= Solution
For noisy data, exact equality $Ax=y$ may be impossible: the data can have a component outside the <operator range>. A <least-squares solution of a linear inverse problem> instead minimizes
$$
J(x)=\frac12\|Ax-y\|_Y^2.
$$
For any variation $h\in X$, the <adjoint operator> gives
$$
DJ(x)[h]=\operatorname{Re}\langle Ax-y,Ah\rangle_Y
=\operatorname{Re}\langle A^*(Ax-y),h\rangle_X.
$$
Thus <statistical homogeneity> in every direction is the <normal equation for a linear inverse problem>, $A^*Ax=A^*y$. It is also sufficient for a global minimum: when it holds,
$$
\boxed{J(x+h)-J(x)=\frac12\|Ah\|_Y^2\geq0.}
$$
The residual is orthogonal to the closure of the range, since it lies in $\ker A^*$. A normal-equation solution therefore fits the component of the data representable by the forward map, while leaving the perpendicular component unfitted. If there is more than one such solution, select the <minimum-norm least-squares solution>, which lies in $(\ker A)^\perp$.
These facts explain the formulation, but do not make the inverse problem well posed. A <compact operator> of infinite rank typically has nonclosed <operator range>, so even the least-squares infimum need not be attained. A <least-squares solution> exists exactly when the <orthogonal projection> of $y$ onto $\overline{\operatorname{Ran}A}$ belongs to $\operatorname{Ran}A$. In a <singular system of a compact operator>, this becomes the <Picard criterion>. Also, the nonzero <eigenvalues> of $A^*A$ are the squares of the <singular values> of $A$: solving the <operator normal equation> directly can worsen numerical conditioning. \b[The <operator normal equation> expresses least squares; it does not itself regularize the inverse problem.] The subsequent finite <Landweber iteration> and its stopping rule provide the <inverse-problem regularization>.
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