Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 70 4 c ii Solution Created 2026-10-03 Updated 2026-10-14
For , minimize . Its first variation in every direction givesThe self-adjoint operator on the left has lower bound , so it is invertible and the minimizer is unique. The regularized solution has the singular expansionSince , its gain is at most . ConsequentlyThis proves existence, uniqueness and continuous data dependence at fixed positive parameter. For the particular perturbation from part (i), the error is instead of . To recover a minimum-norm exact solution as noise tends to zero, choose while ; dominated convergence of the exact-data singular coefficients removes the bias.
To connect this with the provided inverse expansion, put and . The exact intertwining identity isIt follows by multiplying on either side by the appropriate resolvent. A useful exact inverse identity isExpanding to two terms gives the printed approximation, namelyThis is the truncated Tikhonov resolvent. Multiplication gives , so the expression is not an exact inverse. For example and give the approximate value instead of the exact . Its exact remainder isIn the regime it is a controlled Neumann expansion. Using it to reconstruct givesFor each fixed this is a bounded polynomial in bounded operators, withThus the supplied approximation also has continuous data dependence in its domain of use. It must not be extrapolated to with fixed: its gain on any nonzero singular mode diverges, so it would not define a convergent regularization family there. The exact Tikhonov filter supplies the well-posed regularized problem; the truncated formula is only a controlled computational approximation.