APPLICATION OF A RIESZ-TYPE FORMULA TO WEIGHTED BERGMAN SPACES

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1 PROCEEINGS OF HE AMERICAN MAHEMAICAL SOCIEY Volume 131, Number 1, Pages S ) Article electronically published on May 13, 00 APPLICAION OF A RIESZ-YPE FORMULA O WEIGHE BERGMAN SPACES ALI ABKAR Communicated by Juha M. Heinonen) Abstract. Let denote the unit disk in the complex plane. We consider a class of superbiharmonic weight functions w : R + whose growth are subject to the condition 0 wz) C1 z ) forsomeconstantc. We first establish a Reisz-type representation formula for w, and then use this formula to prove that the polynomials are dense in the weighted Bergman space with weight w. 1. Introduction We denote by the unit disk and by the unit circle in the complex plane. A weight function, orsimplyaweight, in is any continuous positive function w : [0, + [. he problem. A real-valued function w defined on the unit disk is said to be superbiharmonic provided that w is locally integrable and the bi-laplacian w is a positive distribution on ; here stands for the Laplace operator defined by = z = z z = 1 ) 4 x + y, z = x + iy. We consider a class of superbiharmonic weight functions w satisfying condition A) 0 wz) C1 z ), for every z, for some constant C. We shall prove that if w is a superbiharmonic function satisfying condition A), then w enjoys a Riesz-type representation formula in terms of the biharmonic Green function for the operator, and the harmonic compensator. In the following we shall briefly introduce these notions. he biharmonic Green function for the operator in the unit disk is the function Γz,ζ) which solves, for fixed ζ, the following boundary value problem: z Γz,ζ) =δ ζz) z, Γz,ζ) =0, z, nz) Γz,ζ) =0, z, Received by the editors August 16, Mathematics Subject Classification. Primary 31A30; Secondary 30E10, 30H05, 46E10. his research was supported in part by a grant from the Institute for heoretical Physics and Mathematics IPM), ehran, Iran. 155 c 00 American Mathematical Society

2 156 ALI ABKAR where the symbol δ ζ stands for the unit point mass at ζ, and nz) denotes the inward normal derivative in the sense of distributions. It is well-known [4] that the biharmonic Green function has the form Γz,ζ) = z ζ log z ζ 1 ζz + 1 z ) 1 ζ ), z,ζ). We define the harmonic compensator Hζ,z)by Hζ,z)= 1 z ) P z,ζ) = 1 z ), ζ,z), 1 zζ where P z,ζ) denotes the Poisson kernel for the unit disk. It turns out that superbiharmonic functions w satisfying condition A) can be represented by a Riesz-type formula which takes into account the growth of w as z approaches the boundary. Indeed, we shall see that 1 1) wz) = Γz,ζ) wζ) daζ)+ Hζ,z) nζ) wζ) dσζ), z, where da = π 1 dx dy is the normalized area measure in and dσ =π) 1 dθ denotes the normalized arc-length measure on. Moreover, the normal derivative in the second term should be understood in a generalized sense: nζ) wζ) dσζ) is the weak-star limit of a family of positive continuous functions on the unit circle, so that it can be identified with a positive measure on the unit circle. Application to weighted Bergman spaces. We say that a function f, analytic in, belongs to the weighted Bergman space L p a,w), 0 <p<+, provided that the following integral is finite: f p L p a,w) = fz) p wz) daz) < +. If 1 p<+, it follows that L p a,w) is a Banach space of analytic functions with norm L p a,w), andfor0<p<1, it is a quasi-banach space. For p =, the space L a,w) is a Hilbert space with the inner product f,g L a,w) = fz)gz)wz) daz), f,g L a,w). In the case where the weight function w is identically 1, the corresponding space is called the Bergman space and is denoted by L p a). Note that condition A) entails wz) daz) < +, which guarantees that the polynomials are contained in the weighted Bergman space L p a,w). We devote the last section to an application of the representation formula 1 1) to the weighted Bergman space L p a,w) whose non-radial) weight w is superbiharmonic and satisfies condition A). It is shown that the polynomials are dense in such weighted Bergman spaces. We remark that if the weight function w is radial w depends only on z ), the result is well-known [9]. he question of weighted polynomial approximation for such weights was raised by H. Hedenmalm in [5], p We answer this question in the positive.

3 APPLICAION OF A RIESZ-YPE FORMULA 157 If we weaken condition A) above, it is possible to find a representation formula for w with one more term. his is done in [1], where the author joint with Hedenmalm succeeded to establish a Riesz representation formula for w.. A Riesz-type representation formula In this section we will find a representation formula for a superbiharmonic function w satisfying the condition 0 wz) C1 z ) forsomeconstantc. he following lemma asserts that wz) is integrable against the area measure 1 z ) daz) in the unit disk. he rather long) proof uses the assumption A) on w, twice the application of Green s formula together with some sharp estimates of the laplacians and the normal derivatives involved. We omit the details and refer the reader to [1], Lemma 3.1. Lemma.1. Let w be a superbiharmonic function satisfying 0 wz) C1 z ). hen 1 z ) wz) daz) < +. Proof. he assertion is a consequence of Lemma 3.1 in [1]. In the following lemma we collect some facts on the biharmonic Green function for the unit disk. he proof of parts c) and d) can be found in [], Lemma.3. Part a) can be verified by a direct argument using the defining formula for Γz,ζ), and finally, b) is an immediate consequence of a). Lemma.. Let Γz,ζ) denote the biharmonic Green function for the unit disk. hen a) Γz,ζ) has the power series representation [ ) 1 z 1 ζ )] n+ Γz,ζ) =, z,ζ), n+1) n=0 n +1)n +) 1 zζ b) Γz,ζ) > 0, for every z,ζ), c) for every z,ζ) we have 1 ) 1 z ) 1 ζ ) 1 z ) 1 ζ Γz,ζ), 1 ζz 1 ζz d) for every z,ζ) we have ) 1 z 1 ζ ) ) Γz,ζ) 1+ z 1 ζ ). We now combine the preceding lemmas and obtain the following proposition. Proposition.3. Let w be a superbiharmonic weight function satisfying the condition 0 wz) C1 z ). LetΓz,ζ) denote the biharmonic Green function for the unit disk. hen Γz,ζ) wζ)daζ) < +, z. Proof. According to Lemma.d), for every z,ζ), ) 1 z 1 ζ ) ) Γz,ζ) 1+ z 1 ζ ).

4 158 ALI ABKAR Hence Γz,ζ) 1 ζ ), z,ζ), from which follows Γz,ζ) wζ) daζ) 1 ζ ) wζ) daζ) < +, in accordance with Lemma.1. Proposition.4. Let w be a superbiharmonic weight function satisfying the condition 0 wz) C1 z ). efine the function v by the formula: vz) = Γz,ζ) wζ) daζ), z. hen for every integer k we have lim r 1 z k vrz) dσz) =0. 1 r In fact, sup z k vrz) k 1 r dσz) vrz) 1 r dσz) 0, as r 1. Proof. For an integer k and 0 <r<1 we define 1) z k vrz) Ck, r) = 1 r dσz). We first make the following simple, but important, observation. Since the function vrz)/1 r ) is nonnegative, it follows that for every integer k we have 0 Ck, r) z k vrz) 1 r dσz) =C0,r). Hence it suffices to verify the validity of the statement of the proposition for k =0; that is to show that C0,r) 0asr 1. o this end, we note that Γrz, ζ) 0 C0,r)= 1 r wζ) daζ) dσz). It now follows from Fubini s theorem and Lemma.c) that 0 C0,r) 1 r z ) ) 1 ζ ) wζ) dσz) daζ) 1 r 1 rzζ ) = 1 ζ ) ) wζ) daζ) 1 r 1 rzζ dσz). But 1 1 rzζ dσz) = 1 1 rζ 1 rζ z rζ dσz) = 1 1 rζ. his together with ) yields 3) 0 C0,r) 1 r 1 r ζ 1 ζ ) wζ) daζ).

5 APPLICAION OF A RIESZ-YPE FORMULA 159 o deal with the area integral appearing in 3) we write dλζ) = 1 ζ ) wζ) daζ), ζ. We note that dλζ) is a positive measure on the unit disk, and according to Lemma.1 the integral of the constant function 1 against the measure dλ is finite. Hence 3) can be written as 4) 0 C0,r) 1 r 1 r ζ dλζ). Since the integrand in 4) is nonnegative and bounded from above by 1, we can apply the dominated convergence theorem to the right-hand side of 4) to obtain 1 r 0 lim C0,r) lim r 1 r 1 1 r dλζ) =0, ζ from which the proposition follows. Corollary.5. With the same conditions as in the preceding proposition sup z k vrz) dσz) = o1 r) as r 1. k We have prepared the ground for the main result of this section. heorem.6. Let w be a superbiharmonic weight function satisfying the condition 0 wz) C1 z ). hen wz) = Γz,ζ) wζ) daζ)+ Hζ,z) nζ) wζ) dσζ), z, where Hζ,z) is the harmonic compensator and n w is the generalized normal derivative of w; in the sense that nζ) wζ)dσζ) is a positive measure on the unit circle. Proof. We define the real-valued function v by vz) = Γz,ζ) wζ) daζ), z. According to Proposition.3 the function v is well-defined. Moreover, v is nonnegative, because Γz,ζ) is positive in, and w 0 by our assumption. We shall first verify that the weight function w can be written as wz) =vz)+ 1 z ) hz), z, where h is a harmonic function in the unit disk. It is clear that w v is a biharmonic function, hence according to the Almansi representation formula for biharmornic functions see [6], Lemma 3.1) there exist two real-valued harmonic functions h and u such that wz) vz) =uz)+ 1 z ) hz), z. We fix a real number r, 0 <r<1, and an integer n. henwehave wrz) vrz) =urz)+ 1 r ) hrz), z,

6 160 ALI ABKAR from which it follows that z n wrz) vrz) ) dσz) = z n urz) dσz)+1 r ) z n hrz) dσz) = û r n)+1 r )ĥrn). Here ûn) stands for the n-th Fourier coefficient of u; moreover,u r z) =urz). Since u and h are harmonic, it follows that 5) z n wrz) dσz) z n vrz) dσz) =r n ûn)+1 r )r n ĥn). Letting now r 1, we see that the right-hand side of 5) tends to ûn). As for the left-hand side of -5) we see that the first integral tends to zero, as r 1. Indeed, we know from our assumption on w that 0 wrz) C1 r z ) =C1 r), for z, and hence 0 z n wrz) dσz) wrz) dσz) C1 r) 0, as r 1. he fact that the second integral on the left-hand side of 5) approaches zero as r 1 follows from Corollary.5. Hence for any integer n, wehaveûn) =0, from which it follows that u is identically zero. his implies that 6) wz) vz) =1 z )hz), z, where h is some harmonic function in the unit disk. We fix once again 0 <r<1 and use 6) to write 7) wrz) 1 r vrz) = hrz), z. 1 r It then follows from Proposition.4 that for every integer k we have 8) lim r 1 z k wrz) dσz) = lim 1 r r 1 z k hrz) dσz) =ĥk). Note that the limit on the right-hand side of 8) exists, since h is a harmonic function. In fact the equality 8) holds for every trigonometric polynomial p on the unit circle, that is, 9) lim r 1 pz) wrz) dσz) = lim pz)hrz) dσz). 1 r r 1 We shall see that 10) sup 0<r<1 h r L1 ) < +. o see this, we first use 7) to write 11) hrz) dσz) wrz) 1 r dσz)+ vrz) 1 r dσz). he first integral in 11) is bounded by the constant 1, as was observed earlier. he uniform boundedness of the second integral in 11) follows from Proposition.4 for k = 0. Hence 10) holds. his L 1 -boundedness of the functions h r implies

7 APPLICAION OF A RIESZ-YPE FORMULA 161 that there exists a unique real-valued Borel measure µ on the unit circle such that h is the Poisson integral of this measure see [10], heorem 11.30); or 1 z 1) hz) = dµζ), z. 1 ζz Moreover, the measure µ is the weak-star limit of the measures dµ r = h r dσ see [8], p. 33). his means that for every trigonometric polynomial p on the unit circle we have pz) dµz) = lim pz)h r z) dσz) r 1 = lim pz) wrz) dσz) lim pz) vrz) r 1 1 r r 1 1 r dσz). It is implicit in 9) that the last limit in the above displayed formula is zero from which it follows that 13) pz) dµz) = lim r 1 pz) wrz) 1 r dσz). Now, 13) implies that for every nonnegative trigonometric polynomial p on the unit circle pz) dµz) 0, proving that µ is a positive measure on. his together with 1) implies that h is a positive harmonic function in the unit disk. herefore we obtain from 6) 0 wz) vz) = 1 z ) hz) wz) C1 z ), z, so that 0 hz) C, thatis,his a nonnegative bounded harmonic function in the unit disk. It follows from 6) and 1) that for every z, wz) vz) = 1 z ) ) 1 z hz) = 1 ζz dµζ) 14) = Hζ,z) dµζ), where Hζ,z) is the harmonic compensator defined by Hζ,z)= ) 1 z ) 1 z P z,ζ) =, ζ,z), 1 ζz in which P z,ζ) denotes the Poisson kernel for the unit disk. he measure µ in the representation formula 14) was obtained as a weakstar limit of the functions w r z)/1 r ), for z. his weak-star limit can be regarded as the generalized normal derivative of the function w on the boundary. he proof of the theorem is now complete. 3. Application to weighted Bergman spaces We start with a lemma which says that the harmonic compensator has some kind of monotonicity property with respect to its second argument. he proof uses the method of [], p. 85 see also [7], p. 64, where the result is attributed to the current author).

8 16 ALI ABKAR Lemma 3.1. Let z = e iθ be fixed, and consider ) 1 ζ He iθ,ζ)= 1 ζe iθ, ζ. hen for 0 <r<1 and ζ <r, we have d rh e iθ, ζ ) ) > 0. dr r Proof. For 0 <r<1and ζ <rwe have H e iθ, ζ ) r ζ ) = r r r ζe iθ. A computation reveals that d e dr H iθ, ζ ) =4 r ζ r rr ζe iθ and therefore r d dr H e iθ, ζ ) r It then follows that d dr r ζ ) r ζe iθ = 4 r ζ ) r r ζe iθ ζ ) r ζe iθ rh e iθ, ζ ) ) = H e iθ, ζ ) + r d r r dr H e iθ, ζ ) r r ζ ) = rr ζe iθ ) r r r ζe iθ) + 1 r r ζe iθ), ) r + 1 r r ζe iθ) + 1 r r ζe iθ). r ζ r r + ζ ) + ) r ζ 1 r ζe 1 iθ r ζe iθ r ζ ) r ζ ) r r ζe iθ r + ζ ) > 0, because for ζ <rwe have 1 r ζe iθ + 1 =Re r ζeiθ he proof is now complete. ) 1 r ζe iθ r ζe iθ r ζ. In the next lemma we prove that rγ z, ζ r ) has the same monotonicity property as well. Lemma 3.. For 0 <r<1, z, and ζ <r, the function rγ z, ζ r ) is an increasing function of r.

9 APPLICAION OF A RIESZ-YPE FORMULA 163 Proof. We shall use the following formula, due to Hadamard, which represents the biharmonic Green function Γz,ζ) in terms of the harmonic compensator Hz,ζ) see [6], p. 74): Γz,ζ) = 1 1 π H e iθ, z ) H e iθ, ζ ) sdθds, z,ζ). π s s Let us denote by max{ z, ζ } π Kz,ζ)=χ [ max{ z, ζ }, 1], z,ζ), where χ I stands for the characteristic function of the interval I. Assuming that ζ <r,weobtain r Γ z, ζ ) = 1 1 π K z, ζ ) H e iθ, z ) rs H e iθ, ζ ) dθ ds. r π 0 π r s rs It is easy to see that the function K z, ζ ) r increases with r. On the other hand, since ζ r <s,wehave ζ <rs, and hence the function rs H e iθ, ζ ) rs increases with r, according to Lemma 3.1. his completes the proof. Proposition 3.3. Let w be a superbiharmonic weight function satisfying the condition 0 wz) C1 z ). hen the function rw z r ),for0 <r<1 and z <r, is increasing in r. Proof. his is an immediate consequence of heorem.6, Lemma 3.1, and Lemma 3.. Lemma 3.4. Suppose that µ is a finite positive measure on a measure space X, 0 < p<+, andf n,fare measurable functions such that lim sup f n p dµ f p dµ < +, n + X X and f n f almost everywhere with respect to the measure µ. hen f f n p dµ =0. lim n + X Proof. his is a well-known statement; for a proof the reader is referred to [3], p. 1, or [7], p. 66. We can now state the main result of this section. heorem < p < + ) Let w be a superbiharmonic weight function satisfying the condition 0 wz) C1 z ). hen the polynomials are dense in the weighted Bergman space L p a,w). Proof. Let f be a function in the weighted Bergman space L p a,w), 0 <p<+. We intend to prove that the function f can be approximated by the polynomials in norm. Since the dilation f r for every 0 <r<1 can be approximated in norm by the polynomials, it suffices to show that f r f in norm as r 1. It is easy to see that f r f pointwise, hence what we need to verify is f r L p a,w) f L p a,w), as r 1.

10 164 ALI ABKAR his together with the standard Lemma 3.4 implies that f r f L p a,w) 0, as r 1, from which the result follows. We start by writing f r p L p a,w) = f r z) p wz)daz). We now replace z with z/r on the right side of the above formula to obtain f r p L p a,w) = 1 r 3 fz) p rw z r )daz). r According to Proposition 3.3 the integrand is an increasing function of r, hence we can apply the monotone convergence theorem to conclude that f r p L p a,w) f p L p a,w), as r 1, from which the theorem follows. Acknowledgement I wish to thank Professor H. Hedenmalm of Lund University, Sweden, for the discussions we had. References 1. A. Abkar, H. Hedenmalm, A Riesz representation formula for super-biharmonic functions, Ann. Acad. Sci. Fenn. Math ), MR 00c: A. Aleman, S. Richter and C. Sundberg, Beurling s theorem for the Bergman space, Acta Math ), MR 98a: P. L. uren, heory of H p spaces, Academic Press, New York, MR 4: P. R. Garabedian, Partial differential equations, John Wiley and Sons, New York, London, and Sydney, MR 8: V. P. Havin and N. K. Nikolski, Linear and complex analysis. Problem book 3. Part II, Lecture Notes in Mathematics, 1574, Springer Verlag, Berlin, MR 96c:00001b 6. H. Hedenmalm, A computation of Green functions for the weighted biharmonic operators z α,withα> 1, uke Math. J ), MR 95k: H. Hedenmalm, B. Korenblum, K. Zhu, heory of Bergman spaces, Graduate exts in Mathematics, 199, Springer Verlag, New York, 000. MR 001c: K. Hoffman, Banach spaces of analytic functions, over Publications, Inc., New York, MR 9d: S. N. Mergelyan, On completeness of systems of analytic functions, Amer. Math. Soc. ranslations ), W. Rudin, Real and complex analysis, third edition, McGraw Hill Book Company, Singapore, MR 88k:0000 epartment of Mathematics, Imam Khomeini International University, P.O. Box 88, Qazvin 34194, Iran Current address: epartment of Mathematics, Institute for Studies in heoretical Physics and Mathematics, P.O. Box , ehran, Iran address: abkar@ipm.ir