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

Computational Methods and Function Theory
Volume 5 (2005), No. 2, 445–457
On the Solution of the Dirichlet Problem
with Rational Holomorphic Boundary Data
Peter Ebenfelt and Michael Viscardi
(Communicated by Dmitry Khavinson)
Abstract. We study the Dirichlet problem for the Laplace operator in a simply
connected bounded domain Ω in R 2 ∼ = C with boundary data that are
rational functions of one complex variable. Our main result is a characterization
of those domains Ω for which all solutions are rational in terms of the
Riemann mapping to the unit disk and in terms of the Bergman kernel of the
domain.
Keywords. Dirichlet problem, Bergman kernel, Riemann mapping.
2000 MSC. 30E25, 35J25.
1. Introduction
Let Ω be a simply connected bounded domain in R 2 ∼ = C. Then it is well known
that the Dirichlet problem
⎧
⎪⎨ ∆u := ∂2 u
(1)
∂x + ∂2 u
2 ∂y = 0 in Ω
2
⎪⎩
u(z, z) = h(z, z)
on ∂Ω,
where h(z, ¯z) is a continuous function in a neighborhood of the boundary ∂Ω,
is uniquely solvable under quite general conditions (see e.g. [1, p. 251]). In the
paper [11], the corresponding Dirichlet problem was studied in the unit ball B n
in R n , n ≥ 2, and the following question was asked: Assuming that the boundary
data is rational without poles on the unit sphere S n−1 , is the resulting solution of
the Dirichlet problem also rational? In the case n = 2, it has been shown (see e.g.
[10, 11]) that the answer is yes; that is, when Ω is the unit disk and the boundary
data h(z, ¯z) is rational (as a function of z and ¯z), the resulting solution u(z, ¯z)
must also be rational (as a function of z and ¯z). In [11] it was also shown that the
answer is no in higher dimensions, namely, for all odd n ≥ 3 and all even n with
Received October 28, 2005.
The first author is supported in part by DMS-0401215.
ISSN 1617-9447/$ 2.50 c○ 2005 Heldermann Verlag

446 P. Ebenfelt and M. Viscardi CMFT
4 ≤ n ≤ 270 (although, presumably, this holds also for all even n ≥ 4). In other
papers (e.g. [2, 3, 8]), the analogous question with polynomials replacing rational
functions was investigated on annuli, ellipsoids, and other quadratic surfaces. We
mention in particular the paper [13] in which it was shown that the solutions to
the Dirichlet problem (in any number of dimensions n ≥ 2) with polynomial data
are always polynomial if and only if Ω is an ellipsoid. This resolves a conjecture
by Khavinson and Shapiro [12].
It is natural to extend these studies by considering the question in more general
domains. In this paper, we let Ω be any simply connected bounded domain
in C and take the boundary data h(z, ¯z) to be a rational function of z without
poles on ∂Ω. Note, in particular, that the boundary data is holomorphic in a
neighborhood of ∂Ω. The case where h(z, ¯z) is allowed to be a rational function
of both z and ¯z is quite different, and is the subject of the forthcoming paper [7].
We shall be concerned with characterizing those Ω for which the solution to
the Dirichlet problem (1) is rational whenever the boundary data h(z, ¯z) is a
rational function of z with no poles on ∂Ω. Clearly, if this happens, then the
boundary of Ω is contained in a real-analytic (indeed, real-algebraic) variety. We
shall here further assume that ∂Ω is smooth and, hence, that the boundary is a
real-analytic (non-singular) curve. Observe that when ∂Ω is a real-analytic curve
and h(z, ¯z) is real-analytic near ∂Ω, then there is a unique holomorphic function
R(z) near ∂Ω such that h(z, ¯z) = R(z) on ∂Ω. Hence, another way of formulating
the main hypothesis in this paper, under the a priori assumption that ∂Ω is a
real-analytic curve and the boundary data is real-analytic, is that the unique
holomorphic extension R(z) of h(z, ¯z) is a rational function. Since any simply
connected bounded domain Ω is conformally equivalent to the unit disk D, we are
naturally led to results involving the Riemann maps of Ω onto the unit disk D.
Furthermore, in [5] it is shown that the Bergman kernel K(z, w) associated with Ω
is rational (as a function of z and ¯w) if and only if any Riemann map is rational.
Thus, we also obtain connections between the Dirichlet problem and the Bergman
kernel, all of which are contained in our main theorem, Theorem 1.
2. Main results
Our main results are summarized in the following theorem.
Theorem 1. Let Ω be a simply connected bounded domain in C with smooth
real-analytic boundary, and take distinct points a, a 1 , a 2 ∈ Ω. Let K(z, w) denote
the Bergman kernel associated with Ω. Then the following are equivalent.
(i) The solution to the Dirichlet problem (1) is rational for
h(z, ¯z) = 1
z − a .

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 447
(ii) The solution to the Dirichlet problem (1) is rational for h(z, ¯z) = R(z),
where R(z) is any rational function without poles on ∂Ω.
(iii) The Riemann map f : Ω → D is rational.
(iv) K(z, a 1 ) and K(z, a 2 ) are rational functions of z.
(v) K(z, w) is a rational function of z and ¯w.
We mention here a result by S. Bell [5] stating that the Bergman kernel on a
multiply connected domain is rational if and only if the domain is simply connected
and a Riemann map f : Ω → D is rational. This implies the equivalence
(iii) ⇐⇒ (v) above. However, the proof we give is independent of Bell’s result.
The proof proceeds as follows: We first prove a preliminary lemma, then prove
the chain of implications (v) ⇒ (iv) ⇒ (iii) ⇒ (ii) ⇒ (i) ⇒ (v). For clarity, we
have written out some of these implications as theorems.
We conclude this section with a few remarks concerning notation. Throughout
the proof of Theorem 1 we shall use the notation
g ∗ (z) := g(z),
and shall use O(Ω) to denote the space of analytic functions in Ω. (Note that g
is holomorphic in some domain Ω if and only if g ∗ is holomorphic in the domain
Ω ∗ := {z : ¯z ∈ Ω}.) We shall use the notation P 1 for the Riemann sphere (a.k.a.
the extended complex plane). We also use the notation C(z) for the field of
rational functions in z.
3. Proof of Theorem 1
We start by proving the following simple lemma.
Lemma 1. Suppose f is an algebraic function whose derivative is rational.
Then f itself is rational.
Proof. Let a 1 , a 2 , . . . , a n be the distinct poles of f ′ . Expand f ′ in terms of its
singular parts:
n∑
f ′ (z) = p(z) + s k (z)
(2)
= p(z) +
= p(z) +
k=1
n∑
p k
∑
k=1 j=1
p n∑ ∑ k
k=1 j=2
c k,j
(z − a k ) j
c k,j
n∑
(z − a k ) + j
k=1
c k,1
z − a k
,
where the p k and c k,j are fixed constants, and p(z) is a polynomial.

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.

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 449
Since K(z, a 2 ) and f ′ (z) are rational, this implies that f(z) is algebraic. Thus,
by Lemma 1, f is rational, as desired.
For ease of reading, we shall write the implication (iii) ⇒ (ii) as a theorem.
Theorem 2. Let Ω be a domain as in Theorem 1, and suppose that a Riemann
map f : Ω → D is rational. Then the solution to the Dirichlet problem
{
∆u = 0 in Ω
u(z, z) = R(z)
on ∂Ω
is rational for every rational function R(z) without poles on ∂Ω.
Proof. Fix a rational function R(z), and let a 1 , a 2 , . . . , a n be the poles of R in Ω.
Expanding R in terms of its singular parts, we have
(5) R(z) =
n∑
s k (z) + ˜R(z) =
k=1
n∑
p k
∑
k=1 j=1
c k,j
(z − a k ) j + ˜R(z),
where the p k and c k,j are fixed constants, and the function ˜R(z) is holomorphic
in Ω and rational. We now propose the following lemma.
Lemma 2. For any fixed k, 1 ≤ k ≤ n, there exist constants b k,j
rational function Q k (z) ∈ O(Ω) such that
(6)
p k
∑
j=1
p
b k,j
[f(z) − f(a k )] = ∑ k
c k,j
j (z − a k ) + Q k(z).
j
j=1
∈ C and a
Proof of Lemma 2. Expanding f(z) as a power series about the point a k yields
i.e.
Thus, we have
(7)
f(z) = f(a k ) + f ′ (a k )(z − a k ) + O(z − a k ) 2 ,
f(z) − f(a k ) = (z − a k )[f ′ (a k ) + O(z − a k )].
b k,j
[f(z) − f(a k )] = b k,j
1
j (z − a k ) j [f ′ (a k ) + O(z − a k )] j
b k,j
1
=
f ′ (a k ) j (z − a k ) j [1 + O(z − a k )] . j
Recall that f is one-to-one in Ω, so that f ′ (a k ) ≠ 0 (since a k ∈ Ω). Hence,
it is permissible to have this term in the denominator of the right-hand side of

450 P. Ebenfelt and M. Viscardi CMFT
equation (7). We now Taylor expand the second term in the right-hand side of
equation (7) about the point a k to obtain
(8)
b k,j
[f(z) − f(a k )] j
b k,j
=
f ′ (a k ) j (z − a k ) [1 + e 1,j(z − a k ) + e 2,j (z − a k ) 2 + · · · ]
j
= b [
k,j 1
f ′ (a k ) j (z − a k ) + e 1,j
j (z − a k ) + · · · + e ]
j−1,j
+ ψ j−1 k (z) ,
z − a k
where the e n,j are constants in C and ψ k is rational (since we are assuming that f
is rational).
Now, b k,j /[f(z) − f(a k )] j clearly has a pole at a k , as well as at all other points z
such that f(z) = f(a k ). However, since a k ∈ Ω and f is one-to-one inside Ω, a k
is the only pole of the b k,j /[f(z) − f(a k )] j in Ω. Thus, a k is the only pole in Ω
of the right-hand side of equation (8). But note that, by construction, ψ k does
not have a pole at a k ; hence, ψ k only has poles outside of Ω. So ψ k is a rational
function and holomorphic in Ω.
We now sum both sides of equation (8) as j goes from 1 to p k to get
(9)
Define
p k
∑
j=1
b k,j
[f(z) − f(a k )] j
=
p
∑ k
[
b k,j
f
j=1
′ (a k ) j
p
∑ k
+ψ k (z)
j=1
1
(z − a k ) + e 1,j
j (z − a k ) + · · · + e ]
j−1,j
j−1 z − a k
b k,j
f ′ (a k ) j .
Q k (z) := ψ k (z)
p k
∑
j=1
b k,j
f ′ (a k ) j .
Since ψ k ∈ O(Ω) and is rational, we conclude that Q k ∈ O(Ω) and that Q k is
rational. Thus, Q k satisfies the conditions of Lemma 2.
For ease of notation, let us also define the following:
Note that
e 0,j := 1,
(10) d j,j =
d l,j := e (j−l),j
f ′ (a k ) j , 1 ≤ l ≤ j, 1 ≤ j ≤ p k.
1 ≠ 0 for all j.
f ′ (a k )
j

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 451
Equation (9) can now be written as
(11)
p k
∑
j=1
p
b k,j
[f(z) − f(a k )] = ∑ k j
j=1
j∑
l=1
d l,j b k,j
(z − a k ) l + Q k(z).
Reversing the order of summation in the right-hand side of equation (11) yields
(12)
p k
∑
j=1
j=1
b k,j
[f(z) − f(a k )] j
= p k
∑
=
p k
∑
l=1 j=l
p
∑ k
l=1
j=1
d l,j b k,j
(z − a k ) l + Q k(z)
∑ pk
j=l d l,jb k,j
(z − a k ) l + Q k (z).
We now swap the indices l and j in the right-hand side of equation (12) to get
p k p
∑ b
k
∑
k,j
(13)
[f(z) − f(a k )] = ∑ pk
l=j d j,lb k,l
+ Q j (z − a k ) j k (z).
Thus, by equation (6) we want to show that there exist b k,j ∈ C such that
p
∑ k
∑ pk
l=j d p
j,lb k,l
∑ k
c k,j
+ Q
(z − a k ) j k (z) =
(z − a k ) + Q k(z),
j
i.e.
i.e.
j=1
p k
∑
l=j
p k
∑
j=1
∑ pk
l=j d j,lb k,l
(z − a k ) j =
j=1
p k
∑
j=1
c k,j
(z − a k ) j ,
d j,l b k,l = c k,j for all j, 1 ≤ j ≤ p k .
Writing this as a system of equations, we get
d 1,1 b k,1 + d 1,2 b k,2 + · · · + d 1,pk b k,pk = c k,1
d 2,2 b k,2 + · · · + d 2,pk b k,pk = c k,2
.
d pk ,p k
b k,pk = c k,pk .
The resulting coefficient matrix
⎛
⎞
d 1,1 d 1,2 . . . d 1,pk
0 d
A = ⎜ 2,2 . . . d 2,pk
⎝ .
⎟
. . .. .
⎠
0 0 . . . d pk ,p k
is triangular, so
det A = d 1,1 d 2,2 . . . d pk ,p k ≠ 0

452 P. Ebenfelt and M. Viscardi CMFT
by equation (10). By Cramer’s Rule, we may solve uniquely for the b k,j so that
they satisfy the above system of equations, and therefore satisfy equation (6).
Thus, we have found the desired b k,j and Q k (z), proving Lemma 2.
We shall now use the results of Lemma 2 to construct the desired function u.
Sum both sides of equation (6) as k goes from 1 to n to get
p n∑ k p
∑ b k,j
n∑
[f(z) − f(a
k=1 j=1
k )] = ∑ k
c k,j
n∑
j (z − a
k=1 j=1 k ) + Q j k (z)
k=1
(14)
= R(z) − ˜R(z)
n∑
+ Q k (z),
where the second equality follows from equation (5). Define
S(z) := ˜R(z)
n∑
− Q k (z),
ˆT (w) :=
n∑
k=1 j=1
k=1
p
∑ k
k=1
b k,j
[w − f(a k )] j .
By Lemma 2 and the paragraph preceding it, S(z) ∈ O(Ω) and is rational. Also,
since f is a bijective rational map from Ω to D and a k ∈ Ω for all k, we know
that ˆT (w) is rational and ˆT (w) ∈ O(P 1 \D).
Substituting these new functions into equation (14), we obtain
i.e.
ˆT (f(z)) = R(z) − S(z),
(15) R(z) = S(z) + ˆT (f(z)).
We now make one final substitution
( )
T (z) := ˆT 1 ∗ .
f(z)
Note that T is rational since ˆT and f are rational. Furthermore, since f : Ω → D
and I : D → P 1 \D (where I denotes the inversion in the plane) are both bijective,
T (z) ∈ O(Ω) since ˆT (z) ∈ O(P 1 \D).
Finally, for z ∈ ∂Ω, we know that f(z) ∈ ∂D = {w : |w| 2 = ww = 1}, which
gives
( ) ( )
T (z) = ˆT 1 ∗ =
f(z)
ˆT 1
= ˆT (f(z)),
f(z)
so that equation (15) becomes
(16) R(z) = S(z) + T (z) for z ∈ ∂Ω.

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 453
We now define
(17) u(z, z) := S(z) + T (z).
Since S, T ∈ O(Ω), u is harmonic in Ω, and for z ∈ ∂Ω, equation (16) yields
u(z, z) = R(z).
Hence, u is the solution to the Dirichlet problem posed in Theorem 2. Since S
and T are rational, equation (17) implies that u is rational. This completes the
proof of Theorem 2.
(ii) ⇒ (i). This is trivial.
We shall also write the implication (i) ⇒ (v) as a theorem.
Theorem 3. Let Ω be as in Theorem 1, and let a ∈ Ω. Suppose the solution u
to the Dirichlet problem
⎧
⎨∆u = 0
in Ω
⎩u(z, z) = 1 on ∂Ω
z − a
is rational. Then the Bergman kernel K(z, w) associated with Ω is also rational.
Proof. Since u is harmonic in Ω, we may write u(z, z) = s(z) + t(z), where
s, t ∈ O(Ω). Furthermore, since u is rational, s and t are also rational. Define
R(z) := 1 − s(z) + t(z).
z − a
Note that R is rational. Since u(z, z) = s(z) + t(z) = 1/(z − a) on ∂Ω, we have
R(z) = 1 − s(z) + t(z) = t(z) + t(z) ∈ R
z − a
on ∂Ω.
Thus, we have constructed a rational function R such that R(∂Ω) ⊂ R.
Now, since the boundary of Ω is a smooth real-analytic closed curve, it is well
known that u can be extended as a harmonic function to an open neighborhood
of Ω (this follows from a version of the Schwarz Reflection Principle; see
e.g. [9, 14]). Thus, s and t can be extended as holomorphic functions past the
boundary of Ω; in particular, they will be continuous on ∂Ω, so R will also be
continuous on ∂Ω. Since Ω is bounded and simply connected, ∂Ω is a compact,
connected subset of C. So by the above, R(∂Ω) is a compact, connected subset
of R, i.e. a closed interval L = [z 1 , z 2 ] ⊂ R.
Lemma 3. R maps Ω bijectively to P 1 \L ′ , where L ′ is some closed subinterval
of L, i.e. L ′ = [z 3 , z 4 ] ⊆ [z 1 , z 2 ] with z 3 < z 4 .

454 P. Ebenfelt and M. Viscardi CMFT
Proof of Lemma 3. Consider the function
F (w) := 1 ∫
R ′ (z)
2πi R(z) − w dz,
∂Ω
w ∈ C\L.
Note that F is analytic on C\L since R(∂Ω) = L. Let Z S (g) and P S (g) denote the
number of zeros and poles, respectively, of a function g in a domain S (counting
multiplicity). The Argument Principle implies
(18) F (w) = Z Ω [R(z) − w] − P Ω [R(z) − w] for all w ∈ C\L.
Since the right-hand side of equation (18) is always an integer, F must be constant
(since F (w) is analytic on C\L and C\L is connected). Also, note that
lim w→∞ F (w) = 0, so in fact F ≡ 0. Since R has exactly one (simple) pole in Ω
at the point a by definition, equation (18) now yields
Z Ω [R(z) − w] = P Ω [R(z) − w] = 1
for all w ∈ C\L.
So the equation R(z) = w has exactly one solution in Ω for any w ∈ C\L. Since
R −1 (L) is at most a proper real-analytic variety in Ω, we conclude that R is a
one-to-one mapping onto R(Ω), and we have
(19) P 1 \L ⊆ R(Ω) ⊆ P 1 .
We now choose a suitable Möbius transformation ϕ such that ϕ(R(Ω)) does not
contain the point ∞, i.e.
(20) P 1 \M ⊆ ϕ(R(Ω)) ⊆ C,
where M = ϕ(L). Then, since Ω is simply connected and ϕ(R(z)) is one-toone
and analytic in Ω, ϕ(R(Ω)) is a simply connected subset of C. It is well
known that this is equivalent to saying that M ′ := P 1 \ϕ(R(Ω)) is connected
(see [9, p. 202]). Also, note that ϕ(R(Ω)) is an open subset of P 1 by the open
mapping theorem (see [9, p. 99]), so M ′ is a closed subset of P 1 . By equation (20),
we therefore have ϕ(R(Ω)) = P 1 \M ′ for some closed, connected, possibly empty
subset M ′ ⊆ M. Thus, R(Ω) = P 1 \L ′ for some closed, possibly empty subinterval
L ′ ⊆ L.
If L ′ = ∅, then R(Ω) = P 1 , which is impossible since P 1 is compact while Ω is
not.
If L ′ = {z 0 } for some z 0 , then R(Ω) = P 1 \{z 0 }, which is also impossible since this
would imply that Ω is conformally equivalent to C, violating Liouville’s Theorem.
Hence, L ′ = [z 3 , z 4 ] with z 3 < z 4 , as desired.
Lemma 4. A Riemann map f : Ω → D is algebraic.
Proof of Lemma 4. First, note that by composing R with an appropriate Möbius
transformation, we may assume, without loss of generality, that L ′ = [−1, 1].
It is well known that the function φ(w) = 1/2(w + 1/w) maps the unit disk D

5 (2005), No. 2 The Dirichlet Problem with Rational Holomorphic Data 455
conformally and bijectively to the Riemann sphere P 1 minus the segment [−1, 1],
so by Lemma 3, the function
(21) f(z) := (φ −1 ◦ R)(z)
is a biholomorphism from Ω to D. Since φ and R are rational, we conclude that f
is algebraic, as desired.
Lemma 5. The function f ′ (z) is rational.
Proof of Lemma 5. We first establish a relation between the solution to the
Dirichlet problem in Theorem 3 and the Bergman kernel K of Ω. This result
has already been proven before (see [4, p. 97]), but we include a proof here for
completeness.
Let h be any function in the Bergman space H 2 (Ω) = L 2 (Ω)∩O(Ω). By Cauchy’s
Integral Formula,
h(a) = 1 ∫
h(w)
2πi ∂Ω w − a dw = 1 ∫
u(w, w)h(w) dw,
2πi ∂Ω
since u(w, w) = 1/(w − a) on ∂Ω. Applying Stokes’ Theorem, and remembering
that h is holomorphic in Ω, we have
h(a) = − 1 ∫ [
2i ∂
]
∂
u(w, w) · h(w) + u(w, w) · 2i
2πi Ω ∂w ∂ w h(w) dA
(22)
= − 1 ∫
∂
u(w, w) · h(w) dA.
π Ω ∂ w
But from the definition of the Bergman kernel, we also have
∫
(23) h(a) = K(a, w) · h(w) dA.
Ω
Combining equations (22) and (23), we obtain
∫
K(a, w) · h(w) dA = − 1 ∫
∂
u(w, w) · h(w) dA,
Ω π Ω ∂ w
i.e.
∫ [
(24) K(a, w) + 1 ]
∂
π ∂ w u(w, w) h(w) dA = 0 for all h ∈ H 2 (Ω).
Ω
Note that the function K(a, w) + (1/π) · (∂/∂ w)u(w, w) is anti-holomorphic,
being the sum of two anti-holomorphic functions. Thus, we may write
B(w) := K(a, w) + 1 ∂
u(w, w),
π ∂ w
where B is holomorphic in Ω. Note that B is also in L 2 (Ω); thus, we have
B(w) ∈ H 2 (Ω). Equation (24) now becomes
∫
B(w) · h(w) dA = 0 for all h ∈ H 2 (Ω).
Ω

456 P. Ebenfelt and M. Viscardi CMFT
This is equivalent to saying that B is orthogonal to every function h ∈ H 2 (Ω).
But since H 2 (Ω) is a Hilbert space, it follows that B ≡ 0, i.e.
(25) K(a, w) = − 1 ∂
u(w, w).
π ∂ w
This is the desired formula relating K and u(w, w).
Since u(w, w) is rational, (∂/∂ w)u(w, w) is also rational. Thus, equation (25)
implies that the Bergman kernel K(a, w) is a rational function of w.
Now, by composing the Riemann map f found in Lemma 4 with an appropriate
Möbius transformation, we may assume without loss of generality that f(a) = 0.
Thus, substituting z = a into the formula for the Bergman kernel (equation (4)),
we have
(26) K(a, w) = f ′ (a)
· f
π
′ (w).
Since K(a, w) is a rational function of w, we conclude that f ′ (w) is also rational,
as desired.
Combining Lemmas 1, 4, and 5, we conclude that f is rational, i.e. statement (iii)
in our Main Theorem. Now, the identity (4) clearly implies that that the
Bergman kernel K(z, w) is a rational function of z and ¯w. Thus, the conclusion
of Theorem 3 immediately follows, and the proof is complete.
This completes the proof of Theorem 1.
Acknowledgement. The authors would like to thank S. Bell, D. Khavinson,
and H. S. Shapiro for helpful comments and suggestions.
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Peter Ebenfelt
E-mail: pebenfel@math.ucsd.edu
Address: Department of Mathematics, University of California, San Diego, La Jolla, CA
92093-0112, U.S.A.
Michael Viscardi
E-mail: mviscardi@earthlink.net
Address: Josan Academy, 5524 Rabbit Ridge Rd., San Diego, CA 92130, U.S.A.