christopher A. Wong Feb 17 '16 at 1:08
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Hint: $\sum_{n=1}^{\infty}a_{n} < \infty \implies \lim_{n\to \infty} a_{n} = 0 \implies a_{N} < 1$ for some $N$. Hence, $g(x) = f^{N}(x)$ is a contraction by the hypothesis and hence by the contraction mapping theorem, there exists $x_{0}$ such that $f^{N}(x_{0}) = x_{0}$. $f(x_{0})=f(f^{N}(x_{0}))=f^{N}(f(x_{0}))$ so $f(x_{0})$ is a fixed point of $f^{N}$ so by the uniqueness $f(x_{0})=x_{0}$.

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  • $\begingroup$ What exactly is the contradiction? $\endgroup$ – clocktower Feb 17 '16 at 1:06
  • $\begingroup$ @clocktower. He meant to say "contraction." $\endgroup$ – DanielWainfleet Feb 17 '16 at 1:24
  • $\begingroup$ (1) Since $x_0$ is the UNIQUE fixed point of $f^N$, and $f(x_0)$ is also a fixed point of $f^N$, we have $x_0=f(x_0).$...(2) If $y$ is any fixed point of $f$ then $y$ is a fixed point of $f^N$ so $y=x_0$.....(3) All that is needed is that $a_N<1$ for some $N$. $\endgroup$ – DanielWainfleet Feb 17 '16 at 1:31
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The usual proof of Banach's theorem applies to the generalization: For every $x_0\in A$ the sequence defined either recursively by $x_{n+1}=f(x_n)$ or explicitly as $x_n=f^n(x_0)$ is Cauchy, since for $n<m $ $$ d(x_n,x_m) \le \sum_{k=n}^{m-1} d(x_{k+1},x_k) \le \sum_{k=n}^{m-1} a_k d(x_1,x_0). $$ For the limit $x_\infty$ this leads to a better error estimate for $d(x_n,x_\infty)$ than the plain application of classical case to $f^N$ as explained in the other answers.

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