Conditional convergence of double-discrepancy composite inversion

Because $F$ and $G$ are compact, hence bounded, $A=G\circ F$ is compact: the image under $F$ of any bounded sequence is bounded, and compactness of $G$ gives a subsequence whose images under $G$ converge. Put $a_j={\lambda }_j{\mu }_j$ and $w_j=⟨w,x_j⟩$. The adjoint identity [F6] and the singular-vector equations show $N(F)=\{x_j:j\in J{\}}^{\perp }$ and $N(G)=\{y_j:j\in J{\}}^{\perp }$: each kernel vector is orthogonal to the indicated family, and conversely a vector orthogonal to the family is orthogonal to the full range of the adjoint, hence is killed by the operator. Since $G$ is injective on $\overline{R(F)}=\overline{R(G^{\ast })}$ (its coefficients there are the positive ${\mu }_j$), $\ker A=\ker F$; hence $(x_j)$ is a basis of $N(A{)}^{\perp }$. By [F1], $A{x}_j=a_j{z}_j$. The adjoint identity [F6] gives $A^{\ast }=F^{\ast }{G}^{\ast }$, so [F5] yields $A^{\ast }{z}_j=a_j{x}_j$, $A^{\ast }A{x}_j=a_j^2{x}_j$, and $\mathrm{\mid }A\mathrm{\mid }{x}_j=a_j{x}_j$. The vectors $z_j$ span $\overline{R(A)}$, since the range of $A$ is generated by $A{x}_j=a_j{z}_j$ and $a_j>0$. Compact spectral calculus and [F3] therefore give $x^{†}={\sum }_{j\in J}{a}_j^{\nu }{w}_j{x}_j$ and $z=A{x}^{†}={\sum }_{j\in J}{a}_j^{\nu +1}{w}_j{z}_j$. The source element is perpendicular to $\ker A$, so [F4] gives $A^{†}z=x^{†}$. Thus $y^{†}=F{x}^{†}$ lies in $N(G{)}^{\perp }$ and $x^{†}$ lies in $N(F{)}^{\perp }$; since $G{y}^{†}=z$ and $F{x}^{†}=y^{†}$, the minimum-norm property [F4] gives $G^{†}z=y^{†}$ and $F^{†}{y}^{†}=x^{†}$. Finally, $v:={\sum }_j{\lambda }_j^{\nu +1}{w}_j{y}_j$ satisfies $y^{†}=\mathrm{\mid }G{\mathrm{\mid }}^{\nu }v$ and $\mathrm{\parallel }v{\mathrm{\parallel }}^2={\sum }_j{\lambda }_j^{2\nu +2}\mathrm{\mid }{w}_j{\mathrm{\mid }}^2\le \mathrm{\parallel }w{\mathrm{\parallel }}^2\le {\rho }^2$.

Let $e_G:=R_{\alpha }^{G}z-y^{†}$. On each $y_j$, $I_Y-R_{\alpha }^{G}G$ has multiplier $p_{\alpha }^G({\mu }_j^2)$, whose absolute value is at most $S_G$ because $I_Z-G{R}_{\alpha }^G$ has that same multiplier on $z_j$. Thus $e_G=-\mathrm{\mid }G{\mathrm{\mid }}^{\nu }{p}_{\alpha }^G(G^{\ast }G)v=\mathrm{\mid }G{\mathrm{\mid }}^{\nu }{v}^{\mathrm{\prime }}$ with $\mathrm{\parallel }{v}^{\mathrm{\prime }}\mathrm{\parallel }\le S_G\mathrm{\parallel }v\mathrm{\parallel }\le S_G\rho$. Also $G{e}_G=G{R}_{\alpha }^{G}z-z$. The difference between the exact and noisy first residuals is $(G{R}_{\alpha }^G-I_Z)(z-z^{\delta })$, of norm at most $S_G\delta$; the stopping inequality therefore gives $\mathrm{\parallel }G{e}_G\mathrm{\parallel }\le ({\tau }_G+S_G)\delta$. Applying [F8] to $e_G=\mathrm{\mid }G{\mathrm{\mid }}^{\nu }{v}^{\mathrm{\prime }}$ proves $\mathrm{\parallel }{R}_{\alpha }^{G}z-y^{†}\mathrm{\parallel }\le (({\tau }_G+S_G)\delta {)}^{\nu \mathrm{/}(\nu +1)}(S_G\rho {)}^{1\mathrm{/}(\nu +1)}$.

Put $e_A:=x^{†}-R_{\gamma }^F{R}_{\alpha }^{G}z$. On the common singular system, $R_{\gamma }^F{R}_{\alpha }^{G}A$ has multiplier $(1-p_{\gamma }^F({\lambda }_j^2))(1-p_{\alpha }^G({\mu }_j^2))$. The bounds on $F{R}_{\gamma }^F$ and $G{R}_{\alpha }^G$ bound the two factors by $C_F,C_G$; on $N(A)$ the product with $A$ is zero. Therefore $e_A=\mathrm{\mid }A{\mathrm{\mid }}^{\nu }u$ with $u=(I-R_{\gamma }^F{R}_{\alpha }^{G}A)w$ and $\mathrm{\parallel }u\mathrm{\parallel }\le (1+C_F{C}_G)\rho$. The same coefficient formula, now on $Z$, gives $\mathrm{\parallel }A{R}_{\gamma }^F{R}_{\alpha }^G\mathrm{\parallel }\le C_F{C}_G$. The exact composite residual decomposes as $(z-z^{\delta })+(z^{\delta }-G{R}_{\alpha }^G{z}^{\delta })+(G{R}_{\alpha }^G{z}^{\delta }-A{R}_{\gamma }^F{R}_{\alpha }^G{z}^{\delta })+A{R}_{\gamma }^F{R}_{\alpha }^G(z^{\delta }-z)$. These four terms have norms at most $\delta ,{\tau }_G\delta ,{\tau }_F\delta ,C_F{C}_G\delta$, respectively, by the noise bound, the two stopping rules, and the composite operator bound. Since $A{e}_A=z-A{R}_{\gamma }^F{R}_{\alpha }^{G}z$, it follows that $\mathrm{\parallel }A{e}_A\mathrm{\parallel }\le (1+{\tau }_G+{\tau }_F+C_F{C}_G)\delta$. Applying [F8] to $e_A=\mathrm{\mid }A{\mathrm{\mid }}^{\nu }u$ proves $\mathrm{\parallel }{e}_A\mathrm{\parallel }\le C_1{\rho }^{1\mathrm{/}(\nu +1)}{\delta }^{\nu \mathrm{/}(\nu +1)}$.

Replace $D_H(\nu +1)$ by $\max \{D_H(\nu +1),0\}+1$ for $H=F,G$; the qualification bounds remain valid and these constants are positive. Define $q_{\gamma }(h):=A{R}_{\gamma }^F{R}_{\alpha }^{G}h-G{R}_{\alpha }^{G}h$. Its common-system multiplier on $z_j$ is $-p_{\gamma }^F({\lambda }_j^2)(1-p_{\alpha }^G({\mu }_j^2))$. Since $a_j^{\nu +1}={\lambda }_j^{\nu +1}{\mu }_j^{\nu +1}$, qualification at $s=\nu +1$, ${\mu }_j\le 1$, and $\mathrm{\mid }1-p_{\alpha }^G({\mu }_j^2)\mathrm{\mid }\le C_G$ give $\mathrm{\parallel }{q}_{\gamma }(z)\mathrm{\parallel }\le D_F(\nu +1)C_G\rho {\gamma }^{(\nu +1)\mathrm{/}2}$. On arbitrary noise coordinates, $\mathrm{\mid }{p}_{\gamma }^F({\lambda }_j^2)\mathrm{\mid }\le S_F$ and $\mathrm{\mid }1-p_{\alpha }^G({\mu }_j^2)\mathrm{\mid }\le C_G$, so $\mathrm{\parallel }{q}_{\gamma }(z^{\delta }-z)\mathrm{\parallel }\le S_F{C}_G\delta$. Let ${\mathrm{\ell }}_0$ denote the first index of $({\gamma }_{\mathrm{\ell }})$. If $\mathrm{\ell }>{\mathrm{\ell }}_0$, the preceding noisy discrepancy is greater than ${\tau }_F\delta$; subtracting the noise bound gives $\mathrm{\parallel }{q}_{{\gamma }_{\mathrm{\ell }-1}}(z)\mathrm{\parallel }>({\tau }_F-S_F{C}_G)\delta$. The exact-data bound just proved then yields ${\gamma }_{\mathrm{\ell }-1}>[({\tau }_F-S_F{C}_G)\mathrm{/}(D_F(\nu +1)C_G){]}^{2\mathrm{/}(\nu +1)}(\delta \mathrm{/}\rho {)}^{2\mathrm{/}(\nu +1)}$, and ${\gamma }_{\mathrm{\ell }}\ge \theta {\gamma }_{\mathrm{\ell }-1}$ gives the same lower bound with factor $\theta$. Hence $\mathrm{\parallel }{R}_{{\gamma }_{\mathrm{\ell }}}^F\mathrm{\parallel }\le K_F\mathrm{/}\sqrt{{\gamma }_{\mathrm{\ell }}}\le C^{\mathrm{\prime }\mathrm{\prime }}{\rho }^{1\mathrm{/}(\nu +1)}{\delta }^{-1\mathrm{/}(\nu +1)}$. If $\mathrm{\ell }={\mathrm{\ell }}_0$, then ${\gamma }_{\mathrm{\ell }}={\gamma }_{{\mathrm{\ell }}_0}$ is fixed. The common system and ${\lambda }_j,{\mu }_j\le 1$ give $\mathrm{\parallel }A\mathrm{\parallel }\le 1$, so $0<\delta <\mathrm{\parallel }z\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\mid }A{\mathrm{\mid }}^{\nu +1}w\mathrm{\parallel }\le \rho$ for all sufficiently small $\delta$. Thus $(\rho \mathrm{/}\delta {)}^{1\mathrm{/}(\nu +1)}\ge 1$, and $\mathrm{\parallel }{R}_{{\gamma }_{{\mathrm{\ell }}_0}}^F\mathrm{\parallel }\le K_F\mathrm{/}\sqrt{{\gamma }_{{\mathrm{\ell }}_0}}\le C^{\mathrm{\prime }\mathrm{\prime }}(\rho \mathrm{/}\delta {)}^{1\mathrm{/}(\nu +1)}$ after enlarging $C^{\mathrm{\prime }\mathrm{\prime }}$.

Let $k_0$ denote the first index of $({\alpha }_k)$. If $k>k_0$, the preceding noisy residual is greater than ${\tau }_G\delta$. Its difference from the exact-data residual is $(G{R}_{{\alpha }_{k-1}}^G-I_Z)(z-z^{\delta })$, of norm at most $S_G\delta$, so $\mathrm{\parallel }G{R}_{{\alpha }_{k-1}}^{G}z-z\mathrm{\parallel }>({\tau }_G-S_G)\delta$. Expanding $z=G\mathrm{\mid }G{\mathrm{\mid }}^{\nu }v$ in the common basis and applying the qualification bound at $s=\nu +1$ gives $\mathrm{\parallel }G{R}_{{\alpha }_{k-1}}^{G}z-z\mathrm{\parallel }\le D_G(\nu +1)\rho {\alpha }_{k-1}^{(\nu +1)\mathrm{/}2}$. Thus ${\alpha }_{k-1}>[({\tau }_G-S_G)\mathrm{/}{D}_G(\nu +1){]}^{2\mathrm{/}(\nu +1)}(\delta \mathrm{/}\rho {)}^{2\mathrm{/}(\nu +1)}$, and ${\alpha }_k\ge \theta {\alpha }_{k-1}$ gives the same lower bound for ${\alpha }_k$ with the extra factor $\theta$. Consequently $\mathrm{\parallel }{R}_{{\alpha }_k}^G(z^{\delta }-z)\mathrm{\parallel }\le K_G\delta \mathrm{/}\sqrt{{\alpha }_k}\le C{\rho }^{1\mathrm{/}(\nu +1)}{\delta }^{\nu \mathrm{/}(\nu +1)}$, where $C$ is independent of $\rho ,\delta$. If $k=k_0$, then ${\alpha }_k={\alpha }_{k_0}$ is fixed. Since $\rho >0$, we have $\delta \le \rho$ for all sufficiently small $\delta$, and hence $\mathrm{\parallel }{R}_{{\alpha }_{k_0}}^G(z^{\delta }-z)\mathrm{\parallel }\le K_G\delta \mathrm{/}\sqrt{{\alpha }_{k_0}}\le K_G{\alpha }_{k_0}^{-1\mathrm{/}2}{\rho }^{1\mathrm{/}(\nu +1)}{\delta }^{\nu \mathrm{/}(\nu +1)}$. Combining this noise bound with step 2.1 and the triangle inequality gives $\mathrm{\parallel }{R}_{{\alpha }_k}^G{z}^{\delta }-y^{†}\mathrm{\parallel }\le C^{\mathrm{\prime }}{\rho }^{1\mathrm{/}(\nu +1)}{\delta }^{\nu \mathrm{/}(\nu +1)}$.

The exact/noisy decomposition is $x^{†}-R_{{\gamma }_{\mathrm{\ell }}}^F{R}_{{\alpha }_k}^G{z}^{\delta }=e_A+R_{{\gamma }_{\mathrm{\ell }}}^F{R}_{{\alpha }_k}^G(z-z^{\delta })$. By linearity, $\mathrm{\parallel }{R}_{{\alpha }_k}^G(z-z^{\delta })\mathrm{\parallel }\le \mathrm{\parallel }{R}_{{\alpha }_k}^{G}z-y^{†}\mathrm{\parallel }+\mathrm{\parallel }{R}_{{\alpha }_k}^G{z}^{\delta }-y^{†}\mathrm{\parallel }$, and these terms are bounded by steps 2.1 and 3.1. Combining those estimates with the exact composite bias from step 2.2 and the outer norm bound from step 2.3 gives the two terms in the Statement, with constants independent of $\delta$ and $\rho$. Since $\nu >1$, both powers of $\delta$ are positive, so the error tends to zero for fixed $\rho$.