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

Write $u^{†}=K^{†}f$, $e=R_{\alpha }f-u^{†}$, and $z={\beta }^{-2}{B}^{\ast }B{u}^{†}$, with $\beta >0$ as required by coercivity and by the printed inverse power. The Tikhonov regularization with a coercive penalty operator normal equation givesThis also holds for inconsistent data in $\mathcal{D}(K^{†})$, because $f-K{u}^{†}\perp \mathcal{R}(K)$ and therefore $K^{\ast }f=K^{\ast }K{u}^{†}$. Taking the inner product with $e$ givesFor any $w$ with $\Vert w\Vert \le r$, put $d=\Vert z-K^{\ast }w\Vert$ and use Cauchy-Schwarz inequality to bound the right-hand side by $\alpha {\beta }^2(d\Vert e\Vert +r\Vert Ke\Vert )$. Since $\Vert Be\Vert \ge \beta \Vert e\Vert$, completing the square in $\Vert Ke\Vert$ yieldsThe positive root of this quadratic inequality is at most $d+\sqrt{\alpha }\beta r/2$. Infimizing over $w$ proves the approximate source bound for coercive Tikhonov regularization, in fact slightly more strongly:Injectivity gives $\overline{\mathcal{R}(K^{\ast })}=\mathcal{N}(K{)}^{\perp }=U$. Given any positive tolerance, approximate $z$ by $K^{\ast }w$ for some finite-norm $w$; for all sufficiently large $r$ it is admissible in the infimum. ThusNo attainment of that infimum is needed.