Question

State and prove contraction lemma

Answer 1

$\textsf{The contraction lemma: Suppose that X is a complete metric space and that } \\ f : X \to X \textsf{ is a contraction mapping on X. Then there is a unique } z \isin X \textsf{such that } \\ f(z) = z. \textsf{ Furthermore, if } x_0\textsf{ is any point in X,} \textsf{then } f^n(x_0)\to z \textsf{ as } n\to \infin. \\ \hspace{2cm}\\ Proof. \textsf{ Let } x_0 \in X.\textsf{ We will show that the sequence } \lbrace x_R= f^n(x_0)\rbrace _{n-1}^\infin \textsf{ is a Cauchy}\\ \textsf{sequence in X. Let } D=d(x_0,f(x_0)) \textsf{and let }0<c<1 \textsf{ be the contraction constant for }f. \textsf{ By the definition of contraction mapping } \\ d(f(x_0),f^2(x_0)) \leq c . d(x_0,f(x_0))=c.D. \textsf{ By induction one can establish that }\\ d(f^n(x_0),f^{n+1}(x_0)) \leq c^n.D.\textsf{ Thus,}\\ \hspace{1cm}\\ \hspace{2cm} d(x_0,f^{n+1}(x_0)) \leq \displaystyle\sum_{k=0}^nc^k .D<\frac D{1-c}\\ \textsf{Similarly, } d(f^n(x_0),f^m(x_0)) \leq D.c^n. \displaystyle\sum_{k=0}^{m-n-1}c^k<\frac{c^n.D}{1-c} \textsf{ for all } n<m. \textsf{ So, to see}\\ \textsf{that the sequence is Cauchy, let }\epsilon>0 \textsf{ and choose N such that } \frac{c^N.D}{1-c}<\epsilon. \\ \textsf{ then for } N\leq n \leq m, d(f^n(x_0),f^m(x_0)) < \epsilon. \textsf{ So the sequence is Cauchy.}\\ \textsf{since } \lbrace x_R= f^n(x_0)\rbrace _{n-1}^\infin \textsf{ is Cauchy}, \textsf{it converges to a point } z \in X. \textsf{ But for this } z\\ \lim_{n\to\infin}f^n(x_0)=z= \lim_{n\to\infin}f^{n+1}(x_0)=f(z). \\ \textsf{ So, } z \textsf{ is a fixed point for } f. \textsf{ On the other hand, if there were another fixed point } z' \ne z,\\ \textsf{then } d(f(z),f(z'))=d(z,z')>c.d(z,z'). \\ \textsf{ This last inequality contradicts the assumption that } f \textsf{is a contraction mapping.}\\ \textsf{So, there is only one fixed point.}$