Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 3

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator, withThe Moore–Penrose inverse of an operator is the generally unbounded mapdefined when the Picard criterion holds, with the component in $\ker A^{\ast }$ sent to zero. It is the minimum-norm least-squares solution of $Af=g$.

A regularization of an inverse problem consists of bounded maps $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,g^{(\delta )})$ such thatwhenever $\Vert g^{(\delta )}-g\Vert \le \delta$ and $g$ is in the domain of $A^{†}$.

For Tikhonov regularization, minimizinggivesThe scalar spectral filter satisfiesand consequentlyFor exact data, each filter factor ${\sigma }_j^2/({\sigma }_j^2+\alpha )$ tends to one, so $R_{\alpha }g\to A^{†}g$. Choosingtherefore makes both the approximation error and propagated data error vanish. For example, $\alpha (\delta )=\delta$ is an admissible a priori regularization parameter choice.

For the Volterra integration operatorthe adjoint operator isSince $1/{\sigma }_n=(2n-1)\pi /2$, direct integration givesSimilarly,because $\cos ((2n-1)\pi /2)=0$. The half-integer sine and cosine families are orthonormal bases of $L^2([0,1])$, so $({\sigma }_n,u_n,v_n)$ is a singular system of a compact operator.

The Tikhonov solution for noisy data $g^{(\delta )}$ is thereforeWriting out the inner product and the singular functions makes this explicit:The factors ${\sigma }_n/({\sigma }_n^2+\alpha )$ suppress the unstable reciprocal growth $1/{\sigma }_n$ at high index.