 Solution

For $\alpha >0$, put $N(\alpha )=\lfloor {\alpha }^{-1}\rfloor$ and define the bounded operatorPart (e) shows $R_{\alpha }f\to A^{†}f$ for every $f\in \mathrm{dom}(A^{†})$ as $\alpha \downarrow 0$. Since $\Vert I-\tau A^{\ast }A\Vert \le 1$,Choose the a priori ruleThen $N(\alpha (\delta ))\to \infty$ whileFor $\Vert f^{\delta }-f\Vert \le \delta$,Thus $\{R_{\alpha }\}$ with this parameter rule is a regularization of an inverse problem.