Underestimated noise can put the Morozov target below every residual

Relabelling coordinates by $k\mapsto k+1$ is an isometry from the complete space ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{R})$ in [F2] to ${\mathrm{\ell }}^2({\mathbb{N}}_{+},\mathbb{R})$, so the latter is Hilbert. Truncate $A$ after coordinate $M\ge 1$. Each diagonal multiplier has modulus at most one, so both $A$ and its finite-rank truncation are bounded by one; the operator-norm error is $1\mathrm{/}(M+1)$. By [F3]–[F4] and [F2], [F10], $A$ is compact on the Hilbert space [F1]. Its range is contained in $e_1^{\perp }$. Given $z\in e_1^{\perp }$ and $\epsilon >0$, choose a finite $F\subset \{2,3,\dots \}$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{z}_n{\mathrm{\mid }}^2<{\epsilon }^2$ using the finite-subsum definition in [F1]. The truncation $z_F={\sum }_{n\in F}{z}_n{e}_n$ lies in $\mathrm{Ran}A$, since $A({\sum }_{n\in F}n{z}_n{e}_n)=z_F$, and $\mathrm{\parallel }z-z_F\mathrm{\parallel }<\epsilon$. Thus $\overline{\mathrm{Ran}A}=e_1^{\perp }$.

The datum $y_{\delta }=2\delta e_1$ is orthogonal to $\overline{\mathrm{Ran}A}$, so $\mathrm{dist}(y_{\delta },\overline{\mathrm{Ran}A})=2\delta$ and $\mathrm{\parallel }{y}_{\delta }\mathrm{\parallel }=2\delta$. For every $x$, the first coordinate of $Ax$ is zero, so $⟨Ax,e_1⟩=0$. By [F8], $⟨x,A^{\ast }{e}_1⟩=0$ for all $x$; taking $x=A^{\ast }{e}_1$ and using [F9] gives $A^{\ast }{e}_1=0$. Thus [F5] gives $x_{\lambda }=0$ and the residual is exactly $2\delta$ for every $\lambda >0$.

The exact endpoint inputs are $d=G=2\delta$, while the declared target is $\tau \delta =3\delta \mathrm{/}2$. The compatibility condition in [F7] fails, and the target is below the residual floor from step 2.1. Thus no exact discrepancy crossing exists and the finite routine reports incompatible-endpoints, not a parameter.