Unanchored Levenberg–Marquardt step with explicit early stopping

Assume Countable Choice, as required for the Hilbert-space adjoints in the normal equation.

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 $\mathbb{F}$. In the complex case, each $F^{\mathrm{\prime }}(x)$ is a bounded complex-linear operator and the Fréchet remainder is measured over complex increments, so ${\lim }_{\begin{array}{c}h\to 0,h\mathrm{\ne }0\\ x+h\in U\end{array}}\frac{\mathrm{\parallel }F(x+h)-F(x)-F^{\mathrm{\prime }}(x)h\mathrm{\parallel }}{\mathrm{\parallel }h\mathrm{\parallel }}=0.$ Inputs are $x_0\in U$, noisy data $y_{\delta }\in Y$, a bound $\delta \ge 0$, a discrepancy factor $\tau >1$, a finite budget $N\in \mathbb{N}$, and parameters ${\alpha }_0>0$ and $q\in (0,1)$. The output is a tuple $(x_k,k,\text{status})$. Starting with $k=0$, repeat the following checks and update. If $x_k\mathrm{\notin }U$ return $(x_k,k,\text{domain-failure})$. Otherwise put $A_k=F^{\mathrm{\prime }}(x_k)$ and ${\alpha }_k={\alpha }_0{q}^k$. Before computing an update, if $\mathrm{\parallel }F(x_k)-y_{\delta }\mathrm{\parallel }\le \tau \delta$, return $(x_k,k,\text{discrepancy})$. If $k=N$, return $(x_k,k,\text{budget-exhausted})$.

For $k<N$, solve the unique Hilbert-space Tikhonov normal equation $(A_k^{\ast }{A}_k+{\alpha }_k{I}_X)s_k=A_k^{\ast }(y_{\delta }-F(x_k))$ and set $x_{k+1}=x_k+s_k$, then replace $k$ by $k+1$ and repeat the checks. The shifted normal operator is invertible by The shifted Hilbert-space normal operator is invertible with inverse norm at most $1\mathrm{/}\lambda$. If the new iterate leaves $U$, the next check returns domain-failure. The procedure asserts no convergence theorem; its prescribed step is the unanchored Levenberg–Marquardt form used in the assigned contract.