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An error bound $A\delta /\alpha +Bc\alpha$, with positive $A,B,c$, is minimized at ${\alpha }_{\ast }=\sqrt{A\delta /(Bc)}$, with value $2\sqrt{ABc\delta }$. Differentiate the scalar bound, or apply the arithmetic-geometric mean inequality. This is an a priori regularization parameter choice for a linear regularization with inverse-parameter noise amplification and first-order approximation bias; the parameter must also lie in the method's allowed interval.

The Landweber spectral filter progressively admits smaller singular-value components. Its approximation bias tends to zero on exact admissible data, but its noise amplification grows. With $\Vert A\Vert \le 1$, a sufficient a priori regularization parameter choice is $n(\delta )\to \infty$ and $\sqrt{n(\delta )}\delta \to 0$. Taking regularization parameter $\alpha =1/n$ expresses this as $\alpha \to 0$ and $\delta /\sqrt{\alpha }\to 0$; the noise-bias decomposition for linear regularization proves convergence.

For $\alpha >0$, Tikhonov regularization minimizes the strictly convex quadraticDifferentiating in every direction gives the Tikhonov normal equation $(A^{T}A+\alpha I)v=A^T{y}_{\delta }$. Symmetry of $A$ therefore givesThe spectral filter suppresses unstable division by small eigenvalues. Put $e_i=(y_{\delta }-y,u_i)$ and $x_i=(x,u_i)$. The exact noise-bias decomposition for linear regularization isBy orthonormality and the triangle inequality,The scalar gain satisfies $\lambda /({\lambda }^2+\alpha )\le 1/(2\sqrt{\alpha })$, because $(\lambda -\sqrt{\alpha }{)}^2\ge 0$. Hence the Tikhonov stability bound givesThis explicitly separates data noise from regularization bias. For a known bound $X\ge \Vert x\Vert >0$, minimizing this last bound gives an a priori regularization parameter choiceOne data-based bound is $X=(\Vert y_{\delta }\Vert +\delta )/{\lambda }_1$. More generally, $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$ make the robust estimate converge to zero.

For this fixed finite matrix, the sharper gain bound ${\max }_i{\lambda }_i/({\lambda }_i^2+\alpha )\le 1/{\lambda }_1$ is also available. Thus the ${\delta }^{2/3}$ rate is a consequence of the robust bound, not an optimality claim: choosing $\alpha X/{\lambda }_1^2\le C\delta /{\lambda }_1$ gives an $O(\delta /{\lambda }_1)$ estimate. No unique statistically optimal parameter is specified without further assumptions on the signal or noise.

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 $A$.

Suppose each $R_{\alpha }:\mathcal{V}\to \mathcal{U}$ is a bounded linear operator and $R_{\alpha }f\to K^{†}f$ for every $f\in \mathcal{D}(K^{†})$ as $\alpha \downarrow 0$. A standard sufficient a priori regularization parameter choice satisfiesFor every datum with $\Vert f^{\delta }-f\Vert \le \delta$, the noise-bias decomposition for linear regularization givesThus the parameter rule gives a convergent regularization of an inverse problem, uniformly over data in the prescribed noise ball for each fixed admissible exact datum. For Tikhonov regularization, $\Vert R_{\alpha }\Vert \le 1/(2\sqrt{\alpha })$, so $\alpha \to 0$ and $\delta /\sqrt{\alpha }\to 0$ suffice. The same sufficient noise scaling holds for spectral cutoff regularization.

A regularization of an inverse problem is a family of continuous maps $R_{\alpha }:\mathcal{V}\to \mathcal{U}$, $\alpha >0$, such thatfor every $f$ in the domain of the Moore–Penrose inverse of an operator. A linear regularization has each $R_{\alpha }$ bounded and linear. For example, Tikhonov regularization givesA nonlinear example is quartic-norm variational regularization:Its strictly convex norm penalty selects the same minimum-norm least-squares solution as $\alpha \downarrow 0$, but the map from data to minimizer is generally nonlinear.

A convergent regularization of an inverse problem additionally has a parameter rule $\alpha (\delta ,f_{\delta })>0$ such that, for every admissible exact $f$,An a priori regularization parameter choice depends only on $\delta$ and fixed information about the problem.

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator, withThe Moore–Penrose inverse of an operator is the generally unbounded mapdefined when the Picard criterion holds, with the component in $\ker A^{\ast }$ sent to zero. It is the minimum-norm least-squares solution of $Af=g$.

A regularization of an inverse problem consists of bounded maps $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,g^{(\delta )})$ such thatwhenever $\Vert g^{(\delta )}-g\Vert \le \delta$ and $g$ is in the domain of $A^{†}$.

For Tikhonov regularization, minimizinggivesThe scalar spectral filter satisfiesand consequentlyFor exact data, each filter factor ${\sigma }_j^2/({\sigma }_j^2+\alpha )$ tends to one, so $R_{\alpha }g\to A^{†}g$. Choosingtherefore makes both the approximation error and propagated data error vanish. For example, $\alpha (\delta )=\delta$ is an admissible a priori regularization parameter choice.