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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 minimizesFor any variation $h\in X$, the adjoint operator givesThus statistical homogeneity in every direction is the normal equation for a linear inverse problem, $A^{\ast }Ax=A^{\ast }y$. It is also sufficient for a global minimum: when it holds,The residual is orthogonal to the closure of the range, since it lies in $\ker A^{\ast }$. 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 $\overline{\mathrm{Ran}A}$ belongs to $\mathrm{Ran}A$. In a singular system of a compact operator, this becomes the Picard criterion. Also, the nonzero eigenvalues of $A^{\ast }A$ are the squares of the singular values of $A$: solving the operator normal equation directly can worsen numerical conditioning. 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.

Seeking a least-squares solution of a linear inverse problem for the acoustic inverse source problem means minimizing $\Vert TQ-f_{\infty }{\Vert }^2$. Its normal equation for a linear inverse problem isFor compatible data, the formal minimum-norm solution can be written with the Moore–Penrose inverse of an operator:The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^{\ast }T{)}^{-1}{T}^{\ast }{f}_{\infty }$ is only formal because $T$ has a nontrivial kernel of a linear map.

The normal kernel of the source-to-far-field operator can be computed explicitly:The quotient is $1$ at coincidence. This smooth kernel already exhibits the loss of information.

There are two distinct causes of ill-posedness. First, a fixed-frequency nonradiating source gives exact nonuniqueness. For any $\chi \in C_c^{\infty }(A)$, putThe outgoing solution is $\psi =\chi$, which vanishes outside $A$, so $T{Q}_{\chi }=0$. Equivalently, integration by parts in the source-to-far-field operator at fixed frequency givessince the exponential solves the homogeneous Helmholtz equation and the boundary terms vanish. This provides an infinite-dimensional family of invisible sources. Single-frequency data determine only the spherical restriction of a three-dimensional Fourier transform.

Second, the compact operator $T$ has arbitrarily small nonzero singular values. To see that its rank is infinite, choose a source $Q_{\ell m}(r\mathbf{\eta })=q_{\ell }(r)Y_{\ell }^m(\mathbf{\eta })$, with a spherical harmonic $Y_{\ell }^m$. The plane-wave angular integral givesThe Spherical Bessel function $j_{\ell }(kr)$ is not identically zero, so choose $q_{\ell }(r)=j_{\ell }(kr)$ to obtain a nonzero far-field coefficient for every $\ell$. The mutually orthogonal angular modes prove infinite rank. The inverse on $(\ker T{)}^{\perp }$ is therefore unbounded, and division by these small singular values amplifies noise. Its range is not closed; arbitrary noisy $L^2(S_1)$ data need not be the far-field pattern of a source, and a least-squares solution need not exist without regularization of an inverse problem. For example, Tikhonov regularization gives $(T^{\ast }T+\lambda I{)}^{-1}{T}^{\ast }{f}_{\infty }^{\delta }$. It stabilizes the recovered visible component but cannot identify the missing fixed-frequency nonradiating source component.