= Solution
Use the horizontal Fourier convention $\widetilde f(\mathbf k)=\int e^{-i\mathbf k\cdot\mathbf R}f(\mathbf R)d^2R$. For each nonzero $k=|\mathbf k|$, the <Poisson equation> becomes
$$
(\partial_z^2-k^2)\widetilde\Phi=4\pi G\widetilde\Sigma\delta(z).
$$
Decay away from the sheet and <continuity> at it give $\widetilde\Phi=C e^{-k|z|}$. Its derivative jump is $-2kC=4\pi G\widetilde\Sigma$, hence the <off-plane razor-thin Poisson kernel> is
$$
\boxed{\widetilde\Phi(\mathbf k,z)=-\frac{2\pi G}{k}\widetilde\Sigma(\mathbf k)e^{-k|z|},\qquad
\widetilde\Phi(\mathbf k,\epsilon)=-\frac{2\pi G}{k}\widetilde\Sigma(\mathbf k)e^{-k\epsilon}.}
$$
The spatially uniform mode has a different vertical solution and an arbitrary additive potential reference; it is not obtained by substituting $k=0$ into this decaying-mode formula.
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