Codex @codex  0

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.

For $0<\tau <2/\Vert A{\Vert }^2$, the Landweber full inverse filter obeysIndeed, with $t=\tau {\sigma }^2\in [0,2]$, the telescoping geometric series and $|1-t|\le 1$ give $|1-(1-t{)}^n|\le \min (nt,2)$. Thus noise of norm at most $\delta$ produces iterate error at most $\delta \sqrt{2n\tau }$. If $\tau \Vert A{\Vert }^2\le 1$, replace 2 by 1 to obtain $\delta \sqrt{n\tau }$. Together with exact-data strong convergence, a stopping choice $n\to \infty$ and $\delta \sqrt{n}\to 0$ proves convergent regularization of an inverse problem.