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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.

Interpret the normal-operator source condition as the displayed intersection condition: there is $v\in U$ with $w=K^{\ast }Kv\in \partial J(u^{†})$. Fix any $\alpha >0$ and put $\bar{u}=u^{†}+\alpha v$. The subgradient optimality condition for the artificial-data objective isIt holds because its first term equals $-\alpha K^{\ast }Kv=-\alpha w$. By convexity, this is sufficient for global minimality, proving the forward implication. Conversely, if $u^{†}$ minimizes that objective, the same optimality condition givesTaking $v=(\bar{u}-u^{†})/\alpha$ proves the intersection condition. Thus the two conditions are equivalent when the source vector is allowed to be zero.

The printed additional requirement $v\ne 0$ is not equivalent in general. For $K=I$ and $J\equiv 0$, the displayed source condition holds, and choosing $\bar{u}=u^{†}$ makes $u^{†}$ the minimizer, but $\partial J(u^{†})=\{0\}$ forces $v=0$. A nonzero-source version needs an additional hypothesis, such as requiring $w\ne 0$; it cannot be inferred from the printed intersection alone.

The relevant nonlinear generalized singular vector relation, without imposing unit-data normalization, isThis is a generalized eigenfunction in the forward-operator metric, rather than a distributional generalized eigenfunction of a linear operator. It supplies the normal-operator source condition with $v=\lambda u^{†}$. For a convex positively one-homogeneous functional, the subgradient inequality applied at zero and at $2u^{†}$ gives $⟨p,u^{†}⟩=J(u^{†})$. If $K{u}^{†}\ne 0$, thenAssume the usual positive eigenvalue $\lambda >0$ and $0<\alpha \lambda \le 1$. By positive homogeneity,The shifted-comparator Bregman bound therefore yieldsThe vector $(1-\alpha \lambda )u^{†}$ is itself a minimizer: $p$ remains a subgradient along the nonnegative ray, and substitution satisfies the subgradient optimality condition. If $K$ is injective, this also proves $u_{\alpha }=(1-\alpha \lambda )u^{†}$. Otherwise distinct minimizers can differ in the kernel while sharing the same predicted data.

For a nonnegative convex positively one-homogeneous functional, the branch continues as $(1-\alpha \lambda {)}_{+}{u}^{†}$ above the threshold, since $p/(\alpha \lambda )\in \partial J(0)$ there. Nonnegativity is needed for that clipping argument; bare positive homogeneity allows signed linear functionals.

Zero Bregman distance does not imply exact recovery. For $K=I$, $J(s)=|s|$, $u^{†}=a>0$ and $0<\alpha <a$, the minimizer is $u_{\alpha }=a-\alpha \ne a$, but $D_J^1(a-\alpha ,a)=0$. Both points touch the same supporting line, as described by zero Bregman distance and supporting faces.

Let $w=K^{\ast }Kv\in \partial J(u^{†})$ from the normal-operator source condition, and set $e=u-u^{†}$. Since $u^{†}$ is a least-squares solution, the normal equation for a linear inverse problem gives $K^{\ast }(K{u}^{†}-f)=0$. Expand the objective difference and use the Bregman distance definition:The cross term with the residual of the least-squares solution vanishes by the normal equation; exact data are not required. Compare the minimizing $u_{\alpha }$ with $z=u^{†}-\alpha v$. At $z$, the completed square is zero. Dividing the resulting inequality by $\alpha$ proves the stronger shifted-comparator Bregman boundDropping the nonnegative squared term gives the requested estimate. If the comparator lies outside the effective domain of $J$, its distance is infinite and the inequality is valid but gives no finite error bound.