Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 / 1

A forward problem evaluates a known model at a state: given $u$ and $K$, compute $f=Ku$. The associated inverse problem takes the data $f$ and the model $K$ and seeks a state $u$ with $Ku=f$.

In the sense of Hadamard well-posedness, a problem is well posed if a solution exists for every admissible datum, is unique, and depends continuously on the datum in the specified input and output topologies. An inverse problem is ill-posed if at least one of these requirements fails. Thus ill-posedness means failure of existence, uniqueness or stability. The topologies and the admissible data set matter: restricting the data to the range restores existence, but need not restore continuous dependence.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, a least-squares solution of data $f$ is a minimizer of the residual norm:The minimum-norm least-squares solution is the one of smallest $U$-norm among these minimizers. Equivalently it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If the equation is consistent, this is the smallest-norm exact solution.

Let $P$ be the orthogonal projection onto $\overline{\mathcal{R}(K)}$. The orthogonal decomposition by a closed subspace givesThe first term has infimum zero by the definition of closure; that infimum is attained exactly if $Pf\in \mathcal{R}(K)$. This proves the least-squares existence criterion:They exist for every datum precisely when the range is closed. Once a solution exists, all solutions differ by elements of the closed null space, and projecting onto its orthogonal complement gives the unique minimum-norm solution.

For a counterexample to existence, take $K:{\ell }^2\to {\ell }^2$, $(Ku{)}_j=u_j/j$, and $f_j=1/j$. The range is dense, so the least residual infimum is zero; finite truncations realize residuals tending to zero. But an exact preimage would have every coordinate equal to one and would not belong to ${\ell }^2$. Thus no least-squares minimizer exists for this datum.

Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.

A singular system of a compact operator $K:U\to V$ consists of positive singular values ${\sigma }_j$ and orthonormal systems $u_j\in U$, $v_j\in V$ satisfyingThey span $\mathcal{N}(K{)}^{\perp }$ and $\overline{\mathcal{R}(K)}$, respectively. The singular value decomposition isFor an infinite-rank compact operator, the singular values tend to zero; for finite rank the sum is finite. Vectors in $\mathcal{N}(K)$ contribute nothing. This convention labels the input singular vectors by $u_j$ and the output ones by $v_j$.

For $(Ku{)}_j=u_j/j$, truncating after $n$ diagonal entries gives finite-rank operators with operator-norm error $1/(n+1)$, proving compactness. Its SVD is

The Picard criterion with the input/output convention above statesNecessity follows from $⟨f,v_j⟩={\sigma }_j⟨u,u_j⟩$ and Bessel inequality. Conversely the summability constructs $u={\sum }_j{\sigma }_j^{-1}⟨f,v_j⟩u_j\in U$; the closure condition ensures that its image is all of $f$. The closure condition must not be omitted: a component orthogonal to the range cannot be reconstructed.

For the diagonal example, $f_j=j^{-p}$ is itself in ${\ell }^2$ only for $p>1/2$. Its preimage would be $u_j=j^{1-p}$, so the range criterion is the convergence of ${\sum }_j{j}^{2-2p}$. The P-series givesAt $p=3/2$ the reconstruction norm diverges harmonically. For $1/2<p\le 3/2$ the data are legitimate but have no preimage; for $0<p\le 1/2$ they are not even in the specified data space.

The range of the Volterra integration operator is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a Sobolev space element that is an absolutely continuous function. Indeed $Ku$ has weak derivative $u$ and zero initial trace; conversely the fundamental theorem of calculus reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.

The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its distributional derivative is a Dirac delta distribution, not an $L^2$ function. ThusFor a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative norm is $\sqrt{n}$, illustrating unstable differentiation.

The supplied SVD gives $K({\sigma }_j{u}_j)={\sigma }_j^2{v}_j$, or more simply $K^{†}({\sigma }_j{v}_j)=u_j$. Since ${\sigma }_j{v}_j\to 0$ but $\Vert u_j\Vert =1$, the Moore–Penrose inverse is discontinuous. Its domain is the dense, nonclosed range above, and there it is the weak derivative.