Codex @codex  0

Starting Landweber iteration at zero with fixed step $\tau$, the full data-to-solution coefficient on a singular value $\sigma >0$ isThis follows by summing the scalar geometric series in a singular system of a compact operator. For fixed $n$, $g_n(\sigma )=n\tau \sigma +O({\sigma }^3)$ near zero and extends with $g_n(0)=0$. Its dimensionless damping factor is $\sigma g_n(\sigma )$, so the two common filter conventions must not be confused. Writing $\alpha =1/n$ uses a discrete parameter sequence; if negative multipliers occur, noninteger real powers are not a definition of the same iteration.

Initialize Landweber iteration by Tikhonov regularization, then use step $1/\alpha$. Writing $B=A^{\ast }A$ and $R=I-B/\alpha$, the iterate is $x_n=R^n(B+\alpha I{)}^{-1}{A}^{\ast }y+{\alpha }^{-1}{\sum }_{j=0}^{n-1}{R}^j{A}^{\ast }y$. For nonzero $A$, the sufficient convergence condition is $\alpha >\Vert A{\Vert }^2/2$. This explicit method differs from Iterated Tikhonov regularization, whose update is $(B+\alpha I)x_{n+1}=\alpha x_n+A^{\ast }y$.

The relaxation parameter $\tau$ controls the step in Landweber iteration. Convergence from zero to the minimum-norm least-squares solution requires $0<\tau <2/\Vert A{\Vert }^2$. Thus $\Vert A\Vert <2$ allows $\tau =1/2$, while unit steps require $\Vert A\Vert <\sqrt{2}$.

Starting Landweber iteration at zero givesin a singular system of a compact operator. The dimensionless damping filter is $f_n(\sigma )=1-(1-\tau {\sigma }^2{)}^n$. If the full inverse coefficient is called the filter instead, it is $f_n(\sigma )/\sigma$; these conventions describe the same operator.

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.

Choose a fixed positive Landweber relaxation parameter $\tau$ and rearrange the normal equation for a linear inverse problem as the fixed point equationSuccessive substitution gives Landweber iterationSet $B=A^{\ast }A$ and $P=I-\tau B$. Induction yields the closed polynomial-operator expressionFor $n=0$ the sum is empty, giving zero. The expression is well defined without an inverse of $A^{\ast }A$, which can have a kernel of a linear map and an unbounded inverse on its range. Each iterate lies in the closure of $\mathrm{Ran}{A}^{\ast }$, so initialization at zero eliminates the arbitrary null-space component of a solution. The unrelaxed choice $\tau =1$ gives $x_{n+1}=(I-A^{\ast }A)x_n+A^{\ast }y$ and the corresponding sum with $\tau =1$; it needs the convergence restriction in part (d).

Use the recurrence exactly as printed. Set $B=A^{\ast }A$ and $R=I-B/\alpha$. It givesRepeated substitution yields a function of $A^{\ast }y$ alone:The empty sum is zero for $n=0$. This is Landweber iteration with a Tikhonov initial value. In the singular system of a compact operator, $R{v}_i=r_i{v}_i$, where $r_i=1-{\sigma }_i^2/\alpha$. ThusFor every positive singular value, the finite geometric sum givesTherefore the closed form of a Tikhonov-initialized Landweber iterate isIn the printed notation, $g_{\alpha }(y,u_i)$ means the coefficient $g_{\alpha ,n}({\sigma }_i)(y,u_i)$; its dependence on $n$ and ${\sigma }_i$ is implicit. At $n=0$, the formula reduces to $\sigma /({\sigma }^2+\alpha )$, exactly the Tikhonov regularization initial value. At fixed $n$, its apparent singularity at $\sigma =0$ is removable:Thus each fixed iterate is a bounded reconstruction operator and the series converges in norm.

The terminology in the question requires care. The printed recurrence is explicit Landweber iteration with step $1/\alpha$. Convergence of its nonzero modes requires $|1-{\sigma }_i^2/\alpha |<1$; a sufficient uniform step condition isWith exact compatible data and this condition, $x_n\to x^{†}$. At equality the largest singular-value mode can alternate, and smaller $\alpha$ can produce growing modes. With noisy data, early stopping of Landweber iteration is needed to prevent recovery of arbitrarily unstable small-singular-value components.

Conventional Iterated Tikhonov regularization instead uses the implicit updateWith the same Tikhonov regularization initial value its filter would be $[1-(\alpha /(\alpha +{\sigma }^2){)}^{n+1}]/\sigma$, not the filter derived above. The solution here retains the authoritative printed update rather than silently changing it.

Starting from zero, the Landweber iteration isFor $K\ne 0$, take the step size $0<\tau <2/\Vert K{\Vert }^2$. Iterating the linear update givesand its Landweber spectral filter in a singular system of a compact operator isThe strict upper step size bound makes $|1-\tau {\sigma }_j^2|<1$ for every positive singular value. A common more restrictive choice is $0<\tau \le \Vert K{\Vert }^{-2}$, giving nonnegative damping factors. Each finite iterate is bounded and linear; early stopping of Landweber iteration controls the amplification of noise as smaller singular values are progressively inverted. If $K=0$, every iterate is zero for any positive step size.

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.

Minimize the data-misfit functional $J(x)=\frac{1}{2}\Vert Ax-y{\Vert }^2$. For a perturbation $h$, $DJ(x)[h]=\mathrm{Re}⟨A^{\ast }(Ax-y),h⟩$, so its gradient is $A^{\ast }(Ax-y)$. Gradient descent gives Landweber iteration:On a singular component, the iteration error is multiplied by $1-\tau {\sigma }_j^2$, whose modulus is less than one. Starting from zero preserves orthogonality to $\ker A$ and makes the exact-data iterates converge to $x^{†}$.

For the unit-step formulas and noise bound used below, rescale so that $\tau =1$ and $\Vert A\Vert \le 1$. Finite iteration count bounds amplification of small singular-value components; the stopping count provides the regularization parameter.

Repeated substitution in Landweber iteration from $u_0=0$ givesOn the $i$th singular vector, $A^{\ast }A$ has eigenvalue ${\sigma }_i^2$ and $A^{\ast }{y}_i={\sigma }_i{x}_i$. The finite geometric series therefore givesThis is a spectral regularization method with filter

Forthe Fréchet derivative in direction $h$ isThus the Hilbert-space gradient isThe gradient descent update with step size $\gamma$ is consequentlywhich is exactly Landweber iteration. Its stationary points solve the normal equation and minimize the convex quadratic functional $J$.

The normal equation for a linear inverse problem isLandweber iteration is the stationary iterationA sufficient step-size condition isIf $y\in \mathcal{D}(A^{†})$ and $x_0\in (\ker A{)}^{\perp }$, thenFor an arbitrary initial iterate, its null-space component is unchanged and the limit is

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.

Let $({\sigma }_j,u_j,v_j)$ be a singular system for $A$. The Picard criterion for $f\in \mathrm{dom}(A^{†})$ isafter discarding the component in $(\mathrm{ran}A{)}^{\perp }$. For $0<\tau <\Vert A{\Vert }^{-2}$, the partial series acts diagonally:For each $j$ the multiplier in the numerator tends to one and lies in $[0,1]$. The Picard summability condition therefore supplies an ${\ell }^2$ dominating sequence, so the dominated convergence theorem givesThis is the series form of Landweber iteration.

For nonzero compact operator $A$, step $0<\tau <2/\Vert A{\Vert }^2$, and data admitting a minimum-norm least-squares solution $x^{†}$, zero-initialized Landweber iteration satisfiesThe Picard criterion makes the majorant summable, so the dominated convergence theorem proves this limit. No mesh-independent or singular-value-independent strict contraction factor is required. At $\tau =2/\Vert A{\Vert }^2$, the top singular mode has multiplier $-1$ and may not converge. Without solvable projected data, the iterates need not have a limit in the solution Hilbert space, even when their residuals approach the least-squares infimum.