Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 16F (Analysis II)

The contraction mapping theorem states: if $(X,d)$ is a nonempty complete metric space and $f:X\to X$ satisfies $d(fx,fy)\le qd(x,y)$ for a fixed $0\le q<1$, then it has exactly one fixed point and the iteration from every starting point converges to it.

To prove existence, choose $x_0$ and let $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and geometric sum then imply, for $m>n$,The iterates are a Cauchy sequence; completeness supplies their limit $x_{\ast }$. The contraction is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ forces equality of the points. The same estimate gives geometric convergence from every initial point.

For the particular map, the original PDF gives $X=[\sqrt{a/2},\infty )$, which is a closed, complete subset of the real line. For positive $x$, the arithmetic-geometric mean inequality gives $f(x)=(x+a/x)/2\ge \sqrt{a}$, proving $f(X)\subset X$. FurthermoreThe mean value theorem therefore gives $|f(x)-f(y)|\le |x-y|/2$, a genuine contraction. The fixed-point equation is $x^2=a$, and only the positive root belongs to $X$, soThe printed square-root endpoint is essential: the TeX transcription loses it.