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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.

Starting from zero, the Landweber iteration isFor $K\ne 0$, take the step size $0<\tau <2/\Vert K{\Vert }^2$. Iterating the linear update givesand its Landweber spectral filter in a singular system of a compact operator isThe strict upper step size bound makes $|1-\tau {\sigma }_j^2|<1$ for every positive singular value. A common more restrictive choice is $0<\tau \le \Vert K{\Vert }^{-2}$, giving nonnegative damping factors. Each finite iterate is bounded and linear; early stopping of Landweber iteration controls the amplification of noise as smaller singular values are progressively inverted. If $K=0$, every iterate is zero for any positive step size.

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.

Use early stopping of Landweber iteration. For each finite $n$, the Landweber spectral filter gives a bounded reconstruction operator; as $n\to \infty$, it approaches the generalized inverse on exact admissible data. Taking $\alpha =1/n$, a sufficient rule for a convergent regularization of an inverse problem isor equivalently $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$. For example, $n(\delta )=\lfloor {\delta }^{-1}\rfloor$ satisfies both in the normalized problem. The first condition removes exact-data bias; the second prevents arbitrarily small measurement errors from being amplified by excessive iteration.

On each left singular vector $u_j$, use $A^{\ast }{u}_j={\sigma }_j{v}_j$ and $(I-A^{\ast }A)v_j=(1-{\sigma }_j^2)v_j$. The finite geometric series givesThus with $\alpha =1/n$, the dimensionless Landweber spectral filter isThe 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, $0\le f_n(\sigma )\le \min (n{\sigma }^2,1)$, so the inverse coefficient is at most $\min (n\sigma ,1/\sigma )\le \sqrt{n}$. This verifies directly why finite iteration is stable and why admitting progressively smaller singular values eventually amplifies noise.