REPRESENTATION THEOREM FOR HARMONIC BERGMAN AND BLOCH FUNCTIONS

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1 Tanaka, K. Osaka J. Math. 50 (2013), REPRESENTATION THEOREM FOR HARMONIC BERGMAN AND BLOCH FUNCTIONS KIYOKI TANAKA (Received March 6, 2012) Abstract In this paper, we give the representation theorem for harmonic Bergman functions and harmonic Bloch functions on smooth bounded domains. As an application, we discuss Toeplitz operators. 1. Introduction Let be a smooth bounded domain in the n-dimensional Euclidean space Ê n, i.e., for every boundary point ¾, there exist a neighborhood V of in Ê n and a C ½ - diffeomorphism fï V f (V ) Ê n such that f () 0 and f ( V ) {(y 1,, y n ) ¾ Ê n Á y n 0} f (V ). For 1 p ½, we denote by b p b p () the harmonic Bergman space on, i.e., the set of all real-valued harmonic functions f on such that f p Ï Ê f p dx 1p ½, where dx denotes the usual n-dimensional Lebesgue measure on. As is well-known, b p is a closed subspace of L p L p () and hence, b p is a Banach space (for example see [1]). Especially, when p 2, b 2 is a Hilbert space, which has the reproducing kernel, i.e., there exists a unique symmetric function R(, ) on such that for any f ¾ b 2 and any x ¾, (1) f (x) R(x, y) f (y) dy. The function R(, ) is called the harmonic Bergman kernel of. When is the open unit ball B, an explicit form is known: R(x, y) R B (x, y) (n 4)x4 y 4 (8x y 2n 4)x 2 y 2 n nb(1 2x y x 2 y 2 ) 1n2, where x y denotes the Euclidean inner product in Ê n and B is the Lebesgue measure 2000 Mathematics Subject Classification. Primary 46E15; Secondary 31B05, 47B35. This work is partially supported by the JSPS Institutional Program for Young Researcher Overseas Visits Promoting international young researchers in mathematics and mathematical sciences led by OCAMI.

2 948 K. TANAKA of B. We denote by P the corresponding integral operator (2) P(x) Ï R(x, y)(y) dy for x ¾. It is known that PÏ L p b p is bounded for 1 p ½; see Theorem 4.2 in [6]. The following result is shown in [8]. Theorem A. Let 1 p ½ and let be a smooth bounded domain. Then we can choose a sequence { i } in satisfying the following property: For any f ¾ b p (), there exists a sequence {a i } ¾ l p such that (3) f (x) ½ a i R(x, i )r( i ) (1 1p)n, where r(x) denotes the distance between x and. The equation (3) is called an atomic decomposition of f. The above theorem shows the existence of a sequence { i } permitting an atomic decomposition for every f ¾ b p. Theorem A does not refer to the case p 1. This deeply comes from the fact that PÏ L 1 b 1 is not bounded. In the present paper, we give an atomic decomposition for p 1 by using a modified reproducing kernel R 1 (, ), introduced in [3]. Theorem 1. Let 1 p ½ and let be a smooth bounded domain. Then we can choose a sequence { i } in satisfying the following property: For any f ¾ b p (), there exists a sequence {a i } ¾ l p such that f (x) ½ a i R 1 (x, i )r( i ) (1 1p)n. Also, we consider the harmonic Bloch space. We define the harmonic Bloch space B by where B Ï { f Ï ÊÏ f is harmonic and f B ½}, f B Ï sup{r(x)ö f (x)ï x ¾ } and Ö denotes the gradient operator (x 1,, x n ). Note that B is a seminorm on B. We fix a reference point x 0 ¾. B can be made into a Banach space by introducing the norm f Ï f (x 0 )f B.

3 HARMONIC BERGMAN AND BLOCH SPACE 949 Also, É B denotes the space of all Bloch functions f such that f (x0 )0. Then, ( É B, B ) is a Banach space. Using a kernel we have the following theorem. Theorem 2. ÉR 1 (x, y) R 1 (x, y) R 1 (x 0, y), Let be a smooth bounded domain. Then we can choose a sequence { i } in satisfying the following property: For any f ¾ É B, there exists a sequence {ai } ¾ l ½ such that f (x) ½ j1 a j É R1 (x, j )r( j ) n. In case that a domain is the unit ball or the upper half space, preceding results are obtained in [5] and [4]. We often abbreviate inessential constants involved in inequalities by writing X º Y, if there exists an absolute constant C 0 such that X CY. 2. Preliminaries In this section, we will introduce some results in [6] and [3]. Those results play important roles in this paper. First, we introduce some estimates for the harmonic Bergman kernel. These estimates are obtained by H. Kang and H. Koo [6]. We use the following notations. We put d(x, y) Ï r(x) r(y) x y for x, y ¾, where r(x) denotes the distance between x and. For an n-tuple «Ï («1,, «n ) of nonnegative integers, called a multi-index, we denote «Ï «1 «n and D «x Ï (x 1 ) «1 (x n ) «n. We also use D i Ï x i and D i j Ï 2 x i x j. Theorem B (H. Kang and H. Koo [6]). (1) There exists a constant C 0 such that Let «, be multi-indices. C D x «D y R(x, y) d(x, y) n«for every x, y ¾. (2) There exists a constant C 0 such that for every x ¾. R(x, x) C r(x) n

4 950 K. TANAKA Second, we explain the modified reproducing kernel R 1 (x, y) introduced by B.R. Choe, H. Koo and H. Yi [3]. We call ¾C (Æ ½ ) a defining function if satisfies the conditions that {x ¾ Ê n (x) 0}, {x ¾ Ê n (x) 0} and Ö does not vanish on. Here, we choose a defining function with condition that (4) Ö 2 1 for some ¾ C ½ (Æ ). We can easily construct the above defining function, because is smooth. Remark that r(x) is comparable to (x). We define a differential operator K 1 by (5) K 1 f Ï f 1 2 ½(2 f ) for f ¾ C ½. We also define a kernel R 1 (x, y) by R 1 (x, y) Ï K 1 (R x )(y) for x, y ¾, where R x (y) Ï R(x, y), and denote by P 1 operator the corresponding integral P 1 f (x) Ï R 1 (x, y) f (y) dy. We call R 1 (x, y) the modified reproducing kernel. This kernel satisfies the reproducing property and has the following estimates. Theorem C (B.R. Choe, H. Koo and H. Yi [3]). Let be a smooth bounded domain. Then (1) R 1 has the reproducing property, i.e., P 1 f f for f ¾ b 1. (2) Let «be multi-index. Then there exists C 0 such that for x, y ¾ (6) D x «R r(y) 1(x, y) C d(x, y) n1«and (7) Ö y R 1 (x, y) C d(x, y) n1. (3) P 1 Ï L p b p is bounded for 1 p ½.

5 HARMONIC BERGMAN AND BLOCH SPACE 951 Finally, we prepare some lemmas. Lemma 2.1 (Lemma 4.1 in [6]). Let s be a nonnegative real number and t 1. If s t 0, then there exists a constant C 0 such that for every x ¾. dy C d(x, y) ns r(y) t r(x) st We define the associated integral operator I s by I s f (x) Ï r(y) s f (y) dy. d(x, y) ns Lemma 2.2. If s 0, then I s Ï L p L p is bounded for 1 p ½ and if s 0, then I s Ï L p L p is bounded for 1 p ½. Proof. When s 0 and 1 p ½, the L p -boundedness of I s is shown by Schur s test; see Lemma 2.6 in [8]. We have only to show that I s Ï L 1 L 1 is bounded for s 0. By Lemma 2.1, we have I s f L 1 This completes the proof. f (y)r(y) s C f L 1. r(y) s d(x, y) ns f (y) dy dx 1 dx dy d(x, y) ns 3. Representation theorem for harmonic Bergman functions In this section, we give a proof of Theorem 1. We need to take sequences { i } i with the following property in the same similar way in [8]. Lemma 3.1. There exists a number c 0 such that for each 0 Æ 14, we can choose a sequence { i } i and a disjoint covering { } i of satisfying the following conditions: (a) is measurable for each i ¾ Æ and { } i are mutually disjoint; (b) B( i, cær( i )) B( i, Ær( i )) for each i ¾ Æ.

6 952 K. TANAKA In what follow, { i } i, { } i are taken in Lemma 3.1. We define operators A, U and S as follows: (8) (9) and A{a i }(x) Ï ½ a i R 1 (x, i )r( i ) (1 1p)n, U f Ï { f ( i )r( i ) (1 1p)n } i, (10) S f (x) Ï ½ R 1 (x, i ) f ( i ). Theorem 1 means that AÏ l p b p is onto for 1 ½. First, we show the boundedness of the operators A, U and S. Lemma 3.2. are bounded. Let 1 p ½. Then UÏ b p l p, AÏ l p b p and SÏ b p b p Proof. First, we show that U is bounded. For any f ¾ b p, by using the condition (b) in Lemma 3.1, we have U f p l p ½ f ( i )r( i ) (1 1p)n p º º º ½ ½ ½ f ( i ) p r( i ) n f ( i ) p dy f (y) p dy f p p. Next, we show that A is bounded. For any {a i } ¾ l p and any x ¾, by Theorem C, we have A{a i }(x) º a i r( i ) (1 1p)n r( i ) d(x, i ) n1 i a i r( i ) (1 1p)n 1 i º a i r( i ) (1 1p)n 1 i I 1 g(x), r( i ) dy d(x, i ) n1 r(y) dy d(x, y) n1

7 HARMONIC BERGMAN AND BLOCH SPACE 953 where g(x) Ï È ia i r( i ) (1 1p)n 1 Ei (x) and Ei denotes the characteristic function of. Since B( i, cær( i )), we have (11) r( i ) (1 1p)n 1 1 (cæ) n 1p. Hence, we have g p L p º i a i p 1 Ei (x) dx {a i } p l p. Therefore, by Lemma 2.1, we have A{a i } b p º I 1 g L p º {a i } l p. S is bounded, because S AÆ U. This completes the proof. The next lemma is essential for the proof of main theorem. Lemma 3.3. Let 1 p ½. Then there exist { i } i and { } i such that SÏ b p b p is bijective. Proof. For 0 Æ 14, we take { i } i and { } in Lemma 3.1. We have only to show that I S 1 for a sufficiently small Æ 0. By the condition of { }, for f ¾ b p we have (I S) f (x) ½ ½ f (y) R 1 (x, y) dy R 1 (x, i ) f ( i ) ½ First, we estimate F 1 (x). By (7), we have F 1 (x) º f (y)(r 1 (x, y) R 1 (x, i )) dy ( f (y) f ( i )) R 1 (x, i ) dy Ï F 1 (x) F 2 (x) say. º Æ ½ ½ f (y)y i Ö y R 1 (x, Æy) dy 1 f (y)r( i ) d(x, Æy) n1 dy

8 954 K. TANAKA º Æ ½ ÆI 1 f(x). r(y) d(x, y) n1 f (y) dy Next, we estimate F 2 (x). For any y ¾, by the mean-value property, we have (12) R 1 (y, z) R 1 ( i, z) Ær( i )Ö x R 1 (Æy, z) for some Æy on the line segment between y and i. Therefore, by (12) and (6), we have Hence, by Theorem C, we have f (y) f ( i ) R 1 (y, z) R 1 ( i, z)f(z) dz º Æ F 2 (x) ½ º Æ Æ º ÆI 1 f(y). r( i )r(z) d(æy, z) n2 f (z) dz f (y) f ( i )R 1 (x, i ) dy r(y) d(x, y) n1 I 1 f(y) dy r(y) d(x, y) n1 I 1 f(y) dy ÆI 1 Æ I 1 f(x). ÆC f b p. Remark that this constant C is By Lemma 2.2, we have (I S) f b p independent of Æ. Hence, if we choose Æ C 1, then we obtain (I S) 1. This completes the proof. Proof of Theorem 1. By Lemma 3.3, we choose a sequence { j } such that SÏ b p b p is bijective. Hence, AÏ l p b p is onto, which implies Theorem Representation theorem for harmonic Bloch functions In this section, we give a proof of Theorem 2. estimate for B (see [3]): We need to recall a pointwise (13) f (x) º f B (1log r(x) 1 )

9 HARMONIC BERGMAN AND BLOCH SPACE 955 for any x ¾ and any f ¾B. We need some operators discussed in [3]. Let F 1 denote the class of all differential operators F of form (14) F 0 n for some real functions i ¾ C ½ (Æ ). We put i D i F(x, y) Ï F(R x )(y) for F ¾ F 1. The following theorem is shown [3]. Theorem D. For c 1 0 and F 1 ¾ F 1, we put H 1 Ï c 1 (K 1 G 1 ), where K 1 is the differential operator defined in (5) and G 1 (x)ï Ê (14) (y)f 1(x, y)(y)dy. We can choose a constant c 1 0 and F 1 ¾ F 1 with the following properties: (a) H 1 b p L p is bounded for each 1 p ½; (b) H 1 Ï B L ½ is bounded and H 1 (B 0 ) C 0 B 0 b ½ ; (c) P 1 H 1 f f for f ¾ b 1. REMARK. Recall R 1 (x, y) R 1 (x, y) R 1 (x 0, y) where x 0 is a fixed reference point. Denote by É P1 the corresponding operator É P1 f (x) Ï Ê É R 1 (x, y) f (y) dy. From Theorem D, we easily have (15) É P1 H 1 f f for any f ¾ É B. We give the estimates for H 1. Lemma 4.1. Let 0 Æ 1 and x ¾. Then (16) H 1 f (y) H 1 f (x) º Æ f B for any f ¾ B and y ¾ B(x, Ær(x)). Proof. To obtain the estimate for H 1, we show the properties on F 1 and G 1. First, we give the estimate for F 1. STEP 1. Let F ¾ F 1, f ¾ B and x ¾. Then (17) F f (x) F f (y) º Æ f B for 0 Æ 1 and y ¾ B(x, Ær(x)).

10 956 K. TANAKA Proof of Step 1. we have Let f ¾ B. By the mean-value property, for y ¾ B(x, Ær(x)) F f (y) F f (x) 0 f(y) f (x) n The proof of Step 1 finished. º 0 Ær(Æy)Ö f (Æy) n º Æ f B. i D i ( f (y) f (x)) i Ær(Æy) 2 Ö D i f (Æy) We put É K1 f 2(½ ) f 4Ö Ö f. Then, É K1 ¾ F 1 and K 1 f É K1 f for any harmonic function f. In particular, we have (18) K 1 f (y) K 1 f (x) º Æ f B for any f ¾ B, x ¾ and y ¾ B(x, Ær(x)). STEP 2. Let F 1 and G 1 satisfy the conditions of Theorem D. Then (19) G 1 f (y) G 1 f (x) º Æ f B for any f ¾ B, x ¾ and y ¾ B(x, Ær(x)). Proof of Step 2. we have For f ¾ B, by the mean-value property, for y ¾ B(x, Ær(x)) G 1 f (y) G 1 f (x) º f (z)r(z)y xö F 1 R Æy (z) dz for some Æy on the line segment between x and y. Because r(x) comparable to r(æy), by (6) and (13), we have G 1 f (y) G 1 f (x) º f (z)r(z)y xö F 1 R Æy (z) dz º Æ f B º Æ f B. 1 r(æy) d(æy, z) n1 dz The proof of Step 2 finished. By (18) and Step 2, we obtain Lemma 4.1. Again, for 0 Æ 14 we choose a sequence { j } in and a disjoint covering {E j } of obtained by Lemma 3.1. We define the operators É AÏ l ½ É B, É SÏ É B É B

11 HARMONIC BERGMAN AND BLOCH SPACE 957 and É UÏ É B l ½ by (20) (21) ÉA{a i }(x) Ï ½ ÉS f (x) Ï ½ j1 j1 a j É R1 (x, j )E j, H 1 f ( j ) É R1 (x, j )E j, and (22) É U f Ï {H 1 f ( j )} j. In the similar manner as in the proof of Theorem 1. We begin with showing that ÉA, U É and S É are bounded. AÏ É l É ½ B, UÏ É B É l ½ and SÏ É B É B É are bounded. Lemma 4.2. Proof. It is obvious that É UÏ É B l ½ is bounded by Theorem D. Since É S É A É U, we have only to show that É AÏ l ½ É B is bounded. Taking {ai } ¾ l ½, by (6) and Lemma 2.1, we have r(x)ö( É A{ai })(x) r(x) which implies É AÏ l ½ É B is bounded. ½ j1 ½ º {a i } l ½ {a i } l ½ º {a i } l ½, a j Ö x É R1 (x, j )E j j1 r( j ) r(x) E d(x, j ) n1 j r(x)r(y) dy d(x, y) n2 Finally, we state an important lemma for the representation theorem. Lemma 4.3. There exists { i } i and { } i such that É SÏ É B É B is bijective. Proof. For 0 Æ 14, we take { i } i and { } in Lemma 3.1. We show that I S É 1 for a sufficiently small Æ 0. By Theorem D, we have (I S) É f (x) ½ H 1 f (y) É R1 (x, y) dy E ½ j j1 ½ E j j1 ½ j1 E j j1 E j É R1 (x, y)(h 1 f (y) H 1 f ( j )) dy H 1 f ( j )( É R1 (x, y) É R1 (x, j )) dy. H 1 f ( j ) É R1 (x, j ) dy

12 958 K. TANAKA We calculate r(x)ö(i É S) f (x) to estimate the Bloch norm. r(x)ö(i É S) f (x) r(x) Ö x ½ j1 r(x) Ö x R1 É (x, y)(h 1 f (y) H 1 f ( j )) dy E j ½ j1 E j H 1 f ( j )( R1 É (x, y) R1 É (x, j )) dy. Because Ö x É R1 (x, y) Ö x R 1 (x, y), Lemma 4.1 shows that the first term is bounded by Æ f B r(x) ½ j1 E j Ö x É R1 (x, y) dy º Æ f B r(x) The second term can be estimated by r(x)h 1 f L ½ ½ j1 E j Ö x R 1 (x, y) Ö x R 1 (x, j ) dy º Æ f B r(x) r(y) d(x, y) n2 dy º Æ f B. º Æ f B. r(y) dy d(x, y) n2 Thus, there exist a constant C 0 such that (I É S) f B CÆ f B. Hence, if we take Æ C 1, then É SÏ É B É B is bijective. This completes the proof. Finally, we give a proof of Theorem 2. Proof of Theorem 2. We put f ¾ B. É By Lemma 4.3, we can choose a constant Æ 0 such that SÏ É B É B É is bijective. This implies AÏ É l É ½ B is surjective. Hence, for any f ¾ B, É we can find a sequence {a ¼ i } ¾ l ½ such that A{a É ¼ i } f. Therefore, if we put a j Ï a ¼ j E j r( j ) n, then {a i } is in l ½ and satisfies f (x) È ½ j1 a j R(x, j )r( j ) n. This completes the proof. 5. Application In this section, we analyze positive Toeplitz operators on b 2 by using Theorem 1. We define several operators and functions. M() denotes the space of all complex Berel measures on. For ¾ M(), the corresponding Toeplitz operator T with symbol is defined by (23) T f (x) Ï R(x, y) f (y) d(y) (x ¾ ).

13 HARMONIC BERGMAN AND BLOCH SPACE 959 Let be a finite positive Borel measure. For Æ ¾ (0, 1), the averaging function Ç Æ defined by (24) Ç Æ (x) Ï (B(x, Ær(x))) V (B(x, Ær(x))) (x ¾ ). is We recall Schatten -class operators. A compact operator T on a separable Hilbert space is called Schatten -class operator, if the following norm is finite; (25) T S (X) Ï ½ 1 s m (T ) m1 where {s m (T )} m is the sequence of all singular value of T. Let S be the space of all Schatten -class operators on b 2. In [2], B.R. Choe, Y.J. Lee and K. Na studied conditions that positive Toeplitz operators are bounded, compact and in Schatten - class on b 2 for 1 ½. We would like to discuss a condition that positive Toeplitz operators are in Schatten -class on b 2. Theorem 3. Let 2(n 1)(n 2) and be a finite positive Borel measure. Choose a sequence { j } in Theorem 1. Then, if È ½ j1 Ç Æ ( j ) ½, then T ¾ S. We recall a general property; see for example [7]. Lemma 5.1 ([7]). If T is a compact operator on a Hilbert space H and 0 2, then for any orthonormal basis {e n }, we have (26) T S (H) ½ ½ n1 k1 T e n, e k. Proof of Theorem 3. When 1 ½, the statement of Theorem 3 is shown in [2]. Hence, we assume 1. We put a sequence { j } satisfying the assumption of Theorem 3. We show the following inequality (27) ½ ½ j1 A T Ae i, e j º ½ j1 Ç Æ ( j ), where {e n } is an orthonormal basis for l 2 and A is the operator of the atomic decomposition obtained by Theorem 1. First, we calculate A T Ae i, e j. Ae i (x) R 1 (x, i )r( i ) n2 and T Ae i (x) r( i ) n2 R(x, y) R 1 (y, i ) d(y).

14 960 K. TANAKA Therefore, we have Then, we have ½ ½ j1 ½ º º A T Ae i, e j r( i ) n2 r( j ) n2 A T Ae i, e j ½ j1 r( i) n2 r( j ) n2 R 1 (x, i ) R 1 (x, j ) d(x) ½ ½ ½ r( i ) n2 r( j ) n2 B( j1 k1 k,ær( k )) ½ ½ ½ r( i ) n2 r( j ) n2 k, Ær( k ))) k1(b( j1 By 1, we have ½ ½ ½ r( i ) n2 r( j ) n2 k, Ær( k ))) k1(b( j1 ½ ½ º Ç Æ ( k ) r( k ) n k1 ½ ½ Ç Æ ( k ) k1 2 r( i ) n2 d( k, i ) (n1) r( k ) n2 r( i ) n2 d( k, i ) (n1) 2. By Lemma 2.1 and the condition of, we have ½ r( k ) n2 r( i ) n2 d( k, i ) (n1) º º ½ B( i,æ i ) R 1 (y, i ) R 1 (y, j ) d(y). r( i ) r( j ) d(x, i ) n1 d(x, j ) d(x) n1 r( i ) r( j ). d( k, i ) n1 d( k, j ) n1 r( i ) r( j ) d( k, i ) n1 d( k, j ) n1 r( k ) n2 r( i ) n2 n dy d( k, i ) n(n1) n r( k ) n2 r(y) n2 n d( k, y) n(n1) n dy º 1. By an estimate (26) and Lemma 5.1, we obtained A T A ¾ S. By Lemma 3.3, there exists S 1 Ï b 2 b 2 and S 1 is bounded. Because T (U S 1 ) A T A(U S 1 ), we obtain T ¾ S. This completes the proof. ACKNOWLEDGEMENTS. This work was done during my visit to Korea University. I am grateful to Professor Hyungwoon Koo and Professor Boo Rim Choe for their hospitalities and suggestions.

15 HARMONIC BERGMAN AND BLOCH SPACE 961 References [1] S. Axler, P. Bourdon and W. Ramey: Harmonic Function Theory, Springer, New York, [2] B.R. Choe, Y.J. Lee and K. Na: Toeplitz operators on harmonic Bergman spaces, Nagoya Math. J. 174 (2004), [3] B.R. Choe, H. Koo and H. Yi: Projections for harmonic Bergman spaces and applications, J. Funct. Anal. 216 (2004), [4] B.R. Choe and H. Yi: Representations and interpolations of harmonic Bergman functions on half-spaces, Nagoya Math. J. 151 (1998), [5] R.R. Coifman and R. Rochberg: Representation theorems for holomorphic and harmonic functions in L p : in Representation Theorems for Hardy Spaces, Astérisque 77, Soc. Math. France, Paris, 1980, [6] H. Kang and H. Koo: Estimates of the harmonic Bergman kernel on smooth domains, J. Funct. Anal. 185 (2001), [7] D.H. Luecking: Trace ideal criteria for Toeplitz operators, J. Funct. Anal. 73 (1987), [8] K. Tanaka: Atomic decomposition of harmonic Bergman functions, Hiroshima Math. J. 42 (2012), Department of Mathematics Graduate School of Science Osaka City University Sugimoto, Sumiyoshi-ku, Osaka Japan