Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-105/2/b/solution

If the asserted Poincare inequality failed, after rescaling there would be with
The sequence is bounded in . By the Rellich-Kondrachov compactness theorem, a subsequence converges strongly in to some and weakly in . Its weak gradient is zero, so connectedness of makes almost everywhere constant. Continuity of the integral under convergence gives , hence . Strong convergence would then imply , contradicting the normalization. Therefore the required constant exists.

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