A problem similar to Banach fixed point theorem

Question

a) Let $(X,d)$ be a complete metric space and let $T: X \to X$. Prove that if there exists a natural $n$ such that $T^n(x)$ (composition of $T$ $n$ times) is a contraction then $T(x)$ has a unique fixed point.

($T$ is a contraction iff there exists $0 \leq c <1$ such that for all $x,y \in X$:

$d(T(x),T(y))< c \cdot d(x,y)$)

b)Is the statement true if $T$ is a weak contraction?

(i.e. for all $x,y \in X$: $d(T(x),T(y)) < d(x,y)$)

I have been trying to prove that $T$ is a contraction, but I can't do it. Any help?

Answer 1

There are no more than one fixed point. Indeed, suppose that there are two: $x$ and $y$. Then $$d(x,y)=d(T^n(x),T^n(y))<c\cdot d(x,y)$$ and that implies $x=y$.

Now take any point $x_0$, and define $x_{k+1}=T^n(x_k)$. For any pair of natural numbers $p>q$ we have $$d(x_p,x_q)<c^qd(x_{p-q},x_0)\leq c^q\sum_{j=1}^{p-q}d(x_j,x_{j-1})<c^q\sum_{j=1}^{p-q}c^jd(x_1,x_0)=c^{q+1}\frac{1-c^{p-q}}{1-c}d(x_1,x_0)$$

This shows that the sequnce $\{x_n\}$ is a Cauchy sequence. Can you finish?

Answer 2

A different approach for the first part. Let $p$ be the unique fixed point of $T^n$. Then $$ T^n(T(p))=T(T^n(p))=T(p)\implies T(p)\text{ is a fixed point of }T^n\implies T(p)=p. $$

For the second part, consider $f\colon\mathbb{R}\to\mathbb{R}$ given by $$ f(x)=x-\arctan x+\frac\pi2. $$ It has no fixed points and it is a weak contraction since $$ f'(x)=\frac{x^2}{1+x^2}\implies0\le f'(x)<1\quad\forall x\in\mathbb{R}. $$

Answer 3

Since $T^n$ is a contraction, it has a unique fixed point, say $y$. So, $$T^n(Ty)=T(T^n y)=Ty.$$ Thus, $Ty$ is also a fixed point of $T^n.$ By uniqueness we have, $Ty=y.$

To prove uniqueness of $y,$ suppose $x$ is a fixed point of $y.$ Then $$T^nx=T^{n-1}(Tx)=T^{n-1}x=T^{n-2}(Tx)=\cdots x$$ Thus, $x$ is also a fixed point of $T^n.$ Hence, $x=y.$