= Solution
A <linear regularization> is a family of bounded operators $R_\alpha:Y\to X$ such that $R_\alpha y\to A^\dagger y$ for every $y$ in the domain of the <Moore–Penrose inverse of an operator> as $\alpha\downarrow0$. For noisy data $\|y^\delta-y\|\leq\delta$, a <convergent regularization of an inverse problem> also specifies a parameter choice $\alpha(\delta,y^\delta)$ giving convergence to $A^\dagger y$, uniformly over the allowed noise at each fixed admissible exact datum.
Expand $x_\alpha$ in the input singular vectors. The <Tikhonov normal equation> implies
$$
(\alpha+\sigma_n^2)\langle x_\alpha,u_n\rangle=\sigma_n\langle y,v_n\rangle.
$$
The component in $\ker A$ is zero because $\alpha>0$. Comparing with the given filter convention gives
$$
\boxed{f_\alpha(\sigma)=\frac{\sigma^2}{\alpha+\sigma^2},\qquad R_\alpha y=\sum_n\frac{\sigma_n}{\alpha+\sigma_n^2}\langle y,v_n\rangle u_n.}
$$
The <singular-system Tikhonov filter> is often described by its gain $g_\alpha(\sigma)=\sigma/(\alpha+\sigma^2)$; the requested $f_\alpha$ includes one extra factor of $\sigma$. For admissible data, $f_\alpha\to1$ and $0\leq f_\alpha\leq1$, so <dominated convergence theorem> applied to the squared <Picard criterion> coefficients proves exact-data convergence. Also
$$
\|R_\alpha\|\leq\sup_{\sigma\geq0}\frac{\sigma}{\alpha+\sigma^2}=\frac{1}{2\sqrt\alpha}.
$$
By the <noise-bias decomposition for linear regularization>, $\alpha(\delta)\to0$ and $\delta/\sqrt{\alpha(\delta)}\to0$ suffice for noisy-data convergence; for example $\alpha=\delta$ as $\delta\downarrow0$. This establishes both consistency and stability of <Tikhonov regularization>.
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