Landweber iteration with discrepancy and budget statuses

Assume Countable Choice, as required for the Hilbert-space adjoint in the iteration.

Let $X,Y$ be real or complex Hilbert spaces, let $A:X\to Y$ be a nonzero compact linear operator, and let $\omega$ satisfy $0<\omega <2\mathrm{/}\mathrm{\parallel }A{\mathrm{\parallel }}^2$. The input consists of data $y_{\delta }\in Y$, a noise bound $\delta \ge 0$, a discrepancy factor $\tau >1$, and a finite iteration budget $N\in \mathbb{N}$. Define the associated infinite mathematical path by $x_0=0$ and $x_{k+1}=x_k+\omega A^{\ast }(y_{\delta }-A{x}_k),k\ge 0.$ For the finite procedure, evaluate only the prefix through index $N$, in increasing order, and return at its first stopping condition. Values of the mathematical path after an earlier return remain defined by the recurrence but need not be computed by an implementation.

For $k=0,\dots ,N$ in increasing order, compute $r_k=y_{\delta }-A{x}_k$. If $\mathrm{\parallel }{r}_k\mathrm{\parallel }\le \tau \delta$, return $(x_k,k,\mathtt{\text{discrepancy}})$. If $k=N$, return $(x_k,k,\mathtt{\text{budget-exhausted}})$. Otherwise set $x_{k+1}=x_k+\omega A^{\ast }(y_{\delta }-A{x}_k)\mathrm{.}$

The procedure uses exact Hilbert-space operations. It reports a finite-budget failure to cross as inconclusive; that status is not a convergence conclusion.