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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.

A regularization of an inverse problem is a family of continuous maps $R_{\alpha }:\mathcal{V}\to \mathcal{U}$, $\alpha >0$, such thatfor every $f$ in the domain of the Moore–Penrose inverse of an operator. A linear regularization has each $R_{\alpha }$ bounded and linear. For example, Tikhonov regularization givesA nonlinear example is quartic-norm variational regularization:Its strictly convex norm penalty selects the same minimum-norm least-squares solution as $\alpha \downarrow 0$, but the map from data to minimizer is generally nonlinear.

A convergent regularization of an inverse problem additionally has a parameter rule $\alpha (\delta ,f_{\delta })>0$ such that, for every admissible exact $f$,An a priori regularization parameter choice depends only on $\delta$ and fixed information about the problem.

Use the singular system of a compact operator convention $A{x}_j={\sigma }_j{y}_j$ and $A^{\ast }{y}_j={\sigma }_j{x}_j$. Starting from zero gives no component in $\ker A$. Write $x(t)={\sum }_j{c}_j(t)x_j$. The coefficient equations areSolving these scalar linear ordinary differential equations givesConsequently asymptotic regularization has the spectral regularization method representationThe component of $f$ orthogonal to the closure of the range of $A$ is annihilated by $A^{\ast }$ and contributes nothing. The scalar filter satisfies $(1-e^{-{\sigma }^2/\alpha })/\sigma \le \min (\sigma /\alpha ,1/\sigma )\le {\alpha }^{-1/2}$, so every fixed-$\alpha$ reconstruction is a bounded linear operator. Its long-time limit is the Moore–Penrose inverse of an operator on admissible data.

For admissible data, the Moore–Penrose inverse of an operator selects the minimum-norm least-squares solution:Its domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$. When the equation is consistent it is the solution of $Ax=y$ orthogonal to the kernel; for inconsistent admissible data it solves the normal equation for a linear inverse problem and projects away the component orthogonal to the range.

Take a singular system of a compact operator satisfying $A{v}_j={\sigma }_j{u}_j$, $A^{\ast }{u}_j={\sigma }_j{v}_j$. The generalized inverse isThe Picard criterion requires the squared coefficients to be summable. For an infinite-rank compact operator, ${\sigma }_j\to 0$, so $\Vert A^{†}{u}_j\Vert =1/{\sigma }_j\to \infty$ even though $\Vert u_j\Vert =1$. Hence $A^{†}$ is unbounded. Finite-rank compact operators do not have this obstruction. Data failing the Picard condition need not have a least-squares minimizer at all, even when the infimum residual is zero.

A problem is well posed in the sense of Hadamard when a solution exists for every admissible datum, is unique, and depends continuously on the datum. It is ill posed if any one of these three properties fails.

For the inverse problem $Au=f$, a regularization of an inverse problem is a family of bounded maps $R_{\alpha }:Y\to X$ that approximate the generally unbounded Moore–Penrose inverse of an operator $A^{†}$. It is a convergent regularization of an inverse problem if there is a parameter rule $\alpha =\alpha (\delta ,f^{\delta })$ such thatas $\delta \downarrow 0$ for every $f\in \mathcal{D}(A^{†})$.

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.

The operator range is the image of a linear map $K$. It need not be closed in an infinite-dimensional Hilbert space; that failure makes the Moore–Penrose inverse of an operator unbounded on its domain.

For $Ku(y)={\int }_0^{y}u(x)dx$ on the Hilbert space $L^2(0,1)$, every image has a weak derivative $f^{\prime }=u$ and zero trace at zero. Conversely every such Sobolev space element is this integral of its weak derivative. The range is dense because it contains smooth functions supported in $(0,1)$, but it is not closed: a step function can be approximated in $L^2$ by continuous ramps and is not itself in $H^1$. The Moore–Penrose inverse of an operator has this range as its domain and differentiates there.

Regularization replaces sensitive inversion by a controlled approximation. For a bounded linear operator between Hilbert spaces, a linear regularization is a family of bounded linear operators $R_{\alpha }$ satisfying $R_{\alpha }y\to A^{†}y$ as $\alpha \downarrow 0$ for each datum in the domain of the Moore–Penrose inverse of an operator. Noisy-data convergence additionally requires a suitable regularization parameter choice.^[ref] Nonlinear variational regularization can instead use nonquadratic penalties.

The noise-bias decomposition for linear regularization separates two costs: amplifying measurement noise and changing the exact-data reconstruction. Truncated singular value decomposition discards weak modes; Tikhonov regularization attenuates them smoothly. A quadratic norm penalty expresses a preference for smaller reconstructions. Neither method manufactures a missing null space component: with complete averaging, the minimum-norm least-squares solution for $(0.5,0.5{)}^T$ is $(0.5,0.5{)}^T$, even if the original image was $(1,0{)}^T$. The preference selects a representative; extra observations or justified constraints are needed to identify the original.

For Tikhonov regularization, the Tikhonov stability bound limits transmitted measurement noise of norm at most $\delta$ to $\delta /(2\sqrt{\alpha })$. Taking $\alpha \to 0$ while $\delta /\sqrt{\alpha }\to 0$ gives a sufficient rule for convergent regularization of an inverse problem. Holding $\alpha$ fixed leaves approximation bias; letting it vanish too quickly can admit overwhelming measurement noise.