Reading guide · Proof index

Fixed point theorem and Picard’s theorem again

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L7.6.2: Complete-space contraction principle with finite geometric bound and explicit limit step

Proof.

Pick x0∈X.x_0 \in X\text{.} Define a sequence {xn}n=1∞\{ x_n \}_{n=1}^\infty by xn+1≔φ(xn).x_{n+1} \coloneqq \varphi(x_n)\text{.} Then
d(xn+1,xn)=d(φ(xn),φ(xn−1))≤kd(xn,xn−1).\begin{equation*} d(x_{n+1},x_n) = d\bigl(\varphi(x_n),\varphi(x_{n-1})\bigr) \leq k d(x_n,x_{n-1}) . \end{equation*}
Repeating nn times, we get d(xn+1,xn)≤knd(x1,x0).d(x_{n+1},x_n) \leq k^n d(x_1,x_0)\text{.} For m>n,m > n\text{,}
d(xm,xn)≤∑i=nm−1d(xi+1,xi)≤∑i=nm−1kid(x1,x0)=knd(x1,x0)∑i=0m−n−1ki≤knd(x1,x0)∑i=0∞ki=knd(x1,x0)11−k.\begin{equation*} \begin{split} d(x_m,x_n) & \leq \sum_{i=n}^{m-1} d(x_{i+1},x_i) \\ & \leq \sum_{i=n}^{m-1} k^i d(x_1,x_0) \\ & = k^n d(x_1,x_0) \sum_{i=0}^{m-n-1} k^i \\ & \leq k^n d(x_1,x_0) \sum_{i=0}^{\infty} k^i = k^n d(x_1,x_0) \frac{1}{1-k} . \end{split} \end{equation*}
In particular, the sequence is Cauchy (why?). Since XX is complete, we let x≔lim⁡n→∞xn,x \coloneqq \lim_{n\to\infty} x_n\text{,} and we claim that xx is our unique fixed point.
Fixed point? The function φ\varphi is a contraction, so it is Lipschitz continuous:
φ(x)=φ(lim⁡n→∞xn)=lim⁡n→∞φ(xn)=lim⁡n→∞xn+1=x.\begin{equation*} \varphi(x) = \varphi\Bigl( \lim_{n\to\infty} x_n\Bigr) = \lim_{n\to\infty} \varphi(x_n) = \lim_{n\to\infty} x_{n+1} = x . \end{equation*}
Unique? Let xx and yy be fixed points.
d(x,y)=d(φ(x),φ(y))≤k d(x,y).\begin{equation*} d(x,y) = d\bigl(\varphi(x),\varphi(y)\bigr) \leq k\, d(x,y) . \end{equation*}
As k<1,k < 1\text{,} the inequality means that d(x,y)=0,d(x,y) = 0\text{,} and hence x=y.x=y\text{.} The theorem is proved.