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Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator, withThe Moore–Penrose inverse of an operator is the generally unbounded mapdefined when the Picard criterion holds, with the component in $\ker A^{\ast }$ sent to zero. It is the minimum-norm least-squares solution of $Af=g$.

A regularization of an inverse problem consists of bounded maps $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,g^{(\delta )})$ such thatwhenever $\Vert g^{(\delta )}-g\Vert \le \delta$ and $g$ is in the domain of $A^{†}$.

For Tikhonov regularization, minimizinggivesThe scalar spectral filter satisfiesand consequentlyFor exact data, each filter factor ${\sigma }_j^2/({\sigma }_j^2+\alpha )$ tends to one, so $R_{\alpha }g\to A^{†}g$. Choosingtherefore makes both the approximation error and propagated data error vanish. For example, $\alpha (\delta )=\delta$ is an admissible a priori regularization parameter choice.

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator $A:X\to Y$, soThe Moore–Penrose inverse of an operator has domainand acts bywith the orthogonal component of $y$ sent to zero. Equivalently, its domain consists of the data satisfying the Picard criterion. It obeys

If $Ax=y$ is exactly solvable, every solution is $x^{†}+z$ with $z\in \ker A$, andis the unique minimum-norm least-squares solution. It recovers the component of the original $x$ orthogonal to the null space; no data can determine the null-space component.

Let $({\sigma }_j,u_j,v_j)$ be a singular system for $A$. The Picard criterion for $f\in \mathrm{dom}(A^{†})$ isafter discarding the component in $(\mathrm{ran}A{)}^{\perp }$. For $0<\tau <\Vert A{\Vert }^{-2}$, the partial series acts diagonally:For each $j$ the multiplier in the numerator tends to one and lies in $[0,1]$. The Picard summability condition therefore supplies an ${\ell }^2$ dominating sequence, so the dominated convergence theorem givesThis is the series form of Landweber iteration.

For nonzero compact operator $A$, step $0<\tau <2/\Vert A{\Vert }^2$, and data admitting a minimum-norm least-squares solution $x^{†}$, zero-initialized Landweber iteration satisfiesThe Picard criterion makes the majorant summable, so the dominated convergence theorem proves this limit. No mesh-independent or singular-value-independent strict contraction factor is required. At $\tau =2/\Vert A{\Vert }^2$, the top singular mode has multiplier $-1$ and may not converge. Without solvable projected data, the iterates need not have a limit in the solution Hilbert space, even when their residuals approach the least-squares infimum.