# Poisson equation on half-space

Let $H$ be the open half-space of $\Bbb R^n$ defined by $x_n > 0$. Let $f : \overline H \to \Bbb R$ continuous and harmonic on $H$. Define the function $F : \Bbb R^n \to \Bbb R$ by $$F(x_1,\dotsc,x_n) = \begin{cases} f(x_1,\dotsc,x_n) && \text{if x_n>0}\\ -f(x_1,\dotsc,-x_n) && \text{if not.} \end{cases}$$

$F$ defines a distribution on $\Bbb R^n$ and one computes $$\Delta F = 2 \partial_n( f \sigma),$$ where $\sigma$ is the surface distribution of the hyperplane $x_n = 0$.

The conclusion is that the Cauchy problem $$f\in C(\overline H), \quad \left\{ \begin{gathered} \Delta f_{|H} = 0 \\ f_{|x_n = 0} = g \end{gathered} \right .$$ translates into the distribution equation $$F \in \mathcal D \Bbb R^n\\ \Delta F = 2 \partial_n(g\sigma).$$

If $g$ has compact support, it is well-known that the latter equation has a solution (using a convolution with the elementary solution of the laplacian).

Here is my question.

Assume that $g\in C_c(\partial H)$, and that $\Delta F = 2\partial_n(g\sigma)$. How to show that $F$ comes from a solution of the first system ?

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Do you want to know that $F\in C(\overline{H})$? Or what do you mean with $F$ "coming from the first system"? –  Vobo Jul 5 '12 at 16:17
@Vobo — I do. I want to prove that there exists a $f$ solution of the first system such that F is obtained from $f$ by imparity on the $n$th variable. –  Lierre Jul 5 '12 at 16:26
You can prove directly that $P(x,x_n)\ast g$ (where $P$ is the classical Poisson kernel) has the property $P(x,x_n)\ast g(x)\to g(x)$ as $x_n\to 0$. A little more complicated is the second step, that for any other solution $Q$, $(P\ast g)-Q$ is a solution of the special problem for $g= 0$, which are precisely harmonic functions in the plane with imparity in the last variable. –  Vobo Jul 5 '12 at 17:09
@Vobo — Thanks. I'm a bit disappointed though that we have to use the Poisson kernel, and not only the existence of a distribution solution. –  Lierre Jul 6 '12 at 6:51
This basically follows from the well-known fact, that any harmonic distribution is in fact a harmonic function. You can extend your result to any distributional boundary $g$, you don't need $g\in C_c$. –  Vobo Jul 6 '12 at 19:43