# Show that $T^N$ contraction and $T$ has a unique fixed point in $X$

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)

• Should it be $d(T^nx,T^ny)\leq \alpha_nd(x,y)$? – Dave Apr 23 at 3:27
• @Dave.sorryy i edited now thank you – Inverse Problem Apr 23 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.
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 $$\alpha_N<1$$.