On the Solution of the Dirichlet Problem with Rational Holomorphic ...

448 P. Ebenfelt and M. Viscardi CMFT
For 1 ≤ k ≤ n and j ≥ 2,
∫
c k,j
(z − a k ) dz = c k,j 1
j 1 − j (z − a k ) + d j−1 k,j ∈ C(z), d k,j ∈ C.
For 1 ≤ k ≤ n, ∫
c k,1
z − a k
dz = c k,1 log(z − a k ) + d k , d k ∈ C.
Thus, integrating both sides of equation (2)) yields:
n∑
(3) f(z) = p(z) + r(z) + b k (z),
where r is a rational function and b k (z) := c k,1 log(z−a k ). Suppose c 1,1 ≠ 0. Since
the a k are distinct, it follows that b 1 (z) has infinitely many branches near the
point a 1 while b 2 (z), . . . , b n (z) are analytic there. This is a contradiction, since
we are assuming f is algebraic (and hence has only finitely many branches). We
conclude that c 1,1 = 0. Similarly, we obtain c k,1 = 0 for k = 2, . . . , n. Thus, the
summation in equation (3) is equal to zero, so
as desired.
k=1
f(z) = p(z) + r(z) ∈ C(z),
We now proceed to the proof of Theorem 1.
(v) ⇒ (iv). This is trivial.
(iv) ⇒ (iii). Suppose that K(z, a 1 ) and K(z, a 2 ) are rational functions of z.
By composing f with an appropriate Möbius transformation, we may assume
without loss of generality that f(a 1 ) = 0. We shall use the following standard
formula for the Bergman kernel:
(4) K(z, w) =
Taking w = a 1 , we have
f ′ (z)f ′ (w)
π(1 − f(z)f(w)) 2 .
K(z, a 1 ) = f ′ (a 1 )
π
f ′ (z).
Since K(z, a 1 ) is a rational function of z, this implies that f ′ (z) is rational.
Now take w = a 2 in equation (4) to get
i.e.
K(z, a 2 ) =
f ′ (z)f ′ (a 2 )
π(1 − f(z)f(a 2 )) 2 ,
π(1 − f(z)f(a 2 )) 2 K(z, a 2 ) − f ′ (z)f ′ (a 2 ) = 0.