One-step error recurrence for the assigned nonlinear regularized Gauss–Newton step

Assume Countable Choice. Let $X,Y$ be real or complex Hilbert spaces over the same scalar field and let $U\subseteq X$ be open. Let $F:U\to Y$ be continuously Fréchet differentiable over that field, with bounded complex-linear derivatives in the complex case. Let $x^{†}\in U$ satisfy $F(x^{†})=y$, and suppose $x_k,x^{†}$ lie in a convex subset of $U$ on which $F^{\mathrm{\prime }}$ is $L$-Lipschitz. Let $y_{\delta }$ satisfy $\mathrm{\parallel }{y}_{\delta }-y\mathrm{\parallel }\le \delta$, put $e_k=x_k-x^{†}$ and $A_k=F^{\mathrm{\prime }}(x_k)$, and use the step of Unanchored Levenberg–Marquardt step with explicit early stopping with ${\alpha }_k>0$. Define $R_k:=y-F(x_k)+A_k{e}_k\mathrm{.}$ Then $e_{k+1}={\alpha }_k(A_k^{\ast }{A}_k+{\alpha }_{k}I{)}^{-1}{e}_k+(A_k^{\ast }{A}_k+{\alpha }_{k}I{)}^{-1}{A}_k^{\ast }(y_{\delta }-y+R_k),$ and $\mathrm{\parallel }{e}_{k+1}\mathrm{\parallel }\le \mathrm{\parallel }{e}_k\mathrm{\parallel }+\frac{\delta +(L\mathrm{/}2)\mathrm{\parallel }{e}_k{\mathrm{\parallel }}^2}{2\sqrt{{\alpha }_k}}\mathrm{.}$ If in addition the tangential-cone condition $\mathrm{\parallel }F(x)-F(z)-F^{\mathrm{\prime }}(z)(x-z)\mathrm{\parallel }\le \eta \mathrm{\parallel }F(x)-F(z)\mathrm{\parallel }$ holds for all $x,z$ in that subset, then also $\mathrm{\parallel }{R}_k\mathrm{\parallel }\le \eta (\mathrm{\parallel }F(x_k)-y_{\delta }\mathrm{\parallel }+\delta )$, which may replace $(L\mathrm{/}2)\mathrm{\parallel }{e}_k{\mathrm{\parallel }}^2$ in the recurrence bound. This lemma alone does not assert contraction or discrepancy-stopped convergence.