= Solution
For nonzero $A$, a sufficient step-size condition for <strong convergence of relaxed Landweber iteration> is
$$
\boxed{0<\tau<\frac2{\|A\|^2}.}
$$
For the unrelaxed iteration, this reads $\|A\|^2<2$. The data must also admit a <minimum-norm least-squares solution> $x^\dagger$; for the exact problem, $y\in\operatorname{Ran}A$ is sufficient. More generally, the projected-data condition in part (a), or the <Picard criterion> $\sum_i|\langle y,u_i\rangle|^2/\sigma_i^2<\infty$, ensures the required solution.
Every positive <singular value> satisfies $|1-\tau\sigma_i^2|<1$, so for fixed $\sigma_i>0$,
$$
\boxed{\lim_{\alpha\to0}g_\alpha(\sigma_i)=\frac1{\sigma_i}.}
$$
This is the correct inverse coefficient. For exact data with a <least-squares solution>, the <norm> error is
$$
\|x_n-x^\dagger\|^2=\sum_i
|1-\tau\sigma_i^2|^{2n}\frac{|\langle y,u_i\rangle|^2}{\sigma_i^2}\longrightarrow0.
$$
The 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 $\tau=2/\|A\|^2$, 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 $|1-t|\leq1$, and
$$
|1-(1-t)^n|\leq\min(nt,2),\qquad
|g_{1/n}(\sigma)|\leq\min(n\tau\sigma,2/\sigma)\leq\sqrt{2n\tau}.
$$
This proves the <uniform noise bound for relaxed Landweber iteration>
$$
\|x_n^{(\delta)}-x_n\|\leq\delta\sqrt{2n\tau}.
$$
Together with the exact-data convergence,
$$
\boxed{\|x_n^{(\delta)}-x^\dagger\|
\leq\delta\sqrt{2n\tau}+\|x_n-x^\dagger\|\longrightarrow0}
$$
whenever $n=n(\delta)\to\infty$ and $\delta\sqrt{n(\delta)}\to0$. Equivalently, the <a priori regularization parameter choice> satisfies $\alpha(\delta)\to0$ and $\delta/\sqrt{\alpha(\delta)}\to0$. For example, $n(\delta)=\lfloor\delta^{-1}\rfloor$ works as $\delta\to0$. The sharper familiar bound $\delta\sqrt{n\tau}$ is available when $\tau\|A\|^2\leq1$.
Thus \b[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 $A$.
Back to article page
