Past exam of the mathematics course of the University of Cambridge 2003 ib Paper 3 11F Solution Created 2026-09-24 Updated 2026-10-14
The contraction mapping theorem states that a contraction mapping of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if with , iteration from any converges to that fixed point.
To prove this, put . Induction gives , so for the triangle inequality givesThus is a Cauchy sequence, and completeness supplies a limit . A contraction mapping is continuous, so . Two fixed points satisfy , forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate .
For the function space, take nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of makes the uniform metric finite. Nonnegativity and symmetry follow from ; means at every . Taking suprema in proves the triangle inequality. Hence is a metric.
Suppose is a Cauchy sequence for this uniform metric. At each , , so completeness of supplies a value . Given , choose with for all . Fix and let pointwise; then for every . Thus . The uniform limit theorem makes this limit continuous, so . This proves the completeness of the continuous-map space in the uniform metric.
Finally fix with contraction constant , and write and . Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing proves continuity of at every . The estimate uses the constant of the fixed map only; it does not require a common contraction constant for all nearby .