 Solution

For nonzero $A$, a sufficient step-size condition for strong convergence of relaxed Landweber iteration isFor the unrelaxed iteration, this reads $\Vert A{\Vert }^2<2$. The data must also admit a minimum-norm least-squares solution $x^{†}$; for the exact problem, $y\in \mathrm{Ran}A$ is sufficient. More generally, the projected-data condition in part (a), or the Picard criterion ${\sum }_i|⟨y,u_i⟩{|}^2/{\sigma }_i^2<\infty$, ensures the required solution.

Every positive singular value satisfies $|1-\tau {\sigma }_i^2|<1$, so for fixed ${\sigma }_i>0$,This is the correct inverse coefficient. For exact data with a least-squares solution, the norm error isThe summable majorant is precisely the Picard criterion, so the dominated convergence theorem proves strong convergence, despite the absence of a uniform contraction constant when the singular values approach zero. Initialization at zero selects the solution orthogonal to $\ker A$. At $\tau =2/\Vert A{\Vert }^2$, a top singular mode of a nonzero compact operator has multiplier $-1$ and can oscillate indefinitely; the strict step bound avoids this failure.

Pointwise convergence of filters alone is not a full noisy-data inverse-problem regularization proof. For each finite $n$, put $t=\tau {\sigma }^2$; the step restriction gives $|1-t|\le 1$, andThis proves the uniform noise bound for relaxed Landweber iterationTogether with the exact-data convergence,whenever $n=n(\delta )\to \infty$ and $\delta \sqrt{n(\delta )}\to 0$. Equivalently, the a priori regularization parameter choice satisfies $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$. For example, $n(\delta )=\lfloor {\delta }^{-1}\rfloor$ works as $\delta \to 0$. The sharper familiar bound $\delta \sqrt{n\tau }$ is available when $\tau \Vert A{\Vert }^2\le 1$.

Thus finite iteration followed by an appropriate stopping rule is a convergent regularization of an inverse problem. Taking $n\to \infty$ for fixed noisy data is not the same operation and can amplify the small-singular-value noise without bound. If $A=0$, the zero starting iterate stays zero and is the minimum-norm least-squares solution for every datum; the step restriction is needed only for nonzero $A$.