I want to show that if $u(x,y)$, a twice differentiable function on the upper half plane, $\{(x,y) \in \mathbb{R}^2\mid y \geq 0\},$ obeys the following conditions,
- is harmonic, i.e. $(\partial_x^2 + \partial_y^2)u=0$ everywhere on the upper half plane
- $u(x,0)\geq 0$ on the $x$-axis
- $\frac{\partial u}{\partial y}\bigg|_{y=0} \geq 0$ on the $x$-axis
then it will also obey $u(x,y) \geq 0$ everywhere on the upper half plane.
It feels intuitively obvious, with my loose intuitive grasp of the shape of harmonic functions on the plane: if one starts on the $x$-axis entirely positive with $u$ increasing everywhere as one steps into the upper plane, how could it possibly ever drop below the lowest value on the $x$-axis? Wouldn't that sort of depth have to come from somewhere? But beyond this intuition I've really made no progress. I've attempted some creative contours to use Stokes' and the Divergence theorem, but have come up with nothing. I don't know that much about partial differential equations – only as much as they teach in a physics undergraduate course – so I could be missing something obvious. Can it be proven? Is it true?
Best Answer
If you know that $f(x,y)$ is harmonic and non-negative in the open half-plane $y > 0$, then there is a positive Lebesgue measure $\mu$ on $\mathbb{R}$ and a non-negative real constant $A$ such that $$ f(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-x')^2+y^2}d\mu(x')+Ay, $$ where $\mu$ satisfies $$ \int_{-\infty}^{\infty}\frac{d\mu(t)}{1+t^2} < \infty. $$ The components of this representation are unique. Conversely, any such representation gives a non-negative harmonic function in the open upper half-plane. This is a variant of a theorem of Gustav Herglotz from 1911.
The only significant difference between this case and yours is that you have added differentiability in the normal direction on the real axis. You don't need to impose differentiability in the normal direction in order to end up with a non-negative harmonic function. Differentiability in the normal direction would seem to require that $\mu(-\infty,t]$ is differentiable in $t$ on $\mathbb{R}$, and vice-versa, which would then reduce the integral to a classic Poisson integral of a non-negative continuous function $\rho(t)$ for which $\int_{-\infty}^{\infty}\frac{\rho(t)}{1+t^2}dt < \infty$, plus $Ay$ where $A$ is a non-negative constant.