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A minimum-norm least-squares solution minimizes $\Vert Ax-y\Vert$ and, among all such minimizers, minimizes $\Vert x\Vert$. It lies in $(\ker A{)}^{\perp }$ and equals $A^{†}y$ when the Moore–Penrose inverse of an operator is defined at $y$.

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.

SetStarting from $x_0=0$, repeated substitution in Landweber iteration givesThis is the finite partial sum of a Neumann series. Whenever $A^{\ast }A$ is boundedly invertible on the relevant subspace and $\Vert I-\gamma A^{\ast }A\Vert <1$,so

For a genuinely compact operator on an infinite-dimensional space, nonzero singular values can accumulate at zero, so this inverse is generally unbounded and the Neumann series need not converge in operator norm. Under the condition in part ii it nevertheless converges componentwise on admissible data to the Moore–Penrose inverse of an operator; stopping after finitely many terms suppresses poorly determined small-singular-value components and acts as a regularization of an inverse problem.