Landweber semiconvergence for diagonal noisy data

Reindexing by $n\mapsto n+1$ is a linear isometry from ${\mathrm{\ell }}^2(I,\mathbb{R})$ onto ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{R})$, so [F7] makes $X=Y$ Hilbert spaces. All coordinates below are indexed by $j\ge 1$, and $(Ax{)}_j=2^{-j}{x}_j$. By [F8], $\mathrm{\parallel }Ax{\mathrm{\parallel }}^2={\sum }_{j\ge 1}{4}^{-j}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \frac{1}{4}\mathrm{\parallel }x{\mathrm{\parallel }}^2$, with equality at $e_1$; hence $A$ is bounded linear and [F5] gives $\mathrm{\parallel }A\mathrm{\parallel }=1\mathrm{/}2$. The truncation $(A_{N}x{)}_j=2^{-j}{x}_j$ for $j\le N$ and $0$ otherwise satisfies $\mathrm{\parallel }{A}_{N}x{\mathrm{\parallel }}^2={\sum }_{j=1}^N{4}^{-j}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2\le \frac{1}{4}\mathrm{\parallel }x{\mathrm{\parallel }}^2$. It is bounded linear with range having ordered basis $e_1,\dots ,e_N$ (the empty basis if $N=0$), so [F3] makes it compact. Also $\mathrm{\parallel }A-A_N\mathrm{\parallel }={\sup }_{j>N}{2}^{-j}=2^{-(N+1)}\to 0$, so [F4] makes $A$ compact. Thus $0<\omega =1<2\mathrm{/}\mathrm{\parallel }A{\mathrm{\parallel }}^2=8$. The data norm is $\mathrm{\parallel }{y}^{\delta }{\mathrm{\parallel }}^2={\delta }^2{\sum }_{j=1}^{\mathrm{\infty }}{2}^{-j}={\delta }^2$.

By [F6] and [F8], $A^{\ast }{e}_j=2^{-j}{e}_j$: for every $x$, $⟨Ax,e_j⟩=2^{-j}{x}_j=⟨x,2^{-j}{e}_j⟩$, and the adjoint is unique. Thus $A{A}^{\ast }{e}_j=4^{-j}{e}_j$, and the un-stopped path from [F1] has residual coordinates given by [F2]: $\delta 2^{-j\mathrm{/}2}(1-4^{-j}{)}^k$. Their squared sum is dominated termwise by ${\delta }^2{2}^{-j}$, whose tail tends to zero by the geometric-series formula. Given $ϵ>0$, choose a finite initial segment with omitted bound below $ϵ$; on that finite segment every residual coordinate tends to zero. Hence the full squared residual sum tends to zero. Each coordinate factor also decreases with $k$, so the residual norm is nonincreasing.

Here $A^{\ast }A{e}_j=4^{-j}{e}_j$. Substituting into [F2] gives $x_{k,j}^{\delta }=\delta 2^{j\mathrm{/}2}(1-(1-4^{-j}{)}^k)$. For an integer $n\ge 2$, Bernoulli's inequality gives $(1+1\mathrm{/}(n-1){)}^n\ge 1+n\mathrm{/}(n-1)>2$, hence $(1-1\mathrm{/}n{)}^n<1\mathrm{/}2$. Taking $n=4^j$ shows $\mathrm{\mid }{x}_{k_j,j}^{\delta }\mathrm{\mid }\ge \delta 2^{j\mathrm{/}2}\mathrm{/}2$, which tends to infinity. Thus $\mathrm{\parallel }{x}_{k_j}^{\delta }-x^{†}\mathrm{\parallel }\to \mathrm{\infty }$ although the residual tends to zero; this is semiconvergence and shows why the method needs an explicit stopping rule.