= Solution
Seeking a <least-squares solution of a linear inverse problem> for the <acoustic inverse source problem> means minimizing $\|TQ-f_\infty\|^2$. \b[Its <normal equation for a linear inverse problem> is]
$$
\boxed{T^*TQ=T^*f_\infty.}
$$
For compatible data, the formal minimum-norm solution can be written with the <Moore–Penrose inverse of an operator>:
$$
\boxed{Q^\dagger=T^\dagger f_\infty
=\left[(T^*T)|_{(\ker T)^\perp}\right]^{-1}T^*f_\infty,\qquad
Q=Q^\dagger+h,\quad h\in\ker T.}
$$
The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^*T)^{-1}T^*f_\infty$ 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:
$$
(T^*TQ)(\mathbf r)=\frac1{16\pi^2}\int_AQ(\mathbf r')
\left[\int_{S_1}e^{ik\boldsymbol\theta\cdot(\mathbf r-\mathbf r')}\,dS(\boldsymbol\theta)\right]d^3\mathbf r'
=\frac1{4\pi}\int_A\frac{\sin(k|\mathbf r-\mathbf r'|)}{k|\mathbf r-\mathbf r'|}Q(\mathbf r')\,d^3\mathbf r'.
$$
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 $\chi\in C_c^\infty(A)$, put
$$
Q_\chi=(\Delta+k^2)\chi.
$$
The outgoing solution is $\psi=\chi$, which vanishes outside $A$, so $TQ_\chi=0$. Equivalently, integration by parts in the <source-to-far-field operator at fixed frequency> gives
$$
\int_A e^{-ik\boldsymbol\theta\cdot\mathbf r}(\Delta+k^2)\chi(\mathbf r)\,d^3\mathbf r=0,
$$
since 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 $Q_{\ell m}(r\boldsymbol\eta)=q_\ell(r)Y_\ell^m(\boldsymbol\eta)$, with a <spherical harmonic> $Y_\ell^m$. The plane-wave angular integral gives
$$
(TQ_{\ell m})(\boldsymbol\theta)=-(-i)^\ell Y_\ell^m(\boldsymbol\theta)\int_0^{r_0}q_\ell(r)j_\ell(kr)r^2\,dr.
$$
The <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 $(T^*T+\lambda I)^{-1}T^*f^\delta_\infty$. It stabilizes the recovered visible component but cannot identify the missing <fixed-frequency nonradiating source> component.
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