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In a singular system of a compact operator, Landweber iteration with a Tikhonov initial value has coefficients $g_{\alpha ,n}({\sigma }_i)(y,u_i)$. Summing the finite geometric series gives the displayed filter. For fixed $n$, $g_{\alpha ,n}(\sigma )=(n+1)\sigma /\alpha +O({\sigma }^3)$ at zero, so each finite iterate is bounded. Under $\alpha >\Vert A{\Vert }^2/2$ the filter approaches $1/\sigma$ on every positive singular mode; noisy data require early stopping of Landweber iteration.

Use the recurrence exactly as printed. Set $B=A^{\ast }A$ and $R=I-B/\alpha$. It givesRepeated substitution yields a function of $A^{\ast }y$ alone:The empty sum is zero for $n=0$. This is Landweber iteration with a Tikhonov initial value. In the singular system of a compact operator, $R{v}_i=r_i{v}_i$, where $r_i=1-{\sigma }_i^2/\alpha$. ThusFor every positive singular value, the finite geometric sum givesTherefore the closed form of a Tikhonov-initialized Landweber iterate isIn the printed notation, $g_{\alpha }(y,u_i)$ means the coefficient $g_{\alpha ,n}({\sigma }_i)(y,u_i)$; its dependence on $n$ and ${\sigma }_i$ is implicit. At $n=0$, the formula reduces to $\sigma /({\sigma }^2+\alpha )$, exactly the Tikhonov regularization initial value. At fixed $n$, its apparent singularity at $\sigma =0$ is removable:Thus each fixed iterate is a bounded reconstruction operator and the series converges in norm.

The terminology in the question requires care. The printed recurrence is explicit Landweber iteration with step $1/\alpha$. Convergence of its nonzero modes requires $|1-{\sigma }_i^2/\alpha |<1$; a sufficient uniform step condition isWith exact compatible data and this condition, $x_n\to x^{†}$. At equality the largest singular-value mode can alternate, and smaller $\alpha$ can produce growing modes. With noisy data, early stopping of Landweber iteration is needed to prevent recovery of arbitrarily unstable small-singular-value components.

Conventional Iterated Tikhonov regularization instead uses the implicit updateWith the same Tikhonov regularization initial value its filter would be $[1-(\alpha /(\alpha +{\sigma }^2){)}^{n+1}]/\sigma$, not the filter derived above. The solution here retains the authoritative printed update rather than silently changing it.