A series expansion for generalized harmonic functions

Markus Klintborg; Anders Olofsson

doi:10.1007/s13324-021-00561-w

A series expansion for generalized harmonic functions

Klintborg, Markus; Olofsson, Anders 2021-06-11 00:00:00 We consider a class of generalized harmonic functions in the open unit disc in the complex plane. Our main results concern a canonical series expansion for such func- tions. Of particular interest is a certain individual generalized harmonic function which suitably normalized plays the role of an associated Poisson kernel. Keywords Harmonic function · Power series · Poisson kernel · Hypergeometric function Mathematics Subject Classiﬁcation Primary 31A05; Secondary 33C05 · 35J15 0 Introduction Let D be the open unit disc in the complex plane C and denote by ∂ = ∂/∂z and ∂ = ∂/∂z ¯ the usual complex partial derivatives. This work is concerned with second order partial differential operators of the form ¯ ¯ L = (1 −|z| )∂ ∂ + pz∂ + qz ¯∂ − pq, z ∈ D, (0.1) p,q z z z z where p, q ∈ C are complex parameters. Of particular interest are solutions of the associated homogeneous equation L u =0in D. p,q We say that a function u is (p, q)-harmonic if u is twice continuously differentiable in D (in symbols u ∈ C (D)) and L u = 0in D, where L is as in (0.1). Notice p,q p,q Anders Olofsson anders.olofsson@math.lu.se Markus Klintborg kborgmarkus@protonmail.com Mathematics, Faculty of Science, Centre for Mathematical Sciences, Lund University, P.O. Box 118, 221 00 Lund, Sweden 0123456789().: V,-vol 122 Page 2 of 28 M. Klintborg, A. Olofsson that a function u is (p, q)-harmonic if and only if its complex conjugate u ¯ is (q ¯ , p ¯)- harmonic. Observe also that a (0, 0)-harmonic function is a harmonic function in D in the usual sense. An interesting example of a (p, q)-harmonic function is the function 2 p+q+1 (1 −|z| ) u (z) = , z ∈ D (0.2) p,q p+1 q+1 (1 − z) (1 −¯z) (see Theorem 1.4). Here powers are deﬁned in the usual way using the principal branch of the logarithm, that is, we require that log(1) = 0. Notice that the above functions u have the hermitian symmetry property that u ¯ = u for p, q ∈ C. p,q p,q q ¯ ,p ¯ Recall that elements of the unit circle T = ∂ D act on the unit disc D as rotations. On a function level we consider rotation operators i θ R u(z) = u(e z), z ∈ D, (0.3) i θ i θ for e ∈ T acting on functions u in D. A basic observation concerning the differential operator L is the commutativity relation p,q L R i θ u = R i θ L u, u ∈ C (D), p,q p,q e e i θ for e ∈ T. This latter commutativity relation suggests an analysis of (p, q)-harmonic functions using concepts natural to classical Fourier analysis on the unit circle. Let Z be the set of integers. For a suitably smooth function u in D we deﬁne its m-th homogeneous part by the formula −imθ i θ u (z) = e u(e z) dθ, z ∈ D, (0.4) 2π for m ∈ Z. Notice that the m-th homogeneous part u of u is the m-th Fourier coefﬁcient of the vector-valued function i θ T e → R i θ u, where R i θ is as in (0.3). ∞ n n We set C (D) = C (D), where C (D) is the space of n-times continuously n=0 differentiable functions in D for n ∈ N ={0, 1, 2,... }. We topologize these spaces in the usual way using the semi-norms j k u = max|∂ ∂ u(z)|, j ,k;K z∈K where j , k ∈ N are non-negative integers and K ⊂ D is a compact subset of D. Let us recall the classical hypergeometric function deﬁned by (a) (b) z n n F (a, b; c; z) = , z ∈ D, (0.5) (c) n! n=0 A series expansion for generalized harmonic functions Page 3 of 28 122 for a, b, c ∈ C such that c =−1, −2,... .Here (a) = 1 and (a) = a(a + 1)...(a + n − 1) for n = 1, 2,... are Pochhammer symbols. Let us return to a (p, q)-harmonic function u. We show that the m-th homogeneous part of u has the form 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, m m for some c ∈ C when m ∈ N is a non-negative integer (see Theorem 4.3). Here F is the hypergeometric function (0.5). Hermitian symmetry leads to a similar formula for u when m ∈ Z = Z\N is a negative integer (see Corollary 4.4). Notice that u ∈ C (D) for m ∈ Z. A principal result of the present paper concerns the asymptotic behavior of the m-th homogeneous part u of a (p, q)-harmonic function u. We show that 1/|m| lim supu < 1 j ,k;K |m|→∞ for all j , k ∈ N and K ⊂ D compact (see Theorem 4.6). This result enables us to use the classical root test to establish absolute convergence of the function series u in C (D) (see Corollary 4.7). m=−∞ A further analysis leads to a function series characterization of (p, q)-harmonic functions. A function u in D is (p, q)-harmonic if and only if it has the form 2 m u(z) = c F (−p, m − q; m + 1;|z| )z m=0 2 m + c F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, (0.6) −m m=1 for some sequence {c } of complex numbers such that m=−∞ 1/|m| lim sup|c | ≤ 1 |m|→∞ (see Theorem 5.1). As mentioned above, the sums in (0.6) are absolutely convergent in ∞ ∞ the space C (D). As a consequence we have that u ∈ C (D) if u is (p, q)-harmonic (see Corollary 4.9). This characterization of (p, q)-harmonic functions improves on a result by Ahern et al. [1, Theorem 2.1] when specialized to our setting. We then turn to coefﬁcient formulas for (p, q)-harmonic functions. We show that m m c = ∂ u(0)/m! and c = ∂ u(0)/m! m −m 122 Page 4 of 28 M. Klintborg, A. Olofsson for m ∈ N, where the c ’s are as in (0.6) (see Theorem 5.3). As a consequence, we have that ∂ u(0) 2 m u(z) = F (−p, m − q; m + 1;|z| )z m! m=0 ∂ u(0) 2 m + F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, m! m=1 whenever u is a (p, q)-harmonic function (see Theorem 5.4) as well as a corresponding uniqueness result for such functions (see Corollary 5.5). Let us return to the function u in (0.2). A calculation of partial derivatives at the p,q origin of the function u leads to the series expansion p,q (p + 1) 2 m u (z) = F (−p, m − q; m + 1;|z| )z p,q m! m=0 (q + 1) 2 m + F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D m! m=1 (see Theorem 6.3). This latter series expansion generalizes a well-known partial frac- tion decomposition formula for the classical Poisson kernel for D which is obtained for p = q = 0. A main contribution of this paper concerns series expansion of (p, q)-harmonic functions. Of particular mention is a limit theorem for associated hypergeometric functions: lim F (−p, m − q; m + 1; z) = (1 − z) , z ∈ D m→∞ (see Theorem 2.6). Apart from its intrinsic interest, this limit theorem provides an efﬁcient tool for the study of limit properties of homogeneous parts of (p, q)-harmonic functions. The results of this paper have applications to Poisson integral representations of (p, q)-harmonic functions which is possible when p, q ∈ C\Z are such that Re(p)+ Re(q)> −1. In the ﬁnal section of the paper we comment brieﬂy on the connection to such theory. We have traced the study of (p, q)-harmonic functions back to Daryl Geller [13]. Other signiﬁcant contributions are those of Ahern and collaborators [1,2]. Our interest in those topics [18–21] arose in connection to standard weights and earlier work of Garabedian [12]. Borichev and Hedenmalm [6] have established a connection to poly- harmonic theory. Other recent related papers are [5,8–10,14,16,17,22]. The authors thank the referee for a careful reading of the manuscript. A series expansion for generalized harmonic functions Page 5 of 28 122 1 The function u is (p, q)-harmonic p,q Let u be as in (0.2)for some p, q ∈ C. Observe that the functions u have the p,q p,q hermitian symmetry property u ¯ = u . Indeed, p,q q ¯ ,p ¯ 2 p ¯+¯ q+1 (1 −|z| ) u ¯ (z) = u (z) = = u (z) p,q p,q q ¯ ,p ¯ p ¯+1 q ¯ +1 (1 −¯z) (1 − z) for z ∈ D. We shall discuss in this section some differentiation formulas for the functions u . In particular, we shall show that the function u is (p, q)-harmonic p,q p,q (see Theorem 1.4). Lemma 1.1 Let u be as in (0.2) for some p, q ∈ C. Then p,q |z| z z∂u (z) = − (p + q + 1) + (p + 1) u (z) p,q p,q 1 −|z| 1 − z for z ∈ D. Proof Differentiating using the product rule for differentiation we have that 1 1 2 p+q ∂u (z) = (p + q + 1)(1 −|z| ) (−¯z) p,q q+1 p+1 (1 −¯z) (1 − z) 2 p+q+1 + (1 −|z| ) (−(p + 1)) (−1) p+2 (1 − z) z ¯ 1 = − (p + q + 1) + (p + 1) u (z) p,q 1 −|z| 1 − z for z ∈ D, where the last equality is straightforward to check. This yields the conclusion of the lemma. We next turn to the ∂-derivative of u . p,q Lemma 1.2 Let u be as in (0.2) for some p, q ∈ C. Then p,q |z| z ¯ z ¯∂u (z) = − (p + q + 1) + (q + 1) u (z) p,q p,q 1 −|z| 1 −¯z for z ∈ D. Proof We shall use the hermitian symmetry property u ¯ = u . From Lemma 1.1 p,q q ¯ ,p ¯ we have that |z| z z∂u (z) = − (p ¯ +¯ q + 1) + (q ¯ + 1) u (z) q ¯ ,p ¯ q ¯ ,p ¯ 1 −|z| 1 − z for z ∈ D. A complex conjugation now yields the conclusion of the lemma. 122 Page 6 of 28 M. Klintborg, A. Olofsson Following earlier practice from [20] we denote by A = z∂ −¯z∂ the angular deriva- tive. We next calculate the angular derivative of u . p,q Corollary 1.3 Let u be as in (0.2) for some p, q ∈ C. Then p,q z z ¯ (z∂ −¯z∂)u (z) = (p + 1) − (q + 1) u (z) p,q p,q 1 − z 1 −¯z for z ∈ D. Proof The result is evident from Lemmas 1.1 and 1.2. We mention that Corollary 1.3 generalizes [20, Theorem 1.11]. Recall the partial fraction formula 1 −|z| z z ¯ = + + 1 (1.1) |1 − z| 1 − z 1 −¯z for the classical Poisson kernel for the unit disc. Formula (1.1) is straightforward to check. Theorem 1.4 Let u be as in (0.2) for some p, q ∈ C. Then L u = 0 in D, p,q p,q p,q where L is as in (0.1). p,q Proof Recall Lemma 1.2. Notice that the differential operator z∂ satisﬁes the product rule for differentiation. It is straightforward to check that 2 2 ∂ |z| |z| z = 2 2 2 ∂z 1 −|z| (1 −|z| ) whenever the formula makes sense. Differentiating using the product rule we have that |z| |z| ∂∂u (z) =−(p + q + 1) u (z) p,q p,q 2 2 (1 −|z| ) |z| z ¯ + − (p + q + 1) + (q + 1) z∂u (z) p,q 1 −|z| 1 −¯z for z ∈ D. From Lemma 1.1 we now have that |z| |z| ∂∂u (z) =−(p + q + 1) u (z) p,q p,q 2 2 (1 −|z| ) |z| z ¯ + − (p + q + 1) + (q + 1) 1 −|z| 1 −¯z |z| z × − (p + q + 1) + (p + 1) u (z) p,q 1 −|z| 1 − z A series expansion for generalized harmonic functions Page 7 of 28 122 for z ∈ D. Expanding the above product we see that 2 4 |z| |z| 2 2 |z| ∂∂u (z) = − (p + q + 1) + (p + q + 1) u (z) p,q p,q 2 2 2 2 (1 −|z| ) (1 −|z| ) |z| z z ¯ − (p + q + 1) (p + 1) + (q + 1) u (z) p,q 1 −|z| 1 − z 1 −¯z |z| + (p + 1)(q + 1) u (z) p,q |1 − z| 2 2 for z ∈ D. Multiplying by a factor (1 −|z| )/|z| we see that |z| 1 2 2 (1 −|z| )∂∂u (z) = (p + q + 1) − (p + q + 1) u (z) p,q p,q 2 2 1 −|z| 1 −|z| z z ¯ − (p + q + 1) (p + 1) + (q + 1) u (z) p,q 1 − z 1 −¯z 1 −|z| + (p + 1)(q + 1) u (z) p,q |1 − z| for z ∈ D. Notice an appearance of the classical Poisson kernel in the rightmost term above. Using the partial fraction formula (1.1) we have that |z| (1 −|z| )∂∂u (z) = (p + q + 1) (p + q) − 1 u (z) p,q p,q 1 −|z| + ((p + 1)(q + 1) − (p + q + 1)(p + 1)) u (z) p,q 1 − z z ¯ + ((p + 1)(q + 1) − (p + q + 1)(q + 1)) u (z) p,q 1 −¯z + (p + 1)(q + 1)u (z) p,q for z ∈ D. A simpliﬁcation of terms now leads to the formula |z| (1 −|z| )∂∂u (z) = (p + q + 1)(p + q) u (z) p,q p,q 1 −|z| z z ¯ − p(p + 1) u (z) − q(q + 1) u (z) + pqu (z) (1.2) p,q p,q p,q 1 − z 1 −¯z for z ∈ D. Recall Lemmas 1.1 and 1.2. In view of these two lemmas our latter formula (1.2) says that L u = 0in D. p,q p,q 122 Page 8 of 28 M. Klintborg, A. Olofsson 2 A sequence of hypergeometric functions Let us ﬁrst consider a second order partial differential operator of the form ¯ ¯ L = (1 −|z| )∂ ∂ + pz∂ + qz ¯∂ − r , z ∈ D, (2.1) p,q;r z z z z where p, q, r ∈ C are complex parameters. A principal case is when r = pq. Notice that L = L , where L is as in (0.1). The introduction of an additional p,q;pq p,q p,q parameter r ∈ C allows for more general operators appearing in the study of conduc- tivity problems, see for instance Calderón [7] or Astala and Päivärinta [4]. We shall evaluate the operator L on a complex-valued function u in the punc- p,q;r tured disc D\{0} of the form 2 m u(z) = f (|z| )z , z ∈ D\{0}, (2.2) for some f ∈ C (0, 1) and m ∈ Z. We introduce also the ordinary differential operator d d H = (1 − x )x + (c −[a + b + 1]x ) − ab, (2.3) a,b;c dx dx where a, b, c ∈ C are complex parameters. Notice that the famous hypergeometric ordinary differential equation (1 − x )xy (x ) + (c −[a + b + 1]x )y (x ) − aby(x ) = 0 takes the form H y = 0 using the operator H . a,b;c a,b;c Theorem 2.1 Let L be as in (2.1) for some p, q, r ∈ C. Let u be a function of the p,q;r form (2.2) for some f ∈ C (0, 1) and m ∈ Z. Then m 2 L u(z) = z H f (|z| ), z ∈ D\{0}, p,q;r a,b;c where c = m + 1 and a + b = m − p − q, (2.4) ab = r − pm. Proof A differentiation shows that m−1 2 m 2 ∂u(z) = mz f (|z| ) + z zf ¯ (|z| ) for z ∈ D\{0}, and similarly that m+1 2 ∂u(z) = z f (|z| ) A series expansion for generalized harmonic functions Page 9 of 28 122 for z ∈ D\{0}. Another differentiation gives that m 2 m+1 2 ∂∂u(z) = (m + 1)z f (|z| ) + z zf ¯ (|z| ) for z ∈ D\{0}. A calculation using these formulas gives that 2 m 2 m 2 2 m+1 2 L u(z) = (1 −|z| )((m + 1)z f (|z| ) + z |z| f (|z| )) + qzz ¯ f (|z| ) p,q;r m−1 2 m 2 m 2 + pz(mz f (|z| ) + z zf ¯ (|z| )) − rz f (|z| ) m 2 2 2 2 2 = z |z| (1 −|z| ) f (|z| ) + (m + 1 −[m + 1 − p − q]|z| ) f (|z| ) 2 m 2 − (r − pm) f (|z| ) = z H f (|z| ) a,b;c for z ∈ D\{0}, where c = m + 1, the numbers a and b areasin(2.4) and H is as a,b;c in (2.3). This yields the conclusion of the theorem. Equations (2.4) say that a and b are the zeros of the quadratic polynomial P (λ) = λ − (m − p − q)λ + r − pm,λ ∈ C. p,q;r ;m Notice that P (λ) = (λ + p)(λ + q − m) + r − pq,λ ∈ C. p,q;r ;m In particular, the zeros of P are −p and m − q in the principal case when p,q;r ;m r = pq. Theorem 2.1 suggests a natural construction of (p, q)-harmonic functions. Proposition 2.2 Let p, q ∈ C. Consider the function 2 m u (z) = F (−p, m − q; m + 1;|z| )z , z ∈ D, where m ∈ N and F is the hypergeometric function (0.5). Then u is a (p, q)-harmonic function. Proof Clearly u ∈ C (D). It is well-known that the hypergeometric function y = F (a, b; c;·) satisﬁes the hypergeometric equation H y = 0(see [3, Section 2.3]). a,b;c The result now follows by Theorem 2.1. Corollary 2.3 Let p, q ∈ C. Consider the function 2 |m| u (z) = F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, where m ∈ Z and F is the hypergeometric function (0.5). Then u is a (p, q)- harmonic function. 122 Page 10 of 28 M. Klintborg, A. Olofsson Proof We consider the complex conjugate 2 |m| u (z) = F (−¯ q, |m|− p¯;|m|+ 1;|z| )z , z ∈ D. By Proposition 2.2 we have that u ¯ is a (q ¯ , p ¯)-harmonic function. From hermitian symmetry we conclude that u is a (p, q)-harmonic function. Following earlier practice, a function u in D\{0} is said to be homogeneous of order m ∈ Z with respect to rotations if it has the property that i θ imθ u(e z) = e u(z), z ∈ D\{0}, i θ for e ∈ T. Notice that every function u of the form (2.2) is homogeneous of order m with respect to rotations. Theorem 2.4 Let p, q ∈ C and m ∈ N. Let u ∈ C (D) be homogeneous of order m with respect to rotations. Then u is (p, q)-harmonic if and only if it has the form 2 m u(z) = cF (−p, m − q; m + 1;|z| )z , z ∈ D, (2.5) for some c ∈ C, where F is the hypergeometric function (0.5). Proof From Proposition 2.2 we know that every function u of the form (2.5)is (p, q)- harmonic. Assume next that u is (p, q)-harmonic. Since u is homogeneous of order m,we can put u on the form (2.2)for some f ∈ C (0, 1). By Theorem 2.1 we have that H f (x ) = 0, 0 < x < 1, −p,m−q,m+1 where H is as in (2.3). Below we shall check that a,b;c m+1 (m + 1) f (x ) + p(m − q) f (x ) = o(1/x ) (2.6) as x → 0. Condition (2.6) allows us to apply [18, Proposition 1.3] to conclude that f is a constant multiple of the hypergeometric function F (−p, m − q; m + 1;·).This will then complete the proof of the theorem. We proceed to check (2.6). Recall formula (2.2). Since u is bounded near the origin m/2 we have that f (x ) = O(1/x ) as x → 0. A differentiation of (2.2) gives that m+1 2 ∂u(z) = z f (|z| ), z ∈ D\{0}. (m+1)/2 Since ∂u is bounded near the origin we have that f (x ) = O(1/x ) as x → 0. We have now checked (2.6). Theorem 2.4 and its proof are modeled on [18, Theorem 2.1]. We have merely supplied some details. A series expansion for generalized harmonic functions Page 11 of 28 122 We shall use the fact that the hypergeometric functions are analytic in D.Let H (D) be the space of analytic functions in D. The space H (D) is topologized in the usual manner using the semi-norms f = max| f (z)| (2.7) z∈K for K ⊂ D compact. Convergence in the space H (D) is usually referred to as normal convergence in D. Recall the terminology that a subset F of H (D) is called a normal family if every sequence of functions of F has a subsequence which converges in H (D). The limit function is not required to belong to F. We refer to Conway [11, Chapter VII] for background. Recall also the binomial series: (a) 1 F (a, b; b; z) = z = (2.8) n! (1 − z) n=0 for z ∈ D. Lemma 2.5 Let p, q ∈ C and consider the functions f (z) = F (−p, m − q; m + 1; z), z ∈ D, for m ∈ N, where F is the hypergeometric function (0.5). Then F ={ f : m ∈ N} is a normal family of analytic functions in D. Proof We shall prove that the functions in F are uniformly bounded on compact subsets of D. The conclusion of the lemma then follows by a classical result of Montel (see Conway [11, Theorem VII.2.9]). Let K ⊂ D be compact. Since K ⊂ D is compact there exists 0 < r < 1 such that max |z| < r. Choose N such that z∈K |1 − (q + 1)/(m + 1)|≤ 1/r for m > N . Observe that n−1 n−1 (m − q) m − q + k q + 1 = = 1 − . (m + 1) m + 1 + k m + 1 + k k=0 k=0 Therefore |(m − q) /(m + 1) |≤ 1/r for n ∈ N provided m > N . n n We now estimate the f ’s with m > N.From(0.5) we have that (−p) (m − q) z n n f (z) = , z ∈ D. (m + 1) n! n=0 122 Page 12 of 28 M. Klintborg, A. Olofsson Notice in this sum an appearance of the quotient (m − q) /(m + 1) considered in n n the previous paragraph. From the triangle inequality and the result of the previous paragraph we have that (| p|) |z| 1 | f (z)|≤ = n | p| n! r (1 −|z|/r ) n=0 for |z| < r and m > N , where the last equality follows by the binomial series (2.8). This proves that the functions in F are uniformly bounded on K . The following limit theorem will be much useful. Theorem 2.6 Let p, q ∈ C. Then lim F (−p, m − q; m + 1; z) = (1 − z) , z ∈ D, m→∞ with normal convergence, where F is the hypergeometric function (0.5). Proof Let the f ’s be as in Lemma 2.5. From Lemma 2.5 we have that the set F = { f : m ∈ N} is a normal family. From (0.5) we have that (−p) (m − q) n n (n) f (0) = (m + 1) for m, n ∈ N. Recall that the Pochhammer symbol (·) is a monic polynomial of (n) degree n. Passing to the limit we have that lim f (0) = (−p) for n ∈ N. m→∞ m n (n) From (2.8) we have that f (0) = (−p) for n ∈ N, where f (z) = (1 − z) , z ∈ D. A standard argument now yields that f → f in H (D) as m →∞. Assume to reach a contradiction that there exists a compact set K ⊂ D such that { f } does not m=0 converge uniformly to f on K . Passing to a subsequence we can assume that max| f (z) − f (z)|≥ δ> 0 (2.9) z∈K for k = 1, 2,... . Since the set F is a normal family, we can after passage to another subsequence if necessary, assume that f → g in H (D) as k →∞ for some (n) g ∈ H (D). From the ﬁrst paragraph of the proof we have g (0) = (−p) for n ∈ N, and a uniqueness argument gives that g = f in D. Thus f → f in H (D) as k →∞, which contradicts (2.9). We emphasize that Theorem 2.6 appears much natural in view of the binomial series (2.8). A series expansion for generalized harmonic functions Page 13 of 28 122 3 A generalized power series From the product rule for differentiation we have that (n) (n−k) (k) ( fg) = f g k=0 for, say, f , g ∈ C [0, 1). Lemma 3.1 Let j , k, m ∈ N be such that m ≥ j. Let u be a function of the form (2.2) for some f ∈ C [0, 1). Then j (m + k)! j k m+k− j 2l (k+l) 2 ∂ ∂ u(z) = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D. Proof Recall (2.2). Differentiating with respect to z ¯ we have that k m+k (k) 2 ∂ u(z) = z f (|z| ) for z ∈ D. Another differentiation using the product rule for differentiation gives that j (m + k)! j k m+k−( j −l) (k+l) 2 l ∂ ∂ u(z) = z f (|z| )z ¯ l (m + k − ( j − l))! l=0 j (m + k)! m+k− j 2l (k+l) 2 = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D. This completes the proof of the lemma. Let C (D) be the set of smooth complex-valued functions in the unit disc D.The space C (D) is topologized by means of the family of semi-norms j k u = max|∂ ∂ u(z)|, (3.1) j ,k;K z∈K where j , k ∈ N and K ⊂ D is compact. Recall that u → u in C (D) as m →∞ means that lim u − u = 0 for all j , k ∈ N and K ⊂ D compact. m→∞ m j ,k;K A (formal) series u of functions u ∈ C (D) for m ∈ N is said to be m m m=0 absolutely convergent in C (D) if u < +∞ m j ,k;K m=0 122 Page 14 of 28 M. Klintborg, A. Olofsson whenever j , k ∈ N and K ⊂ D is compact. By completeness of C (D) we have that ∞ ∞ every series absolutely convergent in C (D) is convergent in C (D). ∞ ∞ Theorem 3.2 Let { f } be a sequence in C [0, 1) such that m=0 1/m (n) lim sup max | f (x )| ≤ 1 0≤x ≤r m→∞ for n ∈ N and 0 < r < 1. Set 2 m u (z) = f (|z| )z , z ∈ D, m m 1/m for m = 0, 1, 2,... . Then lim sup u < 1 for all j , k ∈ N and K ⊂ D m→∞ j ,k;K compact, where · is as in (3.1). j ,k;K Proof Fix j , k ∈ N and K ⊂ D compact. Set r = max |z| < 1. From Lemma 3.1 z∈K we have that j (m + k)! j k m+k− j 2l (k+l) 2 ∂ ∂ u (z) = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D provided m ≥ j. We next apply the triangle inequality to see that j (m + k)! m+k− j 2l (n) u ≤ r r max f m j ,k;K m [0,r ] k≤n≤ j +k l (m + k + l − j )! l=0 m+k− j j 2 j (n) ≤ r (m + k) (1 + r ) max f (3.2) m [0,r ] k≤n≤ j +k for m ≥ j, where we have used the notation (2.7). We shall next apply the m-th root to (3.2) and pass to the limit as m →∞. In view of the assumption on the sequence { f } we conclude from (3.2) that m=0 1/m lim supu ≤ r . j ,k;K m→∞ Since 0 < r < 1, this yields the conclusion of the theorem. From the conclusion of Theorem 3.2 we have that u < +∞ when- m j ,k;K m=0 ever j , k ∈ N and K ⊂ D is compact. Thus the series u is absolutely m=0 convergent in C (D). We can think of a function series 2 m f (|z| )z (3.3) m=0 A series expansion for generalized harmonic functions Page 15 of 28 122 as a vector-valued power series. The sequence of coefﬁcients { f } in (3.3)isnow m=0 a sequence of functions in C [0, 1). From this point of view, Theorem 3.2 generalizes a well-known root criteria for power series (see for instance Conway [11, Theorem III.1.3]). Of particular concern to us are sequences { f } of the form m=0 f (x ) = F (−p, m − q; m + 1; x ), 0 ≤ x < 1, for m ∈ N, where p, q ∈ C and F is the hypergeometric function (0.5). We next observe that Theorem 2.6 guarantees that every such sequence { f } satisﬁes the k=0 assumption of Theorem 3.2. Proposition 3.3 Let p, q ∈ C. Then 1/m (n) lim sup max F (−p, m − q; m + 1; x ) ≤ 1 0≤x ≤r m→∞ for n ∈ N and 0 < r < 1, where F is the hypergeometric function (0.5). Proof Recall the fact that the complex derivative f → f is continuous in the topology of normal convergence of analytic functions (see Conway [11, Theorem VII.2.1]). Recall Montel’s theorem characterizing normal families of analytic functions (see Conway [11, Theorem VII.2.9]). In view of these two results, the proposition follows from Theorem 2.6. 4 Analysis of homogeneous parts i θ Notice that the rotation operators T e → R i θ from (0.3) have the group properties that R = I is the identity and R i (θ +τ) = R i θ R i τ e e i θ i τ for e , e ∈ T. Recall the notion of homogeneity with respect to rotations made precise in the paragraph before Theorem 2.4. Observe that a function u in D is homogeneous of imθ i θ order m ∈ Z with respect to rotations if and only if R i θ u = e u for e ∈ T. Proposition 4.1 Let u be the m-th homogeneous part of u ∈ C (D) for some m ∈ Z, where n ∈ N ∪{∞}. Then u ∈ C (D) and u is homogeneous of order m with m m respect to rotations. Proof Differentiations under the integral in (0.4) show that u ∈ C (D).Fromformula (0.4) we have that −imτ u = e R udτ (4.1) i τ 2π T 122 Page 16 of 28 M. Klintborg, A. Olofsson i θ in a vector-valued sense. Let e ∈ T. From the group property of rotation operators we have that 1 1 −imτ −imτ R u = e R R udτ = e R udτ i θ i θ i τ i (θ +τ) e e e e 2π 2π T T in a vector-valued sense. A change of variables now gives that imθ −imτ imθ R i θ u = e R i τ udτ = e u . m m e e 2π i θ Since e ∈ T is arbitrary, this yields that u is homogeneous of order m with respect to rotations. Below we shall make use of the commutativity relation L R i θ u = R i θ L u, u ∈ C (D), (4.2) p,q p,q e e i θ for e ∈ T. In order to prove (4.2) it sufﬁces to check that the differential operators z∂ and z ¯∂ commute with the rotations R i θ which is evident. We now return to (p, q)-harmonic functions. Proposition 4.2 Let p, q ∈ C. Let u be a (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ Z. Then u is (p, q)-harmonic. 2 2 Proof From Proposition 4.1 we have that u ∈ C (D) since u ∈ C (D). Recall formula (4.1). Applying the operator L we have that p,q −imτ L u = e L R i τ udτ p,q m p,q 2π in a vector-valued sense. We now use the commutativity relation (4.2) to conclude that −imτ L u = e R i τ L udτ = 0, p,q m p,q 2π where the last equality is evident since u is (p, q)-harmonic. Let us next calculate the homogeneous parts of a (p, q)-harmonic function. Theorem 4.3 Let p, q ∈ C. Let u be (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ N. Then 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, (4.3) m m for some c ∈ C, where F is the hypergeometric function (0.5). m A series expansion for generalized harmonic functions Page 17 of 28 122 Proof By Proposition 4.1, the function u is homogeneous of order m with respect to rotations. By Proposition 4.2, the function u is (p, q)-harmonic. The result now follows from Theorem 2.4. Corollary 4.4 Let p, q ∈ C. Let u be (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ Z\Z . Then 2 |m| u (z) = c F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, (4.4) m m for some c ∈ C, where F is the hypergeometric function (0.5). Proof The complex conjugate u ¯ is the −m =|m|-th homogeneous part of the func- tion u ¯. By hermitian symmetry, the function u ¯ is (q ¯ , p ¯)-harmonic. From Theorem 4.3 we have that 2 |m| u (z) = a F (−¯ q, |m|− p¯;|m|+ 1;|z| )z , z ∈ D, m m for some a ∈ C. A complex conjugation now yields (4.4) with c =¯a . m m m As a by-product from Theorem 4.3 and Corollary 4.4 we have that u ∈ C (D) for m ∈ Z if u is (p, q)-harmonic. Recall formula (0.4) for the homogeneous parts of a function u in D.Fromthe triangle inequality we have that sup |u (z)|≤ sup |u(z)| (4.5) |z|=r |z|=r for 0 < r < 1 and m ∈ Z. Notice that (4.5) generalizes a well-known bound for Fourier coefﬁcients. We next estimate the constants c appearing in Theorem 4.3 or Corollary 4.4. Lemma 4.5 Let p, q ∈ C. Let u be a (p, q)-harmonic function and consider its m-th homogeneous part u for m ∈ Z. Let c be as in (4.4) or (4.4) depending on whether m m − 1/|m| m ∈ N or m ∈ Z . Then lim sup |c | ≤ 1. |m|→∞ Proof For simplicity we consider the case m ∈ N. The case m ∈ Z is analogous or follows by hermitian symmetry. Let 0 < r < 1. From (4.4) and (4.5) we have that 2 m |c ||F (−p, m − q; m + 1; r )|r ≤ max|u(z)| (4.6) |z|=r for m ∈ N. From Theorem 2.6 we have that 2 2 p lim F (−p, m − q; m + 1; r ) = (1 − r ) = 0. m→∞ Therefore, a passage to the limit in (4.6) gives that 1/m lim sup|c | ≤ 1/r . m→∞ 122 Page 18 of 28 M. Klintborg, A. Olofsson 1/m Since 0 < r < 1 is arbitrary we conclude that lim sup |c | ≤ 1. This yields m→∞ the conclusion of the lemma. The following theorem is our main result about the asymptotic behavior of the homogeneous parts of a (p, q)-harmonic function. Theorem 4.6 Let p, q ∈ C. Let u be a (p, q)-harmonic function and denote by u its m-th homogeneous part for m ∈ Z. Then u ∈ C (D) for m ∈ Z and 1/|m| lim supu < 1 j ,k;K |m|→∞ for all j , k ∈ N and K ⊂ D compact. Proof From Theorem 4.3 and Corollary 4.4 we have that u ∈ C (D) for m ∈ Z. We consider ﬁrst the case m ∈ N. From Theorem 4.3 we have (4.4)for m ∈ N, where c ∈ C. Notice that the function u in (4.4) is constructed from the function m m f (x ) = c F (−p, m − q; m + 1; x ), 0 ≤ x < 1, m m as in Theorem 3.2. From Proposition 3.3 and Lemma 4.5 we have that the assumption of Theorem 3.2 is satisﬁed. An application of Theorem 3.2 yields the conclusion that 1/m lim sup u < 1for j , k ∈ N and K ⊂ D compact. m→∞ j ,k;K We next consider the case m ∈ Z . The complex conjugate u ¯ of u is the (−m)-th m m homogeneous part of the (q ¯ , p ¯)-harmonic function u ¯. From the result of the previous paragraph we thus have that 1/|m| 1/m lim supu = lim sup(u ¯) < 1 m m j ,k;K k, j ;K m→−∞ m→∞ for j , k ∈ N and K ⊂ D compact. This completes the proof of the theorem. We point out that Theorem 2.6 forms an integral part in the proof of Theorem 4.6. Recall the notion of absolute convergence in C (D), see the paragraph just before Theorem 3.2. Corollary 4.7 Let p, q ∈ C. Let u be a (p, q)-harmonic function with m-th homo- geneous part u for m ∈ Z. Then the function series u is absolutely m m m=−∞ convergent in C (D). Proof Theorem 4.6 allows us to apply the root test to conclude that u < +∞ m j ,k;K m=−∞ for j , k ∈ N and K ⊂ D compact. The following lemma is well-known but included for the sake of completeness. A series expansion for generalized harmonic functions Page 19 of 28 122 Lemma 4.8 Let u ∈ C (D) for some n ∈ N and denote by u its m-th homogeneous part for m ∈ Z. Then |m| u = lim 1 − u N →+∞ N + 1 m=−N in C (D). Proof It is straightforward to check that |m| 1 i θ 1 − u = K (e )R i θudθ m N N + 1 2π m=−N in a vector-valued sense, where |m| i θ imθ i θ K (e ) = 1 − e , e ∈ T, N + 1 m=−N is the Fejér kernel. It is well-known that the K ’s are non-negative and form an approximate identity as N →∞. Since u ∈ C (D), it is well-known that the function i θ n T e → R i θ u ∈ C (D) is continuous from T into C (D). The proof is now completed by a standard argument. We refer to Katznelson [15, Section I.2] for details. We think of Lemma 4.8 as a version of Fejér’s theorem adapted to our context. We next deduce that (p, q)-harmonic functions belong to the space C (D). Corollary 4.9 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Then u ∈ C (D) and u = u in C (D), where u is the m-th homogeneous part of u for m m m=−∞ m ∈ Z. Proof From Lemma 4.8 we have that |m| u = lim 1 − u (4.7) N →+∞ N + 1 m=−N in C (D) since u has such regularity. From Corollary 4.7 we know that the function series u is absolutely convergent in C (D). We can thus omit the conver- m=−∞ gence factors in (4.7) and deduce that u = u in C (D). By completeness m=−∞ we have that u ∈ C (D). 122 Page 20 of 28 M. Klintborg, A. Olofsson 5 A series expansion of harmonic functions The analysis from Sect. 4 leads to a natural function series description of (p, q)- harmonic functions. Theorem 5.1 Let p, q ∈ C. Then u is a (p, q)-harmonic function if and only if it has the form 2 m u(z) = c F (−p, m − q; m + 1;|z| )z m=0 2 m + c F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, (5.1) −m m=1 for some sequence {c } of complex numbers such that m=−∞ 1/|m| lim sup|c | ≤ 1, (5.2) |m|→∞ where F is the hypergeometric function (0.5). Moreover, the sums in (5.1) are abso- lutely convergent in the space C (D) when (5.2) holds. Proof Consider ﬁrst a (formal) function series of the form (5.1) with coefﬁcient sequence {c } satisfying (5.2). From Theorem 3.2 and Proposition 3.3 we have m=−∞ that 1/|m| lim supu < 1 j ,k;K |m|→∞ for j , k ∈ N and K ⊂ D compact, where 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, m m for m ∈ N and 2 |m| u (z) = c F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, m m for m ∈ Z . The root test now applies to show that the series u in (5.1)is m=−∞ absolutely convergent in C (D). This deﬁnes a function u = u by (5.1). m=−∞ From Proposition 2.2 and Corollary 2.3 we have that each term u in (5.1)is (p, q)- harmonic. Since the series expansion (5.1) converges in the space C (D),wehave that u ∈ C (D). Applying the operator L we have L u = L u = 0 p,q p,q p,q m m=−∞ in C (D), which shows that u is (p, q)-harmonic. Consider next a (p, q)-harmonic function u. We proceed to derive (5.1). Denote by u the m-th homogeneous part of u for m ∈ Z. From Theorem 4.3 we have (4.4)for m ∈ N and from Corollary 4.4 we have (4.4)for m ∈ Z , where c ∈ C for m ∈ Z. From Corollary 4.9 we have (5.1) in the form that u = u in C (D). m=−∞ A series expansion for generalized harmonic functions Page 21 of 28 122 By Corollary 4.7 we have that the function series u in (5.1) is absolutely m=−∞ convergent in C (D). From Lemma 4.5 we have that (5.2) holds. We emphasize that the function series (5.1) has an interpretation of homogeneous expansion of the (p, q)-harmonic function u. We mention that Theorem 5.1 improves on a result by Ahern et al. [1, Theorem 2.1] when specialized to the present setting. Theorem 5.1 contains also some earlier results of Olofsson and Wittsten. Theorem 5.1 with (p, q) = (0,α), α ∈ R, yields [21, Theorem 1.2]. Theorem 5.1 with (p, q) = (α/2,α/2), α ∈ R, yields [18, Theorem 2.2]. We shall next turn to formulas for the coefﬁcients c appearing in (5.1). Proposition 5.2 Let p, q ∈ C. Let u be a (p, q)-harmonic function of the form (5.1) for some sequence {c } of complex numbers satisfying (5.2). Then m=−∞ i θ −imθ c = lim u(re )e dθ |m| r →0 2πr for m ∈ Z. Proof We consider the case m ∈ N. Using the expansion (5.1) it is straightforward to check that i θ −imθ 2 m u(re )e dθ = c F (−p, m − q; m + 1; r )r 2π for 0 < r < 1. We next divide by r and pass to the limit to see that i θ −imθ lim u(re )e dθ = c F (−p, m − q; m + 1; 0) = c . m m r →0 2πr This yields the conclusion of the proposition when m ∈ N. The remaining case m ∈ Z\N is proved similarly or follows by hermitian symmetry. We omit the details. The integral quantity i θ −imθ I (r ) = u(re )e dθ (5.3) |m| 2πr in Proposition 5.2 is naturally thought of as a line integral. In fact, 1 u(z) 1 u(z) I (r ) = dz and I (r ) =− dz ¯ m −m m+1 m+1 2πi z 2πi z ¯ |z|=r |z|=r for m ∈ N and 0 < r < 1, where the circle of integration is traversed once in positive direction. These formulas are straightforward to check. 122 Page 22 of 28 M. Klintborg, A. Olofsson Theorem 5.3 Let p, q ∈ C. Let u be a (p, q)-harmonic function of the form (5.1) for some sequence {c } of complex numbers satisfying (5.2). Then m=−∞ m m c = ∂ u(0)/m! and c = ∂ u(0)/m! m −m for m ∈ N. Proof We shall calculate the integral limit in Proposition 5.2.Let I (r ) be as in (5.3). Recall from Corollary 4.9 that u ∈ C (D). Consider the Taylor expansion of u at the origin of degree m ≥ 0: j k j k m+1 u(z) = ∂ ∂ u(0)z z ¯ + O(|z| ) j !k! j ,k≥0 j +k≤m as z → 0. We have that 1 1 j k j +k i ( j −k)θ −imθ I (r ) = ∂ ∂ u(0)r e e dθ + O(r ) 2πr j !k! j ,k≥0 j +k≤m = ∂ u(0)/m!+ O(r ) as r → 0, where the last equality follows by cancellation. A passage to the limit as r → 0 now yields that c = ∂ u(0)/m!. We now turn to the formula for c . A similar analysis as in the previous paragraph −m shows that I (r ) = ∂ u(0)/m!+ O(r ) −m as r → 0. A passage to the limit as r → 0 yields that c = ∂ u(0)/m!.This −m completes the proof of the theorem. The coefﬁcient formulas in Theorem 5.3 leads to an addendum to Theorem 5.1. Theorem 5.4 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Then ∂ u(0) 2 m u(z) = F (−p, m − q; m + 1;|z| )z m! m=0 ∂ u(0) 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D, where F is the hypergeometric function (0.5). Proof Recall the series expansion (5.1) and (5.2) established in Theorem 5.1.By m m Theorem 5.3 we have that c = ∂ u(0)/m! and c = ∂ u(0)/m! for m ∈ N.This m −m yields the conclusion of the theorem. A series expansion for generalized harmonic functions Page 23 of 28 122 As a by-product of Theorem 5.4 we have a uniqueness result of classical type. Corollary 5.5 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Assume that u(0) = 0 and m m ∂ u(0) = ∂ u(0) = 0 for m = 1, 2,... . Then u(z) = 0 for all z ∈ D. Proof The result is evident by Theorem 5.4. 6 Further properties of the function u p,q Let u be as in (0.2)for some p, q ∈ C. In this section we derive the series expansion p,q of u (see Theorem 6.3). Of interest are also properties of integral means of u p,q p,q (see Theorems 6.4 and 6.6). Lemma 6.1 Let u be as in (0.2) for some p, q ∈ C. Let m ∈ N. Then p,q (p + 1) ∂ u (z) = u (z) +¯zg (z), z ∈ D, (6.1) p,q p,q m (1 − z) where g ∈ C (D). Proof We prove the lemma by induction on m ∈ N. Clearly (6.1) holds for m = 0 with g = 0. By Lemma 1.1 we have that p + 1 ∂u (z) = u (z) +¯zg (z), z ∈ D, (6.2) p,q p,q 1 1 − z with p + q + 1 g (z) =− u (z), z ∈ D. 1 p,q 1 −|z| This proves (6.1)for m = 1. Assume next that (6.1) holds for some m = n ≥ 1, that is, (p + 1) ∂ u (z) = u (z) +¯zg (z), z ∈ D, p,q p,q n (1 − z) where g ∈ C (D). Differentiating we have that n (p + 1) n+1 ∂ u (z) = (p + 1) u (z) + ∂u (z) +¯z∂g (z) p,q n p,q p,q n n+1 n (1 − z) (1 − z) 122 Page 24 of 28 M. Klintborg, A. Olofsson for z ∈ D.Wenextuse (6.2) to see that n (p + 1) p + 1 n+1 ∂ u (z) = (p + 1) u (z) + u (z) +¯zg (z) p,q n p,q p,q 1 n+1 n (1 − z) (1 − z) 1 − z (p + 1) n+1 +¯z∂g (z) = u (z) +¯zg (z) n p,q n+1 n+1 (1 − z) for z ∈ D, where (p + 1) g (z) = g (z) + ∂g (z), z ∈ D. n+1 1 n (1 − z) This proves (6.1)for m = n + 1. The conclusion of the lemma now follows by the principle of induction. Notice that formula (6.1) interprets naturally as a congruence in C (D) modulo zC ¯ (D). Theorem 6.2 Let u be as in (0.2) for some p, q ∈ C. Then p,q m m ∂ u (0) = (p + 1) and ∂ u (0) = (q + 1) p,q m p,q m for m ∈ N. Proof We shall use Lemma 6.1. A point evaluation at the origin in (6.1) shows that ∂ u (0) = (p + 1) for m ∈ N. Moreover, by hermitian symmetry u ¯ = u p,q m p,q q ¯ ,p ¯ we have that m m ¯ ¯ ∂ u (0) = ∂ u ¯ (0) = ∂ u (0) = (q ¯ + 1) = (q + 1) p,q q ¯ ,p ¯ q ¯ ,p ¯ m m for m ∈ N. We can now derive the series expansion of the function u . p,q Theorem 6.3 Let p, q ∈ C. Then 2 p+q+1 (1 −|z| ) (p + 1) 2 m = F (−p, m − q; m + 1;|z| )z p+1 q+1 (1 − z) (1 −¯z) m! m=0 (q + 1) 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D, where F is the hypergeometric function (0.5). A series expansion for generalized harmonic functions Page 25 of 28 122 Proof Let u be as in (0.2). From Theorem 1.4 we know that the function u is p,q p,q (p, q)-harmonic. By Theorem 5.4 it has the function series expansion ∂ u (0) p,q 2 m u (z) = F (−p, m − q; m + 1;|z| )z p,q m! m=0 ∂ u (0) p,q 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D. The result now follows from Theorem 6.2. We point out that Theorem 6.3 generalizes a well-known partial fraction de- composition formula for the classical Poisson kernel for D, see formula (1.1). Suitably specialized Theorem 6.3 yields [21, Theorem 2.5] and [18, Theorem 3.2]. Notice that the series expansion in Theorem 6.3 is absolutely convergent in C (D). We denote by Re(z) and Im(z) the real and imaginary parts of a complex number z, respectively. The standard Gamma function is deﬁned by z−1 −t (z) = t e dt for Re(z)> 0. It is well-known that continues to a meromorphic function in C with simple poles at the points z = 0, −1, −2,... . We next turn to L means of u . p,q Theorem 6.4 Let u be as in (0.2) for some p, q ∈ C such that Re(p)+Re(q)> −1. p,q Then 1 (Re(p) + Re(q) + 1) i θ |Im( p)−Im(q)| |u (re )| dθ ≤ e p,q Re( p)+Re(q) 2π T ( + 1) for 0 ≤ r < 1, where is the Gamma function. Proof From deﬁnition of powers we have that ( p+q+1) log(1−|z| ) −( p+q+2) log|1−z| −i ( p−q) arg(1−z) u (z) = e e e p,q for z ∈ D. Notice that the complex number 1 − z lies in the open right half-plane when z ∈ D. Passing to absolute values we have that 2 Re( p)+Re(q)+1 (1 −|z| ) |Im( p)−Im(q)| |u (z)|≤ e p,q Re( p)+Re(q)+2 |1 − z| for z ∈ D. We next apply [18, Theorem 3.1] with α = Re(p) + Re(q) to conclude that 1 π (α + 1) i θ |Im( p)−Im(q)| |u (re )| dθ ≤ e p,q 2π (α/2 + 1) T 122 Page 26 of 28 M. Klintborg, A. Olofsson for 0 ≤ r < 1. This yields the conclusion of the theorem. Remark 6.5 A similar analysis as in the proof of Theorem 6.4 shows that i θ lim |u (re )| dθ =+∞ p,q r →1 2π if Re(p) + Re(q) ≤−1. We omit the details. A classical result known as Gauss summation formula says that (c) (c − a − b) lim F (a, b; c; x ) = (6.3) x →1 (c − a) (c − b) when a, b ∈ C and c ∈ C\{0, −1, −2,... } are such that Re(c)> Re(a) + Re(b) (see [3, Theorem 2.2.2]). Theorem 6.6 Let p, q ∈ C be such that Re(p) + Re(q)> −1. Then 1 (p + q + 1) i θ lim u (re ) dθ = , p,q r →1 2π (p + 1) (q + 1) where is the Gamma function. Proof Recall the series expansion of the function u established in Theorem 6.3.By p,q cancellation of terms we have that i θ 2 u (re ) dθ = F (−p, −q; 1; r ) p,q 2π for 0 ≤ r < 1. Since Re(p) + Re(q)> −1, we can apply Gauss summation formula (6.3) to calculate the limit of the above quantity as r → 1. This yields the conclusion of the theorem. 7 The Dirichlet problem: concluding remarks In a restricted range of parameters, the class of (p, q)-harmonic functions can be analyzed in terms of their boundary values. We shall discuss in this section some rudiments of such theory. Let p, q ∈ C\Z be such that Re(p) + Re(q)> −1. The (p, q)-harmonic Poisson kernel is deﬁned by 2 p+q+1 (1 −|z| ) K (z) = c , z ∈ D, p,q p,q p+1 q+1 (1 − z) (1 −¯z) where (p + 1) (q + 1) c = . p,q (p + q + 1) A series expansion for generalized harmonic functions Page 27 of 28 122 Notice that the constant c is a non-zero complex number in this range of parameters. p,q By Theorem 1.4 the function K is (p, q)-harmonic. Theorems 6.4 and 6.6 ensure p,q that the function K satisﬁes some standard properties for approximate identities. p,q The (p, q)-harmonic Poisson integral is deﬁned by −i θ i θ K [ f ](z) = K (ze ) f (e ) dθ, z ∈ D, p,q p,q 2π for integrable functions f ∈ L (T) on T. Let C (T) be the space of continuous functions on T and ﬁx ϕ ∈ C (T).Bythe (p, q)-harmonic Dirichlet problem for ϕ we understand the problem of ﬁnding a (p, q)-harmonic function u such that lim u = ϕ in C (T), where r →1 r i θ i θ i θ u (e ) = u(re ), e ∈ T, for 0 < r < 1. Following usual practice, we formulate this latter Dirichlet problem as L u =0in D, p,q (7.1) u = ϕ on T, where L is as in (0.1). p,q Theorem 7.1 Let p, q ∈ C\Z be such that Re(p) + Re(q)> −1. Let ϕ ∈ C (T). Then a function u in D satisﬁes (7.1) if and only if it has the form u(z) = K [ϕ](z), z ∈ D. p,q The proof of Theorem 7.1 follows a standard scheme for such results and is therefore omitted, see [18,21] for details. The present paper suggests a ﬁner study of (p, q)-harmonic functions. Earlier results of such type concern Poisson integral representations, pointwise boundary limits, Green functions and Lipschitz continuity of generalized harmonic functions, see [18–21]. We mention here also work of Ahern and collaborators [1,2]. Funding Open access funding provided by Lund University. Declarations Conﬂict of interest The authors declare that they have no conﬂict of interest. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted 122 Page 28 of 28 M. Klintborg, A. Olofsson by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. 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Publisher: Springer Journals
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ISSN: 1664-2368
eISSN: 1664-235X
DOI: 10.1007/s13324-021-00561-w
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Abstract

We consider a class of generalized harmonic functions in the open unit disc in the complex plane. Our main results concern a canonical series expansion for such func- tions. Of particular interest is a certain individual generalized harmonic function which suitably normalized plays the role of an associated Poisson kernel. Keywords Harmonic function · Power series · Poisson kernel · Hypergeometric function Mathematics Subject Classiﬁcation Primary 31A05; Secondary 33C05 · 35J15 0 Introduction Let D be the open unit disc in the complex plane C and denote by ∂ = ∂/∂z and ∂ = ∂/∂z ¯ the usual complex partial derivatives. This work is concerned with second order partial differential operators of the form ¯ ¯ L = (1 −|z| )∂ ∂ + pz∂ + qz ¯∂ − pq, z ∈ D, (0.1) p,q z z z z where p, q ∈ C are complex parameters. Of particular interest are solutions of the associated homogeneous equation L u =0in D. p,q We say that a function u is (p, q)-harmonic if u is twice continuously differentiable in D (in symbols u ∈ C (D)) and L u = 0in D, where L is as in (0.1). Notice p,q p,q Anders Olofsson anders.olofsson@math.lu.se Markus Klintborg kborgmarkus@protonmail.com Mathematics, Faculty of Science, Centre for Mathematical Sciences, Lund University, P.O. Box 118, 221 00 Lund, Sweden 0123456789().: V,-vol 122 Page 2 of 28 M. Klintborg, A. Olofsson that a function u is (p, q)-harmonic if and only if its complex conjugate u ¯ is (q ¯ , p ¯)- harmonic. Observe also that a (0, 0)-harmonic function is a harmonic function in D in the usual sense. An interesting example of a (p, q)-harmonic function is the function 2 p+q+1 (1 −|z| ) u (z) = , z ∈ D (0.2) p,q p+1 q+1 (1 − z) (1 −¯z) (see Theorem 1.4). Here powers are deﬁned in the usual way using the principal branch of the logarithm, that is, we require that log(1) = 0. Notice that the above functions u have the hermitian symmetry property that u ¯ = u for p, q ∈ C. p,q p,q q ¯ ,p ¯ Recall that elements of the unit circle T = ∂ D act on the unit disc D as rotations. On a function level we consider rotation operators i θ R u(z) = u(e z), z ∈ D, (0.3) i θ i θ for e ∈ T acting on functions u in D. A basic observation concerning the differential operator L is the commutativity relation p,q L R i θ u = R i θ L u, u ∈ C (D), p,q p,q e e i θ for e ∈ T. This latter commutativity relation suggests an analysis of (p, q)-harmonic functions using concepts natural to classical Fourier analysis on the unit circle. Let Z be the set of integers. For a suitably smooth function u in D we deﬁne its m-th homogeneous part by the formula −imθ i θ u (z) = e u(e z) dθ, z ∈ D, (0.4) 2π for m ∈ Z. Notice that the m-th homogeneous part u of u is the m-th Fourier coefﬁcient of the vector-valued function i θ T e → R i θ u, where R i θ is as in (0.3). ∞ n n We set C (D) = C (D), where C (D) is the space of n-times continuously n=0 differentiable functions in D for n ∈ N ={0, 1, 2,... }. We topologize these spaces in the usual way using the semi-norms j k u = max|∂ ∂ u(z)|, j ,k;K z∈K where j , k ∈ N are non-negative integers and K ⊂ D is a compact subset of D. Let us recall the classical hypergeometric function deﬁned by (a) (b) z n n F (a, b; c; z) = , z ∈ D, (0.5) (c) n! n=0 A series expansion for generalized harmonic functions Page 3 of 28 122 for a, b, c ∈ C such that c =−1, −2,... .Here (a) = 1 and (a) = a(a + 1)...(a + n − 1) for n = 1, 2,... are Pochhammer symbols. Let us return to a (p, q)-harmonic function u. We show that the m-th homogeneous part of u has the form 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, m m for some c ∈ C when m ∈ N is a non-negative integer (see Theorem 4.3). Here F is the hypergeometric function (0.5). Hermitian symmetry leads to a similar formula for u when m ∈ Z = Z\N is a negative integer (see Corollary 4.4). Notice that u ∈ C (D) for m ∈ Z. A principal result of the present paper concerns the asymptotic behavior of the m-th homogeneous part u of a (p, q)-harmonic function u. We show that 1/|m| lim supu < 1 j ,k;K |m|→∞ for all j , k ∈ N and K ⊂ D compact (see Theorem 4.6). This result enables us to use the classical root test to establish absolute convergence of the function series u in C (D) (see Corollary 4.7). m=−∞ A further analysis leads to a function series characterization of (p, q)-harmonic functions. A function u in D is (p, q)-harmonic if and only if it has the form 2 m u(z) = c F (−p, m − q; m + 1;|z| )z m=0 2 m + c F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, (0.6) −m m=1 for some sequence {c } of complex numbers such that m=−∞ 1/|m| lim sup|c | ≤ 1 |m|→∞ (see Theorem 5.1). As mentioned above, the sums in (0.6) are absolutely convergent in ∞ ∞ the space C (D). As a consequence we have that u ∈ C (D) if u is (p, q)-harmonic (see Corollary 4.9). This characterization of (p, q)-harmonic functions improves on a result by Ahern et al. [1, Theorem 2.1] when specialized to our setting. We then turn to coefﬁcient formulas for (p, q)-harmonic functions. We show that m m c = ∂ u(0)/m! and c = ∂ u(0)/m! m −m 122 Page 4 of 28 M. Klintborg, A. Olofsson for m ∈ N, where the c ’s are as in (0.6) (see Theorem 5.3). As a consequence, we have that ∂ u(0) 2 m u(z) = F (−p, m − q; m + 1;|z| )z m! m=0 ∂ u(0) 2 m + F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, m! m=1 whenever u is a (p, q)-harmonic function (see Theorem 5.4) as well as a corresponding uniqueness result for such functions (see Corollary 5.5). Let us return to the function u in (0.2). A calculation of partial derivatives at the p,q origin of the function u leads to the series expansion p,q (p + 1) 2 m u (z) = F (−p, m − q; m + 1;|z| )z p,q m! m=0 (q + 1) 2 m + F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D m! m=1 (see Theorem 6.3). This latter series expansion generalizes a well-known partial frac- tion decomposition formula for the classical Poisson kernel for D which is obtained for p = q = 0. A main contribution of this paper concerns series expansion of (p, q)-harmonic functions. Of particular mention is a limit theorem for associated hypergeometric functions: lim F (−p, m − q; m + 1; z) = (1 − z) , z ∈ D m→∞ (see Theorem 2.6). Apart from its intrinsic interest, this limit theorem provides an efﬁcient tool for the study of limit properties of homogeneous parts of (p, q)-harmonic functions. The results of this paper have applications to Poisson integral representations of (p, q)-harmonic functions which is possible when p, q ∈ C\Z are such that Re(p)+ Re(q)> −1. In the ﬁnal section of the paper we comment brieﬂy on the connection to such theory. We have traced the study of (p, q)-harmonic functions back to Daryl Geller [13]. Other signiﬁcant contributions are those of Ahern and collaborators [1,2]. Our interest in those topics [18–21] arose in connection to standard weights and earlier work of Garabedian [12]. Borichev and Hedenmalm [6] have established a connection to poly- harmonic theory. Other recent related papers are [5,8–10,14,16,17,22]. The authors thank the referee for a careful reading of the manuscript. A series expansion for generalized harmonic functions Page 5 of 28 122 1 The function u is (p, q)-harmonic p,q Let u be as in (0.2)for some p, q ∈ C. Observe that the functions u have the p,q p,q hermitian symmetry property u ¯ = u . Indeed, p,q q ¯ ,p ¯ 2 p ¯+¯ q+1 (1 −|z| ) u ¯ (z) = u (z) = = u (z) p,q p,q q ¯ ,p ¯ p ¯+1 q ¯ +1 (1 −¯z) (1 − z) for z ∈ D. We shall discuss in this section some differentiation formulas for the functions u . In particular, we shall show that the function u is (p, q)-harmonic p,q p,q (see Theorem 1.4). Lemma 1.1 Let u be as in (0.2) for some p, q ∈ C. Then p,q |z| z z∂u (z) = − (p + q + 1) + (p + 1) u (z) p,q p,q 1 −|z| 1 − z for z ∈ D. Proof Differentiating using the product rule for differentiation we have that 1 1 2 p+q ∂u (z) = (p + q + 1)(1 −|z| ) (−¯z) p,q q+1 p+1 (1 −¯z) (1 − z) 2 p+q+1 + (1 −|z| ) (−(p + 1)) (−1) p+2 (1 − z) z ¯ 1 = − (p + q + 1) + (p + 1) u (z) p,q 1 −|z| 1 − z for z ∈ D, where the last equality is straightforward to check. This yields the conclusion of the lemma. We next turn to the ∂-derivative of u . p,q Lemma 1.2 Let u be as in (0.2) for some p, q ∈ C. Then p,q |z| z ¯ z ¯∂u (z) = − (p + q + 1) + (q + 1) u (z) p,q p,q 1 −|z| 1 −¯z for z ∈ D. Proof We shall use the hermitian symmetry property u ¯ = u . From Lemma 1.1 p,q q ¯ ,p ¯ we have that |z| z z∂u (z) = − (p ¯ +¯ q + 1) + (q ¯ + 1) u (z) q ¯ ,p ¯ q ¯ ,p ¯ 1 −|z| 1 − z for z ∈ D. A complex conjugation now yields the conclusion of the lemma. 122 Page 6 of 28 M. Klintborg, A. Olofsson Following earlier practice from [20] we denote by A = z∂ −¯z∂ the angular deriva- tive. We next calculate the angular derivative of u . p,q Corollary 1.3 Let u be as in (0.2) for some p, q ∈ C. Then p,q z z ¯ (z∂ −¯z∂)u (z) = (p + 1) − (q + 1) u (z) p,q p,q 1 − z 1 −¯z for z ∈ D. Proof The result is evident from Lemmas 1.1 and 1.2. We mention that Corollary 1.3 generalizes [20, Theorem 1.11]. Recall the partial fraction formula 1 −|z| z z ¯ = + + 1 (1.1) |1 − z| 1 − z 1 −¯z for the classical Poisson kernel for the unit disc. Formula (1.1) is straightforward to check. Theorem 1.4 Let u be as in (0.2) for some p, q ∈ C. Then L u = 0 in D, p,q p,q p,q where L is as in (0.1). p,q Proof Recall Lemma 1.2. Notice that the differential operator z∂ satisﬁes the product rule for differentiation. It is straightforward to check that 2 2 ∂ |z| |z| z = 2 2 2 ∂z 1 −|z| (1 −|z| ) whenever the formula makes sense. Differentiating using the product rule we have that |z| |z| ∂∂u (z) =−(p + q + 1) u (z) p,q p,q 2 2 (1 −|z| ) |z| z ¯ + − (p + q + 1) + (q + 1) z∂u (z) p,q 1 −|z| 1 −¯z for z ∈ D. From Lemma 1.1 we now have that |z| |z| ∂∂u (z) =−(p + q + 1) u (z) p,q p,q 2 2 (1 −|z| ) |z| z ¯ + − (p + q + 1) + (q + 1) 1 −|z| 1 −¯z |z| z × − (p + q + 1) + (p + 1) u (z) p,q 1 −|z| 1 − z A series expansion for generalized harmonic functions Page 7 of 28 122 for z ∈ D. Expanding the above product we see that 2 4 |z| |z| 2 2 |z| ∂∂u (z) = − (p + q + 1) + (p + q + 1) u (z) p,q p,q 2 2 2 2 (1 −|z| ) (1 −|z| ) |z| z z ¯ − (p + q + 1) (p + 1) + (q + 1) u (z) p,q 1 −|z| 1 − z 1 −¯z |z| + (p + 1)(q + 1) u (z) p,q |1 − z| 2 2 for z ∈ D. Multiplying by a factor (1 −|z| )/|z| we see that |z| 1 2 2 (1 −|z| )∂∂u (z) = (p + q + 1) − (p + q + 1) u (z) p,q p,q 2 2 1 −|z| 1 −|z| z z ¯ − (p + q + 1) (p + 1) + (q + 1) u (z) p,q 1 − z 1 −¯z 1 −|z| + (p + 1)(q + 1) u (z) p,q |1 − z| for z ∈ D. Notice an appearance of the classical Poisson kernel in the rightmost term above. Using the partial fraction formula (1.1) we have that |z| (1 −|z| )∂∂u (z) = (p + q + 1) (p + q) − 1 u (z) p,q p,q 1 −|z| + ((p + 1)(q + 1) − (p + q + 1)(p + 1)) u (z) p,q 1 − z z ¯ + ((p + 1)(q + 1) − (p + q + 1)(q + 1)) u (z) p,q 1 −¯z + (p + 1)(q + 1)u (z) p,q for z ∈ D. A simpliﬁcation of terms now leads to the formula |z| (1 −|z| )∂∂u (z) = (p + q + 1)(p + q) u (z) p,q p,q 1 −|z| z z ¯ − p(p + 1) u (z) − q(q + 1) u (z) + pqu (z) (1.2) p,q p,q p,q 1 − z 1 −¯z for z ∈ D. Recall Lemmas 1.1 and 1.2. In view of these two lemmas our latter formula (1.2) says that L u = 0in D. p,q p,q 122 Page 8 of 28 M. Klintborg, A. Olofsson 2 A sequence of hypergeometric functions Let us ﬁrst consider a second order partial differential operator of the form ¯ ¯ L = (1 −|z| )∂ ∂ + pz∂ + qz ¯∂ − r , z ∈ D, (2.1) p,q;r z z z z where p, q, r ∈ C are complex parameters. A principal case is when r = pq. Notice that L = L , where L is as in (0.1). The introduction of an additional p,q;pq p,q p,q parameter r ∈ C allows for more general operators appearing in the study of conduc- tivity problems, see for instance Calderón [7] or Astala and Päivärinta [4]. We shall evaluate the operator L on a complex-valued function u in the punc- p,q;r tured disc D\{0} of the form 2 m u(z) = f (|z| )z , z ∈ D\{0}, (2.2) for some f ∈ C (0, 1) and m ∈ Z. We introduce also the ordinary differential operator d d H = (1 − x )x + (c −[a + b + 1]x ) − ab, (2.3) a,b;c dx dx where a, b, c ∈ C are complex parameters. Notice that the famous hypergeometric ordinary differential equation (1 − x )xy (x ) + (c −[a + b + 1]x )y (x ) − aby(x ) = 0 takes the form H y = 0 using the operator H . a,b;c a,b;c Theorem 2.1 Let L be as in (2.1) for some p, q, r ∈ C. Let u be a function of the p,q;r form (2.2) for some f ∈ C (0, 1) and m ∈ Z. Then m 2 L u(z) = z H f (|z| ), z ∈ D\{0}, p,q;r a,b;c where c = m + 1 and a + b = m − p − q, (2.4) ab = r − pm. Proof A differentiation shows that m−1 2 m 2 ∂u(z) = mz f (|z| ) + z zf ¯ (|z| ) for z ∈ D\{0}, and similarly that m+1 2 ∂u(z) = z f (|z| ) A series expansion for generalized harmonic functions Page 9 of 28 122 for z ∈ D\{0}. Another differentiation gives that m 2 m+1 2 ∂∂u(z) = (m + 1)z f (|z| ) + z zf ¯ (|z| ) for z ∈ D\{0}. A calculation using these formulas gives that 2 m 2 m 2 2 m+1 2 L u(z) = (1 −|z| )((m + 1)z f (|z| ) + z |z| f (|z| )) + qzz ¯ f (|z| ) p,q;r m−1 2 m 2 m 2 + pz(mz f (|z| ) + z zf ¯ (|z| )) − rz f (|z| ) m 2 2 2 2 2 = z |z| (1 −|z| ) f (|z| ) + (m + 1 −[m + 1 − p − q]|z| ) f (|z| ) 2 m 2 − (r − pm) f (|z| ) = z H f (|z| ) a,b;c for z ∈ D\{0}, where c = m + 1, the numbers a and b areasin(2.4) and H is as a,b;c in (2.3). This yields the conclusion of the theorem. Equations (2.4) say that a and b are the zeros of the quadratic polynomial P (λ) = λ − (m − p − q)λ + r − pm,λ ∈ C. p,q;r ;m Notice that P (λ) = (λ + p)(λ + q − m) + r − pq,λ ∈ C. p,q;r ;m In particular, the zeros of P are −p and m − q in the principal case when p,q;r ;m r = pq. Theorem 2.1 suggests a natural construction of (p, q)-harmonic functions. Proposition 2.2 Let p, q ∈ C. Consider the function 2 m u (z) = F (−p, m − q; m + 1;|z| )z , z ∈ D, where m ∈ N and F is the hypergeometric function (0.5). Then u is a (p, q)-harmonic function. Proof Clearly u ∈ C (D). It is well-known that the hypergeometric function y = F (a, b; c;·) satisﬁes the hypergeometric equation H y = 0(see [3, Section 2.3]). a,b;c The result now follows by Theorem 2.1. Corollary 2.3 Let p, q ∈ C. Consider the function 2 |m| u (z) = F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, where m ∈ Z and F is the hypergeometric function (0.5). Then u is a (p, q)- harmonic function. 122 Page 10 of 28 M. Klintborg, A. Olofsson Proof We consider the complex conjugate 2 |m| u (z) = F (−¯ q, |m|− p¯;|m|+ 1;|z| )z , z ∈ D. By Proposition 2.2 we have that u ¯ is a (q ¯ , p ¯)-harmonic function. From hermitian symmetry we conclude that u is a (p, q)-harmonic function. Following earlier practice, a function u in D\{0} is said to be homogeneous of order m ∈ Z with respect to rotations if it has the property that i θ imθ u(e z) = e u(z), z ∈ D\{0}, i θ for e ∈ T. Notice that every function u of the form (2.2) is homogeneous of order m with respect to rotations. Theorem 2.4 Let p, q ∈ C and m ∈ N. Let u ∈ C (D) be homogeneous of order m with respect to rotations. Then u is (p, q)-harmonic if and only if it has the form 2 m u(z) = cF (−p, m − q; m + 1;|z| )z , z ∈ D, (2.5) for some c ∈ C, where F is the hypergeometric function (0.5). Proof From Proposition 2.2 we know that every function u of the form (2.5)is (p, q)- harmonic. Assume next that u is (p, q)-harmonic. Since u is homogeneous of order m,we can put u on the form (2.2)for some f ∈ C (0, 1). By Theorem 2.1 we have that H f (x ) = 0, 0 < x < 1, −p,m−q,m+1 where H is as in (2.3). Below we shall check that a,b;c m+1 (m + 1) f (x ) + p(m − q) f (x ) = o(1/x ) (2.6) as x → 0. Condition (2.6) allows us to apply [18, Proposition 1.3] to conclude that f is a constant multiple of the hypergeometric function F (−p, m − q; m + 1;·).This will then complete the proof of the theorem. We proceed to check (2.6). Recall formula (2.2). Since u is bounded near the origin m/2 we have that f (x ) = O(1/x ) as x → 0. A differentiation of (2.2) gives that m+1 2 ∂u(z) = z f (|z| ), z ∈ D\{0}. (m+1)/2 Since ∂u is bounded near the origin we have that f (x ) = O(1/x ) as x → 0. We have now checked (2.6). Theorem 2.4 and its proof are modeled on [18, Theorem 2.1]. We have merely supplied some details. A series expansion for generalized harmonic functions Page 11 of 28 122 We shall use the fact that the hypergeometric functions are analytic in D.Let H (D) be the space of analytic functions in D. The space H (D) is topologized in the usual manner using the semi-norms f = max| f (z)| (2.7) z∈K for K ⊂ D compact. Convergence in the space H (D) is usually referred to as normal convergence in D. Recall the terminology that a subset F of H (D) is called a normal family if every sequence of functions of F has a subsequence which converges in H (D). The limit function is not required to belong to F. We refer to Conway [11, Chapter VII] for background. Recall also the binomial series: (a) 1 F (a, b; b; z) = z = (2.8) n! (1 − z) n=0 for z ∈ D. Lemma 2.5 Let p, q ∈ C and consider the functions f (z) = F (−p, m − q; m + 1; z), z ∈ D, for m ∈ N, where F is the hypergeometric function (0.5). Then F ={ f : m ∈ N} is a normal family of analytic functions in D. Proof We shall prove that the functions in F are uniformly bounded on compact subsets of D. The conclusion of the lemma then follows by a classical result of Montel (see Conway [11, Theorem VII.2.9]). Let K ⊂ D be compact. Since K ⊂ D is compact there exists 0 < r < 1 such that max |z| < r. Choose N such that z∈K |1 − (q + 1)/(m + 1)|≤ 1/r for m > N . Observe that n−1 n−1 (m − q) m − q + k q + 1 = = 1 − . (m + 1) m + 1 + k m + 1 + k k=0 k=0 Therefore |(m − q) /(m + 1) |≤ 1/r for n ∈ N provided m > N . n n We now estimate the f ’s with m > N.From(0.5) we have that (−p) (m − q) z n n f (z) = , z ∈ D. (m + 1) n! n=0 122 Page 12 of 28 M. Klintborg, A. Olofsson Notice in this sum an appearance of the quotient (m − q) /(m + 1) considered in n n the previous paragraph. From the triangle inequality and the result of the previous paragraph we have that (| p|) |z| 1 | f (z)|≤ = n | p| n! r (1 −|z|/r ) n=0 for |z| < r and m > N , where the last equality follows by the binomial series (2.8). This proves that the functions in F are uniformly bounded on K . The following limit theorem will be much useful. Theorem 2.6 Let p, q ∈ C. Then lim F (−p, m − q; m + 1; z) = (1 − z) , z ∈ D, m→∞ with normal convergence, where F is the hypergeometric function (0.5). Proof Let the f ’s be as in Lemma 2.5. From Lemma 2.5 we have that the set F = { f : m ∈ N} is a normal family. From (0.5) we have that (−p) (m − q) n n (n) f (0) = (m + 1) for m, n ∈ N. Recall that the Pochhammer symbol (·) is a monic polynomial of (n) degree n. Passing to the limit we have that lim f (0) = (−p) for n ∈ N. m→∞ m n (n) From (2.8) we have that f (0) = (−p) for n ∈ N, where f (z) = (1 − z) , z ∈ D. A standard argument now yields that f → f in H (D) as m →∞. Assume to reach a contradiction that there exists a compact set K ⊂ D such that { f } does not m=0 converge uniformly to f on K . Passing to a subsequence we can assume that max| f (z) − f (z)|≥ δ> 0 (2.9) z∈K for k = 1, 2,... . Since the set F is a normal family, we can after passage to another subsequence if necessary, assume that f → g in H (D) as k →∞ for some (n) g ∈ H (D). From the ﬁrst paragraph of the proof we have g (0) = (−p) for n ∈ N, and a uniqueness argument gives that g = f in D. Thus f → f in H (D) as k →∞, which contradicts (2.9). We emphasize that Theorem 2.6 appears much natural in view of the binomial series (2.8). A series expansion for generalized harmonic functions Page 13 of 28 122 3 A generalized power series From the product rule for differentiation we have that (n) (n−k) (k) ( fg) = f g k=0 for, say, f , g ∈ C [0, 1). Lemma 3.1 Let j , k, m ∈ N be such that m ≥ j. Let u be a function of the form (2.2) for some f ∈ C [0, 1). Then j (m + k)! j k m+k− j 2l (k+l) 2 ∂ ∂ u(z) = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D. Proof Recall (2.2). Differentiating with respect to z ¯ we have that k m+k (k) 2 ∂ u(z) = z f (|z| ) for z ∈ D. Another differentiation using the product rule for differentiation gives that j (m + k)! j k m+k−( j −l) (k+l) 2 l ∂ ∂ u(z) = z f (|z| )z ¯ l (m + k − ( j − l))! l=0 j (m + k)! m+k− j 2l (k+l) 2 = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D. This completes the proof of the lemma. Let C (D) be the set of smooth complex-valued functions in the unit disc D.The space C (D) is topologized by means of the family of semi-norms j k u = max|∂ ∂ u(z)|, (3.1) j ,k;K z∈K where j , k ∈ N and K ⊂ D is compact. Recall that u → u in C (D) as m →∞ means that lim u − u = 0 for all j , k ∈ N and K ⊂ D compact. m→∞ m j ,k;K A (formal) series u of functions u ∈ C (D) for m ∈ N is said to be m m m=0 absolutely convergent in C (D) if u < +∞ m j ,k;K m=0 122 Page 14 of 28 M. Klintborg, A. Olofsson whenever j , k ∈ N and K ⊂ D is compact. By completeness of C (D) we have that ∞ ∞ every series absolutely convergent in C (D) is convergent in C (D). ∞ ∞ Theorem 3.2 Let { f } be a sequence in C [0, 1) such that m=0 1/m (n) lim sup max | f (x )| ≤ 1 0≤x ≤r m→∞ for n ∈ N and 0 < r < 1. Set 2 m u (z) = f (|z| )z , z ∈ D, m m 1/m for m = 0, 1, 2,... . Then lim sup u < 1 for all j , k ∈ N and K ⊂ D m→∞ j ,k;K compact, where · is as in (3.1). j ,k;K Proof Fix j , k ∈ N and K ⊂ D compact. Set r = max |z| < 1. From Lemma 3.1 z∈K we have that j (m + k)! j k m+k− j 2l (k+l) 2 ∂ ∂ u (z) = z |z| f (|z| ) l (m + k + l − j )! l=0 for z ∈ D provided m ≥ j. We next apply the triangle inequality to see that j (m + k)! m+k− j 2l (n) u ≤ r r max f m j ,k;K m [0,r ] k≤n≤ j +k l (m + k + l − j )! l=0 m+k− j j 2 j (n) ≤ r (m + k) (1 + r ) max f (3.2) m [0,r ] k≤n≤ j +k for m ≥ j, where we have used the notation (2.7). We shall next apply the m-th root to (3.2) and pass to the limit as m →∞. In view of the assumption on the sequence { f } we conclude from (3.2) that m=0 1/m lim supu ≤ r . j ,k;K m→∞ Since 0 < r < 1, this yields the conclusion of the theorem. From the conclusion of Theorem 3.2 we have that u < +∞ when- m j ,k;K m=0 ever j , k ∈ N and K ⊂ D is compact. Thus the series u is absolutely m=0 convergent in C (D). We can think of a function series 2 m f (|z| )z (3.3) m=0 A series expansion for generalized harmonic functions Page 15 of 28 122 as a vector-valued power series. The sequence of coefﬁcients { f } in (3.3)isnow m=0 a sequence of functions in C [0, 1). From this point of view, Theorem 3.2 generalizes a well-known root criteria for power series (see for instance Conway [11, Theorem III.1.3]). Of particular concern to us are sequences { f } of the form m=0 f (x ) = F (−p, m − q; m + 1; x ), 0 ≤ x < 1, for m ∈ N, where p, q ∈ C and F is the hypergeometric function (0.5). We next observe that Theorem 2.6 guarantees that every such sequence { f } satisﬁes the k=0 assumption of Theorem 3.2. Proposition 3.3 Let p, q ∈ C. Then 1/m (n) lim sup max F (−p, m − q; m + 1; x ) ≤ 1 0≤x ≤r m→∞ for n ∈ N and 0 < r < 1, where F is the hypergeometric function (0.5). Proof Recall the fact that the complex derivative f → f is continuous in the topology of normal convergence of analytic functions (see Conway [11, Theorem VII.2.1]). Recall Montel’s theorem characterizing normal families of analytic functions (see Conway [11, Theorem VII.2.9]). In view of these two results, the proposition follows from Theorem 2.6. 4 Analysis of homogeneous parts i θ Notice that the rotation operators T e → R i θ from (0.3) have the group properties that R = I is the identity and R i (θ +τ) = R i θ R i τ e e i θ i τ for e , e ∈ T. Recall the notion of homogeneity with respect to rotations made precise in the paragraph before Theorem 2.4. Observe that a function u in D is homogeneous of imθ i θ order m ∈ Z with respect to rotations if and only if R i θ u = e u for e ∈ T. Proposition 4.1 Let u be the m-th homogeneous part of u ∈ C (D) for some m ∈ Z, where n ∈ N ∪{∞}. Then u ∈ C (D) and u is homogeneous of order m with m m respect to rotations. Proof Differentiations under the integral in (0.4) show that u ∈ C (D).Fromformula (0.4) we have that −imτ u = e R udτ (4.1) i τ 2π T 122 Page 16 of 28 M. Klintborg, A. Olofsson i θ in a vector-valued sense. Let e ∈ T. From the group property of rotation operators we have that 1 1 −imτ −imτ R u = e R R udτ = e R udτ i θ i θ i τ i (θ +τ) e e e e 2π 2π T T in a vector-valued sense. A change of variables now gives that imθ −imτ imθ R i θ u = e R i τ udτ = e u . m m e e 2π i θ Since e ∈ T is arbitrary, this yields that u is homogeneous of order m with respect to rotations. Below we shall make use of the commutativity relation L R i θ u = R i θ L u, u ∈ C (D), (4.2) p,q p,q e e i θ for e ∈ T. In order to prove (4.2) it sufﬁces to check that the differential operators z∂ and z ¯∂ commute with the rotations R i θ which is evident. We now return to (p, q)-harmonic functions. Proposition 4.2 Let p, q ∈ C. Let u be a (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ Z. Then u is (p, q)-harmonic. 2 2 Proof From Proposition 4.1 we have that u ∈ C (D) since u ∈ C (D). Recall formula (4.1). Applying the operator L we have that p,q −imτ L u = e L R i τ udτ p,q m p,q 2π in a vector-valued sense. We now use the commutativity relation (4.2) to conclude that −imτ L u = e R i τ L udτ = 0, p,q m p,q 2π where the last equality is evident since u is (p, q)-harmonic. Let us next calculate the homogeneous parts of a (p, q)-harmonic function. Theorem 4.3 Let p, q ∈ C. Let u be (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ N. Then 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, (4.3) m m for some c ∈ C, where F is the hypergeometric function (0.5). m A series expansion for generalized harmonic functions Page 17 of 28 122 Proof By Proposition 4.1, the function u is homogeneous of order m with respect to rotations. By Proposition 4.2, the function u is (p, q)-harmonic. The result now follows from Theorem 2.4. Corollary 4.4 Let p, q ∈ C. Let u be (p, q)-harmonic function and denote by u its m-th homogeneous part for some m ∈ Z\Z . Then 2 |m| u (z) = c F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, (4.4) m m for some c ∈ C, where F is the hypergeometric function (0.5). Proof The complex conjugate u ¯ is the −m =|m|-th homogeneous part of the func- tion u ¯. By hermitian symmetry, the function u ¯ is (q ¯ , p ¯)-harmonic. From Theorem 4.3 we have that 2 |m| u (z) = a F (−¯ q, |m|− p¯;|m|+ 1;|z| )z , z ∈ D, m m for some a ∈ C. A complex conjugation now yields (4.4) with c =¯a . m m m As a by-product from Theorem 4.3 and Corollary 4.4 we have that u ∈ C (D) for m ∈ Z if u is (p, q)-harmonic. Recall formula (0.4) for the homogeneous parts of a function u in D.Fromthe triangle inequality we have that sup |u (z)|≤ sup |u(z)| (4.5) |z|=r |z|=r for 0 < r < 1 and m ∈ Z. Notice that (4.5) generalizes a well-known bound for Fourier coefﬁcients. We next estimate the constants c appearing in Theorem 4.3 or Corollary 4.4. Lemma 4.5 Let p, q ∈ C. Let u be a (p, q)-harmonic function and consider its m-th homogeneous part u for m ∈ Z. Let c be as in (4.4) or (4.4) depending on whether m m − 1/|m| m ∈ N or m ∈ Z . Then lim sup |c | ≤ 1. |m|→∞ Proof For simplicity we consider the case m ∈ N. The case m ∈ Z is analogous or follows by hermitian symmetry. Let 0 < r < 1. From (4.4) and (4.5) we have that 2 m |c ||F (−p, m − q; m + 1; r )|r ≤ max|u(z)| (4.6) |z|=r for m ∈ N. From Theorem 2.6 we have that 2 2 p lim F (−p, m − q; m + 1; r ) = (1 − r ) = 0. m→∞ Therefore, a passage to the limit in (4.6) gives that 1/m lim sup|c | ≤ 1/r . m→∞ 122 Page 18 of 28 M. Klintborg, A. Olofsson 1/m Since 0 < r < 1 is arbitrary we conclude that lim sup |c | ≤ 1. This yields m→∞ the conclusion of the lemma. The following theorem is our main result about the asymptotic behavior of the homogeneous parts of a (p, q)-harmonic function. Theorem 4.6 Let p, q ∈ C. Let u be a (p, q)-harmonic function and denote by u its m-th homogeneous part for m ∈ Z. Then u ∈ C (D) for m ∈ Z and 1/|m| lim supu < 1 j ,k;K |m|→∞ for all j , k ∈ N and K ⊂ D compact. Proof From Theorem 4.3 and Corollary 4.4 we have that u ∈ C (D) for m ∈ Z. We consider ﬁrst the case m ∈ N. From Theorem 4.3 we have (4.4)for m ∈ N, where c ∈ C. Notice that the function u in (4.4) is constructed from the function m m f (x ) = c F (−p, m − q; m + 1; x ), 0 ≤ x < 1, m m as in Theorem 3.2. From Proposition 3.3 and Lemma 4.5 we have that the assumption of Theorem 3.2 is satisﬁed. An application of Theorem 3.2 yields the conclusion that 1/m lim sup u < 1for j , k ∈ N and K ⊂ D compact. m→∞ j ,k;K We next consider the case m ∈ Z . The complex conjugate u ¯ of u is the (−m)-th m m homogeneous part of the (q ¯ , p ¯)-harmonic function u ¯. From the result of the previous paragraph we thus have that 1/|m| 1/m lim supu = lim sup(u ¯) < 1 m m j ,k;K k, j ;K m→−∞ m→∞ for j , k ∈ N and K ⊂ D compact. This completes the proof of the theorem. We point out that Theorem 2.6 forms an integral part in the proof of Theorem 4.6. Recall the notion of absolute convergence in C (D), see the paragraph just before Theorem 3.2. Corollary 4.7 Let p, q ∈ C. Let u be a (p, q)-harmonic function with m-th homo- geneous part u for m ∈ Z. Then the function series u is absolutely m m m=−∞ convergent in C (D). Proof Theorem 4.6 allows us to apply the root test to conclude that u < +∞ m j ,k;K m=−∞ for j , k ∈ N and K ⊂ D compact. The following lemma is well-known but included for the sake of completeness. A series expansion for generalized harmonic functions Page 19 of 28 122 Lemma 4.8 Let u ∈ C (D) for some n ∈ N and denote by u its m-th homogeneous part for m ∈ Z. Then |m| u = lim 1 − u N →+∞ N + 1 m=−N in C (D). Proof It is straightforward to check that |m| 1 i θ 1 − u = K (e )R i θudθ m N N + 1 2π m=−N in a vector-valued sense, where |m| i θ imθ i θ K (e ) = 1 − e , e ∈ T, N + 1 m=−N is the Fejér kernel. It is well-known that the K ’s are non-negative and form an approximate identity as N →∞. Since u ∈ C (D), it is well-known that the function i θ n T e → R i θ u ∈ C (D) is continuous from T into C (D). The proof is now completed by a standard argument. We refer to Katznelson [15, Section I.2] for details. We think of Lemma 4.8 as a version of Fejér’s theorem adapted to our context. We next deduce that (p, q)-harmonic functions belong to the space C (D). Corollary 4.9 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Then u ∈ C (D) and u = u in C (D), where u is the m-th homogeneous part of u for m m m=−∞ m ∈ Z. Proof From Lemma 4.8 we have that |m| u = lim 1 − u (4.7) N →+∞ N + 1 m=−N in C (D) since u has such regularity. From Corollary 4.7 we know that the function series u is absolutely convergent in C (D). We can thus omit the conver- m=−∞ gence factors in (4.7) and deduce that u = u in C (D). By completeness m=−∞ we have that u ∈ C (D). 122 Page 20 of 28 M. Klintborg, A. Olofsson 5 A series expansion of harmonic functions The analysis from Sect. 4 leads to a natural function series description of (p, q)- harmonic functions. Theorem 5.1 Let p, q ∈ C. Then u is a (p, q)-harmonic function if and only if it has the form 2 m u(z) = c F (−p, m − q; m + 1;|z| )z m=0 2 m + c F (−q, m − p; m + 1;|z| )z ¯ , z ∈ D, (5.1) −m m=1 for some sequence {c } of complex numbers such that m=−∞ 1/|m| lim sup|c | ≤ 1, (5.2) |m|→∞ where F is the hypergeometric function (0.5). Moreover, the sums in (5.1) are abso- lutely convergent in the space C (D) when (5.2) holds. Proof Consider ﬁrst a (formal) function series of the form (5.1) with coefﬁcient sequence {c } satisfying (5.2). From Theorem 3.2 and Proposition 3.3 we have m=−∞ that 1/|m| lim supu < 1 j ,k;K |m|→∞ for j , k ∈ N and K ⊂ D compact, where 2 m u (z) = c F (−p, m − q; m + 1;|z| )z , z ∈ D, m m for m ∈ N and 2 |m| u (z) = c F (−q, |m|− p;|m|+ 1;|z| )z ¯ , z ∈ D, m m for m ∈ Z . The root test now applies to show that the series u in (5.1)is m=−∞ absolutely convergent in C (D). This deﬁnes a function u = u by (5.1). m=−∞ From Proposition 2.2 and Corollary 2.3 we have that each term u in (5.1)is (p, q)- harmonic. Since the series expansion (5.1) converges in the space C (D),wehave that u ∈ C (D). Applying the operator L we have L u = L u = 0 p,q p,q p,q m m=−∞ in C (D), which shows that u is (p, q)-harmonic. Consider next a (p, q)-harmonic function u. We proceed to derive (5.1). Denote by u the m-th homogeneous part of u for m ∈ Z. From Theorem 4.3 we have (4.4)for m ∈ N and from Corollary 4.4 we have (4.4)for m ∈ Z , where c ∈ C for m ∈ Z. From Corollary 4.9 we have (5.1) in the form that u = u in C (D). m=−∞ A series expansion for generalized harmonic functions Page 21 of 28 122 By Corollary 4.7 we have that the function series u in (5.1) is absolutely m=−∞ convergent in C (D). From Lemma 4.5 we have that (5.2) holds. We emphasize that the function series (5.1) has an interpretation of homogeneous expansion of the (p, q)-harmonic function u. We mention that Theorem 5.1 improves on a result by Ahern et al. [1, Theorem 2.1] when specialized to the present setting. Theorem 5.1 contains also some earlier results of Olofsson and Wittsten. Theorem 5.1 with (p, q) = (0,α), α ∈ R, yields [21, Theorem 1.2]. Theorem 5.1 with (p, q) = (α/2,α/2), α ∈ R, yields [18, Theorem 2.2]. We shall next turn to formulas for the coefﬁcients c appearing in (5.1). Proposition 5.2 Let p, q ∈ C. Let u be a (p, q)-harmonic function of the form (5.1) for some sequence {c } of complex numbers satisfying (5.2). Then m=−∞ i θ −imθ c = lim u(re )e dθ |m| r →0 2πr for m ∈ Z. Proof We consider the case m ∈ N. Using the expansion (5.1) it is straightforward to check that i θ −imθ 2 m u(re )e dθ = c F (−p, m − q; m + 1; r )r 2π for 0 < r < 1. We next divide by r and pass to the limit to see that i θ −imθ lim u(re )e dθ = c F (−p, m − q; m + 1; 0) = c . m m r →0 2πr This yields the conclusion of the proposition when m ∈ N. The remaining case m ∈ Z\N is proved similarly or follows by hermitian symmetry. We omit the details. The integral quantity i θ −imθ I (r ) = u(re )e dθ (5.3) |m| 2πr in Proposition 5.2 is naturally thought of as a line integral. In fact, 1 u(z) 1 u(z) I (r ) = dz and I (r ) =− dz ¯ m −m m+1 m+1 2πi z 2πi z ¯ |z|=r |z|=r for m ∈ N and 0 < r < 1, where the circle of integration is traversed once in positive direction. These formulas are straightforward to check. 122 Page 22 of 28 M. Klintborg, A. Olofsson Theorem 5.3 Let p, q ∈ C. Let u be a (p, q)-harmonic function of the form (5.1) for some sequence {c } of complex numbers satisfying (5.2). Then m=−∞ m m c = ∂ u(0)/m! and c = ∂ u(0)/m! m −m for m ∈ N. Proof We shall calculate the integral limit in Proposition 5.2.Let I (r ) be as in (5.3). Recall from Corollary 4.9 that u ∈ C (D). Consider the Taylor expansion of u at the origin of degree m ≥ 0: j k j k m+1 u(z) = ∂ ∂ u(0)z z ¯ + O(|z| ) j !k! j ,k≥0 j +k≤m as z → 0. We have that 1 1 j k j +k i ( j −k)θ −imθ I (r ) = ∂ ∂ u(0)r e e dθ + O(r ) 2πr j !k! j ,k≥0 j +k≤m = ∂ u(0)/m!+ O(r ) as r → 0, where the last equality follows by cancellation. A passage to the limit as r → 0 now yields that c = ∂ u(0)/m!. We now turn to the formula for c . A similar analysis as in the previous paragraph −m shows that I (r ) = ∂ u(0)/m!+ O(r ) −m as r → 0. A passage to the limit as r → 0 yields that c = ∂ u(0)/m!.This −m completes the proof of the theorem. The coefﬁcient formulas in Theorem 5.3 leads to an addendum to Theorem 5.1. Theorem 5.4 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Then ∂ u(0) 2 m u(z) = F (−p, m − q; m + 1;|z| )z m! m=0 ∂ u(0) 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D, where F is the hypergeometric function (0.5). Proof Recall the series expansion (5.1) and (5.2) established in Theorem 5.1.By m m Theorem 5.3 we have that c = ∂ u(0)/m! and c = ∂ u(0)/m! for m ∈ N.This m −m yields the conclusion of the theorem. A series expansion for generalized harmonic functions Page 23 of 28 122 As a by-product of Theorem 5.4 we have a uniqueness result of classical type. Corollary 5.5 Let p, q ∈ C. Let u be a (p, q)-harmonic function. Assume that u(0) = 0 and m m ∂ u(0) = ∂ u(0) = 0 for m = 1, 2,... . Then u(z) = 0 for all z ∈ D. Proof The result is evident by Theorem 5.4. 6 Further properties of the function u p,q Let u be as in (0.2)for some p, q ∈ C. In this section we derive the series expansion p,q of u (see Theorem 6.3). Of interest are also properties of integral means of u p,q p,q (see Theorems 6.4 and 6.6). Lemma 6.1 Let u be as in (0.2) for some p, q ∈ C. Let m ∈ N. Then p,q (p + 1) ∂ u (z) = u (z) +¯zg (z), z ∈ D, (6.1) p,q p,q m (1 − z) where g ∈ C (D). Proof We prove the lemma by induction on m ∈ N. Clearly (6.1) holds for m = 0 with g = 0. By Lemma 1.1 we have that p + 1 ∂u (z) = u (z) +¯zg (z), z ∈ D, (6.2) p,q p,q 1 1 − z with p + q + 1 g (z) =− u (z), z ∈ D. 1 p,q 1 −|z| This proves (6.1)for m = 1. Assume next that (6.1) holds for some m = n ≥ 1, that is, (p + 1) ∂ u (z) = u (z) +¯zg (z), z ∈ D, p,q p,q n (1 − z) where g ∈ C (D). Differentiating we have that n (p + 1) n+1 ∂ u (z) = (p + 1) u (z) + ∂u (z) +¯z∂g (z) p,q n p,q p,q n n+1 n (1 − z) (1 − z) 122 Page 24 of 28 M. Klintborg, A. Olofsson for z ∈ D.Wenextuse (6.2) to see that n (p + 1) p + 1 n+1 ∂ u (z) = (p + 1) u (z) + u (z) +¯zg (z) p,q n p,q p,q 1 n+1 n (1 − z) (1 − z) 1 − z (p + 1) n+1 +¯z∂g (z) = u (z) +¯zg (z) n p,q n+1 n+1 (1 − z) for z ∈ D, where (p + 1) g (z) = g (z) + ∂g (z), z ∈ D. n+1 1 n (1 − z) This proves (6.1)for m = n + 1. The conclusion of the lemma now follows by the principle of induction. Notice that formula (6.1) interprets naturally as a congruence in C (D) modulo zC ¯ (D). Theorem 6.2 Let u be as in (0.2) for some p, q ∈ C. Then p,q m m ∂ u (0) = (p + 1) and ∂ u (0) = (q + 1) p,q m p,q m for m ∈ N. Proof We shall use Lemma 6.1. A point evaluation at the origin in (6.1) shows that ∂ u (0) = (p + 1) for m ∈ N. Moreover, by hermitian symmetry u ¯ = u p,q m p,q q ¯ ,p ¯ we have that m m ¯ ¯ ∂ u (0) = ∂ u ¯ (0) = ∂ u (0) = (q ¯ + 1) = (q + 1) p,q q ¯ ,p ¯ q ¯ ,p ¯ m m for m ∈ N. We can now derive the series expansion of the function u . p,q Theorem 6.3 Let p, q ∈ C. Then 2 p+q+1 (1 −|z| ) (p + 1) 2 m = F (−p, m − q; m + 1;|z| )z p+1 q+1 (1 − z) (1 −¯z) m! m=0 (q + 1) 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D, where F is the hypergeometric function (0.5). A series expansion for generalized harmonic functions Page 25 of 28 122 Proof Let u be as in (0.2). From Theorem 1.4 we know that the function u is p,q p,q (p, q)-harmonic. By Theorem 5.4 it has the function series expansion ∂ u (0) p,q 2 m u (z) = F (−p, m − q; m + 1;|z| )z p,q m! m=0 ∂ u (0) p,q 2 m + F (−q, m − p; m + 1;|z| )z ¯ m! m=1 for z ∈ D. The result now follows from Theorem 6.2. We point out that Theorem 6.3 generalizes a well-known partial fraction de- composition formula for the classical Poisson kernel for D, see formula (1.1). Suitably specialized Theorem 6.3 yields [21, Theorem 2.5] and [18, Theorem 3.2]. Notice that the series expansion in Theorem 6.3 is absolutely convergent in C (D). We denote by Re(z) and Im(z) the real and imaginary parts of a complex number z, respectively. The standard Gamma function is deﬁned by z−1 −t (z) = t e dt for Re(z)> 0. It is well-known that continues to a meromorphic function in C with simple poles at the points z = 0, −1, −2,... . We next turn to L means of u . p,q Theorem 6.4 Let u be as in (0.2) for some p, q ∈ C such that Re(p)+Re(q)> −1. p,q Then 1 (Re(p) + Re(q) + 1) i θ |Im( p)−Im(q)| |u (re )| dθ ≤ e p,q Re( p)+Re(q) 2π T ( + 1) for 0 ≤ r < 1, where is the Gamma function. Proof From deﬁnition of powers we have that ( p+q+1) log(1−|z| ) −( p+q+2) log|1−z| −i ( p−q) arg(1−z) u (z) = e e e p,q for z ∈ D. Notice that the complex number 1 − z lies in the open right half-plane when z ∈ D. Passing to absolute values we have that 2 Re( p)+Re(q)+1 (1 −|z| ) |Im( p)−Im(q)| |u (z)|≤ e p,q Re( p)+Re(q)+2 |1 − z| for z ∈ D. We next apply [18, Theorem 3.1] with α = Re(p) + Re(q) to conclude that 1 π (α + 1) i θ |Im( p)−Im(q)| |u (re )| dθ ≤ e p,q 2π (α/2 + 1) T 122 Page 26 of 28 M. Klintborg, A. Olofsson for 0 ≤ r < 1. This yields the conclusion of the theorem. Remark 6.5 A similar analysis as in the proof of Theorem 6.4 shows that i θ lim |u (re )| dθ =+∞ p,q r →1 2π if Re(p) + Re(q) ≤−1. We omit the details. A classical result known as Gauss summation formula says that (c) (c − a − b) lim F (a, b; c; x ) = (6.3) x →1 (c − a) (c − b) when a, b ∈ C and c ∈ C\{0, −1, −2,... } are such that Re(c)> Re(a) + Re(b) (see [3, Theorem 2.2.2]). Theorem 6.6 Let p, q ∈ C be such that Re(p) + Re(q)> −1. Then 1 (p + q + 1) i θ lim u (re ) dθ = , p,q r →1 2π (p + 1) (q + 1) where is the Gamma function. Proof Recall the series expansion of the function u established in Theorem 6.3.By p,q cancellation of terms we have that i θ 2 u (re ) dθ = F (−p, −q; 1; r ) p,q 2π for 0 ≤ r < 1. Since Re(p) + Re(q)> −1, we can apply Gauss summation formula (6.3) to calculate the limit of the above quantity as r → 1. This yields the conclusion of the theorem. 7 The Dirichlet problem: concluding remarks In a restricted range of parameters, the class of (p, q)-harmonic functions can be analyzed in terms of their boundary values. We shall discuss in this section some rudiments of such theory. Let p, q ∈ C\Z be such that Re(p) + Re(q)> −1. The (p, q)-harmonic Poisson kernel is deﬁned by 2 p+q+1 (1 −|z| ) K (z) = c , z ∈ D, p,q p,q p+1 q+1 (1 − z) (1 −¯z) where (p + 1) (q + 1) c = . p,q (p + q + 1) A series expansion for generalized harmonic functions Page 27 of 28 122 Notice that the constant c is a non-zero complex number in this range of parameters. p,q By Theorem 1.4 the function K is (p, q)-harmonic. Theorems 6.4 and 6.6 ensure p,q that the function K satisﬁes some standard properties for approximate identities. p,q The (p, q)-harmonic Poisson integral is deﬁned by −i θ i θ K [ f ](z) = K (ze ) f (e ) dθ, z ∈ D, p,q p,q 2π for integrable functions f ∈ L (T) on T. Let C (T) be the space of continuous functions on T and ﬁx ϕ ∈ C (T).Bythe (p, q)-harmonic Dirichlet problem for ϕ we understand the problem of ﬁnding a (p, q)-harmonic function u such that lim u = ϕ in C (T), where r →1 r i θ i θ i θ u (e ) = u(re ), e ∈ T, for 0 < r < 1. Following usual practice, we formulate this latter Dirichlet problem as L u =0in D, p,q (7.1) u = ϕ on T, where L is as in (0.1). p,q Theorem 7.1 Let p, q ∈ C\Z be such that Re(p) + Re(q)> −1. Let ϕ ∈ C (T). Then a function u in D satisﬁes (7.1) if and only if it has the form u(z) = K [ϕ](z), z ∈ D. p,q The proof of Theorem 7.1 follows a standard scheme for such results and is therefore omitted, see [18,21] for details. The present paper suggests a ﬁner study of (p, q)-harmonic functions. Earlier results of such type concern Poisson integral representations, pointwise boundary limits, Green functions and Lipschitz continuity of generalized harmonic functions, see [18–21]. We mention here also work of Ahern and collaborators [1,2]. Funding Open access funding provided by Lund University. Declarations Conﬂict of interest The authors declare that they have no conﬂict of interest. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted 122 Page 28 of 28 M. Klintborg, A. Olofsson by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. 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A series expansion for generalized harmonic functions

A series expansion for generalized harmonic functions

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A series expansion for generalized harmonic functions

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