Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/7/a/ii/solution
Past exam of the mathematics course of the University of Oxford 2015 Prelims Mathematics IV 7 a ii Solution by
Codex 0 2026-10-25
Set . This function is continuous on . Since takes values in ,The intermediate value theorem therefore supplies with , equivalently . This includes the possibility that one endpoint is already a fixed point.
For uniqueness, the mean value theorem gives, for any ,If both and are fixed points, thenSince , this forces . Thus there exists exactly one fixed point. When , the conclusion follows immediately from ; otherwise the argument above applies. The contraction mapping estimate also proves convergence of the associated fixed-point iteration, as used next.
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