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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 $A$.

Take $f_n(t)=\sin (2\pi nt)$. Its domain norm is $1$, whereas its image norm is $1+2\pi n$. Hence $T$ is unbounded, so it has no finite operator norm and is not a compact operator.

Write $\mathbf{r}=R\mathbf{\theta }$, $|\mathbf{\theta }|=1$. Uniformly for ${\mathbf{r}}^{\prime }$ in the source ball,For fixed $k$ and $r_0$, the outgoing Green function for the three-dimensional Helmholtz equation therefore has the far-field approximation for an outgoing sourceSubstitution into the source integral yieldsThus the source-to-far-field operator at fixed frequency isIt samples the Fourier transform of the source on the sphere of radius $k$. Its square-integrable kernel makes it a Hilbert-Schmidt operator, hence a compact operator. In factUse complex inner products linear in the first argument: $(f,g)=\int f\overline{g}$. Interchanging the volume and surface integrals givesHence the adjoint source-to-far-field operator isThis is a normalized Herglotz wave function; the change in exponential sign comes from the complex conjugation in the adjoint operator.

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.

Let $L=\Delta +c$ with real $c$, and letThe Fredholm alternative for an elliptic Dirichlet problem says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact boundary compatibility in the self-adjoint Fredholm alternative isThese conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward normal derivative fixes the displayed sign.

To prove the alternative, put $X=\{w\in C^{2,\alpha }(\overline{\Omega }):w{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The standard zero-boundary Poisson equation theorem makes $\Delta :X\to Y$ an isomorphism. Multiplication by $c$, followed by ${\Delta }^{-1}$, is a compact operator $K:X\to X$: the embedding $C^{2,\alpha }\hookrightarrow C^{0,\alpha }$ is compact, and multiplication is bounded. Writing $u=\psi +w$ givesThe Fredholm alternative for a compact operator implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.

For the explicit compatibility conditions, use the self-adjoint Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is coercive, its inverse gains two derivatives, and the Rellich-Kondrachov compactness theorem makes that inverse compact. The Fredholm solvability condition for a self-adjoint operator is $F\perp N$. Standard boundary elliptic regularity upgrades the resulting weak solution to $C^{2,\alpha }$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.

Finally Green second identity, with $z=0$ on the boundary and $Lz=0$, givesConsequently $\int zF=\int zf+\int \psi {\partial }_{\nu }z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline{z}$ in the pairings.

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

Apply the spectral theorem for compact Hermitian operators to positive $A=T^{\ast }T$. Its positive eigenvalues ${\mu }_i$ have orthonormal eigenvectors $u_i$ spanning $\ker A^{\perp }$, with a finite or countable list. In the infinite case ${\mu }_i\to 0$. DefineThen $(v_i,v_j)=(u_i,T^{\ast }T{u}_j)/({\lambda }_i{\lambda }_j)={\delta }_{ij}$, so the $v_i$ are orthonormal. Also $\ker A=\ker T$, because $(x,Ax)=\Vert Tx{\Vert }^2$. Decompose $x=x_0+{\sum }_i(u_i,x)u_i$ with $x_0\in \ker T$ and apply bounded $T$ to the norm-convergent expansion. This gives the singular value decompositionThe series converges in norm; its squared tail norm is ${\sum }_{i>m}{\mu }_i|(u_i,x){|}^2\to 0$. No separability of the whole Hilbert space is necessary: the nonzero spectral subspace of a compact operator is separable. For $T=0$ use the empty list $N=0$.

Put $A=T^{\ast }T$. Then $A^{\ast }=T^{\ast }T$ and $(x,Ax)=\Vert Tx{\Vert }^2\ge 0$, with the inner product linear in its second argument as required by the later SVD formula. For nonreal $\lambda$, self-adjointness gives $\Vert (A-\lambda I)x\Vert \ge |\mathrm{Im}\lambda |\Vert x\Vert$. For $\lambda =-a<0$, positivity gives $\Vert (A+aI)x\Vert \ge a\Vert x\Vert$. These bounds give injectivity and closed range. The range is dense because its orthogonal complement is the kernel of the adjoint, to which the same bound applies. Thus the range is all of $H$ and the inverse is bounded in either case. ConsequentlyIf $T$ is compact, composition of the compact operator $T$ with bounded $T^{\ast }$ is compact, so $T^{\ast }T$ is compact too.

First construct the Dirichlet eigensystem. By part (i) and the Lax-Milgram theorem, the inverse $S=A_D^{-1}$ exists from $L^2$ into $H_0^1$. Elliptic regularity gives $Sf\in H^2\cap H_0^1$, and the Rellich-Kondrachov compactness theorem makes $S$ a compact operator on $L^2$. Symmetry givesso $S$ is a self-adjoint operator. It is injective, and $(Sf,f)=B[Sf,Sf]>0$ for $f\ne 0$. The spectral theorem for compact Hermitian operators supplies an orthonormal basis $(w_m)$ with $S{w}_m={\mu }_m{w}_m$, ${\mu }_m>0$, and ${\mu }_m\to 0$. Put ${\lambda }_m={\mu }_m^{-1}$. ThenRepeated elliptic regularity makes each $w_m$ smooth up to the boundary.

As with part (ii), the stated characterization omits boundary compatibility. At $k=0$, its right-hand side is finite for every $u\in L^2$ by Parseval identity, whereas its left-hand side would require $u\in H_0^1$. At higher orders, the preceding polynomial example is also decisive for the Dirichlet eigensystem: on $(0,\pi )$,Although this function belongs to every ordinary $H^k$, the weighted sum diverges at $k=3$.

The correct spectral characterization of elliptic Dirichlet domains iswith $X_k$ defined in part (ii). For finite eigenfunction sums the squared norm $((u,u){)}_k$ is exactly the weighted sum: even orders follow from $L^l{w}_m={\lambda }_m^l{w}_m$, and odd orders also use $B[w_m,w_j]={\lambda }_m{\delta }_{mj}$. These sums are dense in $X_1$: if $v$ is $B$-orthogonal to every $w_m$, then $0=B[v,w_m]={\lambda }_m(v,w_m)$, so $v=0$. They are dense in $X_0$ by construction, and the isomorphisms $A_D:X_{k+2}\to X_k$ propagate density to every $X_k$. Completion in the equivalent norms from part (ii) proves both directions of the corrected equivalence. In particular, the printed characterization is correct for $k=1,2$.

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

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.

To first order in $n_{ϵ}$, replace $q$ by $2n_{ϵ}$ and use the known incident field in the outgoing integral. The Born approximation for scalar wave scattering givesBoth the $n_{ϵ}^2$ contrast term and the internal-field correction are second order for fixed geometry in the perturbative regime established in part i.

Specify the data space before taking an adjoint operator. For example, let $X=L^2(B)$ and $Y=L^2(\Gamma )$, where $\Gamma$ is a bounded measurement surface separated from $B$. Define the linear operatorFor bounded incident field this is a Hilbert-Schmidt operator, hence a compact operator. The equation is $A{n}_{ϵ}=y$, with $y={\psi }_s{|}_{\Gamma }$. Other sampling geometries give corresponding data spaces and weights; the paper does not specify one. Using the usual complex $L^2$ inner products, its adjoint operator isThe complex conjugations are required by the adjoint operator identity, not by wave reciprocity alone.

The Landweber iteration starts from $f_0=0$ and applies gradient descent to $\frac{1}{2}\Vert Af-y{\Vert }^2$:Each step back-propagates the data residual. The Landweber relaxation parameter controls stability, and early stopping of Landweber iteration prevents small singular values from amplifying measurement errors. For an explicitly real-valued index contrast, use the real Hilbert space structure and replace $A^{\ast }$ by $\mathrm{Re}{A}^{\ast }$ in this update. Additional sign or support constraints require corresponding projections; none are assumed here.

A bounded linear operator $T:X\to Y$ is a compact operator when the image of the closed unit ball has compact closure, equivalently when every bounded sequence $(x_n)$ has a subsequence for which $(T{x}_n)$ converges.

Each $T_i$ is compact because its image lies in a finite-dimensional subspace. Let $B_X$ be the unit ball and fix $\epsilon >0$. Choose $i$ with $\Vert T-T_i\Vert <\epsilon /3$. Since $T_i(B_X)$ is relatively compact, it has a finite $\epsilon /3$-net $y_1,\dots ,y_N$. For each $x\in B_X$, choose $y_j$ with $\Vert T_{i}x-y_j\Vert <\epsilon /3$; thenThus $T(B_X)$ is totally bounded. Its closure is complete because $Y$ is a Banach space, so it is compact. This proves the norm-closedness of compact operators and hence

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

As a map into $L^2$, $T$ factors asThe first arrow is bounded by part 3 and the second is the compact embedding from part 2; hence $T:L^2\to L^2$ is a compact operator.

Writing $u=T(f)$ in the defining identity givesIn particular this holds for every test function $v$, so the definition of a distributional derivative yields

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

Choose a nonzero $\varphi \in C_c^{\infty }({\mathbb{R}}^n)$ supported in a ball of radius $1$, and letTranslation preserves both the $L^2$ norm and the weak-derivative norms, sofor every $k$. Thus $({\varphi }_k)$ is bounded in the Sobolev space $H^1({\mathbb{R}}^n)$.

Distinct translates have disjoint supports. Hence, for $j\ne k$,No subsequence is Cauchy in $L^2$, so no subsequence converges there. The embedding $H^1({\mathbb{R}}^n)\hookrightarrow L^2({\mathbb{R}}^n)$ is therefore not a compact operator. This is the noncompactness of a Sobolev embedding by translation caused by loss of compactness at infinity.

A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the reflexive Banach space $X$, which is weakly compact. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.

The compact operator $T$ is weak-to-weak continuous, so $T{x}_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\Vert T{x}_{k_n}-Tz\Vert \ge \epsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence