Uniqueness theorems for weighted harmonic functions in the upper half-plane

We consider a class of weighted harmonic functions in the open upper half-plane known as $\alpha$-harmonic functions. Of particular interest is the uniqueness problem for such functions subject to a vanishing Dirichlet boundary value on the real line and an appropriate vanishing condition at infinity. We find that the non-classical case ($\alpha\neq0$) allows for a considerably more relaxed vanishing condition at infinity compared to the classical case ($\alpha=0$) of usual harmonic functions in the upper half-plane. The reason behind this dichotomy is different geometry of zero sets of certain polynomials naturally derived from the classical binomial series. Our findings shed new light on the theory of harmonic functions, for which we provide uniqueness results under vanishing conditions at infinity along a) geodesics, and b) rays emanating from the origin. The geodesic uniqueness results require vanishing on two distinct geodesics which is best possible. The ray uniqueness results involves an arithmetic condition which we analyze by introducing the concept of an admissible function of angles. We show that the arithmetic condition is to the point and that the set of admissible functions of angles is minimal with respect to a natural partial order.


Introduction
Let H be the open upper half-plane in the complex plane C and consider the weighted Laplace differential operator (1.1) where ∂ and ∂ are the usual complex partial derivatives and α > −1.Here Im z is the imaginary part of the complex number z ∈ C and the function w H;α (z) = (Im z) α for z ∈ H has an interpretation of a standard weight function for H.The study of differential operators of the form (1.1) is suggested by a classical paper of Paul Garabedian [11].
We refer to the differential operator ∆ H;α in (1.1) as the α-Laplacian for H.An α-harmonic function u in H is a twice continuously differentiable function u in H (in symbols: u ∈ C 2 (H)) such that ∆ H;α u = 0 in H.
Notice that ∆ H;0 = ∂ ∂ is the usual Laplacian and that a 0-harmonic function in H is a harmonic function in H in the usual sense.The restriction α > −1 on the weight parameter ensures good supply of α-harmonic functions with well-behaved boundary values.
In this paper we address the uniqueness problem for α-harmonic functions in H, that is, we wish to characterize the identically zero function u ≡ 0 within the class of α-harmonic functions in H.A natural condition is that of a vanishing boundary value (1.2) lim H∋z→x u(z) = 0, x ∈ R, on the real line R.A condition of this type is often referred to as a vanishing Dirichlet boundary value.There are plenty of non-trivial α-harmonic functions in H satisfying (1.2).A simple such example is the function which is α-harmonic in H and satisfies (1.2).In order to obtain satisfactory results on the uniqueness problem above it is thus natural to complement (1.2) with some condition(s) taking into account the behavior of the α-harmonic function u at infinity.Here we think of the point at infinity ∞ as a boundary point of H in the extended complex plane C ∞ = C ∪ {∞}.In the case of usual harmonic functions in H (α = 0) this problem setup is classical.We mention here a recent contribution by Carlsson and Wittsten [9] concerned with a uniqueness result in this flavor for α-harmonic functions in H with α > −1.This result of Carlsson and Wittsten has served as a guidance for the present investigations.We say that a function u in H is of temperate growth at infinity if it satisfies an estimate of the form |u(z)| ≤ C(|z| 2 / Im(z)) N   for z ∈ H with |z| > R, where C, R and N are positive constants.We refer to the parameter N as an order of growth at infinity for the function u.
A first main result concerns α-harmonic functions u in H that are of temperate growth at infinity.We prove that such a function u satisfies (1.2) if and only if it has the form (1.3) u(z) = (α + 1) j j! z k−j zj for k = 0, 1, . . .(see Corollary 5.4).We point out that the polynomials p k,α are naturally derived from the classical binomial series.We also establish (1.3) under a weaker distributional version of (1.2) (see Theorem 5.7).We denote by V α the set of all functions u of the form (1.3) for some n ∈ N and c 0 , . . ., c n ∈ C. The set V α is naturally filtered in the sense that , where V α,n is the set of all functions of the form (1.3) for some c 0 , . . ., c n ∈ C. The set V α,n has a natural structure of a complex vector space of finite dimension n + 1.The space V α,n admits a natural description within the class V α using order of growth at infinity (see Lemma 5.6 and Proposition 5.9).The finer study of behaviors at infinity of functions in the class V α depends on the parameter α > −1 and divides naturally into cases whether α = 0 or α = 0.The reason behind this dichotomy is a different geometry of zero sets for the polynomials p k,α .
We consider next the problem of characterizing the null function in the class V α using a vanishing condition at infinity.For parameters α > −1 with α = 0, we prove that if u ∈ V α is such that (1.4) lim j→∞ u(z j ) (Im(z j )) α+1 = 0 for some sequence {z j } in H with z j → ∞ in C ∞ as j → ∞, then u(z) = 0 for all z ∈ H (see Theorem 6.4).We emphasize the big freedom allowed in the choice of sequence {z j } in (1.4) above.This analysis leads to the following highly flexible uniqueness result for α-harmonic functions in the case α = 0 as well as a distributional version thereof (see Theorem 6.5).
Theorem 1.1.Let α > −1 and α = 0. Let u be an α-harmonic function in H which is of temperate growth at infinity.
(2) Assume that there exists a sequence {z j } in H with z j → ∞ in C ∞ as j → ∞ such that (1.4) holds.Then u(z) = 0 for all z ∈ H.
We now turn our attention to the case α = 0 of usual harmonic functions in H.A function u belongs to the class V 0 if and only if it has the form for some n ∈ Z + = {1, 2, 3, . . .} and c 1 , . . ., c n ∈ C (see Proposition 5.2).Observe that the harmonic polynomial vanishes on a union of k lines passing through the origin.Notice also that the zero set of u k intersects the unit circle at the 2k-th roots of unity.These examples make evident that the flexible uniqueness results for α-harmonic functions in H with α = 0 are no longer true when α = 0 (compare with Theorem 1.1 above).
In order to obtain satisfactory uniqueness results for usual harmonic functions in H we shall restrict condition (1.4) to suitable classes of curves.We consider two such classes of curves, namely, geodesics in H and rays in H emanating from the origin.
By a geodesic in H we understand a ray in H which is parallel to the imaginary axis.We prove that if u ∈ V 0 is such that Theorem 7.1).This analysis leads to corresponding uniqueness results for usual harmonic functions in H (see Theorems 7.2 and 7.3).We emphasize that those geodesic uniqueness results require vanishing on two (2) distinct geodesics which is best possible.Together with the success results for α = 0, this marks a significant advancement from an earlier uniqueness result of Carlsson and Wittsten [9, Corollary 1.9] for α-harmonic functions which required vanishing on an interval of geodesics.By a ray in H emanating from a point a ∈ R we understand a set of the form {a + te iθ : t > 0}, where 0 < θ < π.We shall restrict our attention to rays emanating from the origin (a = 0).In view of translation invariance of the class of harmonic functions this restriction is minor.In order to discuss vanishing of functions along rays we introduce a notion of admissible function of angles which is a function element (E, η) with E ⊂ (0, π) and η : E → Z + having the property that for every k ∈ Z + there exists θ ∈ E such that sin(kθ) = 0 and k ≥ η(θ) (see Definition 8.1).We prove that if u ∈ V 0 and there exists an admissible function of angles (E, η) such that lim t→+∞ u(te iθ )/t η(θ) = 0 for every θ ∈ E, then u(z) = 0 for all z ∈ H (see Theorem 8.2).This analysis leads to corresponding uniqueness results for usual harmonic functions in H (see Theorems 8.3 and 8.4).
The notion of admissible function of angles involves an arithmetic element.In order to illuminate this fact we mention the following result.
Theorem 1.2.Let u be a harmonic function in the open upper half-plane H which is of temperate growth at infinity.
(2) Assume that for some 0 < θ < π which is not a rational multiple of π.Then u(z) = 0 for all z ∈ H.
The examples u k above show that the arithmetic condition on θ in (1.5) is to the point.Behind Theorem 1.2 lies a construction of a one point admissible function of angles (E, η), where E = {θ} and η(θ) = 1.The arithmetic condition on θ in (1.5) ensures that this latter function element (E, η) is an admissible function of angles.More elaborate constructions of admissible functions of angles lead to corresponding uniqueness results for usual harmonic functions in H (see Corollaries 8.5,8.6,8.7 and 8.8).
In the final section we provide constructions of admissible functions of angles (see Theorems 9.1 and 9.2).The set A of admissible functions of angles is structured by a natural partial order.We show that every element in A has a lower bound which is minimal (see Theorem 9.3 and Lemma 9.4).Our constructions of admissible functions of angles yield precisely the minimal elements in A (see Theorem 9.5).
There is a substantial literature on boundary uniqueness problems for (sub-)harmonic functions with contributions of Wolf [26], Shapiro [22], Dahlberg [10], Berman and Cohn [5] and Borichev with collaborators [6,8] to name a few.As far as usual harmonic functions in H are concerned our uniqueness results supersede an earlier uniqueness result of Siegel and Talvila [23,Corollary 3.1].
The α-Laplacian (or rather its symmetric part) is also related to the Laplace-Beltrami equation in the Riemannian space defined by the metric as studied by Weinstein [24] and Huber [15], among others.For historic reasons, solutions of said Laplace-Beltrami equation are referred to as generalized axially symmetric potentials.For a discussion on this connection and more recent applications in this direction we refer to Wittsten [25].Another related area of interest is the recent study of higher order Laplacians initiated by Borichev and Hedenmalm [7].The present paper is rooted in previous investigations.In an earlier paper [20] we initiated a theory of α-harmonic functions in the open unit disc D in the complex plane is the α-Laplacian for D and α ∈ R. A main concern in this theory is the representation of an α-harmonic function in D as a Poisson integral u = P α [f ] in D with respect to the kernel We refer to the function P α as the α-harmonic Poisson kernel for D.
We prove that a Poisson integral representation u = P α [f ] with f a distribution on the unit circle T = ∂D exists if and only if u is α-harmonic in D, u has temperate growth in D and a certain spectral condition is satisfied (see Theorem 3.3).This latter spectral condition is automatically satisfied when α is not a negative integer (see Corollary 3.4).For α > −1, the distribution f ∈ D ′ (T) has a natural interpretation as a (distributional) boundary limit of the function u.On the other hand, for α ≤ −1, existence of a distributional boundary limit of an α-harmonic function u in D, forces u to be analytic in D (see Theorem 3.5).We thus establish Poisson integral representations u = P α [f ] with f ∈ D ′ (T) in situations where a distributional boundary value of u is non-existent.To overcome those difficulties we resort to a study of related hypergeometric functions, found in Section 2.
A link between the settings of the upper half-plane H and the unit disc D is provided by a certain conformal invariance property of α-harmonic functions which was recently studied by the first author [18].Let ϕ : D → H be a biholomorphic map.For a function u in H we consider the weighted pull-back of u by ϕ with respect to the parameter α.Here the power in (1.7) is defined in the usual way using a logarithm of ϕ ′ in D. We shall use the fact that the function v is α-harmonic in D if and only if the function u is α-harmonic in H.This latter fact follows easily from [18,Theorem 1.1].We refer to Geller [13] or Ahern et al. [1,2] for earlier results.
Let us return to an α-harmonic function u in H satisfying some appropriate conditions, notably (1.2) and temperate growth at infinity.We consider a weighted pull-back v of u of the form (1.7).The function v is α-harmonic in D and we propose to study this function by means of its Poisson integral representation v = P α [f ] in D. An ambiguity lies in the choice of biholomorphic map ϕ : D → H which we chose as the Möbius transformation Notice that ϕ(0) = i and ϕ(1) = ∞.
From (1.2) we have that the boundary value f for v vanishes on the set T \ {1}.Standard distribution theory then dictates that the distribution f ∈ D ′ (T) is a finite linear combination of derivatives of a Dirac mass δ 1 located at the point 1 on in D ′ (T) (see Hörmander [14,Theorem 2.3.4]).The Poisson integral P α [δ ′  1 ] is naturally interpreted as an angular derivative iAP α of the Poisson kernel P α , and by iteration we find that v = n k=0 c k (iA) k P α (see Corollary 3.7).In Section 4, those angular derivatives (iA) k P α of P α are carefully investigated using the Möbius transformation ϕ above, and in Section 5 this part of the analysis culminates in the proof of the representation formula (1.3) (see Theorem 5.1).

Series expansion of α-harmonic functions in D
In this section we revisit the series expansion of α-harmonic functions in D. Of particular concern is a characterization of temperate growth of an α-harmonic function in D in terms of polynomial growth of coefficients (see Theorem 2.3).The proof of this latter result depends on properties of related hypergeometric functions.
The hypergeometric function is the function defined by the power series expansion for parameters a, b, c ∈ C with c = 0, −1, −2, . . . .Convergence in (2.1) follows by the standard ratio test.
Recall also the classical binomial series which follows from (2.1) and (2.2).We take as our starting point a certain series expansion of α-harmonic functions in D. It is known that a function u is α-harmonic in D if and only if it has the form where F is the hypergeometric function (2.1) (see [20,Theorem 1.2]).Condition (2.4) ensures that the series expansion (2.3) is absolutely convergent in the space C ∞ (D) of indefinitely differentiable functions in D. As a consequence we have that u ∈ C ∞ (D).We refer to Klintborg and Olofsson [16] for an updated account on these matters.
The property F (a, b; c; 0) = 1 of the hypergeometric function (2.1) yields a normalization of the expansion (2.3).As a consequence we have the coefficient formulas where u is as in (2.3) (see [16,Theorem 5.3]).In particular, an αharmonic function in D is uniquely determined by its germ at the origin.
As an example of a series expansion of the form (2.3) we mention that of the α-harmonic Poisson kernel (2.5) for z ∈ D (see [16,Theorem 6.3]).
We shall make use of a classical result known as Euler's integral formula for the hypergeometric function.This result says that In particular, from (2.6) we have that for k ∈ Z + and α ∈ R. From this latter formula we see that the function F (−α, k; k + 1; •) is nonnegative on the interval [0, 1).Furthermore, the function A passage to the limit in (2.7) shows that where we have used a standard formula for the Beta function.When α ≤ −1, the quantity F (−α, k; k + 1; x) diverges to +∞ as x → 1.
We shall need some more detailed estimates of the hypergeometric functions appearing in (2.3).
Proof.From (2.7) we have that An integration by parts shows that Observe that the logarithm in the rightmost integral is negative.This yields the conclusion of the lemma.
We shall use also another result of Euler which says that (2.9) Proof.We first apply (2.9) to see that for 0 ≤ x < 1.We shall divide into cases depending on whether k Assume first that k + α + 1 ≤ 0. By (2.6) we have that In view of (2.10) this yields the conclusion of the lemma for We next assume that k + α + 1 > 0. By symmetry and (2.6) we have that By monotonicity we have that where the last two equalities follows by standard formulas for the Beta and Gamma functions.In view of (2.10) this yields the conclusion of the lemma for k + α + 1 > 0.
We say that a function u in D is of temperate growth in D if it satisfies an estimate of the form for some positive constants C and N .
Theorem 2.3.Let α ∈ R. Let u be an α-harmonic function in D and consider the expansion (2.3).Then the function u is of temperate growth in D if and only if the 3) has at most polynomial growth.
Proof.Assume that u has temperate growth in D. We first show that the coefficient c k has at most polynomial growth as k → +∞.Let k ∈ N and 0 < r < 1.From (2.3) we have that From the triangle inequality and (2.11) we have that for k ∈ N and 0 < r < 1, where C is as in (2.11).Choosing r = 1 − 1/k with k big in this latter inequality, we see that Let us first consider the case α ≤ 0. Since the function F (−α, |k|; |k| + 1; •) is increasing on [0, 1), we have from (2.12) and the triangle inequality that where C is as in (2.11).Choosing r = 1 − 1/|k| in this latter inequality we see that Let us now consider the case α ≥ 0. Since the function F (−α, |k|; |k| + 1; •) is decreasing on [0, 1), we have from (2.12) and the triangle inequality that (2.13) where we have used (2.8) and C is as in (2.11).Stirling's formula ensures that a quotient of Gamma functions Γ(x)/Γ(x + α) behaves asymptotically as 1/x α when x → +∞ ([3, Section 1.4]).Choosing r = 1 − 1/|k| in (2.13) we see that 3) has at most polynomial polynomial growth, that is, We shall show that the function u has temperate growth in D. We consider first the leftmost sum for k ∈ N. By the triangle inequality we have that for z ∈ D, where in the last equality we have used (2.2).This proves that the function f above has temperate growth in D.
We consider next the function appearing in (2.3).Assume first that α ≥ 0. Then the function F (−α, k; k + 1; •) is decreasing on [0, 1).From the triangle inequality we have that for z ∈ D, where the last two inequalities follows as in the previous paragraph.This proves that the function g above has temperate growth in 1).By the triangle inequality and (2.8) we have that for z ∈ D. Since Γ(x)/Γ(x + α) behaves asymptotically as 1/x α when x → +∞ we deduce as above that the function g has temperate growth in By the triangle inequality and Lemma 2.1 we have that for z ∈ D. We can now deduce as above that the function g has temperate growth in D if α = −1.
Assume finally that α < −1.By the triangle inequality and Lemma 2.2 we have that ).We can then once again deduce as above that the function g has temperate growth in D if α < −1.
The case division in the proof of Theorem 2.3 depends on different behaviors of the hypergeometric functions appearing in (2.3).One can notice that the quantity

Poisson integrals of distributions
A purpose of this section is to further develop a theory of Poisson integral representations of α-harmonic functions in D. Of particular interest is a characterization of Poisson integrals of distributions on T (see Theorem 3.3 and Corollary 3.4).
Along the way we introduce some notation needed later.
We denote by D ′ (T) the space of distributions on T.An integrable function f ∈ L 1 (T) on T is identified with the distribution where C ∞ (T) is the space of indefinitely differentiable test functions on T. The space D ′ (T) is topologized in the usual way using the semi-norms for k ∈ Z be the exponential monomials on T. The Fourier coefficients of a distribution f ∈ D ′ (T) are defined by A distribution on T is uniquely determined by its sequence of Fourier coefficients.
It is well-known that a sequence of complex numbers {c k } ∞ k=−∞ is of at most polynomial growth if and only if it is the sequence of Fourier coefficients for some distribution on T, that is, there exists For a suitably smooth function u in D we set where * denotes convolution, P α,r is defined in accordance with (3.1), r ≥ 0 and e iθ ∈ T. Observe that We next calculate the series expansion of the Poisson integral.
Proof.Let 0 ≤ r < 1.From (2.5) we have that for e iθ ∈ T. Passing to the convolution we have that for z = re iθ ∈ D with r ≥ 0 and e iθ ∈ T. This yields the conclusion of the proposition.
Notice that the expansion in Proposition 3.1 is a series expansion of the form (2.3).As a consequence, we have that the function P α [f ] is α-harmonic in D. We record also that as k → +∞ which follows by Stirling's formula.
Recall the notion of Fourier spectrum of a distribution on T defined by for f ∈ D ′ (T).We shall need a similar notion of spectrum of an α-harmonic function in D. Recall that an α-harmonic function in D is uniquely determined by its sequence of coefficients in (2.3).For such a function u we set where the c k 's are as in (2.3).Observe that Spec(u) = ∅ if and only if u = 0.
Let us denote by Z − the set of negative integers.In view of the series expansion (2.5) of the function P α we have that Spec We next turn our attention to a characterization of Poisson integrals of distributions.Proof.Assume first that u = P α [f ] in D for some f ∈ D ′ (T).From a well-known characterization of Fourier coefficients of distributions on T we know that the sequence of Fourier coefficients { f (k)} ∞ k=−∞ has at most polynomial growth.Proposition 3.1 supplies us with the series expansion of u.Clearly u is α-harmonic in D. From Corollary 3.2 we have that Spec(u) ⊂ Spec(P α ).By Theorem 2.3 we conclude that u has at most temperate growth in D.
Assume next that u is α-harmonic in D, u has temperate growth in D, and Spec(u) ⊂ Spec(P α ).Consider the series expansion (2.3).By Theorem 2.3 we conclude that the sequence of coefficients {c k } ∞ k=−∞ in (2. 3) has at most polynomial growth.We set a k = c k for k ∈ N, for k ∈ Z − and k ∈ Spec(P α ), and a k = 0 for k ∈ Z − and k ∈ Spec(P α ).From (3.2) we have that the sequence {a k } ∞ k=−∞ is of polynomial growth.From a well-known characterization of Fourier coefficients of distributions on T there exists f ∈ D ′ (T) such that f (k) = a k for k ∈ Z. Since Spec(u) ⊂ Spec(P α ), we have by Proposition 3.
The spectral condition in Theorem 3.3 is redundant when α is not a negative integer.
, where u r is as in (3.1).
The boundary behavior of α-harmonic functions in D is conceptually different when α ≤ −1.The next result exemplifies this fact.Theorem 3.5.Let α ≤ −1 and let u be an α-harmonic function in D. Assume that the limit f = lim r→1 u r in D ′ (T) exists, where u r is as in (3.1).Then u is analytic in D.
Proof.Consider the series expansion (2.3).We shall prove that c k = 0 for k < 0. Let k ∈ Z + .From (2.3) we have that By assumption the right hand side in (3.3) has a limit as r → 1. Recall from the discussion following formulas (2.7)-(2.8)that the quantity F (−α, k; k + 1; x) increases to +∞ as x → 1. Passing to the limit in (3.3) as r → 1 we conclude that c −k = 0.This yields the conclusion of the theorem.
We refer to Olofsson [17, Theorem 2.3] for a result analogous to Theorem 3.5 phrased in another setting of generalized harmonic functions in D. Under an assumption of pointwise boundary limit, a similar result also exists for the class of generalized axially symmetric potentials mentioned in the introduction, see Huber [15,Theorem 2].
A traditional approach to the Poisson integral representation if u has the form (2.3).In particular, the function Au is α-harmonic in D if u is.
for z ∈ D, where f ′ is the distributional derivative of f .
Proof.Recall that (f ′ )ˆ(k) = ik f (k) for k ∈ Z.By Proposition 3.1 we have that for z ∈ D, where in the last equality we have used (3.5).
We refer to the differential operator iA = i(z∂ − z ∂) as the angular derivative.Our interest in this operator arose in connection to the paper Olofsson [19].

Angular derivatives of Poisson kernels
This section is devoted to a careful analysis of angular derivatives of Poisson kernels.We shall make good use of the Möbius transformation (4.1) ϕ(z) = i 1 + z 1 − z in our calculations.Notice that ϕ maps D one-to-one onto H, ϕ(0) = i and ϕ(1) = ∞.From standard theory we have that the Möbius transformation ϕ is uniquely determined by these three properties.For the sake of easy reference we record also the formulas which are straightforward to check.
A most natural α-harmonic function in H is the (α+1)-th power of the imaginary part: We first calculate the weighted pull-back u ϕ,α from (1.7) of this function by ϕ.
Notice that Theorem 4.1 with α = 0 yields the well-known formula for the usual Poisson kernel for D.
Our next task is to calculate angular derivatives of Poisson kernels.We begin with a preparatory lemma.Lemma 4.2.Let ϕ be as in (4.1).Then iAϕ = 1 2 (ϕ 2 + 1).Proof.From formula (4.2) we have that Using some elementary algebra we now calculate that where the last equality follows by (4.1).
Let us record some properties of the angular derivative.The differential operator iA satisfies the product rule for differentiation: iA(f g) = giA(f ) + f iA(g) for suitable f and g.Denote by f the pointwise complex conjugate of a complexvalued function f .We shall use the chain rule in the form for suitable f and h.Formula (4.3) is straightforward to check.As a consequence of (4.3) we have that the differential operator iA commutes with the action of complex conjugation of functions: iA( f ) = iAf for suitable f .We now turn to differentiation of powers.Proof.Let u = (Im ϕ) α+1 in D. By a standard rule for differentiation we have that in D, compare with (4.3).Since the differential operator iA commutes with the action of complex conjugation of functions we have that in D. We now use Lemma 4.2 and calculate that in D, where the last equality follows by cancellation.By some elementary algebra we now have that in D. This yields the conclusion of the lemma.
Lemma 4.4.Let ϕ be as in (4.1) and α ∈ R. Then iA((ϕ Proof.We first show that iAϕ ′ = (ϕ − i)ϕ ′ .From (4.2) we have that where the last equality again follows by (4.2).By some elementary algebra we now have that where the last equality follows by (4.1).We now consider the general case.By a well-known differentiation formula we have that iA((ϕ in D, compare with (4.3).We now use the result of the previous paragraph to conclude that We next calculate the angular derivative of the Poisson kernel.
where the last equality is straightforward to check.This yields the conclusion of the theorem.
An earlier version of Theorem 4.5 appears in Olofsson [19,Theorem 1.11]; see also Klintborg and Olofsson [16,Corollary 1.3] for a generalization along similar lines.Here our focus is on applications to the setting of the upper half-plane and we express matters using the Möbius transformation ϕ.
We denote by C[z, z] the algebra of polynomials in z and z with complex coefficients.Define polynomials {h k,α } ∞ k=0 in C[z, z] by h 0,α = 1 and for k ≥ 0. It is straightforward to check that h k,α has degree at most k for k = 0, 1, . . . .Proof.For k = 0 the result is evident.For k = 1 the result follows by Theorem 4.5.We proceed by induction and assume that (iA) k P α = (h k,α • ϕ)P α for some k ≥ 0.
Applying the operator iA we have that where we have used the product rule.From the chain rule (4.3) and Lemma 4.2 we have that We now return to the function (iA) k+1 P α and use Theorem 4.5 to conclude that where in the last equality we have used (4.4).The result now follows by invoking the induction principle.
We shall study the polynomials h k,α in some more detail.Let α ∈ R and let us denote by y the imaginary part.A calculation shows that in the sense of differential operators, where As a consequence, we have that a product y α+1 h is α-harmonic in H if and only if Consider the partial sums for k = 0, 1, . . . of a binomial series (2.2) with a = α + 1.We shall need the associated homogeneous polynomials Proof.We shall evaluate the differential operator D α on a polynomial p of the form for some a 0 , . . ., a k ∈ C. Calculations show that From these formulas we have that for p of the form (4.8).
From the result of the previous paragraph we have that D α p = 0 if and only if (j + 1)a j+1 − (j + α + 1)a j = 0 for j = 0, . . ., k − 1.We conclude that D α p = 0 if and only if p = a 0 p k,α .
We now return to the h k,α 's.
Proof.From Lemma 4.7 we have that the product , where D α is as in (4.5).We next write the polynomial h k,α as a sum of homogeneous polynomials: h k,α = k j=0 p j , where p j is homogeneous of degree j.Clearly p 0 is a constant multiple of p 0,α .Applying the differential operator D α we see that homogeneous of degree j − 1, we have that D α p j = 0 for j = 1, . . .k. From Theorem 4.8 we have that p j is a constant multiple of p j,α for j = 1, . . .k.We conclude that h k,α is a linear combination of p 0,α , . . ., p k,α .

A general representation theorem
In this section we consider α-harmonic functions in H which are of temperate growth at infinity and vanish on the real line.A main task is to establish a representation of such functions.The analysis uses results from Section 4. Theorem 5.1.Let α > −1.Let u be an α-harmonic function in H which is of temperate growth at infinity.Assume that u(z) → 0 as H ∋ z → x for every x ∈ R. Then for some n ∈ N and c 0 , . . ., c n ∈ C, where the p k,α 's are as in (4.7).
Proof.We consider the weighted pull-back where ϕ is as in (4.1).From [18, Theorem 1.1] it follows that v is α-harmonic in D. From temperate growth of u at infinity and vanishing of u on R we have by a compactness argument that v is of temperate growth in D. By Theorem 3.3 we conclude that v = P α [g] for some distribution g ∈ D ′ (T).Since α > −1, we have from general theory that v r → g in D ′ (T) as r → 1, where the v r 's are defined as in (3.1) (see for instance [20,Theorem 5.4]).Since u(z) → 0 as H ∋ z → x for x ∈ R, we see that g vanishes on T \ {1} in a distributional sense.Now since 1 in D ′ (T) for some n ∈ N and complex numbers a 0 , . . ., a n ∈ C (see Hörmander [14,Theorem 2.3.4]).By Corollary 3.7 we have that P α [δ for z ∈ D, where the last equality follows by Theorem 4.6.By Theorem 4.10 the function h k,α is a linear combination of p 0,α , . . ., p k,α .From (5.2) we conclude that for some b 0 , . . ., b n ∈ C. Invoking Theorem 4.1 we see that for z ∈ D, where cϕ ′ (0) −α/2 = 1.A passage back to the function u yields (5.1) with k = cb k for k = 0, . . ., n.This completes the proof of the theorem.
For α > −1 and n ∈ N, we denote by V α,n the set of all functions u of the form (5.1) for some c 0 , . . ., c n ∈ C, where the p k,α 's are as in (4.7).We also set The space V α,n is a complex vector space of finite dimension n + 1.Notice that the set V α is a complex vector space which is naturally filtered by the sets V α,n for n = 0, 1, . . . .
Observe that the conclusion of Theorem 5.1 says that u ∈ V α .We next describe the class V 0 in some more detail.Proposition 5.2.Let n ∈ N. Then a function u belongs to the space V 0,n if and only if it has the form Proof.A calculation using the formula for a finite geometric sum shows that Therefore The result is now evident from definition of the space V 0,n .
We shall next investigate the order of growth of a function of the form for some k ∈ N and α > −1, where p k,α is as in (4.7).From the proof of Proposition 5.2 we have that u k,0 (z) = Im(z k+1 ), z ∈ C, for k ∈ N. Proposition 5.3.Let u k,α be as in (5.4) for some k ∈ N and α > −1.Then for z ∈ H.In particular, if u ∈ V α,n for some n ∈ N, then u has order of growth at most n + α + 1 at infinity.Proof.Let z ∈ H and write z = te iθ with t > 0 and 0 < θ < π.By homogeneity we have that u k,α (z) = sin α+1 (θ)p k,α (e iθ )t k+α+1 .Notice also that |z| 2 / Im(z) = t/ sin(θ).From these two facts we have that u k,α (z) = sin k+2α+2 (θ)p k,α (e iθ ) 2 / Im(z) k+α+1 for z ∈ H, where θ = arg z.This yields the conclusion of the proposition.
The next result points out the significance of the class V α .
Corollary 5.4.Let α > −1.Then a function u belongs to the class V α if and only if it is α-harmonic in H, has temperate growth at infinity and vanishes on the real line in the sense that u(z) → 0 as H ∋ z → x for every x ∈ R.
Proof.The if part is a restatement of Theorem 5.1.From Corollary 4.9 we have that every function u ∈ V α is α-harmonic in H. From Proposition 5.3 we have that every function u ∈ V α is of temperate growth at infinity.It is evident that every function u ∈ V α vanishes on the real line.
Corollary 5.4 explains how the space V α can be thought of as the class of obstructions for the uniqueness problem for α-harmonic functions in H with respect to a vanishing Dirichlet boundary value on the real line and temperate growth at infinity.Lemma 5.5.Let u ∈ V α,n be of the form (5.1) for some α > −1 and n ∈ N. Then lim t→+∞ u(te iθ )/t n+α+1 = c n sin α+1 (θ)p n,α (e iθ ) for 0 < θ < π.
Proof.Let t > 0 and 0 < θ < π.By homogeneity we have that The result now follows by a passage to the limit.
We shall dissect the space V α using orders of growth at infinity.We say that a function u in H has order of growth n + α + 1 at infinity if it satisfies an estimate of the form (5.5) |u(z)| ≤ C(|z| 2 / Im(z)) n+α+1 for z ∈ H with |z| > R, where C ≥ 0 and R > 1 are finite constants.
Proof.From Proposition 5.3 we know that (5.5) holds if u ∈ V α,n .Assume next that u ∈ V α satisfies (5.5).We shall prove that u ∈ V α,n .If u ∈ V α,0 there is nothing to prove.Assume therefore that u ∈ V α,m and u ∈ V α,m−1 for some m ∈ Z + .An application of Lemma 5.5 with 0 < θ < π chosen such that p m,α (e iθ ) = 0 shows that u has order of growth at least m + α + 1. Therefore m ≤ n, so that u ∈ V α,n .
We equip the space V α,n with the semi-norm given by the best constant C in (5.5).Another application of Lemma 5.5 shows that this latter semi-norm is in fact a norm.The space V α,n is thus a normed complex vector space of finite dimension n + 1.Since any two norms on such a space are equivalent, we conclude that We denote by D ′ (R) the space of distributions on the real line R.An integrable where C ∞ 0 (R) is the space of indefinitely differentiable test functions on R with compact support.By u j → u in D ′ (R) as j → ∞ we understand that lim j→∞ u j , ϕ = u, ϕ for every ϕ ∈ C ∞ 0 (R).A standard reference for distribution theory is Hörmander [14].
We shall next extend the validity of Theorem 5.1 to allow for a distributional boundary value.
Proof.The theorem is proved by a regularization argument.Let ψ ∈ C ∞ 0 (R) be a nonnegative compactly supported test function with ∞ −∞ ψ(x) dx = 1 and set ψ ε (x) = ψ(x/ε)/ε for x ∈ R and ε > 0. We consider the regularizations for ε > 0. A differentiation under the integral shows that the function u ε is αharmonic in H.It is straightforward to check that u ε (z) → 0 as H ∋ z → x for every x ∈ R. Using that u is of temperate growth at infinity it is straightforward to check that (5.6) for z ∈ H with |z| > R and 0 < ε < 1, where C ≥ 0 and R > 1 are finite constants and n ∈ N. Notice also that u = lim ε→0 u ε in H in the sense of normal convergence.We now proceed to details.By Theorem 5.1 we have that u ε ∈ V α .From (5.6) and Lemma 5.6 we have that u ε ∈ V α,n .Therefore (5.7) for some c 0 (ε), . . ., c n (ε) ∈ C. Another application of Lemma 5.6 gives that |c k (ε)| ≤ C ′ C for k = 0, 1, . . ., n, where C is as in (5.6) and C ′ = C ′ α,n,R is a positive constant.By a compactness argument we can extract a subsequence ε = ε j → 0 of positive real numbers such that c k (ε j ) → c k as j → ∞ for k = 0, 1, . . ., n.The conclusion of the theorem now follows by setting ε = ε j in (5.7) and letting j → ∞.
We point out that the assumption in Theorem 5.7 is that The case of usual harmonic functions deserves special mention.
Corollary 5.8.Let u be a harmonic function in H which is of temperate growth at infinity.Assume that lim y→0+ u(• + iy) = 0 in D ′ (R).Then u ∈ V 0 .
We record also the following slight improvement of the if part of Lemma 5.6.
Proof.If u ∈ V α,0 there is nothing to prove.Assume therefore that u ∈ V α,m and u ∈ V α,m−1 for some m ∈ Z + .An application of Lemma 5.5 with 0 < θ < π chosen such that p m,α (e iθ ) = 0 shows that u has order of growth at least m + α + 1 at infinity.Therefore m < n + 1, so that u ∈ V α,n .Proposition 5.9 is included for the sake of convenience.We shall prove more refined versions of Proposition 5.9 in later sections.

A uniqueness result for the case α = 0
We now turn our attention to uniqueness results for α-harmonic functions in H.In this section we study the case of parameters α > −1 such that α = 0.The special case of usual harmonic functions is investigated in later sections.We begin our analysis with a study of the zeros of the polynomials p k,α .
A classical result of Eneström-Kakeya says that if Proof.We first show that the zeroes of p are all located in the disc {z ∈ C : |z| ≤ R}.Consider the polynomial g(z) = p(Rz).By assumption the polynomial g satisfies the assumptions of the Eneström-Kakeya theorem quoted above.We conclude that the zeroes of g are all located in D. This yields that the zeroes of p are all located in the disc {z ∈ C : |z| ≤ R}.
We next show that the zeroes of p are all located in the exterior disc {z ∈ C : |z| ≥ r}.Consider the polynomial h(z) = z n p(1/z).Notice that By the result of the previous paragraph, the zeroes of h are all located in the disc {z ∈ C : |z| ≤ 1/r}.Therefore the zeroes of p are all located in the exterior disc {z ∈ C : |z| ≥ r}.
Assume next that −1 < α < 0. In this case the quotients in (6.2) decrease in j and we have that a j (α)/a j+1 (α) ≥ 1/(α + 1) for 0 ≤ j < k.By Corollary 6.1 the zeroes of s k,α are all located in the exterior disc {z ∈ C : |z| ≥ 1/(α + 1)} which is disjoint from D. Remark 6.3.We point out that the zero set of p k,0 intersects the unit circle if k ≥ 1.In fact, from (4.6) we have that We now consider the class V α .Theorem 6.4.Let u ∈ V α for some α > −1 with α = 0. Assume that there is a sequence {z j } in H with |z j | → ∞ as j → ∞ such that Then u(z) = 0 for all z ∈ H. problem for classical harmonic functions along the lines of what we did in Section 6 thus calls for a more demanding vanishing condition at infinity.In this section we shall consider vanishing conditions along geodesics in H, that is, along rays in H that are parallel to the positive imaginary axis.
Let us first consider the class V 0 .
We next turn to uniqueness theorems for harmonic functions.We point out that the case of an empty sequence {θ k } n k=1 (n = 0) in Theorem 9.2 yields the admissible function of angles used in the proof of Corollary 8.5.
We are now in position to calculate the minimal elements in A.
Theorem 9.5.The minimal elements in A are exactly those that belong to the set A 0 .
Proof.By Lemma 9.4 we know that the admissible functions of angles (E, η) from the set A 0 are minimal elements in A. We proceed to show that every minimal element in A has such form.

Theorem 3 . 3 .
Let α ∈ R. Then a function u in D has the form of a Poisson integral u = P α [f ] in D for some f ∈ D ′ (T) if and only if u is α-harmonic in D, u has temperate growth in D, and Spec(u) ⊂ Spec(P α ).

Corollary 3 . 4 .
Let α ∈ R \ Z − .Then a function u in D has the form of a Poisson integral u = P α [f ] in D for some f ∈ D ′ (T) if and only if it is α-harmonic in D and of temperate growth there.Proof.Recall that Spec(P α ) = Z in the present situation.The result follows from Theorem 3.3.The α-harmonic Poisson kernel P α has bounded L 1 -means when α > −1 (see Olofsson [17, Theorem 3.1]).As a consequence, for α > −1, one has is to exhibit the distribution f ∈ D ′ (T) as a limit point of the u r 's as r → 1 and then deduce the Poisson integral representation (3.4) from a uniqueness argument.See for instance Olofsson and Wittsten[20, Theorem 5.5]  for an elaboration on this theme.An interesting point of Theorem 3.3 is that this result establishes Poisson integral representations(3.4)  in cases where boundary limits are non-existent.Structure in (2.3) suggests use of the differential operator A = z∂ − z ∂.Observe that(3.5)

Lemma 4 . 7 .
Let h k,α ∈ C[z, z] be as in(4.4)  for some α ∈ R and k ≥ 0. Then the functionH ∋ z → (Im z) α+1 h k,α (z) is α-harmonic in H.Proof. Recall that the function iAu is α-harmonic in D if u is, see(3.5).From Theorem 4.6 we thus have that the function (h k,α • ϕ)P α is α-harmonic in D. Theorem 4.1 displays the Poisson kernel P α as a weighed pull-back by the function ϕ.From Olofsson[18, Theorem 1.1]  we now conclude that the function z → (Im z) α+1 h k,α (z) is α-harmonic in H, see discussion following formula (1.7).

Corollary 6 . 1 .
k is an analytic polynomial of degree n ≥ 0 such that a k ≤ a k+1 for 0 ≤ k < n, then the zeroes of p are all located in the closed unit disc D. An easy proof of this result can be found in Gardner and Govil[12, Theorem 1.3].Let p be an analytic polynomial of the form (6.1) with positive coefficients:a k > 0 for 0 ≤ k ≤ n.Set r = min 0≤k<n a k /a k+1 and R = max 0≤k<n a k /a k+1 .Then the zeroes of p are all located in the annulus {z ∈ C : r ≤ |z| ≤ R}.