Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 70 / 4 / c

The adjoint operators, orthonormal singular vectors and squared-norm normal equation for a linear inverse problem conditions require Hilbert spaces. For arbitrary normed spaces the Banach-space adjoint takes values in the dual spaces and need not produce this formula. Interpret the given singular value system in Hilbert spaces, with $u_j$ in the unknown space and $v_j$ in the data space, as in the printed identities. Positive singular values tending to zero provide the compact ill-posed case. A general bounded linear operator need not have this property; the identity operator is a well-conditioned counterexample. Components in the kernel must be fixed by a minimum-norm convention, and the unregularized inverse is defined only for data satisfying the corresponding range condition.

Using the singular value system with $A{u}_j={\sigma }_j{v}_j$ and $A^{\ast }{v}_j={\sigma }_j{u}_j$, the formal minimum-norm inverse isIt defines an element of the unknown Hilbert space only if ${\sum }_j|(y,v_j){|}^2/{\sigma }_j^2<\infty$, the Picard criterion. A null-space component is undetermined by the data.

Adding an error along a single left singular vector changes precisely one coefficient:Although this perturbation has data norm $\delta$, its amplification factor is $1/{\sigma }_j$. If ${\sigma }_j\to 0$, these factors are unbounded. More explicitly choose $y^{(j)}={\sigma }_j{v}_j$, tending to zero, while the minimum-norm inverse is $u_j$, of norm one. Thus the inverse is discontinuous at zero. It may also fail to exist for data outside its nonclosed range, so merely minimizing a residual does not guarantee that the formal inverse exists. This is ill-posedness by unbounded small-singular-value amplification, not a claim that every finite-dimensional or bounded forward map has an unstable inverse.

For $\alpha >0$, minimize $J_{\alpha }(x)=\Vert Ax-y{\Vert }^2+\alpha \Vert x{\Vert }^2$. Its first variation in every direction givesThe self-adjoint operator on the left has lower bound $\alpha I$, so it is invertible and the minimizer is unique. The regularized solution has the singular expansionSince ${\sigma }^2+\alpha \ge 2\sigma \sqrt{\alpha }$, its gain is at most $1/(2\sqrt{\alpha })$. ConsequentlyThis proves existence, uniqueness and continuous data dependence at fixed positive parameter. For the particular perturbation from part (i), the error is $\delta {\sigma }_j/({\sigma }_j^2+\alpha )$ instead of $\delta /{\sigma }_j$. To recover a minimum-norm exact solution as noise tends to zero, choose $\alpha \to 0$ while $\delta /\sqrt{\alpha }\to 0$; dominated convergence of the exact-data singular coefficients removes the bias.

To connect this with the provided inverse expansion, put $B=A{A}^{\ast }$ and $C=A^{\ast }A$. The exact intertwining identity isIt follows by multiplying $(C+\alpha I)A^{\ast }=A^{\ast }(B+\alpha I)$ on either side by the appropriate resolvent. A useful exact inverse identity isExpanding $(I+{\alpha }^{-1}C{)}^{-1}$ to two terms gives the printed approximation, namelyThis is the truncated Tikhonov resolvent. Multiplication gives $(\alpha I+B)R_{\alpha ,2}=I+{\alpha }^{-3}{B}^3$, so the expression is not an exact inverse. For example $A=1$ and $\alpha =1$ give the approximate value $1$ instead of the exact $1/2$. Its exact remainder isIn the regime $\Vert C\Vert /\alpha <1$ it is a controlled Neumann expansion. Using it to reconstruct givesFor each fixed $\alpha >0$ this is a bounded polynomial in bounded operators, withThus the supplied approximation also has continuous data dependence in its domain of use. It must not be extrapolated to $\alpha \downarrow 0$ with $A$ fixed: its gain on any nonzero singular mode diverges, so it would not define a convergent regularization family there. The exact Tikhonov filter supplies the well-posed regularized problem; the truncated formula is only a controlled computational approximation.