Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 / 2 / 4 / b / Solution

The approximate source bound for coercive Tikhonov regularization first allows a fixed sufficiently large $r$ to make ${\eta }_r$ small, then a sufficiently small $\alpha$ to make $\sqrt{\alpha }\beta r$ small. This order of limits proves $R_{\alpha }f\to K^{†}f$ for every datum in its domain.

For the noise error let $v=R_{\alpha }g$. Its normal equation for coercive Tikhonov regularization givesCompleting the square shows $\alpha {\beta }^2\Vert v{\Vert }^2\le \Vert g{\Vert }^2/4$, hence $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. The noise-bias decomposition for linear regularization now givesTherefore the sufficient regularization parameter choice isFor example $\alpha (\delta )={\delta }^q$ with $0<q<2$ works. Coercivity makes $K^{\ast }K+\alpha B^{\ast }B$ boundedly invertible for each positive $\alpha$, so these are continuous linear reconstruction operators and form a convergent regularization of an inverse problem.