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Asymptotic regularization stops the gradient flow $x^{\prime }=-A^{\ast }(Ax-f)$ at $t=1/\alpha$, starting from zero. With the singular system of a compact operator convention $A{v}_j={\sigma }_j{u}_j$, its spectral filter givesThe scalar coefficient solves $c_j^{\prime }=-{\sigma }_j^2{c}_j+{\sigma }_j⟨f,u_j⟩$ with zero initial value. Since $1-e^{-s}\le \min (s,1)$ for $s\ge 0$, the operator norm is at most ${\alpha }^{-1/2}$. On the domain of the Moore–Penrose inverse of an operator, dominated convergence theorem of the squared spectral coefficients proves $R_{\alpha }f\to A^{†}f$. The noise-bias decomposition for linear regularization then gives noisy-data convergence when $\delta /\sqrt{\alpha }\to 0$.

The Landweber spectral filter progressively admits smaller singular-value components. Its approximation bias tends to zero on exact admissible data, but its noise amplification grows. With $\Vert A\Vert \le 1$, a sufficient a priori regularization parameter choice is $n(\delta )\to \infty$ and $\sqrt{n(\delta )}\delta \to 0$. Taking regularization parameter $\alpha =1/n$ expresses this as $\alpha \to 0$ and $\delta /\sqrt{\alpha }\to 0$; the noise-bias decomposition for linear regularization proves convergence.

Suppose each $R_{\alpha }:\mathcal{V}\to \mathcal{U}$ is a bounded linear operator and $R_{\alpha }f\to K^{†}f$ for every $f\in \mathcal{D}(K^{†})$ as $\alpha \downarrow 0$. A standard sufficient a priori regularization parameter choice satisfiesFor every datum with $\Vert f^{\delta }-f\Vert \le \delta$, the noise-bias decomposition for linear regularization givesThus the parameter rule gives a convergent regularization of an inverse problem, uniformly over data in the prescribed noise ball for each fixed admissible exact datum. For Tikhonov regularization, $\Vert R_{\alpha }\Vert \le 1/(2\sqrt{\alpha })$, so $\alpha \to 0$ and $\delta /\sqrt{\alpha }\to 0$ suffice. The same sufficient noise scaling holds for spectral cutoff regularization.

Write $y={\psi }_s$ and retain the Landweber relaxation parameter $\tau$. The printed $\Vert A\Vert <2$ allows the fixed choice $\tau =1/2$, since then $\tau \Vert A{\Vert }^2<2$. It does not justify a unit step: the scalar linear operator $A=3/2$ has norm below $2$, but unit-step error is multiplied by $1-|A{|}^2=-5/4$ and diverges. We use $0<\tau <2/\Vert A{\Vert }^2$ throughout.

From the recurrence and $f_0=0$, induction yields the closed formFor the compact operator in part ii, take a singular system of a compact operator with $A{v}_j={\sigma }_j{u}_j$, $A^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. Thus ${\sigma }_j^2$ are the eigenvalues of $A^{\ast }A$; this fixes the notation implicit in the printed hint. In the singular component $j$, the recurrence isSumming the geometric series gives the Landweber spectral filterThere is no contribution from data in $\ker A^{\ast }$ or from $\ker A$ in the reconstruction. Starting at zero is what selects the minimum-norm least-squares solution.

For fixed ${\sigma }_j>0$, $|1-\tau {\sigma }_j^2|<1$, so the numerator tends to $1$. The assumption $y\in D(A^{†})$ is the Picard criterionwith an arbitrary additional component in $\ker A^{\ast }$. The squared reconstruction error isEach term tends to zero and is bounded by a summable term from the Picard criterion. The dominated convergence theorem therefore proves $R_{n}y\to A^{†}y$, where $A^{†}$ is the Moore–Penrose inverse of an operator.

For each finite $n$, the regularization of an inverse problem is stable. Set $t=\tau {\sigma }^2\in (0,2)$. The geometric series and $|1-t|\le 1$ implyConsequently the Landweber noise amplification bound in this general step-size convention is $\Vert R_n\Vert \le \sqrt{2\tau n}$. If additionally $\tau \Vert A{\Vert }^2\le 1$, the sharper numerator bound $\min (nt,1)$ gives $\Vert R_n\Vert \le \sqrt{\tau n}$.

To see the regularization parameter directly, put $\alpha =1/(\tau n)$. Modes with ${\sigma }^2\ll \alpha$ have numerator approximately $n\tau {\sigma }^2$, so their inverse coefficient is approximately $\sigma /\alpha$ rather than $1/\sigma$. Each fixed nonzero mode is eventually restored as $\alpha \downarrow 0$. ThusThe finite iterates suppress unstable small singular values; infinitely many iterations remove this suppression. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, the noise-bias decomposition for linear regularization givesChoosing $n(\delta )\to \infty$ and ${\delta }^{2}n(\delta )\to 0$ proves noisy-data convergence. The reciprocal iteration index is a regularization parameter, and early stopping controls noise amplification.

Let $u^{†}=A^{†}f$. Linearity and the triangle inequality give the noise-bias decomposition for linear regularizationThe first term tends to zero by the assumed noise-amplification bound. The second tends to zero because $\alpha (\delta )\to 0$ and $R_{\alpha }$ is a regularization of an inverse problem. The right side is independent of the particular noisy datum within its allowed ball. Its convergence therefore proves the uniform noisy-data convergence required of a convergent regularization of an inverse problem.

Add and subtract the exact-data iterate. The noise-bias decomposition for linear regularization givesso the supplied Landweber noise amplification bound yieldsWith unit step and zero initial iterate, the exact-data error is $x^{†}-x_n=(I-A^{\ast }A{)}^n{x}^{†}$. Its squared norm isby the dominated convergence theorem. Stopping with $n\to \infty$ but $n{\delta }^2\to 0$ therefore makes both errors vanish.