Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 (Inverse Problems)

Take a minimizing sequence for the variational regularization functionalIts coercivity makes the sequence bounded. Since $X$ is a reflexive Banach space, a subsequence converges weakly to some $u\in X$. A convex norm-lower-semicontinuous functional has weak lower semicontinuity, so this applies to both $J$ and the convex continuous map $u\mapsto \Vert Au-f{\Vert }_Y^2$. Thereforeand $u$ is a minimizer. This is the direct method in the calculus of variations.

An element $u^{†}$ is a $J$-minimizing solution whenIt satisfies the source condition in variational regularization when there is a $w^{†}\in Y^{\ast }$ such thatwhere $\partial J(u^{†})$ is the subdifferential of the convex functional $J$.

Let ${\hat{u}}_{\alpha }$ minimize the nonsquared-residual objective. Comparison with $u^{†}$ givesThe source subgradient inequality givesConsequently,Thus forevery $0<\alpha <{\alpha }_0$ forces $A{\hat{u}}_{\alpha }=f$. The original comparison then gives $J({\hat{u}}_{\alpha })\le J(u^{†})$, so ${\hat{u}}_{\alpha }$ is itself $J$-minimizing. If $J$ is strictly convex, its restriction to the affine solution set has at most one minimizer, hence ${\hat{u}}_{\alpha }=u^{†}$. This is an exact penalty method.

For $p\in \partial J(v)$, the Bregman divergence isPut $r=A{\hat{u}}_{\alpha ,\delta }-f^{\delta }$ and $e=f^{\delta }-f$, so $\Vert e{\Vert }_Y\le \delta$. Comparison with $u^{†}$ givesUsing $p^{†}=A^{\ast }{w}^{†}$,For $0<\alpha <{\alpha }_0$ the last coefficient is negative, soThe estimate holds for any fixed admissible $\alpha$; it does not require $\alpha \to 0$ with the noise level.

The exact solution is feasible becauseThe feasible set $E_{\delta }$ is convex and weakly closed. A minimizing sequence has bounded residual and bounded $J$; the coercivity assumption from part (a), applied to a fixed positive weighted objective, makes it bounded in $X$. Reflexivity gives a weakly convergent subsequence, and weak lower semicontinuity of the residual and $J$ keeps its limit feasible and minimizing.

Since ${\hat{u}}_{\delta }$ minimizes $J$ over $E_{\delta }$,The source condition in variational regularization and feasibility then yieldThus the claimed constant is $\boxed{C=(c+1)\Vert w^{†}{\Vert }_{Y^{\ast }}}$.

The kernel $k(x,y)=e^{-|x-y|}$ is real and symmetric. For $f,g\in L^2[0,1]$, Fubini's theorem givesso $K$ is self-adjoint. It is linear, and because $k\in L^2([0,1{]}^2)$ it is a Hilbert-Schmidt operator, hence bounded, with

Suppose $Kf=\lambda f$. Differentiating the integral expression on the two sides of $y=x$ givesThereforeThe first boundary condition makesSubstitution into the second givesor

Extend $f$ by zero outside $[0,1]$. Since the Fourier transform of $e^{-|x|}$ is $2/(1+{\xi }^2)>0$,for $f\ne 0$. Thus every eigenvalue is positive. From the relation in part (b),The transcendental equation has its successive roots in intervals separated by the poles and zeros of $\tan \omega$, so ${\omega }_n$ grows linearly with $n$. Hence

The compact self-adjoint operator $K$ has an orthonormal basis of eigenvectors $e_n$. Part (c) gives ${\lambda }_n>0$ and ${\sum }_n{\lambda }_n<\infty$, so $K$ is positive and trace class, exactly the required condition for a covariance operator of a Gaussian measure on a Hilbert space.

For independent standard normal variables ${\xi }_n$, define the Hilbert-space Gaussian seriesBecause $\mathbb{E}\Vert U-m{\Vert }^2={\sum }_n{\lambda }_n<\infty$, the series converges in $L^2(\Omega ;X)$ and almost surely. Its law is the Gaussian measure $N(m,K)$.

For observed $v\in {\mathbb{R}}^N$, the Gaussian likelihood isThus the Bayesian inverse problem is to determine the posterior distribution of $U$ given $V=v$. WithBayes' formula gives

The total variation distance isThe problem is a well-posed Bayesian inverse problem in total variation when every $v\in {\mathbb{R}}^N$ determines a unique posterior ${\mu }^v$ and

The heat solution operator at positive time is bounded from $L^2[0,1]$ to $C[0,1]$, so the finite sensor map $G:X\to {\mathbb{R}}^N$ is bounded and continuous. Therefore $\Phi (u;v)$ is jointly continuous and $0<e^{-\Phi }\le 1$. The normalizer satisfies $0<Z(v)\le 1$. If $v_n\to v$, the dominated convergence theorem gives bothand convergence in $L^1({\mu }_0)$ of the normalized posterior densities. Since total variation is one half of this $L^1$ distance for absolutely continuous measures, $d_{\mathrm{TV}}({\mu }^{v_n},{\mu }^v)\to 0$. Existence, uniqueness, and continuous dependence all follow.

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.