Composite double regularization

Let $X,Y,Z$ be real or complex Hilbert spaces over one scalar field, let $F:X\to Y$ and $G:Y\to Z$ be compact linear maps, and set $A:=G\circ F:X\to Z$. For spectral regularization families $R_{\alpha }^G:Z\to Y$ for $G^{†}$ and $R_{\gamma }^F:Y\to X$ for $F^{†}$, and a parameter rule $(\alpha ,\gamma ):(0,\mathrm{\infty })\times Z\to (0,\mathrm{\infty }{)}^2$, define $T_{\delta }(z^{\delta }):=R_{\gamma (\delta ,z^{\delta })}^F(R_{\alpha (\delta ,z^{\delta })}^G{z}^{\delta })$ for $z^{\delta }\in Z$. The order is fixed by the factorization: first regularize $G$, then $F$; no general identity $A^{†}=F^{†}{G}^{†}$ is part of this definition.

Compact Hilbert inverse model

Compact spectral filters and source conditions

Brandt and Horn, §2.1 and Theorem 1, pp. 2–3

Clason, Chapter 5, §5.1, pp. 44–50

The upper row is the forward map: applying $F$ and then $G$ gives $A=G\circ F$. The lower row shows the estimator's domains and codomains: data in $Z$ first pass through $R_{\alpha }^G$ to $Y$, then through $R_{\gamma }^F$ to $X$. Read that row from right to left; the diagram omits the parameter rule's dependence on $\delta$ and $z^{\delta }$.

For noisy data $z^{\delta }$, the lower row is $z^{\delta }\mapsto R_{\alpha }^G{z}^{\delta }\mapsto R_{\gamma }^F{R}_{\alpha }^G{z}^{\delta }$. The diagram records these domains and orders only; no proof about convergence or pseudoinverses relies on it.