Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 13H (Analysis II)

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.