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The outgoing Green function for the three-dimensional Helmholtz equation satisfiesIts normalization follows by integrating over a small ball about $x$: $G_F\sim 1/(4\pi |y-x|)$, so its outward derivative has flux $-1$. Apply Green's second identity in the $y$ variables:The ${\omega }^2$ terms cancel. Thus the surface integral equals $-\Psi (x)$ for $x\in V$, and zero for $x$ outside the closed volume. Equivalently, remove a small source ball and evaluate its flux before taking its radius to zero. A source on the boundary would require a different limiting formula and is not part of this calculation.

For the upper half-space, take a half-ball of radius $\rho$. Interpret the first far-field condition as $\Psi =O({\rho }^{-1})$, and the radiation limit in its usual direction-uniform sense, namely the Sommerfeld radiation conditionThis is the finite-bound interpretation of the PDF's first condition. For fixed $x$ and real $\omega$,Indeed $|y-x|=\rho -\hat{y}\cdot x+O({\rho }^{-1})$, and its radial derivative is $1+O({\rho }^{-2})$. On the large hemisphere the integrand can therefore be rewritten asIts area is $2\pi {\rho }^2$, so its integral tends to zero. The planar integral, taken over expanding disks if needed, consequently equals $-\Psi (x)$ for $x_3>0$ and zero for $x_3<0$.

The outward unit normal vector on the bottom plane is $-e_3$. Denote this planar integral byFor $x_3>0$ reflect the source point to $x^{\ast }=(x_1,x_2,-x_3)$. On the plane,Adding $I(x)=-\Psi (x)$ and $I(x^{\ast })=0$ cancels the boundary values of $\Psi$ and gives $2\int G_F{\partial }_3\Psi =-\Psi (x)$. Thus the Neumann representation for a radiating Helmholtz half-space isThe negative sign is consistent with using the upward derivative ${\partial }_3\Psi$ rather than the outward normal derivative. If that upward derivative is zero everywhere on the plane, the integral is zero. Hence $\Psi =0$ throughout the upper half-space in the specified radiating class; taking a difference also gives uniqueness for prescribed Neumann boundary data.