A CLASS OF INTEGRAL OPERATORS ON THE UNIT BALL OF C n. 1. INTRODUCTION Throughout the paper we fix a positive integer n and let

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1 A CLASS OF INTEGRAL OPERATORS ON THE UNIT BALL OF C n OSMAN KURES AND KEHE ZHU ABSTRACT. For real parameters a, b, c, and t, where c is not a nonpositive integer, we determine exactly when the integral operator T f(z) = (1 z 2 ) a (1 w 2 ) b f(w) dv(w) (1 z, w ) c is bounded on L p (, dv t ), where is the open unit ball in C n, 1 p <, and dv t (z) = (1 z 2 ) t dv(z) with dv being volume measure on. The characterization remains the same if we replace (1 z, w ) c in the integral kernel above by 1 z, w c. 1. INTRODUCTION Throughout the paper we fix a positive integer n and let C n = C C denote the n-dimensional complex Euclidean space. For z = (z 1,, z n ) and w = (w 1,, w n ) in C n we write and The open unit ball in C n is the set The boundary of is the set z, w = z 1 w 1 + z n w n z = z z n 2 = z, z. = {z C n : z < 1}. S n = {ζ C n : ζ = 1}, which will be called the unit sphere in C n. We denote by dv the volume measure on, normalized so that v( ) = 1. For any real parameter t we define dv t (z) = (1 z 2 ) t dv(z). It is well known that dv t is finite if and only if t > 1. 1

2 2 OSMAN KURES AND KEHE ZHU We are going to study the integral operators T = T a,b,c and S = S a,b,c defined by T f(z) = (1 z 2 ) a (1 w 2 ) b f(w) dv(w), (1 z, w ) c and Sf(z) = (1 z 2 ) a Here a, b, and c are real parameters. We can now state our main results. (1 w 2 ) b f(w) dv(w). 1 z, w c Theorem 1. Suppose 1 < p < and c is neither 0 nor a negative integer. Then the following conditions are equivalent. (a) The operator T a,b,c is bounded on L p (, dv t ). (b) The operator S a,b,c is bounded on L p (, dv t ). (c) The parameters satisfy { pa < t + 1 < p(b + 1) c n a + b. Theorem 2. Suppose p = 1 and c is neither 0 nor a negative integer. Then the following conditions are equivalent. (a) The operator T a,b,c is bounded on L 1 (, dv t ). (b) The operator S a,b,c is bounded on L 1 (, dv t ). (c) The parameters satisfy { a < t + 1 < b + 1 c = n a + b. or { a < t + 1 b + 1 c < n a + b. The special case n = 1 and c = 2 + a + b was considered in [3]. The special case a = 0 is especially interesting. We denote the resulting operator by P b,c. Thus (1 w 2 ) b f(w) dv(w) P b,c f(z) =. (1 z, w ) c In particular, P b,c f is holomorphic whenever it is defined. This operator is in some sense similar to the Bergman projection, and Theorems 1 and 2 tell us exactly when P b,c maps L p (, dv t ) into L p (, dv s ), where 1 p <, t is real, and s > 1.

3 INTEGRAL OPERATORS 3 2. AN APPLICATION OF SCHUR S TEST In this section we consider the boundedness of S a,b,c on L p (, dv t ) in the case c = n a + b. Our estimate is based on the following well-known Schur s test. Lemma 3. Let µ be a positive measure on a measure space X, let H(x, y) be a positive measurable function on X X, and let p > 1 with 1 p + 1 q = 1. If there exists a positive measurable function h(x) on X and if there exists a positive constant C such that H(x, y)h(y) q dµ(y) Ch(x) q and X X H(x, y)h(x) p dµ(x) Ch(y) p for all x and y in X, then the integral operator Hf(x) = H(x, y)f(y) dµ(y) is bounded on L p (X, µ) with H C. Proof. See Theorem of [2]. X The following estimate will also be crucial to the proof of our main results. Lemma 4. Suppose α > 1 and s is real. Then the integral (1 w 2 ) α dv(w) I(z) = 1 z, w n+1+α+s has the following asymptotic behavior as z 1. (a) If s < 0, then I(z) 1. (b) If s = 0, then I(z) log(1 z 2 ). (c) If s > 0, then I(z) (1 z 2 ) s. Proof. See Proposition of [1]. The following simple fact will be used many times later in the paper, so we collect it here for convenience of reference. Lemma 5. The measure dv t (z) = (1 z 2 ) t dv(z) is finite on if and only if t > 1. Proof. This follows easily from polar coordinates; see of [1].

4 4 OSMAN KURES AND KEHE ZHU We now prove the main esitmate of this section. Lemma 6. Suppose 1 p <, pa < t + 1 < p(b + 1), and c = n a + b. Then the operator S = S a,b,c is bounded on L p (, dv t ). Proof. If p = 1 and a < t + 1 < b + 1, then for every f L 1 (, dv t ) we can apply Fubini s theorem to obtain Sf(z) dv t (z) (1 z 2 ) a+t (1 w 2 ) b f(w) dv(w) dv(z) 1 z, w n+1+a+b = (1 w 2 ) b (1 z 2 ) a+t dv(z) f(w) dv(w) 1 z, w n+1+a+b = (1 w 2 ) b (1 z 2 ) a+t dv(z) f(w) dv(w) 1 z, w. n+1+a+t+(b t) Since a + t > 1 and b t > 0, we deduce from part (c) of Lemma 4 that there exists a constant C > 0 such that (1 z 2 ) a+t dv(z) 1 z, w C n+1+a+t+(b t) (1 w 2 ) b t for all w. It follows that Sf(z) dv t (z) C f(w) dv t (w), and so S is bounded on L 1 (, dv t ). If 1 < p < and 1/p + 1/q = 1, then the inequalities is equivalent to b + 1 q pa < t + 1 < p(b + 1) < b t p, a + t + 1 p < a q. These two inequalities clearly imply that ( b + 1, a ) ( a + t + 1, b t ) q q p p is nonempty. Pick any number s from the above intersection and let h(z) = (1 z 2 ) s, z. We can write the integral operator S = S a,b,c as Sf(z) = H(z, w)f(w) dv t (w),

5 where INTEGRAL OPERATORS 5 H(z, w) = (1 z 2 ) a (1 w 2 ) b t. 1 z, w n+1+a+b It follows from Lemmas 3 and 4 that the integral operator S is bounded on L p (, dv t ). 3. SUFFICIENCY FOR THE BOUNDEDNESS OF S a,b,c In this section we obtain sufficient conditions for the boundedness of the operator S = S a,b,c on L p (, dv t ). In the next section we shall show that these conditions are also necessary. Lemma 7. If 1 p <, pa < t + 1 < p(b + 1), and c n a + b, then the operator S a,b,c is bounded on L p (, dv t ). Proof. Write Then σ 0 and σ = (n a + b) c. 1 1 z, w c = 1 z, w σ 1 z, w n+1+a+b 2 σ 1 z, w n+1+a+b. Combining this with Lemma 6, we conclude that the operator S is bounded on L p (, dv t ). Lemma 8. If p = 1, a < t + 1 = b + 1, and c < n a + b, then the operator S = S a,b,c is bounded on L 1 (, dv t ). Proof. We apply Fubini s theorem again to obtain Sf(z) dv t (z) B n = (1 w 2 ) b f(w) dv(w) B n f(w) dv t (w) (1 z 2 ) a+t dv(z) 1 z, w c (1 z 2 ) a+t dv(z). 1 z, w c By assumption, a + t > 1 and c = n (a + t) + s, where s = c (n a + t) = c (n a + b) < 0. By Lemma 4, there exists a constant C > 0 such that (1 z 2 ) a+t dv(z) C 1 z, w c for all w. This shows that the operator S is bounded on L 1 (, dv t ).

6 6 OSMAN KURES AND KEHE ZHU 4. NECESSITY FOR THE BOUNDEDNESS OF T a,b,c In this section we obtain necessary conditions for the boundedness of the operator T = T a,b,c on L p (, dv t ). These conditions turn out to be the same as the sufficient conditions we obtained in the previous section for the boundedness of the operator S = S a,b,c on L p (, dv t ). Lemma 9. Suppose 1 p < and T = T a,b,c is bounded on L p (, dv t ). Then pa < t + 1. Proof. Recall that T f(z) = (1 z 2 ) a (1 w 2 ) b f(w) dv(w). (1 z, w ) c Choose a positive number N such that Np + t > 1 and N + b > 1. It follows from Lemma 5 that the function f N (z) = (1 z 2 ) N belongs to L p (, dv t ). Since the kernel (1 z, w ) c is anti-holomorphi in w, an application of the symmetry of shows that there exists a constant C N > 0 such that T f N (z) = C N (1 z 2 ) a, z. Since T f N belongs to L p (, dv t ), it follows from Lemma 5 that pa + t > 1, or pa < t + 1. Lemma 10. Suppose 1 p < and T = T a,b,c is bounded on L p (, dv t ). Then t + 1 p(b + 1), and strict inequality holds when 1 < p <. Proof. Let q be the conjugate of p. Thus 1/p + 1/q = 1 when 1 < p < and q = when p = 1. By duality, the boundedness of T on L p (, dv t ) implies the boundedness of T on L q (, dv t ). It is easy to see that, with respect to the duality L p (, dv t ) = L q (, dv t ), we have T f(z) = (1 z 2 ) b t (1 w 2 ) a+t f(w) dv(w). (1 z, w ) c If 1 < p <, then the boundedness of T on L q (, dv t ) implies q(b t) + t > 1; see Lemma 9 or its proof. It is easy to see that the inequality above is equivalent to t + 1 < p(b + 1). If p = 1, so the operator T is bounded on L ( ), then we can apply T to a bounded function of the form f N (z) = (1 z 2 ) N,

7 INTEGRAL OPERATORS 7 where N is a sufficiently large positive number, to obtain a bounded function T f N. Once again, it is easy to see that there exists a positive constant C N such that T f N (z) = C N (1 z 2 ) b t, z. The boundedness of this function clearly implies that b t 0, or t + 1 1(b + 1). This completes the proof of the lemma. The next lemma shows that, in the special case c = n a + b, strict inequality in t + 1 p(b + 1) must also hold even when p = 1. Lemma 11. Suppose c = n a + b and T = T a,b,c is bounded on L 1 (, dv t ). Then t < b. Proof. By the duality argument used in the proof of the previous lemma, the operator T f(z) = (1 z 2 ) b t (1 w 2 ) a+t f(w) dv(w) (1 z, w ) n+1+a+b is bounded on L ( ). For any point a consider the function f a (z) = (1 a, z )n+1+a+b 1 a, z n+1+a+b, z. It is obvious that f a = 1 for every a. On the other hand, we have T f a T f a (a) = (1 a 2 ) b t (1 w 2 ) a+t dv(w) 1 a, w. n+1+a+b If b = t, then an application of part (b) of Lemma 4 shows that lim T f a =, a 1 a contradiction to the assumption that T is bounded on L ( ). Since Lemma 10 tells us that t > b is impossible, we must have t < b. Lemma 12. Suppose 1 p < and the operator T = T a,b,c is bounded on L p (, dv t ). If c is not a nonpositive integer, then c n a + b. Proof. We consider the functions f N,k defined by f N,k (z) = (1 z 2 ) N z k 1, z, where k is a positive integer and N is large enough so that N + b > 1 and Np + t > 1.

8 8 OSMAN KURES AND KEHE ZHU We first estimate the norm of f N,k in L p (, dv t ): f N,k p t = f N,k (z) p dv t (z) = (1 z 2 ) Np+t z 1 pk dv(z). For z we write z = rζ, where 0 r < 1 and ζ S n, and integrate in polar coordinates (see of [1]) to obtain f N,k p t = 2n 1 0 (1 r 2 ) Np+t r 2n+pk 1 dr ζ 1 pk dσ(ζ), S n where dσ is the normalized surface measure on S n. The radial integral above is equal to n 1 0 (1 r) Np+t r n+ pk 2 1 dr = nb ( Np + t + 1, n + pk ) 2 ) = n Γ(Np + t + 1)Γ ( n + pk 2 Γ ( Np + pk + n + t + 1). 2 We are going to fix N but let k. By Stirling s formula, the radial integral above is comparable to k (Np+t+1) as k. In particular, there exists a constant C 1 > 0, independent of k, such that (1) f N,k p t C 1 ζ k Np+t+1 1 pk dσ(ζ) S n for all k 1. We next compute the norm of T f N,k. Since c is neither 0 nor a negative integer, we use the Taylor expansion for (1 z, w ) c and the multi-nomial expansion for z, w k to obtain T f N,k (z) = (1 z 2 ) a (1 w 2 ) N+b w1 k dv(w) (1 z, w ) c = (1 z 2 ) a (1 w 2 ) N+b w k 1 = = Γ(k + c) k! Γ(c) (1 z 2 ) a Γ(k + c) k! Γ(c) (1 z 2 ) a z1 k m=0 Γ(m + c) m! Γ(c) z, w m dv(w) (1 w 2 ) N+b w1 z, k w k dv(w) B n (1 w 2 ) N+b w 1 2k dv(w).

9 INTEGRAL OPERATORS 9 By polar coordinates and Proposition of [1], the last integral above is equal to 1 2n (1 r 2 ) N+b r 2n+2k 1 dr 0 1 ζ 1 2k dσ(ζ) S n = n (1 r) N+b r n+k 1 dr 0 ζ1 k 2 dσ(ζ) S n = Γ(N + b + 1)Γ(n + k) (n 1)! k! n Γ(N + b n + k) (n 1 + k)!. It follows that n! Γ(N + b + 1)Γ(k + c) T f N,k (z) = Γ(c)Γ(N + b n + k) (1 z 2 ) a z1. k By Stirling s formula, n! Γ(N + b + 1)Γ(k + c) Γ(c)Γ(N + b n + k) 1 k N+b+1+n c as k, and by the arguments in the first paragraph of this proof (with a in place of N), (1 z 2 ) pa z 1 pk 1 dv t (z) ζ k pa+t+1 1 pk dσ(ζ) S n as k. So we can find a constant C 2 > 0, independent of k, such that (2) T f N,k p t C 2 k p(n+b+1+n c)+(pa+t+1) for all k 1. Combining (1) and (2) with the boundedness of T on L p (, dv t ), we obtain a positive constant C, independent of k, such that 1 k pn+pb+p+pn pc+pa+t+1 C k Np+t+1, or 1 C, k 1. kp(n+1+a+b c) This is possible only when c n a + b. 5. COMLETING THE PROOF OF THEOREMS 1 AND 2 We now put all the pieces together to prove the two main theorems stated in the introduction. It is obvious that the boundedness of S a,b,c on L p (, dv t ) implies the boundedness of T a,b,c on L p (, dv t ). So (b) implies (a) in both Theorem 1 and Theorem 2. That (a) implies (c) in Theorem 1 follows from Lemma 9, Lemma 10, and Lemma 12. That (a) implies (c) in Theorem 2 follows from Lemmas 9-12.

10 10 OSMAN KURES AND KEHE ZHU It follows from Lemma 7 that (c) implies (b) in Theorem 1. That (c) implies (b) in Theorem 2 follows from Lemma 7 and Lemma 8. This completes the proof of Theorems 1 and AN APPLICATION In this section we apply the main result proved earlier to characterize a class of Banach spaces of holomorphic functions in, including the weighted Bergman spaces and the holomorphic Besov spaces of. Throughout this section we use m = (m 1, m 2,, m n ) to denote a multi-index of nonnegative integers. It is common practice to write m = m 1 + m m n, m! = m 1!m 2! m n!. If z, we write z m = z m 1 1 z m 2 2 zn mn. Similarly, if f is holomorphic in, we write m m f f(z) = z m 1 1 z m. 2 2 zn mn In this section we are going to modify the definition of dv t as follows: dv t (z) = c t (1 z 2 ) t dv(z), where c t = 1 for t 1, and for t > 1, c t is chosen so that dv t is a probability measure. This slight abuse of notation is clearly harmless, but it will simplify our presentation in many instances. For α > 1 and p > 0 we use A p α = H( ) L p (, dv α ) to denote the weighted Bergman space, where H( ) is the space of all holomorphic functions in. It is well known that A p α is a closed subspace of L p (, dv α ). In particular, there exists an orthogonal projection P α : L 2 (, dv α ) A 2 α. This will be called the (weighted) Bergman projection. It is well known that P α is an integral operator. More specifically, f(w) dv α (w) P α f(z) = (1 z, w ). n+1+α This integral representation can be used to extend the domain of P α to L 1 (, dv α ). The main result of this section is the following.

11 INTEGRAL OPERATORS 11 Theorem 13. Suppose β is real, α > 1, p 1, and N is a positive integer satisfying pn < β + 1 < p(α + 1). If f is holomorphic in, then f P α L p (, dv β ) if and only if the functions (1 z 2 ) N m f(z), where m = N, all belong to L p (, dv β ). Proof. If f = P α (g) for some g L p (, dv β ), then f(z) = c α (1 w 2 ) α g(w) dv(w) (1 z, w ) n+1+α. For each multi-index m with m = N we differentiate under the integral sign to obtain (1 z 2 ) N m f(z) = C m (1 z 2 ) N (1 w 2 ) α w m g(w) dv(w), (1 z, w ) n+1+n+α where C m is a positive constant depending on m. An application of Theorems 1 and 2 shows that the functions (1 z 2 ) N m f(z), m = N, all belong to L p (, dv β ). On the other hand, if the functions (1 z 2 ) N m f(z) are in L p (, dv β ) for every multi-index m with m = N, then the functions (1 z 2 ) N m f(z) belong to L p (, dv β ) for every multi-index m with m N; see Theorem 2.17 of [4]. Consider the function g(z) = C(1 z 2 ) N R α,n f(z), where C is an appropriate constant and R α,n is the radial differential operator defined in [4]. By Proposition 1.15 of [4], R α,n is a linear partial differential operator on H( ) of order N with polynomial coefficients. Therefore, the assumption on f ensures that the function g belongs to L p (, dv β ). Now (1 w 2 ) N+α R α,n f(w) dv(w) P α (g)(z) = CC α. (1 z, w ) n+1+α By Proposition 1.14 of [4], we can choose the constant C such that P α (g)(z) = R α,n R α,n f(z) = f(z), where R α,n is the radial integral operator defined in [4], which is just the inverse of the operator R α,n. This completes the proof of the theorem. An interesting special case is when β = (n + 1). In this case, the theorem above characterizes the holomorphic Besov space B p, 1 p <,

12 12 OSMAN KURES AND KEHE ZHU as the image of the Lebesgue space L p (, dτ) under the weighted Bergman projection P α, where dv(z) dτ(z) = (1 z 2 ) n+1 is the Möbius invariant measure on. See [4] for more information on the Besov spaces B p. REFERENCES [1] W. Rudin, Function Theory in the Unit Ball of C n, Springer-Verlag, New York, [2] K. Zhu, Operator Theory in Function Spaces, Marcel Dekker, New York, [3] K. Zhu, A Forelli-Rudin type theorem with applications, Complex Variables 16 (1991), [4] K. Zhu, Spaces of Holomorphic Functions in the Unit Ball, Springer-Verlag, New York, DEPARTMENT OF MATHEMATICS, SUNY, ALBANY, NY 12222, USA address: kzhu@math.albany.edu

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