= Solution
Use the convention required by the printed expansion:
$$
Av_i=\sigma_i u_i,\qquad A^*u_i=\sigma_i v_i,\qquad \sigma_i>0,
$$
where $u_i\in Y$ and $v_i\in X$ form the positive-singular-value parts of the <singular system of a compact operator>. Consequently $A^*Av_i=\sigma_i^2v_i$. Apply part (b)'s polynomial to each such <eigenvector>, using noisy data:
$$
x_n^{(\delta)}=\sum_i\tau\sigma_i\sum_{j=0}^{n-1}(1-\tau\sigma_i^2)^j
\langle y^{(\delta)},u_i\rangle v_i.
$$
The <geometric series> then gives the full inverse coefficient, or <Landweber full inverse filter>,
$$
\boxed{g_\alpha(\sigma)=\frac{1-(1-\tau\sigma^2)^{1/\alpha}}{\sigma},
\qquad \alpha=1/n,\quad n\in\mathbb N.}
$$
In particular, for the unrelaxed iteration $\tau=1$, $g_\alpha(\sigma)=[1-(1-\sigma^2)^{1/\alpha}]/\sigma$. The dimensionless <Landweber spectral filter> is instead $\sigma g_\alpha(\sigma)$; the printed expression uses the full coefficient and therefore needs the denominator $\sigma$.
For each fixed finite $n$, the filter extends continuously at zero by $g_{1/n}(0)=0$, since $g_{1/n}(\sigma)=n\tau\sigma+O(\sigma^3)$ as $\sigma\to0$. Small <singular values> are suppressed rather than immediately divided into noisy data. A component of $y^{(\delta)}$ in $\ker A^*$ contributes nothing. The parameter here is discrete: $1/\alpha$ is an integer. This matters when $1-\tau\sigma^2<0$; arbitrary real powers are not intended. A family for every $\alpha>0$ can use $n=\lfloor1/\alpha\rfloor$ instead.
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