It seems classically known (and mentioned in several papers without reference) that there exist bounded non-constant harmonic functions on the hyperbolic space $\mathbb{H}^n, n \geq 2$. I am interested in seeing how one can prove this fact. In particular, I can imagine that there could be roundabout proofs (Poisson boundary etc.) or direct constructive proofs (directly solving a pde). In that case, I would be very interested in learning about all of them. Any hints/references would be highly appreciated.
$\begingroup$
$\endgroup$
2
-
4$\begingroup$ Realize $\mathbb{H}^n$ as the unit ball endowed with the hyperbolic metric. A function on the unit ball is harmonic with respect to the hyperbolic metric iff it is harmonic with respect to the Euclidean metric. So you can get harmonic functions on $\mathbb{H}^n$ by integrating a function on the boundary of the unit ball against the Poisson kernel. $\endgroup$– NealCommented Feb 9, 2022 at 17:02
-
1$\begingroup$ The most direct proof is to use the conformal unit ball model and check that a coordinate function is harmonic (if you know the definition of harmonicity, this will be immediate). $\endgroup$– Moishe KohanCommented Feb 9, 2022 at 18:21
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The simplest way is to refer to the solvability of the Dirichlet problem on the hyperbolic space with the boundary data on the sphere at infinity. If you want something more direct, then you can use the following recipe. Begin with the upper half-space model, and notice that the function $y^{n-1}$ is harmonic ($y$ is the "vertical" coordinate; the other $n-1$ coordinates are "horisontal"). Geometrically this function is just an exponent of the Busemann function of the point at infinity. Therefore, in the same way any boundary point produces a harmonic function. These functions are unbounded, but if you integrate them with respect to a sufficiently spread out measure (e.g., the uniform distribution on a hemisphere), then the result will be a non-constant bounded harmonic function (if you take the uniform distribution on the whole sphere, the resulting function will be constant).
A couple of remarks:
There is nothing ``roundabout'' concerning the Poisson boundary as its purpose is precisely to describe the space of bounded harmonic functions.
Both comments above only work in dimension 2 as in higher dimensions the Laplacian is not conformal (a harmonic function need not remain harmonic after a conformal change of metric).
