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When all multipliers are nonzero, the Moore–Penrose inverse of an operator acts by dividing each Fourier coefficient $y_n$ by $c_n$. Its data domain is precisely the square-summability condition displayed. The dense range need not be closed, and writing the formal inverse series does not establish the Picard criterion.

For a bounded operator $A:X\to Y$ between Hilbert spaces, the Moore–Penrose inverse maps admissible data to the unique minimum-norm least-squares solution. If $({\sigma }_j,u_j,v_j)$ is a singular system of a compact operator, thenon the data for which this series converges.

More precisely, its domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$. It vanishes on $(\mathrm{ran}A{)}^{\perp }$ and takes values in $(\ker A{)}^{\perp }$. An infinite-rank compact operator on infinite-dimensional spaces has singular values approaching zero, so its inverse on its range is generally unbounded; Picard criterion specifies admissible data.

Using the singular value system with $A{u}_j={\sigma }_j{v}_j$ and $A^{\ast }{v}_j={\sigma }_j{u}_j$, the formal minimum-norm inverse isIt defines an element of the unknown Hilbert space only if ${\sum }_j|(y,v_j){|}^2/{\sigma }_j^2<\infty$, the Picard criterion. A null-space component is undetermined by the data.

Adding an error along a single left singular vector changes precisely one coefficient:Although this perturbation has data norm $\delta$, its amplification factor is $1/{\sigma }_j$. If ${\sigma }_j\to 0$, these factors are unbounded. More explicitly choose $y^{(j)}={\sigma }_j{v}_j$, tending to zero, while the minimum-norm inverse is $u_j$, of norm one. Thus the inverse is discontinuous at zero. It may also fail to exist for data outside its nonclosed range, so merely minimizing a residual does not guarantee that the formal inverse exists. This is ill-posedness by unbounded small-singular-value amplification, not a claim that every finite-dimensional or bounded forward map has an unstable inverse.

Use the inner product convention linear in its first argument. A singular value system consists of positive singular values ${\sigma }_n$ and orthonormal bases $u_n$ of $(\ker A{)}^{\perp }\subset X$ and $v_n$ of $\overline{\mathrm{ran}A}\subset Y$, withFor an infinite-rank compact operator, ${\sigma }_n\to 0$; a finite-rank operator has a finite system. Applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$ constructs $u_n$, with eigenvalues ${\sigma }_n^2$, and then $v_n=A{u}_n/{\sigma }_n$. This solution uses the input/output naming of this problem, which is the reverse of another common singular-vector convention.

The Moore–Penrose inverse of an operator isIts domain is $\mathrm{ran}A\oplus (\mathrm{ran}A{)}^{\perp }$, equivalently the data satisfying the Picard criterionIt ignores the component in $\ker A^{\ast }$ and lies in $(\ker A{)}^{\perp }$. The series then converges in $X$ by orthonormality and is the minimum-norm least-squares solution. If $y\in \mathrm{ran}A$, it is an exact solution; all other exact solutions are $x^{†}+z$ with $z\in \ker A$. Compactness does not make the inverse bounded when there are infinitely many positive singular values.

A linear regularization is a family of bounded operators $R_{\alpha }:Y\to X$ such that $R_{\alpha }y\to A^{†}y$ for every $y$ in the domain of the Moore–Penrose inverse of an operator as $\alpha \downarrow 0$. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, a convergent regularization of an inverse problem also specifies a parameter choice $\alpha (\delta ,y^{\delta })$ giving convergence to $A^{†}y$, uniformly over the allowed noise at each fixed admissible exact datum.

Expand $x_{\alpha }$ in the input singular vectors. The Tikhonov normal equation impliesThe component in $\ker A$ is zero because $\alpha >0$. Comparing with the given filter convention givesThe singular-system Tikhonov filter is often described by its gain $g_{\alpha }(\sigma )=\sigma /(\alpha +{\sigma }^2)$; the requested $f_{\alpha }$ includes one extra factor of $\sigma$. For admissible data, $f_{\alpha }\to 1$ and $0\le f_{\alpha }\le 1$, so dominated convergence theorem applied to the squared Picard criterion coefficients proves exact-data convergence. AlsoBy the noise-bias decomposition for linear regularization, $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$ suffice for noisy-data convergence; for example $\alpha =\delta$ as $\delta \downarrow 0$. This establishes both consistency and stability of Tikhonov regularization.

A singular value system of a compact operator consists of positive numbers ${\sigma }_j$ and orthonormal families $v_j\in X$, $u_j\in Y$ satisfyingThus $v_j$ are positive-eigenvalue eigenvectors of $A^{\ast }A$ and $u_j$ of $A{A}^{\ast }$, with eigenvalue ${\sigma }_j^2$. The families are complete in $(\ker A{)}^{\perp }$ and $\overline{\mathrm{ran}A}$ respectively. The singular values can be listed nonincreasingly with multiplicities, and tend to zero in the infinite-rank case. For finite rank there are only finitely many positive singular values; zero-kernel directions are handled separately.

Use the convention that $⟨f,g⟩$ is conjugate-linear in its first entry. Then the singular value system givesThe latter series converges precisely on the admissible range component specified by the Picard criterion:with $R^{\perp }$ components annihilated by the inverse. Merely writing a formal singular expansion does not imply it converges in $X$.

Let $y_n=⟨e_n,y⟩$. Using the phase of $u_n$ in the singular value system, the Moore–Penrose inverse of an operator becomesThis gives the admissible data for a periodic convolution inverse. Because the range is dense and the kernel is zero here, the domain of the inverse is exactly the range, not all of $L^2$.

For $y(x)=e^{\alpha x}$ and $\alpha \notin i\mathbb{Z}$,The formal expression would consequently beHowever, the high-frequency obstruction for nonperiodic exponential data prevents this from being a Hilbert-space solution. The numerator is nonzero, $|y_n|$ is asymptotic to a nonzero constant divided by $|n|$, and part (c) proved $|c_n|=o(1/|n|)$. Hence $|y_n/c_n|\to \infty$, so the Picard criterion fails. For $\alpha \notin i\mathbb{Z}$, $A^{†}y$ is not defined in the specified Hilbert space. The formal series is not a convergent generalized solution.

If $\alpha =im$ with $m\in \mathbb{Z}$, the data are a single periodic Fourier mode: $y_n=\sqrt{2\pi }{\delta }_{nm}$. Then there is an exact unique solution,In particular, if the intended parameter is real, only $\alpha =0$ is admissible, giving $f^{†}=1/c_0$.

For the inadmissible cases the inverse problem still has approximate solutions $f_N={\sum }_{|n|\le N}(y_n/c_n)e_n$: their images are Fourier projections converging to $y$ in $L^2$, while their norms diverge. Thus the least-squares residual has infimum zero but no minimizer. This distinguishes an undefined exact inverse from a regularized truncated reconstruction; it is the necessary qualification to the question's unrestricted constant $\alpha$.

For noisy data, exact equality $Ax=y$ may be impossible: the data can have a component outside the operator range. A least-squares solution of a linear inverse problem instead minimizesFor any variation $h\in X$, the adjoint operator givesThus statistical homogeneity in every direction is the normal equation for a linear inverse problem, $A^{\ast }Ax=A^{\ast }y$. It is also sufficient for a global minimum: when it holds,The residual is orthogonal to the closure of the range, since it lies in $\ker A^{\ast }$. A normal-equation solution therefore fits the component of the data representable by the forward map, while leaving the perpendicular component unfitted. If there is more than one such solution, select the minimum-norm least-squares solution, which lies in $(\ker A{)}^{\perp }$.

These facts explain the formulation, but do not make the inverse problem well posed. A compact operator of infinite rank typically has nonclosed operator range, so even the least-squares infimum need not be attained. A least-squares solution exists exactly when the orthogonal projection of $y$ onto $\overline{\mathrm{Ran}A}$ belongs to $\mathrm{Ran}A$. In a singular system of a compact operator, this becomes the Picard criterion. Also, the nonzero eigenvalues of $A^{\ast }A$ are the squares of the singular values of $A$: solving the operator normal equation directly can worsen numerical conditioning. The operator normal equation expresses least squares; it does not itself regularize the inverse problem. The subsequent finite Landweber iteration and its stopping rule provide the inverse-problem regularization.

For nonzero $A$, a sufficient step-size condition for strong convergence of relaxed Landweber iteration isFor the unrelaxed iteration, this reads $\Vert A{\Vert }^2<2$. The data must also admit a minimum-norm least-squares solution $x^{†}$; for the exact problem, $y\in \mathrm{Ran}A$ is sufficient. More generally, the projected-data condition in part (a), or the Picard criterion ${\sum }_i|⟨y,u_i⟩{|}^2/{\sigma }_i^2<\infty$, ensures the required solution.

Every positive singular value satisfies $|1-\tau {\sigma }_i^2|<1$, so for fixed ${\sigma }_i>0$,This is the correct inverse coefficient. For exact data with a least-squares solution, the norm error isThe summable majorant is precisely the Picard criterion, so the dominated convergence theorem proves strong convergence, despite the absence of a uniform contraction constant when the singular values approach zero. Initialization at zero selects the solution orthogonal to $\ker A$. At $\tau =2/\Vert A{\Vert }^2$, a top singular mode of a nonzero compact operator has multiplier $-1$ and can oscillate indefinitely; the strict step bound avoids this failure.

Pointwise convergence of filters alone is not a full noisy-data inverse-problem regularization proof. For each finite $n$, put $t=\tau {\sigma }^2$; the step restriction gives $|1-t|\le 1$, andThis proves the uniform noise bound for relaxed Landweber iterationTogether with the exact-data convergence,whenever $n=n(\delta )\to \infty$ and $\delta \sqrt{n(\delta )}\to 0$. Equivalently, the a priori regularization parameter choice satisfies $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$. For example, $n(\delta )=\lfloor {\delta }^{-1}\rfloor$ works as $\delta \to 0$. The sharper familiar bound $\delta \sqrt{n\tau }$ is available when $\tau \Vert A{\Vert }^2\le 1$.

Thus finite iteration followed by an appropriate stopping rule is a convergent regularization of an inverse problem. Taking $n\to \infty$ for fixed noisy data is not the same operation and can amplify the small-singular-value noise without bound. If $A=0$, the zero starting iterate stays zero and is the minimum-norm least-squares solution for every datum; the step restriction is needed only for nonzero $A$.

Choose a singular system of a compact operator with $K{v}_j={\sigma }_j{u}_j$, $K^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. The Picard criterion, together with range orthogonality, givesFor necessity, expand a solution's component in $(\ker K{)}^{\perp }$ in the orthonormal vectors $v_j$; its coefficients must be $⟨f,u_j⟩/{\sigma }_j$. Their square sum is finite by Bessel's inequality. Conversely, the displayed square sum defines a convergent Hilbert space seriesApplying the bounded linear operator $K$ to the partial sums yields the expansion of $f$ in $\overline{\mathrm{ran}K}=(\ker K^{\ast }{)}^{\perp }$. Thus its limit solves $Ku=f$. The perpendicularity condition is essential here: the Picard criterion alone characterizes the domain of the generalized inverse, which also includes components in $\ker K^{\ast }$.

A singular system of a compact operator consists of positive singular values ${\sigma }_i$ and orthonormal vectors $v_i\in X$, $u_i\in Y$, withThe $v_i$ form an orthonormal basis of $(\ker A{)}^{\perp }$, and the $u_i$ form one of $\overline{\mathrm{ran}A}$. There are finitely many positive singular values for finite-rank $A$; otherwise ${\sigma }_i\to 0$. Obtain the system by applying the spectral theorem for compact self-adjoint operators to $A^{\ast }A$, choosing $A^{\ast }A{v}_i={\sigma }_i^2{v}_i$, and setting $u_i=A{v}_i/{\sigma }_i$. The inner product identities verify orthonormality and the displayed relations.

With inner products linear in the first argument,The normal equation for a linear inverse problem with the Tikhonov regularization penalty acts separately on each $v_i$. Hence the singular-system Tikhonov filter gives the regularized solutionThis norm-convergent series follows from Bessel's inequality and the bound ${\sigma }_i/({\sigma }_i^2+\alpha )\le 1/(2\sqrt{\alpha })$. Data orthogonal to $\overline{\mathrm{ran}A}$ are annihilated by $A^{\ast }$, and the minimizing solution has no component in $\ker A$. The same formula with $y^{\delta }$ reconstructs noisy data. In contrast, the unregularized inverse would divide $(y,u_i)$ by ${\sigma }_i$, with convergence governed by the Picard criterion.

The Picard criterion with the input/output convention above statesNecessity follows from $⟨f,v_j⟩={\sigma }_j⟨u,u_j⟩$ and Bessel inequality. Conversely the summability constructs $u={\sum }_j{\sigma }_j^{-1}⟨f,v_j⟩u_j\in U$; the closure condition ensures that its image is all of $f$. The closure condition must not be omitted: a component orthogonal to the range cannot be reconstructed.

For the diagonal example, $f_j=j^{-p}$ is itself in ${\ell }^2$ only for $p>1/2$. Its preimage would be $u_j=j^{1-p}$, so the range criterion is the convergence of ${\sum }_j{j}^{2-2p}$. The P-series givesAt $p=3/2$ the reconstruction norm diverges harmonically. For $1/2<p\le 3/2$ the data are legitimate but have no preimage; for $0<p\le 1/2$ they are not even in the specified data space.

Fix a singular system of a compact operator, with $K{v}_j={\sigma }_j{w}_j$ and $K^{\ast }{w}_j={\sigma }_j{v}_j$, ${\sigma }_j>0$. A spectral regularization method has the formIt annihilates $\mathcal{R}(K{)}^{\perp }$. Standard sufficient spectral filter conditions are boundedness of ${\sup }_{0<\sigma \le \Vert K\Vert }|\sigma g_{\alpha }({\sigma }^2)|$ for each $\alpha$, pointwise $\lambda g_{\alpha }(\lambda )\to 1$ as $\alpha \downarrow 0$ for $\lambda >0$, and a uniform bound on $|\lambda g_{\alpha }(\lambda )|$. These give bounded linear operators and convergence on the domain of the Moore–Penrose inverse of an operator by the Picard criterion and dominated convergence theorem.

For Tikhonov regularization,This can be computed by solving the Tikhonov normal equation, without knowing the singular value decomposition.

For spectral cutoff regularization, takewith value zero at $\lambda =0$. Only singular values ${\sigma }_j\ge \sqrt{\alpha }$ are inverted. Here $g_{\alpha }$ multiplies $\sigma$; conventions calling the full coefficient $\sigma g_{\alpha }({\sigma }^2)$ the spectral filter are equivalent.

For Iterated Tikhonov regularization, set $A=K^{\ast }K$ and $B=(I+\tau A{)}^{-1}$. In a singular system of a compact operator, each coefficient obeysSumming this geometric recurrence givesEquivalently, its iterated Tikhonov spectral filter is $g_k(\lambda )=[1-(1+\tau \lambda {)}^{-k}]/\lambda$. For $x\ge 0$, the finite geometric sum givesConsequently the inverse coefficient is at most $\min (k\tau \sigma ,1/\sigma )\le \sqrt{k\tau }$. Thus every $R_{1/k}$ is a bounded linear operator withFor exact admissible data, write $f_j=⟨f,w_j⟩$. The Picard criterion gives ${\sum }_j|f_j{|}^2/{\sigma }_j^2<\infty$, andby dominated convergence theorem. Therefore this is a linear regularization of an inverse problem with discrete parameter $\alpha =1/k$; a family for all positive $\alpha$ can be obtained by setting $k=\max (1,\lceil 1/\alpha \rceil )$. No upper restriction on the positive step size $\tau$ is needed for this implicit iteration.

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.

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.

It is sufficient thatand that the stopping rule obeyFor example, $k(\delta )=\lfloor {\delta }^{-1}\rfloor$ works.

For exact data $f\in \mathcal{D}(A^{†})$, each factor $(1-\tau {\sigma }_i^2{)}^k$ tends to zero. The Picard criterion makes $⟨f,y_i⟩/{\sigma }_i$ square summable, while $|1-(1-\tau {\sigma }_i^2{)}^k|$ is uniformly bounded. The dominated convergence theorem on the resulting series yields $u_k(f)\to A^{†}f$.

For noisy data with $\Vert f^{\delta }-f\Vert \le \delta$, the filter representation givesIndeed, when $\tau {\sigma }^2\le 1$, Bernoulli's inequality gives $1-(1-\tau {\sigma }^2{)}^k\le \min (k\tau {\sigma }^2,1)$; when $\tau {\sigma }^2>1$, the quotient is uniformly bounded because $\tau \Vert A{\Vert }^2<2$. Hence the triangle inequality giveswhich proves that early-stopped Landweber iteration is a convergent regularization of an inverse problem.