# Ngô Quốc Anh

## December 29, 2008

### A fixed point theorem

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 17:06

Let $(X,d)$ be a complete metric space and let $F: X\to X$ be such that $F^N : X \to X$ is a contraction for some positive integer $N$. Show that $F$ has a unique fixed point $u \in X$ and that for each $x \in X$, $\mathop {\lim }\limits_{n \to \infty } {F^n}\left( x \right) = u$.

Proof.
Since $F^N : X \to X$ is a contraction, then $F^N$ has a fixed point, say $u_0$, i.e., $F^N u_0=u_0$. Note that $\displaystyle d\left( {F{u_0},{u_0}} \right) = d\left( {{{\left( {{F^N}} \right)}^n}F{u_0},{{\left( {{F^N}} \right)}^n}{u_0}} \right) \leq {k^n}d\left( {F{u_0},{u_0}} \right)$.

Since $k<1$ then $d\left( {F{u_0},{u_0}} \right)=0$. In other word, $u_0$ is a fixed point of $F$.

To prove the uniqueness, assume $u’_0$ is also a fixed point of $F$. Then both $u_0$ and $u’_0$ are fixed points of $F^N$ which implies that $u_0 \equiv u'_0$ due to the uniqueness of fixed point of a contractive mapping.