Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 76 / 3 / c / Solution

Use the convention required by the printed expansion:where $u_i\in Y$ and $v_i\in X$ form the positive-singular-value parts of the singular system of a compact operator. Consequently $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$. Apply part (b)'s polynomial to each such eigenvector, using noisy data:The geometric series then gives the full inverse coefficient, or Landweber full inverse filter,In particular, for the unrelaxed iteration $\tau =1$, $g_{\alpha }(\sigma )=[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. The dimensionless Landweber spectral filter is instead $\sigma g_{\alpha }(\sigma )$; the printed expression uses the full coefficient and therefore needs the denominator $\sigma$.

For each fixed finite $n$, the filter extends continuously at zero by $g_{1/n}(0)=0$, since $g_{1/n}(\sigma )=n\tau \sigma +O({\sigma }^3)$ as $\sigma \to 0$. Small singular values are suppressed rather than immediately divided into noisy data. A component of $y^{(\delta )}$ in $\ker A^{\ast }$ contributes nothing. The parameter here is discrete: $1/\alpha$ is an integer. This matters when $1-\tau {\sigma }^2<0$; arbitrary real powers are not intended. A family for every $\alpha >0$ can use $n=\lfloor 1/\alpha \rfloor$ instead.