= Solution
For $\alpha>0$, put $N(\alpha)=\lfloor\alpha^{-1}\rfloor$ and define the bounded operator
$$
R_\alpha
=\tau\sum_{n=0}^{N(\alpha)}
(I-\tau A^*A)^nA^*.
$$
Part (e) shows $R_\alpha f\to A^\dagger f$ for every $f\in\operatorname{dom}(A^\dagger)$ as $\alpha\downarrow0$. Since $\|I-\tau A^*A\|\leq1$,
$$
\|R_\alpha\|
\leq\tau(N(\alpha)+1)\|A\|.
$$
Choose the a priori rule
$$
\boxed{\alpha(\delta)=\sqrt\delta}.
$$
Then $N(\alpha(\delta))\to\infty$ while
$$
\delta\|R_{\alpha(\delta)}\|
\leq\tau\|A\|\delta(N(\alpha(\delta))+1)
\longrightarrow0.
$$
For $\|f^\delta-f\|\leq\delta$,
$$
\|R_{\alpha(\delta)}f^\delta-A^\dagger f\|
\leq
\delta\|R_{\alpha(\delta)}\|
+\|R_{\alpha(\delta)}f-A^\dagger f\|
\longrightarrow0.
$$
Thus $\{R_\alpha\}$ with this parameter rule is a <regularization of an inverse problem>.
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