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I come across this problem in my functional analysis exercise book:

Let X be complete metric space and g be a function, If $g^q$ contracts then $g$ has a fixed point.

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up vote 2 down vote accepted

HINT: Let $x_0\in X$. Given $x_n$, let $x_{n+1}=f^p(x_n)$. Use the hypothesis that $f^p$ is a contraction to show that the sequence of distances $\langle\rho(x_n,x_{n+1}):n\in\Bbb N\rangle$ shrinks exponentially fast towards $0$. Deduce from this that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy; since $X$ is complete, it converges to some point $y$. Show that $f(y)=y$.

Now let $x_0'$ be another point of $X$, and construct the sequence $\langle x_n':n\in\Bbb N\rangle$ similarly. Show that $\langle\rho(x_n,x_n'):n\in\Bbb N\rangle$ coverges to $0$, and conclude that $\langle x_n':n\in\Bbb N\rangle$ also converges to $y$. Since $x_0'$ was arbitrary, conclude that $y$ is the unique fixed point of $f$.

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How to know f(y)=y.It is the key,thanks. – 89085731 Oct 14 '12 at 16:35
$y$ and $f(y)$ are fixed points of $f^p$. Thus $d(y,f(y)) = d(f^p(y),f^p(f(y)))$. Assume that $f(y)\neq y$. Since $f^p$ is a contraction $d(f^p(y),f^p(f(y))) < d(y,f(y))$. We get a contradiction: $d(y,f(y)) = d(f^p(y),f^p(f(y))) < d(y,f(y))$. – Yury Oct 14 '12 at 16:42
@89085731: See Yury’s comment just above this. (He used $d$ instead of $\rho$ for the metric, but that shouldn’t pose any problem.) – Brian M. Scott Oct 14 '12 at 16:45
@Yury Thanks a lot. – 89085731 Oct 14 '12 at 16:45

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