2
$\begingroup$

I came into two definitions of harmonic measure on a Riemann surface. The first is defined on p.180 of Riemann surfaces, 2nd by Kra and Farkas, which read as follows.

Theorem. Let $M$ be a hyperbolic Riemann surface and $K$ a compact subset with $\partial(M-K)$ regular and $M-K$ connected. Then there exists a function $\omega \in C(\overline{M-K})$ such that

  1. $\omega$ is harmonic on $M-K$.
  2. $\omega=1$ on $\partial (M-K)$.
  3. $0 < \omega < 1$ on $M-K$.

Definition 1The smallest $\omega$ above, is called the harmonic measure of $K$.

The second definition of harmonic measure is taken from p.301, Functions of one complex variable II by Conway.

Definition 2 Let $G$ be a hyperbolic open set and $a \in G$. The unique probability measure $\omega_a$ supported on $\partial_\infty G$ and satisfying \begin{equation} \hat{u}(a)=\int_{\partial_\infty G} u d\omega_a, \forall u \in C_{\mathbb{R}}(\partial_\infty G). \end{equation} is called the harmonic measure for $G$ at $a$. For each continuous function $u \in C_{R}(\partial_\infty G)$, $\hat{u}$ means the Perron solution with respect to boundary value $u$, and $\partial_\infty G$ is the boundary of $G$ on the Riemann sphere.

Obviously, each hyperbolic open set can be viewed as a hyperbolic Riemann surface. My question is , what's the relationship between these two definitions? I think they are not equivalent, since the first definition is intrinsic, but the second is not.

$\endgroup$

1 Answer 1

1
$\begingroup$

For the case of a plane domain, the first definition is a special case of the second. Assuming that $M$ is a plane domain, take $G=M\backslash K$ in the second definition. Then, if $\omega$ is the harmonic measure from the second definition, then $u$ is the first definition is $$u=\int_{\partial{G}\cap K} d\omega.$$ In other words, $u$ is the harmonic function in $G\backslash K$ whose boundary values are $1$ on $\partial G\cap K$ and $0$ on the rest of the boundary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.