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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.

Asymptotic regularization stops the gradient flow $x^{\prime }=-A^{\ast }(Ax-f)$ at $t=1/\alpha$, starting from zero. With the singular system of a compact operator convention $A{v}_j={\sigma }_j{u}_j$, its spectral filter givesThe scalar coefficient solves $c_j^{\prime }=-{\sigma }_j^2{c}_j+{\sigma }_j⟨f,u_j⟩$ with zero initial value. Since $1-e^{-s}\le \min (s,1)$ for $s\ge 0$, the operator norm is at most ${\alpha }^{-1/2}$. On the domain of the Moore–Penrose inverse of an operator, dominated convergence theorem of the squared spectral coefficients proves $R_{\alpha }f\to A^{†}f$. The noise-bias decomposition for linear regularization then gives noisy-data convergence when $\delta /\sqrt{\alpha }\to 0$.

A regularization $R_{\alpha }$ is convergent when its parameter can be chosen from the noise level and data so that the regularized reconstruction converges to the exact Moore–Penrose inverse of an operator solution as the noise level tends to zero.

For $f\in C^2[0,1]$ and $\Vert f^{\delta }-f{\Vert }_2\le \delta$, two translated half-interval integrals have overlap multiplicity at most two. The inequality $|a-b{|}^2\le 2|a{|}^2+2|b{|}^2$ gives $\Vert D_h\Vert \le \sqrt{6}/h$. Averaging $f^{\prime }$ over the relevant interval and applying $|f^{\prime }(x+t)-f^{\prime }(x)|\le \Vert f^{\prime \prime }{\Vert }_{\infty }|t|$ gives the second term. For this to estimate a Moore–Penrose inverse of an operator, also require $f(0)=0$; otherwise the derivative is still approximated but $K^{†}f$ is not defined.

On the Hilbert space ${\ell }^2(\mathbb{N})$, the left shift deletes the first coordinate. It is the adjoint operator of the unilateral shift operator $S$. The identities $LS=I$ and $SLf=f-f_1{e}_1$ show that it is also the Moore–Penrose inverse of an operator for $S$; both maps have operator norm one.

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 $y$.

The orthogonal projection of a prescribed point $u_0$ onto the affine subspace of least-squares solutions is unique. It retains $u_0$'s null space component and chooses the determined component from the Moore–Penrose inverse of an operator.

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.

For a bounded linear operator $A:X\to Y$ between Hilbert spaces, let $N=\ker A$ and $R=\mathrm{ran}A$. The Moore–Penrose inverse of an operator is defined onFor $y=Ax+z$, with $z\in R^{\perp }$, defineThis is independent of the chosen preimage $x$, since two preimages differ by a vector in $N$. Equivalently, it is the inverse of the restriction of $A$ to $N^{\perp }$, applied to the component of the data in $R$, and is zero on $R^{\perp }$.

The generalized solution $x^{†}=A^{†}y$ is the unique minimum-norm least-squares solution. Its residual is orthogonal to the range, giving the operator normal equationThe operator normal equation alone leaves an arbitrary null-space component; the second condition fixes the minimum-norm representative. The identities $A^{†}A=P_{N^{\perp }}$ and $A{A}^{†}=P_{\overline{R}}$ hold on their appropriate domains.

For an infinite-rank compact operator, the range need not be closed, and $A^{†}$ is generally unbounded. Data outside $R\oplus R^{\perp }$ need not have any least-squares solution at all, even though they can be approximated by range elements. This domain qualification is crucial in part (d); the Moore–Penrose notation does not turn an ill-posed inverse into an everywhere-defined bounded operator.

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 $\alpha$.

The given map is the unilateral shift operator. Its adjoint operator is the left shift operator, sinceThe shift is an isometry with trivial kernel and closed range $\{f:f_1=0\}$. Its range has orthogonal complement $\mathrm{span}\{e_1\}$, so every square-summable datum is admissible. Consequently,Indeed $K^{†}K=I$, while $K{K}^{†}f=f-f_1{e}_1$ is the required orthogonal projection. The discarded first component is the residual of the least-squares solution, not an obstruction to the existence of the Moore–Penrose inverse of an operator.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, the Moore–Penrose inverse of an operator assigns to each admissible datum its unique minimum-norm least-squares solution. Its domain isWrite $f=Ku+h$, where $u\in (\ker K{)}^{\perp }$ and $h\in (\mathrm{ran}K{)}^{\perp }=\ker K^{\ast }$. Then $K^{†}f=u$. Restricting to $(\ker K{)}^{\perp }$ removes the ambiguity among solutions differing by an element of the kernel. The orthogonal decomposition also shows that $K{K}^{†}f$ is the orthogonal projection of $f$ onto $\overline{\mathrm{ran}K}$, and $K^{†}Ku$ is the orthogonal projection of $u$ onto $(\ker K{)}^{\perp }$. The inverse is defined on a dense domain and need not be a bounded linear operator when the range is not closed.

The Volterra integration operator is injective and has dense range. Its range of the Volterra integration operator is $\{f\in H^1(0,1):f(0)=0\}$, and its Moore–Penrose inverse of an operator is differentiation on this domain. Thus the error estimate for $K^{†}f$ implicitly requires $f(0)=0$, in addition to the printed $H^2$ assumption. For an arbitrary $H^2$ function the same calculation bounds error against $f^{\prime }$, but that function need not be in $\mathcal{D}(K^{†})$.

Put $a=(1-\alpha )/2$, $b=(1+\alpha )/2$ and $h=\alpha /2$. The three stencils are forward, central finite differences, and backward on $[0,a)$, $[a,b)$ and $[b,1]$, respectively. All shifted evaluation points lie in $[0,1]$. For $g\in L^2(0,1)$, apply $|A-B{|}^2\le 2(|A{|}^2+|B{|}^2)$ and change variables in each of the six evaluation terms. The resulting integration intervals arePair $[0,a]$ with $[a,1-\alpha ]$, $[\alpha ,b]$ with $[b,1]$, and the two middle intervals. Each pair has disjoint interiors, so almost every point is counted at most three times. This overlap multiplicity bound for piecewise difference operators provesFor the bias on the middle interval, write the central finite difference as an average of $f^{\prime }$. The fundamental theorem of calculus giveswhere $\Vert k_{\alpha }{\Vert }_1=\alpha /4$. Extend $f^{\prime \prime }$ by zero to the real line. Young's convolution inequality bounds the middle-interval $L^2$ error by $\alpha c/4$. Combining its squared error with the supplied outer-interval estimate gives the central-difference noise-bias boundThe triangle inequality and the operator norm estimate now giveFor $c>0$, balancing noise and approximation bias minimizes this upper bound atfor sufficiently small $\delta$ that $\alpha <1/2$. Capping the parameter at $1/4$ supplies an admissible rule for all positive noise levels. More generally, $\alpha \to 0$ and $\delta /\alpha \to 0$ suffice; $\alpha =\sqrt{\delta }$ avoids needing $c$ or dividing by $c=0$.

To prove convergence for every admissible datum, not just $H^2$ data, let $f=Ku$ with $u\in L^2(0,1)$ and extend $u$ by zero. Each stencil is a forward, centred or backward average of $u$. Translation continuity in Lp makes each averaged difference from $u$ tend to zero in $L^2(\mathbb{R})$. Restricting those three bounds to their respective intervals proves $R_{\alpha }Ku\to u$. The noise-bias decomposition for linear regularization, with $\delta /\alpha \to 0$, finishes the convergent regularization of an inverse problem.

Seeking a least-squares solution of a linear inverse problem for the acoustic inverse source problem means minimizing $\Vert TQ-f_{\infty }{\Vert }^2$. Its normal equation for a linear inverse problem isFor compatible data, the formal minimum-norm solution can be written with the Moore–Penrose inverse of an operator:The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^{\ast }T{)}^{-1}{T}^{\ast }{f}_{\infty }$ is only formal because $T$ has a nontrivial kernel of a linear map.

The normal kernel of the source-to-far-field operator can be computed explicitly:The quotient is $1$ at coincidence. This smooth kernel already exhibits the loss of information.

There are two distinct causes of ill-posedness. First, a fixed-frequency nonradiating source gives exact nonuniqueness. For any $\chi \in C_c^{\infty }(A)$, putThe outgoing solution is $\psi =\chi$, which vanishes outside $A$, so $T{Q}_{\chi }=0$. Equivalently, integration by parts in the source-to-far-field operator at fixed frequency givessince the exponential solves the homogeneous Helmholtz equation and the boundary terms vanish. This provides an infinite-dimensional family of invisible sources. Single-frequency data determine only the spherical restriction of a three-dimensional Fourier transform.

Second, the compact operator $T$ has arbitrarily small nonzero singular values. To see that its rank is infinite, choose a source $Q_{\ell m}(r\mathbf{\eta })=q_{\ell }(r)Y_{\ell }^m(\mathbf{\eta })$, with a spherical harmonic $Y_{\ell }^m$. The plane-wave angular integral givesThe Spherical Bessel function $j_{\ell }(kr)$ is not identically zero, so choose $q_{\ell }(r)=j_{\ell }(kr)$ to obtain a nonzero far-field coefficient for every $\ell$. The mutually orthogonal angular modes prove infinite rank. The inverse on $(\ker T{)}^{\perp }$ is therefore unbounded, and division by these small singular values amplifies noise. Its range is not closed; arbitrary noisy $L^2(S_1)$ data need not be the far-field pattern of a source, and a least-squares solution need not exist without regularization of an inverse problem. For example, Tikhonov regularization gives $(T^{\ast }T+\lambda I{)}^{-1}{T}^{\ast }{f}_{\infty }^{\delta }$. It stabilizes the recovered visible component but cannot identify the missing fixed-frequency nonradiating source component.

Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

A regularization of an inverse problem is a family of continuous maps $R_{\alpha }:V\to U$, $\alpha >0$, such that $R_{\alpha }f\to K^{†}f$ for each $f\in \mathcal{D}(K^{†})$ as $\alpha \downarrow 0$. A linear regularization requires each $R_{\alpha }$ to be a bounded linear operator.

A regularization parameter choice assigns $\alpha =\alpha (\delta ,f^{\delta })>0$ from the noise bound and possibly the measured data. A convergent regularization of an inverse problem satisfies, for every fixed admissible exact datum,In the usual definition the chosen parameter also tends to zero uniformly over this noise ball. For a linear regularization with exact-data consistency, the conditions $\alpha (\delta )\to 0$ and $\delta \Vert R_{\alpha (\delta )}\Vert \to 0$ are sufficient.

For example, Tikhonov regularization with identity penalty has $R_{\alpha }=(K^{\ast }K+\alpha I{)}^{-1}{K}^{\ast }$ and $\Vert R_{\alpha }\Vert \le 1/(2\sqrt{\alpha })$. Its spectral filter converges to the Moore–Penrose inverse of an operator on its domain. Taking $\alpha (\delta )=\delta$ for small positive noise levels gives both required limits.

The Moore–Penrose inverse of an operator is the linear mapthat assigns each admissible datum its minimum-norm least-squares solution. If $f=y+z$ with $y\in \mathcal{R}(K)$ and $z\in \mathcal{R}(K{)}^{\perp }$, then $K^{†}f$ is the unique $u\in \mathcal{N}(K{)}^{\perp }$ satisfying $Ku=y$. In particular it vanishes on $\mathcal{R}(K{)}^{\perp }$, andThe continuity criterion, with the inherited data norm on its domain, isFor a closed operator range, the restriction of $K$ from $\mathcal{N}(K{)}^{\perp }$ to $\mathcal{R}(K)$ is a bounded bijection between Banach spaces, so its inverse is bounded by the bounded inverse theorem. Conversely, if $K^{†}$ is bounded and $K{u}_n\to y$, replace each $u_n$ by $P_{\mathcal{N}(K{)}^{\perp }}{u}_n=K^{†}K{u}_n$. This is a Cauchy sequence, hence converges to $u$, and $Ku=y$. Thus every limit of range elements remains in the operator range.

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.

Let $\eta =f^{\delta }-f$, so $\Vert \eta \Vert \le \delta$. By linearity, $u_{\delta }^{(k)}-u^{(k)}=R_{1/k}\eta$. The preceding operator norm estimate givesMoreover, $K{R}_{1/k}$ has factors $1-(1+\tau {\sigma }_j^2{)}^{-k}$ on the data singular vectors, and is zero on $\mathcal{R}(K{)}^{\perp }$. Every factor is in $[0,1]$, henceThus, for any fixed $k$, one can take $C=\sqrt{k\tau }$ and $\gamma =1$.

The first constant cannot generally be chosen independently of the iteration count. On the l2 sequence space, take $(Ku{)}_n=u_n/n$, $\tau =1$, and noise $\eta =\delta e_n$. At $k=n^2$ the solution perturbation has normThis disproves a uniform-in-$k$ interpretation of the stated first estimate. For a closed operator range, boundedness of the Moore–Penrose inverse of an operator instead supplies the uniform constant $\Vert K^{†}\Vert$.