Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 335 / 3

A regularization of an inverse problem consists of bounded operators $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,y^{\delta })$ such that, whenever $\Vert y^{\delta }-y\Vert \le \delta$ and $y$ lies in the domain of $A^{†}$,as $\delta \to 0$. It is needed because a compact operator on an infinite-dimensional space has singular values tending to zero, so direct inversion divides noisy data by arbitrarily small numbers and is generally discontinuous.

Tikhonov regularization definesand its normal equation givesFor every fixed $\alpha >0$, $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$, and the data-to-solution operator is bounded. Small changes in $y^{\delta }$ therefore produce small changes in $x_{\alpha }^{\delta }$.

A singular system of a compact operator is a family $({\sigma }_i,u_i,v_i)$ with ${\sigma }_i>0$, ${\sigma }_i\to 0$, orthonormal $u_i\in Y$ and $v_i\in X$, andOn each singular vector, $A^{\ast }A+\alpha I$ acts by multiplication by ${\sigma }_i^2+\alpha$. HenceThe bounded filter $\sigma /({\sigma }^2+\alpha )$ replaces the unstable inverse factor $1/\sigma$.

WriteThe stated iteration is $x_{n+1}=B{x}_n+{\alpha }^{-1}{A}^{\ast }y$, so repeated substitution and the geometric series identity giveFor the singular component $v_i$, put $r_i=1-{\sigma }_i^2/\alpha$. Since the initial Tikhonov solution has coefficient ${\sigma }_i(y,u_i)/({\sigma }_i^2+\alpha )$, one obtainsTherefore the requested spectral filter isThus equation (3) reads $x_n={\sum }_i{g}_{\alpha ,n}({\sigma }_i)(y,u_i)v_i$. Convergence of this stationary iteration requires $|1-{\sigma }_i^2/\alpha |<1$ on the nonzero spectrum, for example $\alpha >\Vert A{\Vert }^2/2$.