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To illustrate Tikhonov regularization precisely, take the unknown to be the trace $g$ on a chosen enclosing sphere of radius $R$. Use Hilbert spaces $X=Y=L^2(S^2)$ and the forward linear operatorAt real positive $kR$, $d_n\ne 0$, because the real Spherical Bessel functions $j_n,y_n$ have a nonzero Wronskian and cannot vanish simultaneously. Moreover $|d_n|\to \infty$. Thus $A$ is an injective compact operator: truncating to finitely many harmonic orders approximates it in operator norm, since the remaining diagonal entries tend uniformly to zero. Its inverse is unbounded. This is a linear field reconstruction; estimating the obstacle from the reconstructed field remains a further nonlinear step.

Given coefficients $b_n^m$ of measured data $y^{\delta }$, minimize the quadratic functionalVarying $g$ in every direction gives $(A^{\ast }A+\alpha I)g_{\alpha }^{\delta }=A^{\ast }{y}^{\delta }$. The operator $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$, so it is invertible and the minimizer is unique. In coefficients the minimization is separable: $A^{\ast }y$ has coefficient $b_n^m/\overline{d_n}$ and $A^{\ast }A$ has multiplier $|d_n{|}^{-2}$. Hence the Tikhonov regularization of spherical far-field continuation isThe implied regularized far-field coefficients are $b_n^m/(1+\alpha |d_n{|}^2)$. This suppresses precisely the high-order coefficients that make backward continuation unstable.

Let $g^{†}$ be the exact trace and $y=A{g}^{†}$, with $\Vert y^{\delta }-y\Vert \le \delta$. The noise-bias decomposition for linear regularization isFor a singular value $s=|d_n{|}^{-1}$, the noise multiplier is $s/(s^2+\alpha )$. Since $s^2+\alpha \ge 2s\sqrt{\alpha }$, its supremum is at most $1/(2\sqrt{\alpha })$. ConsequentlyThis Tikhonov stability bound measures the trace norm, not a distance between obstacle boundaries. The bias coefficient is $\alpha /(|d_n{|}^{-2}+\alpha )$ times the exact coefficient. It tends to zero for each fixed $n,m$, is bounded by one, and $\sum |g_n^{†,m}{|}^2<\infty$. Splitting off a finite sum and bounding its tail therefore proves $B(\alpha )\to 0$ as $\alpha \to 0$.

A convergent regularization parameter choice thus satisfiesIn nondimensionalized variables, $\alpha ={\delta }^p$ with $0<p<2$ is one such rule. Without a smoothness assumption on the exact trace there is no universal bias rate and no universal optimal value of $\alpha$. For a concrete normal-operator source condition, suppose $g^{†}=A^{\ast }Aw$ and $\Vert w\Vert \le C$. Each bias multiplier on $w$ is $\alpha s^2/(s^2+\alpha )\le \alpha$, so $B(\alpha )\le C\alpha$. Minimizing the resulting bound gives, for $C>0$,This optimizes the displayed worst-case bound under that additional source condition, rather than guaranteeing an exact optimum for every datum.

If only the effect of measurement error is minimized, the multiplier bound decreases when $\alpha$ increases. Indeed $\Vert (A^{\ast }A+\alpha I{)}^{-1}{A}^{\ast }\Vert \le \Vert A\Vert /\alpha \to 0$ as $\alpha \to \infty$. But the reconstruction then approaches zero and its bias approaches $\Vert g^{†}\Vert$. Suppressing noise alone does not optimize reconstruction: $\alpha$ must balance noise amplification against regularization bias. A geometrically accurate obstacle estimate additionally needs justified shape constraints and a stable procedure for extracting a boundary from the regularized field.

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.

Substitution into the Tikhonov regularization series gives an explicit reconstruction for every $y\in L^2(0,1)$:Equivalently the summand is ${\sigma }_n⟨y,v_n⟩u_n/(\alpha +{\sigma }_n^2)$. The Tikhonov stability bound makes the series converge in $L^2$ for every positive $\alpha$. It suppresses high-frequency components of unstable differentiation, rather than differentiating arbitrary noisy data pointwise.

The range of the Volterra integration operator is $\{y\in H^1(0,1):y(0)=0\}$. On exactly these data, $A^{†}y=y^{\prime }$ and the series converges to $y^{\prime }$ as $\alpha \downarrow 0$. For arbitrary $L^2$ data outside that range, a finite regularized solution still exists, but a finite-norm unregularized solution need not exist. This is Spectral Tikhonov differentiation for the Volterra operator.

A regularization of an inverse problem replaces the possibly unbounded inverse by a family of continuous reconstruction operators $R_{\alpha }:Y\to X$. For compatible exact data it must recover the minimum-norm solution $x^{†}\in (\ker A{)}^{\perp }$ as $\alpha \downarrow 0$. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, a parameter rule $\alpha =\alpha (\delta )$ must make both bias and amplified noise tend to zero. The target is the minimum-norm solution because data cannot determine a component in the kernel of a linear map.

Tikhonov regularization minimizesThe quadratic norm penalty makes this functional strictly convex and a coercive function. Its first variation, applied to arbitrary real and imaginary perturbations, givesThe positive operator $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$ and has a bounded inverse. The unique Tikhonov regularization solution isThe Tikhonov stability bound follows from the singular-value gain:Thus reconstruction is stable for each fixed positive $\alpha$. Stability is not uniform as $\alpha \downarrow 0$: that limit restores the unbounded inverse. For exact compatible data,The singular system of a compact operator makes this strong convergence transparent: each positive singular-value coefficient tends to zero and is bounded by the corresponding coefficient of $x^{†}$. Consequentlywhenever $\alpha (\delta )\downarrow 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$, for example $\alpha (\delta )=\delta$. This is a convergent regularization of an inverse problem, not merely a bounded formula.

Regularization replaces sensitive inversion by a controlled approximation. For a bounded linear operator between Hilbert spaces, a linear regularization is a family of bounded linear operators $R_{\alpha }$ satisfying $R_{\alpha }y\to A^{†}y$ as $\alpha \downarrow 0$ for each datum in the domain of the Moore–Penrose inverse of an operator. Noisy-data convergence additionally requires a suitable regularization parameter choice.^[ref] Nonlinear variational regularization can instead use nonquadratic penalties.

The noise-bias decomposition for linear regularization separates two costs: amplifying measurement noise and changing the exact-data reconstruction. Truncated singular value decomposition discards weak modes; Tikhonov regularization attenuates them smoothly. A quadratic norm penalty expresses a preference for smaller reconstructions. Neither method manufactures a missing null space component: with complete averaging, the minimum-norm least-squares solution for $(0.5,0.5{)}^T$ is $(0.5,0.5{)}^T$, even if the original image was $(1,0{)}^T$. The preference selects a representative; extra observations or justified constraints are needed to identify the original.

For Tikhonov regularization, the Tikhonov stability bound limits transmitted measurement noise of norm at most $\delta$ to $\delta /(2\sqrt{\alpha })$. Taking $\alpha \to 0$ while $\delta /\sqrt{\alpha }\to 0$ gives a sufficient rule for convergent regularization of an inverse problem. Holding $\alpha$ fixed leaves approximation bias; letting it vanish too quickly can admit overwhelming measurement noise.