Let $M$ be complete and let $f:M\to M$ be continuous .If $f^k$ is a strict contraction for some integer $k>1$ ,show that $f$ has a unique fixed point.
My try:
Let us fix $x_0$.Let $g=f^k$.
Put $x_{n+1}=g^{n+1}(x_0)\implies d(x_{n+1},x_{n})=d(g^{n+1}(x_0),g^{n}(x_0))\le \alpha d(g^{n}(x_{0}),g^{n-1}(x_{0}))\le\ldots\le \alpha ^nd(g(x_0),(x_0)) $. where $0\le \alpha<1$
$d(x_n,x_m)\le d(x_n,x_{n-1})+d(x_{n-1},x_{n-2})+\ldots +d(x_{m+1},x_m)\le \{\alpha^{n-1}+\alpha^{n-2}+\ldots \alpha ^{m}\}d(g(x_0),(x_0))\to 0 $
as $n,m\to \infty$
So,$x_{n}$ is Cauchy $\implies y=\lim x_n$ for some $y$[Since $g$ is continuous] $\implies g(y)=\lim g(x_n)=\lim g^{n+1}(x_0)=\lim x_{n+1}=y\implies y$ is a fixed point of $g\implies f^k(y)=y$.
If $f(y)\neq y\implies d(f(y),y)>0$ .
I am having trouble proving $f(y)=y$.
Will you please help me in doing it?
First: You proved the fix point theorem again, I do not think there is need of it, just apply it to $g := f^k$. Hence $f^k$ has an unique fixed point, say $y \in M$.
It follows that $f$ cannot have a fixed point $x \ne y$, as $f(x) = x$ for some $x \in M$ implies $f^k(x) = x$ and hence $y = x$. Therefore, $f$ has at most one fixed point.
To prove that $y$ is fixed by $f$, note that $f(y)$ is a fixed point of $g$, too: We have $$ g\bigl(f(y)\bigr) = f^{k+1}(y) = f\bigl(g(y)\bigr) = f(y) $$ As $g$ has only one fixed point, we must have $f(y) = y$.