Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/7/a/ii/solution

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, then
Since , 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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