Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 (Inverse Problems in Imaging)

A forward problem evaluates a known model at a state: given $u$ and $K$, compute $f=Ku$. The associated inverse problem takes the data $f$ and the model $K$ and seeks a state $u$ with $Ku=f$.

In the sense of Hadamard well-posedness, a problem is well posed if a solution exists for every admissible datum, is unique, and depends continuously on the datum in the specified input and output topologies. An inverse problem is ill-posed if at least one of these requirements fails. Thus ill-posedness means failure of existence, uniqueness or stability. The topologies and the admissible data set matter: restricting the data to the range restores existence, but need not restore continuous dependence.

For a bounded linear operator $K:U\to V$ between Hilbert spaces, a least-squares solution of data $f$ is a minimizer of the residual norm:The minimum-norm least-squares solution is the one of smallest $U$-norm among these minimizers. Equivalently it is the unique least-squares solution in $\mathcal{N}(K{)}^{\perp }$. If the equation is consistent, this is the smallest-norm exact solution.

Let $P$ be the orthogonal projection onto $\overline{\mathcal{R}(K)}$. The orthogonal decomposition by a closed subspace givesThe first term has infimum zero by the definition of closure; that infimum is attained exactly if $Pf\in \mathcal{R}(K)$. This proves the least-squares existence criterion:They exist for every datum precisely when the range is closed. Once a solution exists, all solutions differ by elements of the closed null space, and projecting onto its orthogonal complement gives the unique minimum-norm solution.

For a counterexample to existence, take $K:{\ell }^2\to {\ell }^2$, $(Ku{)}_j=u_j/j$, and $f_j=1/j$. The range is dense, so the least residual infimum is zero; finite truncations realize residuals tending to zero. But an exact preimage would have every coordinate equal to one and would not belong to ${\ell }^2$. Thus no least-squares minimizer exists for this datum.

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.

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.

A singular system of a compact operator $K:U\to V$ consists of positive singular values ${\sigma }_j$ and orthonormal systems $u_j\in U$, $v_j\in V$ satisfyingThey span $\mathcal{N}(K{)}^{\perp }$ and $\overline{\mathcal{R}(K)}$, respectively. The singular value decomposition isFor an infinite-rank compact operator, the singular values tend to zero; for finite rank the sum is finite. Vectors in $\mathcal{N}(K)$ contribute nothing. This convention labels the input singular vectors by $u_j$ and the output ones by $v_j$.

For $(Ku{)}_j=u_j/j$, truncating after $n$ diagonal entries gives finite-rank operators with operator-norm error $1/(n+1)$, proving compactness. Its SVD is

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.

The range of the Volterra integration operator is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a Sobolev space element that is an absolutely continuous function. Indeed $Ku$ has weak derivative $u$ and zero initial trace; conversely the fundamental theorem of calculus reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.

The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its distributional derivative is a Dirac delta distribution, not an $L^2$ function. ThusFor a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative norm is $\sqrt{n}$, illustrating unstable differentiation.

The supplied SVD gives $K({\sigma }_j{u}_j)={\sigma }_j^2{v}_j$, or more simply $K^{†}({\sigma }_j{v}_j)=u_j$. Since ${\sigma }_j{v}_j\to 0$ but $\Vert u_j\Vert =1$, the Moore–Penrose inverse is discontinuous. Its domain is the dense, nonclosed range above, and there it is the weak derivative.

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.

For a linear regularization, add and subtract the exact-data reconstruction:The triangle inequality gives the noise-bias decomposition for linear regularization:The two summands are noise amplification and approximation bias. Increasing the strength of regularization typically decreases noise amplification and increases bias; reducing regularization has the opposite effect. These are qualitative tendencies and bounds, not a claim that each realized noise error is monotone for every noise vector.

For the one-sided differentiation example, the bounds are $\sqrt{6}\delta /h$ and $\Vert f^{\prime \prime }{\Vert }_{\infty }h/2$. Their sum has an interior balance point when the second-derivative bound is positive and the optimum lies in the permitted interval. The following original sketch plots these bounds against $h$:

For the Volterra operator, the identity $K^{†}f=f^{\prime }$ requires $f(0)=0$ in addition to the printed $C^2$ hypothesis. The constant function $f=1$ is a counterexample to the unrestricted domain assertion: it is not in $\mathcal{D}(K^{†})$. The estimate below is valid for derivatives of every $C^2$ function, and estimates the Moore–Penrose reconstruction when this missing boundary condition holds.

Let $e=f^{\delta }-f$. Apply $|a-b{|}^2\le 2|a{|}^2+2|b{|}^2$ separately to the two halves defining differentiation by one-sided difference quotients:The two translated intervals cover each point at most twice; this proves the factor six. The shifts act on almost-everywhere equivalence classes, so no undefined pointwise data sampling is involved.

For exact data, write the forward difference as $h^{-1}{\int }_0^h{f}^{\prime }(x+t)dt$ and the backward difference as $h^{-1}{\int }_0^h{f}^{\prime }(x-t)dt$. If $M=\Vert f^{\prime \prime }{\Vert }_{\infty }$, either differs from $f^{\prime }(x)$ by at most $h^{-1}{\int }_0^{h}Mtdt=Mh/2$. The interval has length one, so this also bounds the $L^2$ approximation error. The triangle inequality proves the error bound for one-sided differentiation:

For $\delta >0$ and $M=\Vert f^{\prime \prime }{\Vert }_{\infty }>0$, differentiate the bound $F(h)=\sqrt{6}\delta /h+Mh/2$. Its derivative vanishes at $h_{\ast }^2=2\sqrt{6}\delta /M$, and $F^{\prime \prime }(h)>0$. Consequently, when $h_{\ast }<1/2$,If $h_{\ast }\ge 1/2$, the infimum on the open allowed interval is approached as $h\uparrow 1/2$, rather than attained at an admissible endpoint. If $M=0$ and $\delta >0$, the bound decreases throughout the interval and has the same boundary behavior. At zero noise with $M>0$, its infimum is approached as $h\downarrow 0$.

A practical regularization parameter choice independent of $M$ isIt satisfies $h\to 0$ and $\delta /h\to 0$, hence gives a convergent regularization of an inverse problem on the domain of the Volterra operator inverse. This extends beyond $C^2$ exact data: if $f=Ku$ with $u\in L^2$, the differences are local averages of $u$. Extend $u$ by zero outside the interval; continuity of translations in $L^2$ makes these averages converge to $u$ on both halves. Together with $\Vert D_h(f^{\delta }-f){\Vert }_2\le \sqrt{6}\delta /h$ this proves convergence uniformly over the noise ball for every admissible exact datum.

Write $u^{†}=K^{†}f$, $e=R_{\alpha }f-u^{†}$, and $z={\beta }^{-2}{B}^{\ast }B{u}^{†}$, with $\beta >0$ as required by coercivity and by the printed inverse power. The Tikhonov regularization with a coercive penalty operator normal equation givesThis also holds for inconsistent data in $\mathcal{D}(K^{†})$, because $f-K{u}^{†}\perp \mathcal{R}(K)$ and therefore $K^{\ast }f=K^{\ast }K{u}^{†}$. Taking the inner product with $e$ givesFor any $w$ with $\Vert w\Vert \le r$, put $d=\Vert z-K^{\ast }w\Vert$ and use Cauchy-Schwarz inequality to bound the right-hand side by $\alpha {\beta }^2(d\Vert e\Vert +r\Vert Ke\Vert )$. Since $\Vert Be\Vert \ge \beta \Vert e\Vert$, completing the square in $\Vert Ke\Vert$ yieldsThe positive root of this quadratic inequality is at most $d+\sqrt{\alpha }\beta r/2$. Infimizing over $w$ proves the approximate source bound for coercive Tikhonov regularization, in fact slightly more strongly:Injectivity gives $\overline{\mathcal{R}(K^{\ast })}=\mathcal{N}(K{)}^{\perp }=U$. Given any positive tolerance, approximate $z$ by $K^{\ast }w$ for some finite-norm $w$; for all sufficiently large $r$ it is admissible in the infimum. ThusNo attainment of that infimum is needed.

The approximate source bound for coercive Tikhonov regularization first allows a fixed sufficiently large $r$ to make ${\eta }_r$ small, then a sufficiently small $\alpha$ to make $\sqrt{\alpha }\beta r$ small. This order of limits proves $R_{\alpha }f\to K^{†}f$ for every datum in its domain.

For the noise error let $v=R_{\alpha }g$. Its normal equation for coercive Tikhonov regularization givesCompleting the square shows $\alpha {\beta }^2\Vert v{\Vert }^2\le \Vert g{\Vert }^2/4$, hence $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. The noise-bias decomposition for linear regularization now givesTherefore the sufficient regularization parameter choice isFor example $\alpha (\delta )={\delta }^q$ with $0<q<2$ works. Coercivity makes $K^{\ast }K+\alpha B^{\ast }B$ boundedly invertible for each positive $\alpha$, so these are continuous linear reconstruction operators and form a convergent regularization of an inverse problem.

For a real Banach space $X$ and a proper convex function $E:X\to \mathbb{R}\cup \{+\infty \}$, the convex conjugate is the extended-valued function on the continuous dual spaceHere proper means that $E$ never takes $-\infty$ and is finite somewhere. The pairing is the duality pairing, or the inner product after identification in a Hilbert space. The conjugate is convex and lower semicontinuous because it is a supremum of continuous affine functions. Its definition gives the Fenchel–Young inequality $E(x)+E^{\ast }(p)\ge ⟨x,p⟩$; the Fenchel-Moreau theorem recovers $E$ as $E^{\ast \ast }$ under the stated closed-convex hypotheses.

For the absolute value function, $px-|x|\le 0$ for all $x$ if $|p|\le 1$, with equality at $x=0$. If $|p|>1$, taking $x=t\mathrm{sign}(p)$ with $t\to \infty$ makes the expression diverge. Thus its convex conjugate isThis is the indicator functional of a constraint set for the dual unit interval, not a zero-one indicator function.

The indicator functional of a constraint set restricts the conjugate supremum to $[-1,1]$. Complete the square:Its maximizer is $x=p$ when $|p|\le 1$, and the nearer endpoint $x=\mathrm{sign}(p)$ otherwise. The convex conjugate of a constrained quadratic is thereforeThis is the unit-threshold Huber loss; both branches and their first derivatives agree at the transition points. The original PDF constrains $|x|\le 1$, including both signs; the damaged TeX loses that absolute value.

In an infinite-dimensional Hilbert space, every bounded sequence has a weakly convergent subsequence, but it need not have a norm-convergent one. For example an orthonormal sequence $e_n$ has pairwise distance $\sqrt{2}$ and therefore no strongly convergent subsequence, although $e_n\rightharpoonup 0$ by Bessel inequality.

More generally, weak sequential compactness of bounded sequences in a reflexive Banach space ensures a weakly convergent subsequence with its limit in that space. Conversely, this property for all bounded sequences characterizes reflexivity, using the Eberlein-Šmulian theorem and weak compactness of the closed unit ball. Thus reflexivity replaces finite-dimensional norm compactness by weak sequential compactness.

In a nonreflexive Banach space there is no such general weak-subsequence guarantee. The coordinate unit vectors in ${\ell }^1$ give an example: by the Schur property, weak convergence would imply norm convergence, impossible for their pairwise distance two. Banach-Alaoglu theorem supplies weak-star compactness of a dual unit ball, which gives convergent subnets, not automatically convergent subsequences. For a separable predual, the weak-star compact dual ball is metrizable and subsequences do suffice. For an arbitrary Banach space one can also regard a bounded sequence in its bidual and extract a weak-star convergent subnet there; the limit need not belong to the original space. These qualifications distinguish weak from weak-star compactness.

On a real Hilbert space, for a proper lower-semicontinuous convex function $E$, the proximal operator isThe quadratic makes this objective strongly convex and coercive; the affine lower bound available for a proper closed convex function and weak lower semicontinuity give existence. The minimizer is unique. The subdifferential optimality condition isFor a positive parameter $t$, ${\mathrm{prox}}_{tE}$ uses $tE(x)$ in the same formula. On a general Banach space, the analogous norm-squared minimization is Banach-space proximal minimization; it can be set-valued and the identity-valued Hilbert derivative must be replaced by the duality mapping.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.

By the definition of the subdifferential, $p\in \partial J(u)$ meansRearrangement bounds $⟨v,p⟩-J(v)$ by $⟨u,p⟩-J(u)$, and equality is attained at $v=u$. Taking the supremum in the definition of the convex conjugate gives $J^{\ast }(p)=⟨u,p⟩-J(u)$. Conversely this equality bounds every member of that supremum and rearranges to the subgradient inequality. ThereforeApply the same argument to $J^{\ast }$ and use the Fenchel-Moreau theorem $J^{\ast \ast }=J$. With the canonical Hilbert identification of the bidual, the equality is also equivalent to $u\in \partial J^{\ast }(p)$. This proves subgradient inversion under convex conjugacy:The conditions include finiteness at the points in question; expressions involving $+\infty$ are not subgradients merely by formal subtraction.

Let $v={\mathrm{prox}}_J(u)$ and $p=u-v$. The proximal operator optimality condition gives $p\in \partial J(v)$. By subgradient inversion under convex conjugacy, $v\in \partial J^{\ast }(p)$. Since $v=u-p$, this is exactly the optimality condition defining $p={\mathrm{prox}}_{J^{\ast }}(u)$. Uniqueness of both Hilbert proximal minimizers proves Moreau decomposition:Both terms belong to the same Hilbert space after dual identification. No orthogonality of the two terms is asserted for a general convex $J$; that stronger property pertains to special indicator/cone cases.

For a convex functional $J:U\to \mathbb{R}\cup \{+\infty \}$ on a real Banach space, its subdifferential at $v$ is empty unless $J(v)$ is finite; otherwiseGiven $p_v\in \partial J(v)$, the Bregman distance isIt can be infinite if $J(u)$ is infinite. If $J$ is differentiable at $v$, the choice is its derivative; otherwise the base subgradient must be specified. Given also $p_u\in \partial J(u)$, add both directional distances to obtain the symmetric Bregman distanceThe symmetric version here is the sum, not half the sum. These nonnegative quantities are not generally metrics and can vanish at distinct points.

Assume the displayed variational minimizer exists and $K:U\to V$ is bounded and linear. The smooth data-fidelity term and the subdifferential sum rule giveUse the supplied source condition in variational regularization $p^{†}=K^{\ast }w\in \partial J(u^{†})$. Put $e=f^{\delta }-f$ and $q=K(u_{\alpha }-u^{†})$, so $\Vert e\Vert \le \delta$. By the symmetric Bregman distance identity,Dropping the last square and using $\Vert e-\alpha w{\Vert }^2\le 2\Vert e{\Vert }^2+2{\alpha }^2\Vert w{\Vert }^2$ proves the source-condition estimate for symmetric Bregman distance:The source condition also makes $u^{†}$ a $J$-minimizing exact solution: for $Ku=f$, its subgradient inequality has zero pairing with $u-u^{†}$. Existence of $u_{\alpha }$ is not implied by properness and lower semicontinuity alone; it is a hypothesis when no coercivity/compactness condition is provided.

Because $w_k\ge c>0$, a finite sum ${\sum }_k{w}_k|u_k|$ implies $u\in {\ell }^1$. Thus the printed weighted one-norm on l2 equals the extended nonnegative sum on every $u\in {\ell }^2$. Unbounded weights may make that sum infinite for some $u\in {\ell }^1$; membership in ${\ell }^1$ alone is not a finiteness assertion.

For each $N$, $J_N(u)={\sum }_{k=1}^N{w}_k|u_k|$ is continuous in the l2 sequence space, and $J={\sup }_N{J}_N$. If $u^{(n)}\to u$ in norm, then for every $N$,Taking the supremum over $N$ provesIt is also weakly lower semicontinuous by the same finite-coordinate argument, and convex and proper because zero and finite sequences have finite penalty.

At a finite-penalty base point $v$, coordinate variations give the subdifferential of a weighted one-norm on l2:These conditions are sufficient by summing the coordinate supporting inequalities. Since $w_k\ge c$, an infinitely supported $v$ would require infinitely many $|p_k|\ge c$, contradicting $p\in {\ell }^2$. Thus a finite-penalty point can have empty subdifferential; a Bregman base with an actual subgradient must be finitely supported.

For any such $p$, $⟨p,v⟩=J(v)$ and the Bregman distance simplifies toEach summand is nonnegative; the sum may be infinite. The pairing converges by Cauchy-Schwarz inequality. This definition does not silently assume a subgradient exists at every $v\in {\ell }^1$.

Use the standard inverse-problem setting of a bounded linear $K:{\ell }^2\to {\ell }^2$ and exact data having a finite-penalty solution. Let $u^{†}$ denote the unique $J$-minimizing exact solution for $J(u)=\Vert u{\Vert }_1+\beta \Vert u{\Vert }_2^2$, $\beta >0$. This is the limiting target of elastic net regularization in Hilbert sequence space; it need not be the Moore–Penrose minimum-norm solution unless additional hypotheses identify them, for example injectivity of $K$.

For each $\alpha >0$, the reconstruction objective is weakly lower semicontinuous, coercive by $\alpha \beta \Vert u{\Vert }_2^2$, and strictly convex. The direct method on the reflexive l2 sequence space gives existence, and strict convexity gives uniqueness. For two data sets, monotonicity of the one-norm subgradients and the quadratic penalty give, with $d=u_1-u_2$ and $g=f_1-f_2$,Completing the square yields $\Vert d\Vert \le \Vert g\Vert /(2\sqrt{2\alpha \beta })$, so each reconstruction map is continuous.

Minimality against $u^{†}$ givesChoose $\alpha \to 0$ and ${\delta }^2/\alpha \to 0$, for example $\alpha =\delta$. This bounds the Hilbert norms, drives the data residual to zero and gives $\mathrm{limsup}J(u_{\alpha }^{\delta })\le J(u^{†})$. Every weakly convergent subsequence has an exact solution as its limit, by boundedness of $K$; weak lower semicontinuity and uniqueness of the penalty minimizer identify that limit with $u^{†}$. Separately applying weak lower semicontinuity to the one-norm givesWeak convergence and convergence of Hilbert norms imply strong convergence by the Radon-Riesz theorem. This proves norm convergence of elastic net regularization:The argument applies to every sequence of noise levels and admissible data, hence gives the usual uniform noise-ball convergence for a fixed admissible exact datum.

For nonlinearity of elastic net reconstruction, zero is the minimizer exactly when $\Vert K^{\ast }{f}^{\delta }{\Vert }_{\infty }\le \alpha$. This includes a neighborhood of zero. If $K\ne 0$, choose $e_j$ with $K{e}_j\ne 0$ and $f^{\delta }=tK{e}_j$ for $t\Vert K{e}_j{\Vert }^2>\alpha$; zero then fails its optimality condition. A linear map vanishing on a neighborhood would vanish everywhere, so the elastic-net reconstruction is nonlinear whenever $K\ne 0$. For $K=I$ the explicit formula is $[R_{\alpha }f{]}_j=\mathrm{sign}(f_j)(|f_j|-\alpha {)}_{+}/(1+2\alpha \beta )$, exhibiting soft thresholding.

A concrete distinction from Moore–Penrose convergence is $K(u)=u_1+2u_2$ (embedded in the first output coordinate of ${\ell }^2$), exact datum one, and $\beta =1$. Its minimum-norm solution is $(1/5,2/5,0,\dots )$, with penalty $4/5$, while the unique penalty-minimizing exact solution is $(0,1/2,0,\dots )$, with penalty $3/4$. For $0<\alpha <2$ the exact-data elastic-net reconstruction is $(0,(2-\alpha )/(4+2\alpha ),0,\dots )$, converging to the latter. Thus it is not a Moore–Penrose-consistent regularization for every noninjective operator under the earlier definition.

The PDF supplies neither a nonzero assumption on $K$ nor a target/domain qualification here. Its unconditional nonlinearity claim is false for $K=0$, whose reconstruction is the zero linear map. The convergence proof above states the finite-penalty exact-solution hypotheses explicitly; it establishes convergence to the penalty-minimizing solution, not an automatically identical minimum-norm least-squares solution.

Use the standard bounded linear operator setting for $K:U\to V$, so the adjoint operator $K^{\ast }:V\to U^{\ast }$ in the updates is defined on every datum. Put $q=K^{\ast }K{u}_{\lambda }$. The generalized singular vector relation is $\lambda q\in \partial J(u_{\lambda })$ with $\Vert K{u}_{\lambda }\Vert =1$ and $\lambda =J(u_{\lambda })\ge 0$. For an absolutely one-homogeneous functional, a subgradient $p$ at $u$ is characterized by $⟨p,u⟩=J(u)$ and $⟨p,v⟩\le J(v)$ for all $v$. Consequently $\lambda q$ is also a subgradient at every positive multiple of $u_{\lambda }$, and $bq\in \partial J(0)$ for every $0\le b\le \lambda$.

Use the initialization $u^0=p^0=0$ and start the first update at $k=0$; the PDF lists $k=1,2,\dots$ despite specifying only index-zero initial data. Suppose $p^k=b_{k}q$ with $0\le b_k\le \lambda$. DefineIf it is positive, $u^{k+1}=c_{k+1}{u}_{\lambda }$ and $p^{k+1}=\lambda q$ satisfy the optimality equation and the dual update. If it is zero, take $u^{k+1}=0$ and $p^{k+1}=(b_k+\gamma /\alpha )q\in \partial J(0)$. Thus this compatible branch hasInduction proves the finite termination of Bregman iteration on a singular vector formulasOnce the dual coefficient reaches $\lambda$, the next reconstruction is $\gamma u_{\lambda }$. Therefore a guaranteed stopping index for this branch isThe formulas also cover $\lambda =0$, where exact reconstruction occurs at the first update. The subgradient conditions are sufficient for global minimization because each objective is convex.

There is a genuine uniqueness issue in the printed claim about arbitrary iterates. Strict convexity of data fidelity in $Ku$ ensures that any two minimizers at a given step have the same image under $K$. Induction therefore makes every minimizer branch have the same dual variables and attain exact data at the stated finite index. It does not make their states equal to the specified $\gamma u_{\lambda }$ if $K$ has a null space.

For a counterexample take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $K(x,y)=x$, $J(x,y)=|x|$, $u_{\lambda }=(1,0)$, and $\lambda =1$. Every step leaves $y$ completely unconstrained. Choosing the minimizing state's second coordinate to be $k$ gives valid updates and subgradients but never reaches $(\gamma ,0)$ and does not converge as a state sequence. Thus finite exact-data recovery holds for all minimizer branches; exact recovery of the stated vector needs uniqueness or the aligned branch. If $K$ is injective, equal reconstructed data identify the state, and the literal vector conclusion follows.