Is Sommerfeld radiation condition invariant under translations? A smooth function $U:\mathbb{R}^3\setminus B_{r_0}(0)\to\mathbb{C}$ (for some $r_0>0$) satisfies the Sommerfeld Radiation Condition with index $k$, denoted $U\in \texttt{SRC}$, whenever
$$
\lim_{r\to\infty}\;\max_{|x|=r}r\,\Big|\partial_r U(x)-ikU(x)\Big|\,=\,0.
$$
Equivalently, if $U(x)=e^{ik|x|}V(x)$, then
$$
U\in \texttt{SRC} \iff \lim_{r\to\infty}\;\max_{|x|=r}r\,\Big|\partial_r V(x)\Big|\,=\,0.
$$
Here the radial derivative $\partial_r$ is defined as usual by $\partial_r U(x)=\nabla U(x)\cdot x/|x|$.
Question: does $U\in\texttt{SRC}$ imply that $U(x-x_0)\in \texttt{SRC}$, for all $x_0\in\mathbb{R}^3$?
So far, I only know a positive answer when $U$ is radial, that is $U(x)=e^{ik|x|}h(|x|)$.
Any suggestion or reference for the general case? Thanks in advance...
 A: $\newcommand{\x}{\mathbf{x}}
\renewcommand{\a}{\mathbf{a}}
\renewcommand{\b}{\mathbf{b}}
\renewcommand{\d}{\mathbf{d}}
\newcommand{\0}{\mathbf{0}}
\newcommand{\n}{\nabla}
\newcommand{\R}{\mathbb R}
\newcommand{\tth}{\theta}
\newcommand{\thh}{\theta}$
The question can be restated as follows:

Suppose $U$ is a smooth complex-valued function defined outside a neighborhood of $\0\in\R^3$ such that for some real $k>0$
\begin{equation*}
    \n U(\x)\cdot\frac\x{|\x|}-ikU(\x)=o\Big(\frac1{|\x|}\Big) \label{1}\tag{1}
\end{equation*}
(as $|\x|\to\infty$). Does it then follow that
\begin{equation*}
    \n U(\x)\cdot\frac{\x+\a}{|\x+\a|}-ikU(\x)=o\Big(\frac1{|\x|}\Big) \label{2}\tag{2}
\end{equation*}
for each $\a\in\R^3$?

The answer to this question is no.
Indeed, suppose that
\begin{equation*}
    U=e^{i\tth},
\end{equation*}
where $\tth$ is a smooth real-valued function defined outside a neighborhood of $\0\in\R^3$. Conditions \eqref{1} and \eqref{2} can then be rewritten as
\begin{equation*}
    \n\thh(\x)\cdot\x-k|\x|=o(1) \label{1a}\tag{1a}
\end{equation*}
and
\begin{equation*}
    \n\thh(\x)\cdot\d(\x,\a)\,|\x|=o(1), \label{2a}\tag{2a}
\end{equation*}
where
\begin{equation*}
    \d(\x,\a):=\frac{\x+\a}{|\x+\a|}-\frac\x{|\x|}. 
\end{equation*}
We need to construct a function $\tth$ satisfying \eqref{1a} but not \eqref{2a}. For $\x=(x,y,z)\ne\0$ and $u:=\sqrt{x^2+y^2}$, let
\begin{equation*}
    \tth_0(\x):=\frac{1}{u^2}\,\exp\Big\{-\frac{\ln^2 |z|}{u^2}\Big\} \label{*}\tag{*}
\end{equation*}
if $uz\ne0$, with $\tth_0(\x):=0$ otherwise. The function $\tth_0$ is smooth outside any neighborhood of $\0\in\R^3$. Also,
\begin{equation*}
    \n\thh_0(\x)\cdot\x=\frac{2 e^{-t^2} t^2}{u^2}-\frac{2 e^{-t^2} t}{u^3}-\frac{2 e^{-t^2}}{u^2}, \label{**}\tag{**}
\end{equation*}
where $t:=(\ln |z|)/u$. So, if $u\to\infty$, then $\n\thh_0(\x)\cdot\x=o(1)$. If $u$ is bounded but $|\x|^2(=u^2+z^2)\to\infty$, then $|z|\to\infty$ and hence $\ln |z|\to\infty$, so that
\begin{equation*}
    \n\thh_0(\x)\cdot\x=\frac{2 e^{-t^2} t^4}{\ln^2 |z|}-\frac{2 e^{-t^2} t^4}{\ln^3 |z|}
    -\frac{2 e^{-t^2}t^2}{\ln^2 |z|}=o(1). 
\end{equation*}
So, $\n\thh_0(\x)\cdot\x=o(1)$ whenever $|\x|\to\infty$. That is, \eqref{1a} holds with $\tth_0$ and $0$ in place of $\tth$ and $k$, respectively.
However, with $\tth_0$ in place of $\tth$, \eqref{2a} does not hold for $\a=(0,0,1)$ and $\x=(x,0,e^{-x})$ as $x\to\infty$, because then
\begin{equation*}
    \n\thh_0(\x)=(0,0,2e^{x-1}/x^3),
\end{equation*}
\begin{equation*}
    \d(\x,\a)=\Big(-\frac{1+o(1)}{2 x^2},0,\frac{1+o(1)}{x}\Big), 
\end{equation*}
and $|\x|\sim x$.
On the other hand, letting
\begin{equation*}
    \tth_1(\x):=k|\x|,
\end{equation*}
we get $\n\tth_1(\x)=k\x/|\x|$, so that \eqref{1a} obviously holds with $\tth_1$ in place of $\tth$. Moreover, it is not hard to see that for any $\a$ \eqref{2a} holds as well with $\tth_1$ in place of $\tth$.
Letting finally
\begin{equation*}
    \tth:=\tth_0+\tth_1,
\end{equation*}
we see that $\tth$ satisfies \eqref{1a} but not \eqref{2a}. This concludes the proof.

The initial idea behind the construction of the function $\tth_0$ is to make it a nonzero constant on the surface $z=e^{-u}$ (say), which is fast converging to the plane $z=0$ as $u\to\infty$, which will create a large gradient in a direction almost perpendicular to the plane $z=0$ at points of this surface with large $u$, provided that $\tth_0=0$ on the plane $z=0$. On the other hand, $\d(\x,\a)$ is almost perpendicular to $\x$ if $|\x|$ is large and $\x$ almost perpendicular to a fixed vector $\a$. So, by letting $\a$ be a normal vector to the plane $z=0$, we will hopefully get a violation of condition \eqref{2a}. So, one could try -- cf. \eqref{*} --
\begin{equation*}
    \tth_0(\x):=\exp\Big\{-\frac{\ln^2 |z|}{u^2}\Big\} 
\end{equation*}
if $uz\ne0$, with $\tth_0(\x):=0$ otherwise.
However, then instead of \eqref{**} we would have
\begin{equation*}
    \n\thh_0(\x)\cdot\x=2 e^{-t^2} t^2-\frac{2 e^{-t^2} t}{u},
\end{equation*}
and therefore would be unable to say that $\n\thh_0(\x)\cdot\x=o(1)$ whenever $u\to\infty$.
The final fix is then to introduce a slowly enough decreasing factor $\frac1{u^2}$, as was done in \eqref{*}. Then everything will work as desired.
