Rudin orthogonality problem on the Bergman space

Journal of Functional Analysis 261 211) 51 68 www.elsevier.com/locate/jfa Rudin orthogonality problem on the Bergman space Kunyu Guo a, echao Zheng b,c, a School of Mathematical Sciences, Fudan University, Shanghai, 2433, PR China b Center of Mathematics, Chongqing University, Chongqing 41331, PR China c epartment of Mathematics, Vanderbilt University, Nashville, TN 3724, United States Received 3 August 21; accepted 11 March 211 Available online 23 March 211 Communicated by G. Godefroy Abstract In this paper, we study the Rudin orthogonality problem on the Bergman space, which is to characterize those functions bounded analytic on the unit disk whose powers form an orthogonal set in the Bergman space of the unit disk. We completely solve the problem if those functions are univalent in the unit disk or analytic in a neighborhood of the closed unit disk. As a consequence, it is shown that an analytic multiplication operator on the Bergman space is unitarily equivalent to a weighted unilateral shift of finite multiplicity n if and only if its symbol is a constant multiple of the n-th power of a Möbius transform, which was obtained via the Hardy space theory of the bidisk in Sun et al. 28) [1]. 211 Elsevier Inc. All rights reserved. Keywords: Rudin s conjecture; Bergman space; Multiplication operators; Counting functions; Orthogonal functions 1. Introduction Let da denote the area measure in the complex plane C, normalized so that the measure of the open unit disk equals 1. The Bergman space L 2 a ) is the Hilbert space consisting of those holomorphic functions on that are also in the space L 2, da) of square integrable functions on. In this paper, we consider the Rudin orthogonality problem on the Bergman space: For what functions φ bounded and analytic on the unit disk does {φ n : n =, 1,...} form an orthogonal set in L 2 a )? * Corresponding author at: epartment of Mathematics, Vanderbilt University, Nashville, TN 3724, United States. E-mail addresses: kyguo@fudan.edu.cn K. Guo), dechao.zheng@vanderbilt.edu. Zheng). 22-1236/$ see front matter 211 Elsevier Inc. All rights reserved. doi:1.116/j.jfa.211.3.6

52 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 One motivation comes from the Rudin problem on the Hardy space. In a 1988 conference at the Mathematical Sciences Research Institute in Berkeley, California, Rudin noted that if φ is an inner function with φ) =, then {φ n : n =, 1, 2,...} is orthogonal in the Hardy space H 2 ), and asked if the converse is true, that is, are multiples of inner functions the only orthogonal bounded holomorphic functions on the disk? Under some mild assumption on the function φ, Cima, Korenblum and Stessin [3] and Bourdon [2] confirmed the Rudin conjecture. In general, Sundberg [11] and Bishop [1] disproved the Rudin conjecture. However, methods in papers [1, 11] imply a lot of interesting results in operator theory on function spaces and measure theory. Some results on Rudin s orthogonal functions were contained in [5,6]. Another motivation comes from the problem when an analytic multiplication operator on the Bergman space is unitarily equivalent to a weighted unilateral shift of finite multiplicity n. In [1], via the Hardy space of the bidisk, the problem is completely solved. Naturally we would like to find an approach of the Bergman function theory on the problem. Our main result is the following theorem: Theorem 1.1. Let φ be analytic on a neighborhood of the closed unit disk. If {φ k : k =, 1,...} forms an orthogonal set of L 2 a ), then for an integer n and some constant c. φz)= cz n By means of the induced measure on the unit disk and counting functions of analytic functions on the unit disk, we will reduce the Rudin problem on the Bergman space to the problem when general counting functions are the essentially radial. Our main result suggests the following conjecture. Conjecture. Let φ be a bounded and analytic function on the unit disk. If {φ k : k =, 1,...} forms an orthogonal set in L 2 a ), then for an integer n and some constant c. φz)= cz n As a consequence of Theorem 1.1, we obtain a new proof of the main result in [1] that an analytic multiplication operator on the Bergman space is unitarily equivalent to a weighted unilateral shift of finite multiplicity n if and only if its symbol is a constant multiple of the n-th power of a Möbius transform. This was obtained via the theory of the Hardy space of the bidisk in [1]. 2. Counting functions and essentially radial functions First we introduce some notation. For an analytic self-map φ from to, define the induced measure μ φ of φ by μ φ E) = area[φ 1 E)]. π

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 53 For a Borel measure μ on the unit disk, μ is called to be radial if μ satisfies μe) = μ e iθ E ) for <θ< and every measure set E of the unit disk. On the Hardy space, in [1,2,11], it is shown that there is an important connection between induced measures and the classical Nevanlinna counting function N φ of φ. OntheBergman space, we need a general counting function Ñ φ w). Nevanlinna counting function N φ is defined on \{φ)} and given by N φ w) = φz)=w log 1 z, where multiplicities are counted and N φ w) istakentobezeroifw is not in the range of φ.the general counting function Ñ φ w) of φ is defined by Ñ φ w) = 2N φ w) φz)=w 1 z 2 ). The Nevanlinna counting function of φ is lower semi-continuous and satisfies the sub-meanvalue property. A function f on is essentially radial if for almost all r [, 1), fre iθ ) is independent from θ. The following lemma is contained in [2]. Proposition 2.1. Let φ be an analytic self-map of and φ) =. The following are equivalent: a) the set {φ k : k =, 1, 2,...} is orthogonal in H 2 ); b) N φ w) is essentially radial. On the Bergman space we have the following characterization of the radial induced measures. Theorem 2.2. Let φ be an analytic self-map of the unit disk, and φ) =. The following are equivalent: a) {φ k : k =, 1,...} forms an orthogonal set in L 2 a ); b) the measure μ φ is radial; c) Ñ φ w) is essentially radial. Proof. First we show that b) implies a). Using the induced measure, we have the following formula: f φz) ) daz) = fz)dμ φ z)

54 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 for any bounded continuous f on. This immediately gives φz) n φz) m daz) = z n z m dμ φ z). 2.1) Thus φz) n φz) m daz) = for n m. Conversely, assuming a), by 2.1), we have z n z m dμ φ z) = if n m. Using a similar argument to the proof of Lemma 2.4 in [1], we have that for any polynomial Pz,z) of z and z, Pz,z)μ φ z) = Pλz,λz)μ φ z) for each unimodular constant λ. Since polynomials in z and z are dense in the continuous functions on the closed unit disk, we deduce that gλz)μ φ z) = gz)μ φ z) for each continuous function g on the unit disk and any unimodular constant λ. This gives b). Next we show that a) implies c). First we derive the following formula that for n, m 1, φ n z)φ m z) daz) = mn 4π w n 1 w m 1 Ñ φ w) daw). 2.2) Using polar coordinates and then the Littlewood Paley formula for the area measure Lemma 3.1 on p. 228 in [4]) gives π φ n z)φ m z) daz) = 1 [ 2π = mn 1 ] φ n re iθ) φ m re iθ) dθ rdr [ φ n 1 rw)φ m 1 rw) φ rw) ] 2 log 1 w 2 daw) r 3 dr.

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 55 Letting u = rw, we obtain 1 [ = 2 = 2 = 2 φ n 1 rw)φ m 1 rw) φ rw) ] 2 log 1 w 2 daw) r 3 dr 1 [ r 2π [ 1 2π dθ φ n 1 u)φ m 1 u) φ u) ] 2 log r u dau) rdr rdr 1 r φ n 1 se iθ) φ m 1 se iθ) φ se iθ) 2 s log r s ds ] dθ φ n 1 se iθ) φ m 1 se iθ) φ [ 1 se iθ) 2 In the third equality we use polar coordinates for the area measure Using the identity dau) = sdsdθ. s r log r s dr ]sds. we obtain 1 φ n z)φ m z) daz) = mn 2π = mn 2π = mn 2π 2π dθ 1 s r log r s dr = 1 2 log 1 s 1 4 1 s 2 ), φ n 1 se iθ) φ m 1 se iθ) φ se iθ) 2 [log 1 s 2 1 s 2)] sds φ n 1 z)φ m 1 z) φ z) [ 2 log 1 z 2 1 z 2) ] daz) w n 1 w m 1 Ñ φ w) daw). Thus we establish 2.2). Assuming a), 2.2) gives that for n m and m, n 1, = φ n z)φ m z) daz)

56 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 = mn 2π = mn 2π 1 Let k = n m. The above equality gives = 1 w n 1 w m 1 Ñ φ w) daw) [ 2π r n+m 1 e n m)iθ Ñ φ re iθ ) ] dθ dr. [ 2π r 2n k 1 e kiθ Ñ φ re iθ ) ] dθ dr for each integer k and n 1. The Littlewood inequality gives that Thus N φ re iθ ) log 1 r. Letting we obtain Ñ φ re iθ ) 2log 1 r. f k r) = 2π e kiθ Ñ φ re iθ ) dθ, fk r) 4π log 1 r and hence f k r) is in L 2, 1). By Müntz Szasz Lemma [7, p. 336], we have 2π e kiθ Ñ φ re iθ ) dθ = for allmost all r [, 1] and each k. Thus Ñ φ re iθ ) is essentially radial and hence c) holds. Assuming c), by 2.2), we have φ n z)φ m z) daz) = to establish a). This completes the proof.

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 57 3. Radial functions In this section, we will characterize some radial functions. To do so, we need the Berezin transform. For S a bounded operator on L 2 a ), theberezin transform of S is the function S on defined by Sz) = Sk z,k z where k z is the normalized reproducing kernel at z. Then the mapping S S is injective [9]. For f,g L 2 a ), letf g denote the rank one operator on L2 a ) defined by for each h in L 2 a ). f gh) = h, g f Lemma 3.1. Let φ 1,...,φ m be analytic functions on.if m φ k z) 2 is radial, then each φ k must be a polynomial of z. Proof. Since φ k is analytic on the unit disk, write φ k z) as a power series φ k z) = a nk z n, n= where a nk are complex numbers. Since m φ k z) 2 is radial, i.e., for each θ<2π and <r<1 m φ k rz) 2 = m φ k e iθ rz ) 2, taking integration both sides of the above equality with respect to θ gives m φ k rz) 2 = 1 2π 2π = 1 2π 2π = = = m n= m φ k e iθ rz ) 2 dθ m 2 a nk r n e inθ z n dθ n a nk 2 rz 2n m ) a nk 2 rz 2n n= a n rz 2n, n=

58 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 where a n = m a nk 2.Letφk rz) denote the function φ krz). Then φk r isafunctioninthe Bergman space. Letting K z be the reproducing kernel of L 2 a ) at z, we have that the Berezin transform of m φk r φr k is given by and n= [ m φk r φr k ] z) = m 1 z 2) 2 φk r φr k )K z,k z = 1 z 2) 2 m φ r k,k z φ r k,k z = 1 z 2) 2 m φ k rz) 2, [ n)] a n r n z n r n z z) = ) 1 z 2) 2 a n r n z n r n z n K z,k z for each z in the unit disk. Hence [ m φk r φr k ] n= = 1 z 2) 2 an r n z n,k z r n z n,k z n= = 1 z 2) 2 a n rz 2n z) = [ n= As the Berezin transform is injective [9], we obtain that n= a n r n z n r n z n)] z). m φ r k φk) r = a n r n z n r n z n). n= Since {z n : n =, 1,...} is orthogonal in L 2 a ) but the left side of the above identity has finite rank, we conclude that there are only finitely many nonzero a n. This implies that each φk r is a polynomial and so is φ k. Let φ be an analytic self-map of the unit disk. Recall the general counting function Ñ φ w) Ñ φ w) = 2N φ w) φz)=w 1 z 2 ). 3.1)

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 59 Since log x>1 1 x x > 1),wehave Ñ φ w) = φz)=w log 1 z 2 1 z 2) ) > 3.2) for each w φ)\{φ)}. Theorem 3.2. Let φ be a bounded analytic function on the unit disk which vanishes at. If Ñ φ w) is essentially radial and φ = 1, then φ has the following properties: a) The image of under φ equals. b) If φ is analytic on a neighborhood of the closed unit disk, then φ must be a finite Blaschke product. Proof. First we prove a). Assume that φ = 1. Let Ñ φ w) is essentially radial and vanishes at and 3.2) gives that if w E, E = { r [, 1): r = φz) for some z }. Ñ φ w). Since φ) is a simply connected open subset of the unit disk, Ñ φ w) is essentially radial and φ = 1, we have that E must be equal to the interval [, 1). Thus we have that φ)=. This gives a). To get b), by a), we have that the unit circle is contained in the closure of φ) and hence in the image φ) of the closed unit disk under φ. Since φ is continuous on the closure of the unit disk, the maximal modulus principle shows that the image φ ) of the unit circle under φ contains the unit circle. Consider the analytic function ψz)= φz)φ ) 1 z defined on some open annulus containing. Since there are infinitely many points z in such that φz) = 1, hence ψz)= 1.

6 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 This gives ψz) 1 on the annulus. Therefore φz) is an inner function. Since φ is continuous on the unit circle, we conclude that φ is a finite Blaschke product. This completes the proof. 4. Univalent functions In this section, using Proposition 2.1 and Theorem 2.2, we completely solve the Rudin problem on the Bergman space if φ is univalent on the unit disk. Theorem 4.1. Suppose that φ is a univalent self-mapping of the unit disk, and φ) =. If {φ k } k= is orthogonal in L2 a ), then φz)= cz for some constant c. Proof. By Theorem 2.2, we have that the general counting function Ñ φ w) is essentially radial. We may assume that φ = 1. It follows from Theorem 3.2 that φ) equals the unit disk. As φ is univalent, the inverse function φ 1 w) exists on φ) and let ψw) denote the inverse function. We have the precise formula of Ñ φ w) that for w = φ) and w, Ñ φ w) = log 1 ψw) 2 1 ψw) 2 ). By Theorem 2.2, Ñ φ w) is essentially radial. Since ψw) is continuous in the unit disk, we have that Ñ φ w) is radial. Let Fθ,w)depends only on w. Thus Fθ,w)= Ñ φ e iθ w ). F θ =. Taking partial derivative of Fθ,w)with respect to θ, using the chain rule and then evaluating at θ = give F θ [ = i wψ w) wψ + i w) θ= ψw) ψw) Thus we obtain the following equation wψ w) ψw) ] + iwψ w)ψw) iwψ w)ψw). [ wψ ] w) = wψ w)ψw) wψ ψw) w)ψw). Since the left-hand side of the above equality is harmonic, applying the Laplace operator to both sides of the above equation gives [ wψ w)ψw) wψ w)ψw) ] =.

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 61 Noting that = 4 w w, we have wψ w) ) ψ w) wψ w) ) ψ w) =. This gives wψ w)) ψ w) = wψ w)). ψ w) Since wψ w)) ψ w) is analytic in the unit disk except for and wψ w)) is co-analytic in the unit ψ w) disk except for, both sides of the above equation must be constant. Hence we have that there is a constant λ such that wψ w) ) = λψ w). Using the Fourier series expansion of ψ ψw)= a n w n to solve the above equation, we obtain the following equations on the Fourier coefficients a n, n= n 2 a n = λna n for each n. These equations immediately give that a n = ifn λ. Thus λ must be some integer n and ψw)= a n w n. Since ψ is univalent, we conclude that n = 1 and ψw) = a 1 w and hence φz) = cz. This completes the proof. 5. Proof of the main result In this section, we will prove our main result, Theorem 1.1. To do so, first we need the following proposition. Proposition 5.1. Let φ be a finite Blaschke product with order n, and φ) =. If the function 1 z 2 ) φz)=w

62 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 is essentially radial, then for some constant c. φ = cz n Proof. For λ in the unit disk,letφ λ z) denote the Möbius transform φ λ z) = z λ 1 λz. Let n be the order of the finite Blaschke product φz). Then the finite Blaschke product φz) can be uniquely written as φz)= ξφ m λ φ m 1 λ 1 φ m l λ l, 5.1) where λ i are complex numbers in the unit disk, ξ is the unimodular constant and m i are positive integers. We may assume that λ =, and λ i are distinct. Taking ɛ to be enough small, we may assume that I. φ 1 O,ɛ) ) = l O λj, j= where O,ɛ)={ z <ɛ}, O λj is an open neighborhood of λ j, and for all i, j; II. on each O λj, φz) can be expressed as O λi O λj = φz)= φ λj z)ψ j z) ) m j for some analytic function ψ j z) having nonzero point in O λj. For z in O λ, we write φz)= zψ z) ) m and solve the following equation u m = w m m 1 m l, to get m solutions {u j } in the following forms

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 63 where j =, 1,...,m 1, and Then solve the following equations u j = ζ j w m 1 m l, ζ = e 2πi m. zψ z) = ζ j w m 1 m l 5.2) on O λ. Since ψ does not vanish on O λ, the Inverse Function Theorem gives that there are m functions φ ) w),...,φ) m 1 w) on O,ɛ)which are analytic functions of wm 1 m l such that for j =, 1,...,m 1. Similarly, for other zero point λ k of φz), we write and solve the equation φ ) j w)ψ φ ) j w) = ζ j w m 1 m l 5.3) φz)= φ λk z)ψ k z) ) m k, φz)= w m m 1 m l, to get m k functions φ k) w),..., φk) m k 1 w) on Oλ k,ɛ) which are analytic functions of w m m k 1 m k+1 m l such that φ λk φ k) j for j =, 1,...,m k 1, where ς = e 2πi m k. Let and w)ψ k φ k) j w) = ς j w m m k 1 m k+1 m l Fw)= If w <ɛ, then the above construction gives Gw) = φz)=w m m 1 m l φz)=w 1 z 2 ) Gw) = F w m m 1 m l ). 1 z 2 ) = l k= m k 1 j= 1 φ k) j w) 2 ).

64 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 For each unimodular constant η, Fηw)= Fw),wehave Gηw) = F η m m l w m ) m l = F w m ) m l = Gw). This gives that Gw) is radial since Fw) is essentially radial and continuous in {w: w <ɛ}. Let G ɛ w) = Gɛw). Since Gw) is radial in ɛ, applying Lemma 3.1 to the function G ɛ we obtain that each φ k) j w) is a polynomial in w. Noting that φ k) j w) is an analytic function of w m m k 1 m k+1 m l,wehave that φ k) j w) is a polynomial in w m m k 1 m k+1 m l. On the other hand, write hw) = φ ) w). Observing that h is a solution of the following equation by 5.1), we have 5.4) gives φ hw) ) = w m m 1 m l, 5.4) φ hw) ) = hm w)hw) λ 1 ) m 1 hw) λ l ) m l 1 λ 1 hw)) m 1 1 λ l hw)) m l h m w) hw) λ 1 ) m1 hw) λ l ) ml = w m m 1 m l 1 λ1 hw) ) m1 1 λ l hw) ) m l. 5.5) Let s be the degree of the polynomial hw). Using deg pw) to denote the degree of polynomial pw) and computing degrees of both sides of 5.5), we have and deg h m w) hw) λ 1 ) m1 hw) λ l ) ml ) = sm + m 1 + +m l ) deg w m m 1 m l 1 λ1 hw) ) m1 1 λ l hw) ) m l ) = m m 1 m l + sm 1 + +m l ). 5.5) gives that sm = m m 1 m l..

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 65 Since m is a positive integer, we have s = m 1 m l. Since hw) is a polynomial of w m 1 m l and vanishes at, we obtain that there exists a constant γ such that If w <ɛ, 5.2) gives and hence hw) = γw m 1 m l. hw)ψ hw) ) = w m 1 m l, ψ hw) ) = 1 γ. This implies that there is a positive δ such that when z <δ, ψ z) is a constant. From the equality we conclude that φz)= zψ z) ) m, φz)= cz m for some constant c, as desired. Now we are ready to prove Theorem 1.1. Proof of Theorem 1.1. Since {φ k : k =, 1,...} forms an orthogonal set of L 2 a ), by Theorem2.2,wehavethatÑ φ w) is essentially radial. Theorem 3.2 gives that φ is a constant multiple of a finite Blaschke product. Let n be the order of the Blaschke product φ. On the other hand, {φ n : n =, 1, 2,...} forms an orthogonal set in the Hardy space H 2 ). By a theorem in [1,2], N φ w) is essentially radial. Recall Ñ φ w) = 2N φ w) φz)=w 1 z 2 ). Thus the function Fw)= 1 z 2 ) φz)=w is essentially radial. Proposition 5.1 gives that

66 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 φz)= cz n for some constant c. This completes the proof. 6. An application to multiplication operators Given two bounded operators on the Bergman space L 2 a ), we say that T and S are unitarily equivalent if there is a unitary operator U on L 2 a ) such that T = U SU. As an application of Theorem 1.1, we obtain a new proof of the following result obtained in [1] by using the Hardy space theory of the bidisk. For a function in H ), define the multiplication operator M g on the Bergman space L 2 a ) by M g h)z) = gz)hz) for each z in and each function h in L 2 a ). Clearly, M g is a bounded operator on L 2 a ) and M g = g since gh is still in L 2 a ) if h is in L2 a ). Moreover, M g is the restriction of the multiplication operator by g on L 2, da) to the Bergman space L 2 a ). Noting that the multiplication operator on L 2, da) is normal, we have that M g is subnormal on L 2 a ). Theorem 6.1. Let φ be in H ). If the multiplication operator M φ is unitarily equivalent to a weighted unilateral shift of finite multiplicity n, then φz)= cφ n λ z) for a constant c and some Möbius transform φ λ z) = z λ 1 λz. Proof. For each w, define a unitary operator U w : L 2 a ) L2 a ) by U w f = f φ w k w, where k w is the normalized reproducing kernel of L 2 a ), i.e., For any g in H ), it is easy to verify k w = K w / K w. U w M gu w = M g φw.

K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 67 Therefore, we may assume and φ) = φ = 1. We will show that there is a constant c such that φz)= cz n. In the general case, composing φ with a Möbius transform φ λ we can reduce it to the above case. If M φ is unitarily equivalent to a weighted unilateral shift W of finite multiplicity n, W is subnormal since M φ is subnormal. Thus by a theorem in [8], the essential spectrum σ e W ) of W consists of a union of n circles with the center. Let M denote the maximal ideal space of H ). The Corona Theorem [4] shows that σ e M φ ) = φm \ ). 6.1) Since M \ is connected [4], σ e M φ ) must be connected. This gives σ e M φ ) = σ e W ) =. 6.2) By the Corona Theorem [4] that is dense in M, thus φ is an inner function. Combining 6.1) and 6.2) gives that there exist <δ,ɛ<1 such that φz) >ɛ if z >δ. Hence φz) is a finite Blaschke product. Let M denote the wandering subspace of the contraction M φ : L 2 a ) φl2 a ). Noting that M φ is a weighted unilateral shift, we have M j φ M Mi φ M if i j. Thus the set { M,φM,...,φ k M,... } = { M,M φ M,...,Mφ k M,...} is an orthogonal set in L 2 a ). Since φ) =, we have that 1 is in M. Thus for each k, φk is in φ k M and so the set {1,φ,φ 2,...} is orthogonal in L 2 a ). Theorem 1.1 gives φz)= cz n for some constant c. This completes the proof.

68 K. Guo,. Zheng / Journal of Functional Analysis 261 211) 51 68 Acknowledgments This work is partially supported by NSFC 152516), and NKBRPC 26CB8595), and Laboratory of Mathematics for Nonlinear Science, Fudan University. References [1] C. Bishop, Orthogonal functions in H, Pacific J. Math. 22 25) 1 31. [2] P. Bourdon, Rudin s orthogonality problem and the Nevanlinna counting functions, Proc. Amer. Math. Soc. 125 1997) 1187 1192. [3] J. Cima, B. Korenblum, M. Stessin, On Rudin s orthogonality and independence problem, preprint. [4] J. Garnett, Bounded Analytic Functions, Academic Press Inc., 1981. [5] T. Nakazi, The Nevanlinna counting functions for Rudin s orthogonal functions, Proc. Amer. Math. Soc. 131 23) 1267 1271. [6] T. Nakazi, T. Watanabe, Properties of a Rudin s orthogonal function which is a linear combination of two inner functions, Sci. Math. Jpn. 57 23) 413 418. [7] W. Rudin, Real and Complex Analysis, 2nd edition, McGraw Hill, New York, 1974. [8] A. Shields, Weighted shift operators and analytic function theory, in: Math. Surveys, vol. 13, American Mathematical Society, Providence, 1974, pp. 49 128. [9] K. Stroethoff, The Berezin transform and operators on spaces of analytic functions, in: J. Zemanek Ed.), Linear Operators, in: Banach Center Publ., vol. 38, Polish Academy of Sciences, Warsaw, 1997, pp. 361 38. [1] S. Sun,. Zheng, C. Zhong, Multiplication operators on the Bergman space and weighted shifts, J. Operator Theory 59 28) 435 454. [11] C. Sundberg, Measure induced by analytic functions and a problem of Walter Rudin, J. Amer. Math. Soc. 16 23) 69 9.