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)

  • $\begingroup$ Should it be $d(T^nx,T^ny)\leq \alpha_nd(x,y)$? $\endgroup$ – Dave Apr 23 '19 at 3:27
  • $\begingroup$ @Dave.sorryy i edited now thank you $\endgroup$ – Inverse Problem Apr 23 '19 at 3:28


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:

to show existence of a fixed point, consider $$d(Tx_0,x_0)=d(T(T^Nx_0),T^Nx_0)=d(T^N(Tx_0),T^N(x_0))\leq \alpha_N d(Tx_0,x_0)$$ and thus $$(1-\alpha_N)d(Tx_0,x_0)\leq 0$$ but $\alpha_N<1$.


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