Let $(X,d)$ be a complete metric space and $T:X\to X$ be a mapping such that for some sequence $(\alpha_n)\in (0,\infty), d(T^nx,T^ny)\le \alpha _n d(x,y) $, for $x,y\in X$. If $\liminf_{n\to \infty} \alpha_n <1$ , then show that
(a) there exists a $N \in \mathbb{N}$ such that $T^N$ is a contraction mapping and
(b) $T$ has a unique fixed point in $X$
how to prove that $T^N$ is contraction ?can we directly say from $(\alpha_n)\in (0,\infty), d(T^nx,T^ny)\le d(x,y) $
since $|T^nx-T^ny|\le \alpha_n |x-y|$
and how to prove (b)
Best Answer
Hints:
For (a): since $\liminf_{n\to\infty} \alpha_n<1$, there exists $N$ such that $\alpha_N<1$.
For (b): since $T^N$ is a contraction it has a unique fixed point, call it $x_0$. Try getting a unique fixed point of $T$ from this.
For an additional (b) hint: