Landweber noise amplification bound

For $\tau =1$ and $\Vert A\Vert \le 1$, $0\le 1-(1-{\sigma }^2{)}^n\le \min (n{\sigma }^2,1)$ givesThus a data error of norm at most $\delta$ produces an iterate error at most $\sqrt{n}\delta$.

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.