= Solution
On each <left singular vector> $u_j$, use $A^*u_j=\sigma_jv_j$ and $(I-A^*A)v_j=(1-\sigma_j^2)v_j$. The finite <geometric series> gives
$$
x_n=\sum_j\sigma_j\left[\sum_{m=0}^{n-1}(1-\sigma_j^2)^m\right]\langle y,u_j\rangle v_j=\sum_j\frac{1-(1-\sigma_j^2)^n}{\sigma_j}\langle y,u_j\rangle v_j.
$$
Thus with $\alpha=1/n$, the dimensionless <Landweber spectral filter> is
$$
\boxed{f_\alpha(\sigma)=1-(1-\sigma^2)^{1/\alpha}.}
$$
The full inverse-coefficient convention instead uses $[1-(1-\sigma^2)^{1/\alpha}]/\sigma$. With general step $\tau$, replace $\sigma^2$ inside the power by $\tau\sigma^2$.
In the normalized problem, $0\leq f_n(\sigma)\leq\min(n\sigma^2,1)$, so the inverse coefficient is at most $\min(n\sigma,1/\sigma)\leq\sqrt n$. This verifies directly why finite iteration is stable and why admitting progressively smaller <singular values> eventually amplifies noise.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-335-landweber-filters.png]
{title=Landweber filters and progressive noise amplification}
{description=More iterations admit smaller singular-value components. The damping factor tends toward one, while the coefficient applied to measured data approaches the unstable reciprocal singular value.}
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