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.
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.
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 $f$.