= Solution
Let the evaluation point be $\mathbf r_0=(x,y,z)$, $z>0$, and reflect it to $\mathbf r_0^*=(x,y,-z)$. The <method of images> constructs
$$
\boxed{G(\mathbf r,\mathbf r_0)=\frac1{4\pi}\left[\frac1{|\mathbf r-\mathbf r_0|}-\frac1{|\mathbf r-\mathbf r_0^*|}\right].}
$$
The two <Euclidean distances> coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-\nabla_{\mathbf r}^2G=\delta^{(3)}(\mathbf r-\mathbf r_0)$. The difference also decays at infinity.
On the plane write $\mathbf r=(\xi,\eta,s)$ and set $s=0$. The outward <unit normal> of the upper half-space is $n=-e_z$. Direct differentiation gives
$$
\left.\partial_sG\right|_{s=0}=\frac{z}{2\pi[(x-\xi)^2+(y-\eta)^2+z^2]^{3/2}},\qquad\partial_nG=-\partial_sG.
$$
Substituting into the boundary representation from part (c) yields the <Poisson kernel for the upper half-space>:
$$
\boxed{\phi(x,y,z)=\frac{z}{2\pi}\int_{\mathbb R^2}\frac{f(\xi,\eta)}{[(x-\xi)^2+(y-\eta)^2+z^2]^{3/2}}\,d\xi\,d\eta.}
$$
Its sign is positive, as expected from the <maximum principle> for nonnegative boundary data. Its normalization can be checked in <polar coordinates>:
$$
\int_{\mathbb R^2}P_z\,d\xi\,d\eta=z\int_0^\infty\frac{r\,dr}{(r^2+z^2)^{3/2}}=1.
$$
As $z\downarrow0$ the kernel is an <approximate identity>, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.
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