Landweber converges for exact data and balances source bias with noise

For every nonzero eigenvalue $t$ of $T$, [F3]–[F4] give $0<t\le M$, so $0<\omega t<2$ and $\mathrm{\mid }1-\omega t\mathrm{\mid }<1$. Therefore $\mathrm{\mid }{r}_k(t)\mathrm{\mid }$ decreases with $k$ and tends to $0$. Since $x^{†}\perp \ker T$ by [F3] and [F5], [F6] expands it using only positive-eigenvalue components. By [F1], each component of $x_k-x^{†}$ is multiplied by $-r_k(t)$; its magnitude decreases to zero and is bounded by the corresponding component of $x^{†}$. Given $ϵ>0$, choose a finite set of eigenvalues whose omitted squared components of $x^{†}$ sum to less than $ϵ$; on the chosen finite set every component tends to zero. The squared error sum therefore tends to at most $ϵ$, and arbitrariness of $ϵ$ proves norm convergence. The componentwise decrease also proves monotonicity of the norm.

Choose an integer $m>s$. For $v\ge 0$ and an integer $n\ge 0$, $(1+v{)}^n\ge 1+nv$: induction proves this, since multiplying the bound at $n$ by $1+v$ gives $1+(n+1)v+n{v}^2\ge 1+(n+1)v$. For $k\ge 2m$, let $n=\lfloor k\mathrm{/}m\rfloor$, so $mn\le k$ and $n\ge k\mathrm{/}(2m)$. If $0<t\le \min \{M,{\omega }^{-1}\}$, put $u=\omega t\in (0,1]$. At $u=1$ the filter is zero. If $0<u\le 1\mathrm{/}k$, then $u^s(1-u{)}^k\le k^{-s}$. If $k\ge 2m$ and $1\mathrm{/}k<u<1$, the identity $(1-u)(1+u)=1-u^2\le 1$ gives $(1-u{)}^{-k}\ge (1+u{)}^k\ge (1+u{)}^{mn}=((1+u{)}^n{)}^m\ge (nu{)}^m$. Consequently $u^s(1-u{)}^k\le n^{-m}{u}^{s-m}\le n^{-m}{k}^{m-s}\le (2m{)}^m{k}^{-s}\mathrm{.}$ For $k<2m$ the same product is at most $1\le (2m{)}^s{k}^{-s}$. Thus on this range $t^s\mathrm{\mid }{r}_k(t)\mathrm{\mid }={\omega }^{-s}{u}^s(1-u{)}^k\le C_{s,\omega }{k}^{-s}$. If ${\omega }^{-1}<t\le M$, a range that exists only when $\omega M>1$, then $\mathrm{\mid }1-\omega t\mathrm{\mid }=\omega t-1\le \omega M-1=:q\in (0,1)$. Put $a=q^{-1}-1>0$. For $k\ge 2m$ and the same $n$, $q^{-k}=(1+a{)}^k\ge (1+a{)}^{mn}=((1+a{)}^n{)}^m\ge (na{)}^m,$ so $k^s{q}^k\le (2m{)}^m{a}^{-m}{k}^{s-m}\le (2m{)}^m{a}^{-m}$. For $k<2m$, $k^s{q}^k\le (2m{)}^s$. Hence ${\sup }_{k\ge 1}{k}^s{q}^k<\mathrm{\infty }$ and $t^s\mathrm{\mid }{r}_k(t)\mathrm{\mid }\le M^s{q}^k\le C_{s,M,\omega }{k}^{-s}$. Combining the two ranges proves ${\sup }_{t\in \sigma (T)\setminus \{0\}}{t}^s\mathrm{\mid }{r}_k(t)\mathrm{\mid }\le C_{s,A,\omega }{k}^{-s}$.

For $k\ge 1$, the scalar geometric identity gives $g_k(t)=[1-(1-\omega t{)}^k]\mathrm{/}t$ for $t>0$. Since $\mathrm{\mid }1-\omega t\mathrm{\mid }\le 1$, both the bound by $2$ and the finite-sum bound by $k\omega t$ hold for its numerator. Hence $\sqrt{t}\mathrm{\mid }{g}_k(t)\mathrm{\mid }\le \min \{2\mathrm{/}\sqrt{t},\omega k\sqrt{t}\}\le \sqrt{2\omega k}\mathrm{.}$ The last inequality follows by splitting at $t=2\mathrm{/}(\omega k)$. For $Q_k=g_k(T)A^{\ast }$, adjoint multiplication gives $Q_k{Q}_k^{\ast }=g_k(T)T{g}_k(T{)}^{\ast }$, the compact spectral multiplier with scalar values $t\mathrm{\mid }{g}_k(t){\mathrm{\mid }}^2$ and value zero on $\ker T$. [F7] bounds its norm by the supremum of these values; [F4] and $\mathrm{\parallel }{Q}_k{\mathrm{\parallel }}^2=\mathrm{\parallel }{Q}_k{Q}_k^{\ast }\mathrm{\parallel }$ then give $\mathrm{\parallel }{Q}_k\mathrm{\parallel }\le \sqrt{2\omega k}$. Applying [F1] to $y_{\delta }-y$ gives $\mathrm{\parallel }{x}_k^{\delta }-x_k\mathrm{\parallel }=\mathrm{\parallel }{Q}_k(y_{\delta }-y)\mathrm{\parallel }\le \sqrt{2\omega k}\delta$.

Under $x^{†}=T^{s}w$, [F1] gives $x_k-x^{†}=-r_k(T)T^{s}w$. On every eigenspace the product multiplier is $t^s{r}_k(t)$, bounded by [step 1.2]; [F7] therefore yields $\mathrm{\parallel }{x}_k-x^{†}\mathrm{\parallel }\le {\sup }_{t>0}{t}^s\mathrm{\mid }{r}_k(t)\mathrm{\mid }\mathrm{\parallel }w\mathrm{\parallel }$. Substituting the bound in [step 1.2] proves the source-bias estimate.

The triangle inequality gives $\mathrm{\parallel }{x}_k^{\delta }-x^{†}\mathrm{\parallel }\le \mathrm{\parallel }{x}_k^{\delta }-x_k\mathrm{\parallel }+\mathrm{\parallel }{x}_k-x^{†}\mathrm{\parallel }$; substituting the bounds from [step 2.1] and [step 1.3] proves the final estimate.