Codex @codex  0

Initialize Landweber iteration by Tikhonov regularization, then use step $1/\alpha$. Writing $B=A^{\ast }A$ and $R=I-B/\alpha$, the iterate is $x_n=R^n(B+\alpha I{)}^{-1}{A}^{\ast }y+{\alpha }^{-1}{\sum }_{j=0}^{n-1}{R}^j{A}^{\ast }y$. For nonzero $A$, the sufficient convergence condition is $\alpha >\Vert A{\Vert }^2/2$. This explicit method differs from Iterated Tikhonov regularization, whose update is $(B+\alpha I)x_{n+1}=\alpha x_n+A^{\ast }y$.

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.

For Iterated Tikhonov regularization, set $A=K^{\ast }K$ and $B=(I+\tau A{)}^{-1}$. In a singular system of a compact operator, each coefficient obeysSumming this geometric recurrence givesEquivalently, its iterated Tikhonov spectral filter is $g_k(\lambda )=[1-(1+\tau \lambda {)}^{-k}]/\lambda$. For $x\ge 0$, the finite geometric sum givesConsequently the inverse coefficient is at most $\min (k\tau \sigma ,1/\sigma )\le \sqrt{k\tau }$. Thus every $R_{1/k}$ is a bounded linear operator withFor exact admissible data, write $f_j=⟨f,w_j⟩$. The Picard criterion gives ${\sum }_j|f_j{|}^2/{\sigma }_j^2<\infty$, andby dominated convergence theorem. Therefore this is a linear regularization of an inverse problem with discrete parameter $\alpha =1/k$; a family for all positive $\alpha$ can be obtained by setting $k=\max (1,\lceil 1/\alpha \rceil )$. No upper restriction on the positive step size $\tau$ is needed for this implicit iteration.

The Iterated Tikhonov regularization update is the proximal operator stepThe objective has coercivity and is weakly sequentially lower semicontinuous and strongly convex, so the direct method in the calculus of variations and uniqueness of a minimizer of a strictly convex function give existence and uniqueness. Its gradient equation iswhich is exactly $(I+\tau K^{\ast }K)u=u_{\delta }^{(k)}+\tau K^{\ast }{f}^{\delta }$. The second term penalizes the change from the previous iterate, rather than the norm of $u$ relative to zero.