Anchored iteratively regularized Gauss-Newton with finite discrepancy stopping

Assume Countable Choice. Let $X,Y$ be Hilbert spaces over the same field $\mathbb{F}\in \{\mathbb{R},\mathbb{C}\}$, let $U\subseteq X$ be open, and let $F:U\to Y$ be continuously Fréchet differentiable over $\mathbb{F}$: for each $x\in U$, $F^{\mathrm{\prime }}(x)$ is bounded $\mathbb{F}$-linear, the map $x\mapsto F^{\mathrm{\prime }}(x)$ is operator-norm continuous, and ${\lim }_{\begin{array}{c}h\to 0,h\mathrm{\ne }0\\ x+h\in U\end{array}}\frac{\mathrm{\parallel }F(x+h)-F(x)-F^{\mathrm{\prime }}(x)h\mathrm{\parallel }}{\mathrm{\parallel }h\mathrm{\parallel }}=0,$ where $h$ ranges over increments in $\mathbb{F}$. In the complex case the derivative is complex-linear and the limit uses complex increments. Inputs are anchor and initial point $x_0\in U$, datum $y_{\delta }$, noise bound $\delta \ge 0$, discrepancy factor $\tau >1$, finite update cap $N\in {\mathbb{N}}_0$, ${\alpha }_0>0$ and $q\in (0,1)$. For $k=0,\dots ,N$, first check $x_k\in U$; otherwise return $(x_k,k,\mathtt{\text{domain-failure}})$. The output of every return is a tuple $(x_k,k,\text{status})$. Use an exact-call interface with these outcomes: an evaluation call returns either $\mathtt{\text{ok}}(v_k)$ with the exact value $v_k=F(x_k)$ or $\mathtt{\text{failure}}$; a normal-solve call returns either $\mathtt{\text{ok}}(s_k)$ with the exact solution of the displayed equation or $\mathtt{\text{unverified}}$. The interface never returns an incorrect value. If the evaluation call returns $\mathtt{\text{failure}}$, return $(x_k,k,\mathtt{\text{evaluation-failure}})$. If the evaluation returns $\mathtt{\text{ok}}(v_k)$ and $\mathrm{\parallel }{v}_k-y_{\delta }\mathrm{\parallel }\le \tau \delta$, return $(x_k,k,\mathtt{\text{discrepancy}})$. If $k=N$, return $(x_k,k,\mathtt{\text{budget-exhausted}})$. Otherwise set $A_k=F^{\mathrm{\prime }}(x_k)$, ${\alpha }_k={\alpha }_0{q}^k$, and call the normal solver on $(A_k^{\ast }{A}_k+{\alpha }_{k}I)s_k=A_k^{\ast }(y_{\delta }-v_k)+{\alpha }_k(x_0-x_k),$ whose exact solution exists uniquely by The shifted Hilbert-space normal operator is invertible with inverse norm at most $1\mathrm{/}\lambda$. If the solve call returns $\mathtt{\text{unverified}}$, return $(x_k,k,\mathtt{\text{solve-unverified}})$; otherwise set $x_{k+1}=x_k+s_k$. The mathematical evaluation and solve succeed because $F$ is defined on $U$ and the shifted normal equation has a unique solution; the failure outcomes record only a finite implementation's inability to return the exact value or certify the exact solve. No success or convergence claim is attached to these failure statuses.

The mathematical extended recurrence means the same exact updates with the discrepancy and finite-cap tests suppressed, for as long as the states lie in $U$. It is used only for comparison in proofs. A finite actual run follows its prefix and stops at its first test crossing. This anchored update is distinct from the retained unanchored Levenberg-Marquardt step; its extra term is ${\alpha }_k(x_0-x_k)$.