= Solution
Fix a <singular system of a compact operator>, with $Kv_j=\sigma_jw_j$ and $K^*w_j=\sigma_jv_j$, $\sigma_j>0$. A <spectral regularization method> has the form
$$
\boxed{R_\alpha f=\sum_j\sigma_jg_\alpha(\sigma_j^2)
\langle f,w_j\rangle v_j.}
$$
It annihilates $\mathcal R(K)^\perp$. Standard sufficient <spectral filter> conditions are boundedness of $\sup_{0<\sigma\leq\|K\|}|\sigma g_\alpha(\sigma^2)|$ for each $\alpha$, pointwise $\lambda g_\alpha(\lambda)\to1$ as $\alpha\downarrow0$ for $\lambda>0$, and a uniform bound on $|\lambda g_\alpha(\lambda)|$. These give bounded linear operators and convergence on the domain of the <Moore–Penrose inverse of an operator> by the <Picard criterion> and <dominated convergence theorem>.
For <Tikhonov regularization>,
$$
g_\alpha(\lambda)=\frac1{\lambda+\alpha},\qquad
\boxed{R_\alpha=(K^*K+\alpha I)^{-1}K^*.}
$$
This can be computed by solving the <Tikhonov normal equation>, without knowing the <singular value decomposition>.
For <spectral cutoff regularization>, take
$$
\boxed{g_\alpha(\lambda)=\frac{\mathbf1_{[\alpha,\infty)}(\lambda)}{\lambda},}
$$
with value zero at $\lambda=0$. Only <singular values> $\sigma_j\geq\sqrt\alpha$ are inverted. Here $g_\alpha$ multiplies $\sigma$; conventions calling the full coefficient $\sigma g_\alpha(\sigma^2)$ the <spectral filter> are equivalent.
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