Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 / 3

Hadamard well-posedness requires existence, uniqueness, and continuous dependence of $u$ on $f$. A compact operator with infinite-dimensional range cannot have closed range: otherwise its inverse on the orthogonal complement of its kernel would be bounded, making the identity on an infinite-dimensional space compact. Hence the inverse on $\mathrm{ran}A$ is unbounded. If $\ker A\ne \{0\}$ uniqueness also fails, and data outside the range have no exact solution. In every case at least stability fails, so the inverse problem is ill posed.

If $\Vert K\Vert <1$, the partial Neumann series satisfiesSince $\Vert K^{N+1}\Vert \le \Vert K{\Vert }^{N+1}\to 0$, the series converges in operator norm andIf $\tau \ne 0$ and $\Vert I-\tau K\Vert <1$, apply this identity to $I-\tau K$:

The normal equation for a linear inverse problem isIt has a solution exactly whenWhen solutions exist they formThey are unique exactly when $\ker A=\{0\}$, while $A^{†}f$ is always the unique solution in $(\ker A{)}^{\perp }$ and the solution of minimum norm. Here $A^{†}$ is the Moore-Penrose inverse.

The operator $A^{\ast }A$ is positive, self-adjoint, and compact. Because $\mathrm{ran}A$ is infinite dimensional, the spectral theorem for compact Hermitian operators gives positive spectral values tending to zero. Even if $\ker A=\{0\}$, zero remains in the spectrum as a limit point. Hence for every $\tau$,The strict inequality $\Vert I-\tau K\Vert <1$ needed for the operator-norm Neumann series is impossible, so formula (3) cannot be applied directly to invert $A^{\ast }A$.

Let $({\sigma }_j,u_j,v_j)$ be a singular system for $A$. The Picard criterion for $f\in \mathrm{dom}(A^{†})$ isafter discarding the component in $(\mathrm{ran}A{)}^{\perp }$. For $0<\tau <\Vert A{\Vert }^{-2}$, the partial series acts diagonally:For each $j$ the multiplier in the numerator tends to one and lies in $[0,1]$. The Picard summability condition therefore supplies an ${\ell }^2$ dominating sequence, so the dominated convergence theorem givesThis is the series form of Landweber iteration.

For $\alpha >0$, put $N(\alpha )=\lfloor {\alpha }^{-1}\rfloor$ and define the bounded operatorPart (e) shows $R_{\alpha }f\to A^{†}f$ for every $f\in \mathrm{dom}(A^{†})$ as $\alpha \downarrow 0$. Since $\Vert I-\tau A^{\ast }A\Vert \le 1$,Choose the a priori ruleThen $N(\alpha (\delta ))\to \infty$ whileFor $\Vert f^{\delta }-f\Vert \le \delta$,Thus $\{R_{\alpha }\}$ with this parameter rule is a regularization of an inverse problem.