Special Toeplitz operators on a class of bounded Hartogs domains
aa r X i v : . [ m a t h . C V ] A ug Special Toeplitz operators on a class of bounded Hartogs domains
Yanyan Tang & Zhenhan Tu ∗ School of Mathematics and Statistics, Wuhan University, Wuhan, Hubei 430072, P.R. ChinaEmail: [email protected] (Y. Tang), [email protected] (Z. Tu)
Abstract
We introduce a wider class of bounded Hartogs domains, which contains some generaliza-tions of the classical Hartogs triangle. A sharp criteria for the L p − L q boundedness of the Toeplitzoperator with symbol K − t is obtained on these domains, where K is the Bergman kernel on diagonaland t ≥
0. It generalizes the results by Chen and Beberok in the case 1 < p < ∞ . Key words:
Hartogs domains · L p − L q boundedness · Toeplitz operator
Mathematics Subject Classification (2010): · · C n and A (Ω) be the Bergman space of holomorphic functions in the squareintegrable space L (Ω). It is well known that the Bergman projection P Ω : L (Ω) → A (Ω)is an integral operator represented by P Ω ( f )( z ) = Z Ω K Ω ( z, w ) f ( w ) dA ( w ) , where K Ω ( · , · ) is the Bergman kernel on Ω × Ω, and dA is the Lebesgue measure on Ω. The study of the L p boundedness of the Bergman projection is a fact of interesting and fundamental importance. Oneof the most common object is the bounded domain with various boundary conditions. For example,if Ω is a bounded strongly pseudoconvex domain or a smoothly bounded pseudoconvex domain offinite type in C or some bounded Reinhardt domains, then the Bergman projection P Ω is boundedfrom L p (Ω) to itself for all p ∈ (1 , ∞ ) (see Huo [10], Lanzani-Stein [16], McNeal [17], and Phong-Stein[18]). When the domain Ω has serious singularities at the boundary, in general, the L p boundednessof P Ω will no longer hold for all p ∈ (1 , ∞ ) (see, e.g., Chakrabarti-Zeytuncu [4], Edholm-McNeal [7],Krantz-Peloso [15]). There are also smoothly bounded pseudoconvex domains where the Bergmanprojection has a restriction on p for L p boundedness (see Barrett-S¸ahuto˘glu [1]).Given a function ψ ∈ L ∞ (Ω), the Toeplitz operator with symbol ψ is defined by T ψ ( f )( z ) := P Ω ( ψf )( z ) = Z Ω K Ω ( z, w ) f ( w ) ψ ( w ) dA ( w ) . When Ω is a bounded domain and ψ is constant 1, then the Toeplitz operator will degenerate to theBergman projection. Moreover, it is easy to see that T ψ : L p (Ω) → L q (Ω) is bounded if P Ω : L p (Ω) → L q (Ω) is bounded for ψ ∈ L ∞ (Ω), where p, q ∈ (1 , ∞ ). It is natural to consider the following problem. Question
How does the symbol ψ affect the boundedness of the Toeplitz operator T ψ from L p (Ω)to L q (Ω)? Can we obtain a larger range of p for the boundedness of T ψ in comparison with thecorresponding Bergman projection? ∗ Corresponding author.
Y. Tang & Z. TuThe above question is the so-called “gain” L p − L q estimate properties of the Toeplitz operator. In˘Cu˘ckovi´c and McNeal [6], they gave an affirmative answer to the above question on smoothly boundedstrongly pseudoconvex domain Ω in C n by choosing a special symbol ψ := δ η , where η ≥ δ ( · ) = d ( · , ∂ Ω) is the Euclidean distance from the boundary. Later, Khanh-Liu-Thuc [11] also consideredthe same problem on certain classes of smoothly bounded pseudoconvex domains of finite type withsymbol ψ := K − α , where α ≥ K is the Bergman kernel on diagonal. Recently, they continuedto work on the same question on fat Hartogs triangle Ω k = { ( z , z ) ∈ C : | z | k < | z | < } ( k ∈ Z + ),which has singular boundary points (see Khanh-Liu-Thuc [12]).1.2. Generalizations of the Hartogs triangleBesides the power-generalized Hartogs triangles Ω γ = { ( z , z ) ∈ C : | z | γ < | z | < } ( γ ∈ R + )investigated by Chakrabarti-Zeytuncu [4] and Edholm-McNeal [7, 8], Beberok [2] also considered the L p boundedness of the Bergman projection on the following generalization of the Hartogs triangle H n +1 k := { ( z, w ) ∈ C n × C : k z k < | w | k < } , k ∈ Z + , (1.1)where k · k is the Euclidean norm in C n . He proved the following result: Theorem 1.1. (see Beberok [ , Theorem 2 . Let p ∈ (1 , ∞ ) . Then the Bergman projection is abounded operator on L p ( H n +1 k ) if and only if p ∈ (cid:0) nk +2 nk +2 , nk +2 nk (cid:1) . Therefore, the restricted range of p is affected by the dimension and the “camber” of the domain H n +1 k .In recent paper [5], Chen introduced a family of bounded Hartogs domains generalizing the Hartogstriangle as follows. For j = 1 , · · · , l , let Ω j be a bounded smooth domain in C k j , φ j : Ω j → B k j be a biholomorphic mapping, m j = P js =1 k s with m = 0 and m l = k + · · · + k l := k < n ,˜ z j = ( z m j − +1 , . . . , z m j ), and z = (˜ z , · · · , ˜ z l , z k +1 , · · · , z n ) ∈ C n . For 1 ≤ k < n , define H n { k j ,φ j } = { z ∈ C n : max ≤ j ≤ l k φ j ( e z j ) k < | z k +1 | < · · · < | z n | < } , (1.2)when l = k = 1 , n = 2, and φ is the identity map, we obtain the classical Hartogs triangle. Theauthor proved the following result: Theorem 1.2. (see Chen [ , Theorem 1 . For ≤ p < ∞ and ≤ k < n , the Bergman projection P H n { kj,φj } for H n { k j ,φ j } is bounded on L p ( H n { k j ,φ j } ) if and only if p is in the range ( nn +1 , nn − ) . Thus, the L p boundedness of the Bergman projection on H n { k j ,φ j } is only dependent on the dimen-sion n but not on { k j , φ j } .Now, we consider a slightly wider class of non-smooth bounded pseudoconvex domains whichcontains the above two domians (1.1) and (1.2). It is defined by H n { k j ,φ j ,b } = { z ∈ C n : max ≤ j ≤ l k φ j ( e z j ) k < | z k +1 | b < · · · < | z n | b < } , (1.3)where the notations are same as those in (1.2), and b ∈ Z + . When k = n − , l = 1, and φ isthe identity map, we obtain H nb . There exists a biholomorphism Θ : H n { k j ,φ j ,b } → H n { k j ,φ j } , which isdefined by Θ( z ) = ( φ − ( φ (˜ z ) z − bk +1 ) , · · · , φ − l ( φ l (˜ z l ) z − bk +1 ) , z k +1 , · · · , z n ) . When b = 1, Θ is just the identity map.It is well known that even though two domains are biholomorphic equivalence, the corresponding L p behavior of the Bergman projection on these two domains may be totally different from each other,let alone the Toeplitz operator constructed by the Bergman projection. Therefore, it would be ofoeplitz Operator 3interest to consider what will happen for the boundedness of the Bergman projection in this moregeneral setting.In this paper, inspired by the idea in Chen [5] and Khanh-Liu-Thuc [12], we mainly focus on the L p − L q boundedness of the Toeplitz operator with symbol K − t ( z, z ) ( t ≥
0) on H n { k j ,φ j ,b } , where K − t ( z, z ) := ( K ( z, z )) − t , and K ( z, z ) is the Bergman kernel on diagonal for H n { k j ,φ j ,b } . We concludethat the parameter b plays an interesting role in the restricted range of p for the boundedness of theBergman projection on the domain considered in this paper.1.3. Main results Theorem 1.3.
Let < p ≤ q < ∞ and T K − t : L p ( H n { k j ,φ j ,b } ) → L q ( H n { k j ,φ j ,b } ) be the Toeplitz operatorwith symbol K − t ( z, z ) . (1) If q ∈ (cid:2) n +2 k ( b − n − k ( b − , ∞ (cid:1) , then T K − t is unbounded for any t ∈ [0 , ∞ ) . (2) If q ∈ (cid:0) n − k ( b − n +1+ k ( b − − /p , n +2 k ( b − n − k ( b − (cid:1) , then T K − t is bounded if and only if t ≥ p − q . (3) If q ∈ (cid:2) p, n − k ( b − n +1+ k ( b − − /p (cid:3) , then T K − t is bounded if and only if t > p + (1 − p )2 p n +1+ k ( b − n − k ( b − . Setting t = 0 and p = q in Theorem 1.3, we obtain a sharp range of p for the boundedness of theBergman projection on H n { k j ,φ j ,b } as follows. Corollary 1.4.
Let < p < ∞ . Then the Bergman projection from L p ( H n { k j ,φ j ,b } ) to itself is boundedif and only if p ∈ (cid:0) n +2 k ( b − n +1+ k ( b − , n +2 k ( b − n − k ( b − (cid:1) . Remark 1.1. (1)
When b = 1 , Corollary 1.4 is the known result in Chen [5, Theorem 1.1] in the case < p < ∞ . (2) When l = 1 , k = n − , and φ is the identity map, Corollary 1.4 is just the result obtained byBeberok [2, Theorem 2.1] for H nb . It is shown that, influenced by the parameter b , the L p -boundedness of the Bergman projections onthese domains present an interesting phenomenon. More precisely, comparing with the result obtainedby Chen on H n { k j ,φ j } (see Theorem 1.2 mentioned above), the boundedness range of p for the Bergmanprojection on H n { k j ,φ j ,b } is not only dependent on the dimension n but also k and b unless b = 1, where k = k + · · · + k l (see Corollary 1.4). Moreover, the L p -boundedness range (cid:0) n +2 k ( b − n +1+ k ( b − , n +2 k ( b − n − k ( b − (cid:1) becomes smaller with the increase of b and approaches { } as b → ∞ .Through out this paper, we write A . B to mean that there exists a constant C > A ≤ C B , and use A ≈ B for B . A . B . For the multi-index α ∈ Z n , we write α =( ˜ α , · · · , ˜ α l , α k +1 , · · · , α n ) ∈ Z k ×· · ·× Z k l × Z × · · · × Z := Z k × Z n − k , where ˜ α j = ( α m j − +1 , . . . , α m j ) ∈ Z k j with m j = P js =1 k s , m = 0 and m l = k + · · · + k l := k < n , j = 1 , · · · , l . Let | α | = α + · · · + α n .In addition, the volume measures mentioned below are all normalized. H n { k j ,φ j ,b } and H n { k j ,b } When φ j ’s are all identity maps in (1.3), we denote this special domain by H n { k j ,b } = { z ∈ C n : max ≤ j ≤ l k e z j k < | z k +1 | b < · · · < | z n | b < } . Y. Tang & Z. TuIt is easy to verify that H n { k j ,b } is biholomorphic to H n { k j ,φ j ,b } by the biholomorphismΦ( z ) := ( φ − ( e z ) , · · · , φ − l ( e z l ) , z k +1 , · · · , z n ) , z ∈ H n { k j ,b } . (2.1)The careful reader will note that the biholomorphism (2.1) is same as the biholomorphism between H n { k j } and H n { k j ,φ j } , two domains considered by Chen [5]. Applying Bell’s extension Theorem, Chenobtain the equivalence of the L p boundedness of the Bergman projections on these two domains. Here,this method is also adapted to the special Toeplitz operators on the two domains considered by us. Lemma 2.1.
Let < p, q < ∞ , t ≥ , K and K be the Bergman kernels on diagonal for H n { k j ,b } and H n { k j ,φ j ,b } , respectively. Then the following statements are equivalent.(1) T K − t is bounded from L p (cid:0) H n { k j ,b } (cid:1) to L q (cid:0) H n { k j ,b } (cid:1) .(2) T K − t is bounded from L p (cid:0) H n { k j ,φ j ,b } (cid:1) to L q (cid:0) H n { k j ,φ j ,b } (cid:1) . Proof. (1) ⇒ (2). We put H := H n { k j ,b } and H := H n { k j ,φ j ,b } for short. From Chen [5, Section 6], wecould find two real numbers c and d such that0 < c ≤ | det Φ ′ ( z ) | ≤ d, z ∈ H . (2.2)Assume that T K − t is bounded from L p ( H ) to L q ( H ). Let f ∈ L p ( H ), ζ := Φ( z ), and η := Φ( w ) inthe following integral. By the transformation rule for the Bergman kernel under biholomorphism, wehave k T K − t f k qL q ( H ) = Z H (cid:12)(cid:12)(cid:12) Z H K ( ζ, η ) f ( η ) K − t ( η, η ) dv ( η ) (cid:12)(cid:12)(cid:12) q dv ( ζ )= Z H (cid:12)(cid:12)(cid:12) T K − t (cid:0) ( f ◦ Φ) · det Φ ′ · | det Φ ′ | t (cid:1) ( z ) (cid:12)(cid:12)(cid:12) q | det Φ ′ ( z ) | − q dv ( z ) . k T K − t (cid:0) ( f ◦ Φ) · det Φ ′ · | det Φ ′ | t (cid:1) k qL q ( H ) . k ( f ◦ Φ) · det Φ ′ · | det Φ ′ | t k pL p ( H ) . k f k pL p ( H ) . For the last three lines, we apply the boundedness of T K − t and estimate (2.2).(2) ⇒ (1). Same argument for Φ − will show the desired result. This finishes the proof.Therefore, it is sufficient to investigate the L p − L q boundedness of the Toeplitz operator withsymbol K − t ( t ≥
0) on H n { k j ,b } . In the rest of the note, we will focus on the domain H n { k j ,b } .1.2 The orthogonal basis of A ( H n { k j ,b } )Define a biholomorphic map Ψ : H n { k j ,b } → Π n { k j } given byΨ(˜ z , · · · , ˜ z l , z k +1 , z n ) = (cid:16) ˜ z z bk +1 , · · · , ˜ z l z bk +1 , z k +1 z k +2 , · · · , z n − z n , z n (cid:17) , (2.3)where Π n { k j } := B k × · · · × B k l × D ∗ × · · · × D ∗ | {z } n − k . We denote its inverse by G . Then the determinant ofthe complex Jacobian of G is given bydet G ′ ( η ) = n Y j = k +1 η j − b − kj , η ∈ Π n { k j } , (2.4)oeplitz Operator 5and the Bergman kernel on diagonal for H n { k j ,b } is K ( z, z ) = K ( G ( η ) , G ( η )) = 1det | G ′ ( η ) | Q lj =1 (1 − k ˜ η j k ) k j +1 Q nj = k +1 (1 − | η j | ) . (2.5) Lemma 2.2.
Let A := { α ∈ Z n : α ∈ N k × Z n − k } . Then B ( H n { k j ,b } ) := (cid:8) z α : α ∈ A , X mj =1 α j + ( b − X kj =1 α j > (1 − b ) k − m, m = k + 1 , · · · , n (cid:9) is a complete orthogonal basis for A ( H n { k j ,b } ) . Proof. We first consider the following operator Γ Ψ : A (Π n { k j } ) → A ( H n { k j ,b } ) defined byΓ Ψ f := ( f ◦ Ψ) · det Ψ ′ . Since Ψ is biholomorphic from H n { k j ,b } to Π n { k j } , it is easy to verify that Γ Ψ : A (Π n { k j } ) → A ( H n { k j ,b } ) isunitary. Together with the fact that { e β ( w ) := w β = ˜ w ˜ β · · · ˜ w ˜ β l l w β k +1 k +1 · · · w β n n , β ∈ N n } is a completeorthonormal basis in A (Π n { k j } ). Thus { ( e β ◦ Ψ)( z ) · [det G ′ (Ψ( z ))] − , β ∈ N n } (2.6)forms a complete orthonormal basis in A ( H n { k j ,b } ). Substituting (2.3) and (2.4) into (2.6), we couldobtain Lemma 2.2 after uniting similar terms.1.3. Other key LemmasWe first give the generalised version of Schur’s test introduced by Khanh-Liu-Thuc [11], which isan important tool of studying the L p − L q boundedness for Toeplitz operator with symbol ψ . Lemma 2.3. (see [ , Theorem 5 . Let ( X, µ ) , ( Y, υ ) be measure spaces with σ -finite, positive mea-sures; let < p ≤ q < ∞ and r ∈ R . Let K : X × Y → C and ψ : Y → C be measurable functions.Assume that there exist positive measurable functions h , h on Y and f on X such that h − h ψ ∈ L ∞ ( Y, dυ ) and the inequalities Z Y | K ( x, y ) | rp ∗ h p ∗ ( y ) dυ ( y ) ≤ C f p ∗ ( x ) , (2.7) Z X | K ( x, y ) | (1 − r ) q f q ( x ) dµ ( x ) ≤ C h q ( y ) , (2.8) hold for almost every x ∈ ( X, µ ) and y ∈ ( Y, υ ) , where p + p ∗ = 1 and C , C are positive constants.Then the Toeplitz operator T ψ associated to the kernel K and the symbol ψ defined by T ψ ( f )( x ) := Z Y K ( x, y ) f ( y ) ψ ( y ) dυ ( y ) is bounded from L p ( Y, υ ) into L q ( X, µ ) . Furthermore, k T ψ k L p ( Y,υ ) → L q ( X,µ ) ≤ C p − p C q k h − h ψ k L ∞ ( Y,υ ) . Y. Tang & Z. Tu
Lemma 2.4. (see Herbort [ ] , Blocki [ ]) Let Ω be a bounded pseudoconvex domain in C n and let s be any positive number. Then for any holomorphic function f on Ω and any w ∈ Ω , Z { G ( · ,w ) < − s } | f ( z ) | dz ≥ e − ns | f ( w ) | K ( w, w ) , where G ( · , w ) is the pluricomplex Green function on Ω with a pole w . For the definition and properties of the pluricomplex Green function, see Klimek [13, Chapter 6].Next, using approach as in Khanh-Liu-Thuc [12, Lemma 3.2], we obtain a similar estimate asfollows. The difference is that we need to extend their estimates on the unit disc D to the unit ballduring the process of the proof. Lemma 2.5.
Fix w ∈ H n { k j ,b } and s ∈ R + . Then for any z ∈ { z ∈ H n { k j ,b } : G H n { kj,b } ( z, w ) < − s } , wehave K ( z, z ) K ( w, w ) ≈ (cid:12)(cid:12)(cid:12) det Ψ ′ ( z )det Ψ ′ ( w ) (cid:12)(cid:12)(cid:12) . Proof. We divide into two steps to prove Lemma 2.5.
Step 1.
Assume that w ∈ B k , ϕ w is the automorphism of B k taking 0 to w , and k ϕ w ( z ) k < e − s for all z ∈ B k . Then 1 − k z k ≈ − k w k . When k = 1, it will degenerate to the result in [12, Lemma3.2].Indeed, employing the properties of the automorphism of the unit ball (see Rudin [19, Theorem2.2.2]), we have 1 − k z k − k w k = 1 − k ϕ w ◦ ϕ w ( z ) k − k w k = 1 − k ϕ w ( z ) k | − h ϕ w ( z ) , w i| . Hence 1 − k ϕ w ( z ) k k ϕ w ( z ) k ≤ − k z k − k w k ≤ k ϕ w ( z ) k − k ϕ w ( z ) k . Since k ϕ w ( z ) k < e − s , then there exist positive constants C ( s ) = e − s − e − s , depending only on s , suchthat C − ( s )(1 − k w k ) . − k z k . C ( s )(1 − k w k ). Step 2.
Since H n { k j ,b } is holomorphic equivalent to the product domain Π n { k j } via map Ψ (seeformula (2.3)), then associating with the biholomorphic invariant and the product-property of thepluricomplex Green function for pseudoconvex domains (see Klimek [14, Theorem 2.3, Theorem 4.2]),we obtain that G H n { kj,b } ( z, w ) = G Π n { kj } (Ψ( z ) , Ψ( w ))= max n G B k (cid:16) ˜ z z bk +1 , ˜ w w bk +1 (cid:17) , · · · , G D ∗ ( z n , w n ) o for all z, w ∈ H n { k j ,b } . Then for any z ∈ { z ∈ H n { k j ,b } : G H n { kj,b } ( z, w ) < − s } , we havelog (cid:13)(cid:13)(cid:13) ϕ ˜ wjwbk +1 (cid:16) ˜ z j z bk +1 (cid:17)(cid:13)(cid:13)(cid:13) < − s, j = 1 , · · · , l ;log (cid:12)(cid:12)(cid:12) ϕ wjwj +1 (cid:16) z j z j +1 (cid:17)(cid:12)(cid:12)(cid:12) < − s, j = k + 1 , · · · , n − | ϕ w n ( z n ) | < − s. Here we use the fact that G B k ( z, w ) = log k ϕ w ( z ) k (see Klimek [14, Example 2.2]). By the discussionof Step 1, we get 1 − k ˜ z j k | z k +1 | b ≈ − k ˜ w j k | w k +1 | b , j = 1 , · · · , l ;oeplitz Operator 71 − (cid:12)(cid:12)(cid:12) z j z j +1 (cid:12)(cid:12)(cid:12) ≈ − (cid:12)(cid:12)(cid:12) w j w j +1 (cid:12)(cid:12)(cid:12) , j = k + 1 , · · · , n − − | z n | ≈ − | w n | . Thus, applying the explicit formula (2.5) of the Bergman kernel on diagonal for H n { k j ,b } , we couldderive the estimate in Lemma 2.5. The proof is completed. Lemma 2.6. (see Khanh − Liu − Thuc [ , Lemma 2 . Let a ≥ , − < u < , and c > − . Then Z D (1 − | w | ) u | w | c | − z ¯ w | a dv ( w ) . (1 − | z | ) − a + u +2 , z ∈ D . Lemma 2.7.
Let a ≥ , − < u < . Then Z B k (1 − k w k ) u | − h z, w i| ( k +1) a dv ( w ) . (1 − k z k ) u +( k +1)(1 − a ) , z ∈ B k . Proof. When a = 1, it is the result in Chen [5, Lemma 3.2]. Since | − h z, w i| ≥ (1 − k z k ) for z, w ∈ B k , when a ≥
1, we have | − h z, w i| ( k +1) a & (1 − k z k ) ( k +1)( a − | − h z, w i| k +1 . Then Z B k (1 − k w k ) u | − h z, w i| ( k +1) a dv ( w ) . (1 − k z k ) ( k +1)(1 − a ) Z B k (1 − k w k ) u | − h z, w i| k +1 dv ( w ) . (1 − k z k ) ( k +1)(1 − a )+ u , which proves Lemma 2.7. L p − L q boundedness of the Toeplitz operator withsymbol K − t ( z, z ) ( t ≥
0) on H n { k j ,b } .Here and in the sequel, we set C b,k := k ( b − q ≥ n +2 C b,k n − C b,k , then z − n − C b,k n / ∈ A q ( H n { k j ,b } ).(b) If t ≥ z β ∈ B ( H { k j ,b } ), then h K − t ( z, z )¯ z n − C b,k n , z β i H n { kj,b } = 0 unless β = (0 , . . . , , − n − C b,k ).Indeed, for (a), by Lemma 2.2, we learn that z − n − C b,k n ∈ A ( H n { k j ,b } ), and Z H n { kj,b } | z − n − C b,k n | q dv ( z )= n − Y j = k +1 Z D ∗ | w j | j − C b,k ) dv ( w j ) · Z D ∗ | w n | (2 − q )( n − C b,k ) dv ( w n ) Y. Tang & Z. Tu < + ∞ if and only if (2 − q )( n − C b,k ) > −
2. This proves (a).For the second fact (b), let β b,k := ( b − β + · · · + β k ), making the change of variables z = G ( η ),we have h K − t ( z, z )¯ z n − C b,k n , z β i H n { kj,b } = l Y j =1 Z B kj (1 − k ˜ η j k ) t ( k j +1) ¯˜ η ˜ β j j dv (˜ η j ) · n − Y j = k +1 Z D ∗ | η j | (2+2 t )( j − C b,k ) (1 − | η j | ) − t ¯ η β + ··· + β j + β b,k j dv ( η j ) · Z D ∗ | η n | (2+2 t )( n − C b,k ) (1 − | η n | ) − t ¯ η n − C b,k + | β | + β b,k n dv ( η n ) = 0if and only if β = (0 , · · · , , − n − C b,k ). This proves part (b).Then by the fact ( b ), for all t ≥ z β ∈ B ( H { k j ,b } ), we learn that h T K − t (¯ z n − C b,k n ) , z β i H n { kj,b } = h P H n { kj,b } ( K − t ( z, z )¯ z n − C b,k n ) , z β i H n { kj,b } = h K − t ( z, z )¯ z n − C b,k n , z β i H n { kj,b } = 0 unless β = (0 , . . . , , − n − C b,k ) . Note that the complete orthogonal basis for A ( H n { k j ,b } ) in Lemma 2.2, then there exists a non-zeroconstant C such that T K − t (¯ z n − C b,k n ) = Cz − n − C b,k n . We finish the proof of Theorem 1.3 (1) bycombining with the above fact (a).3.2. Proof of Theorem 1.3 (2). Sufficiency.
Suppose now that n − C b,k n +1+ C b,k − /p < q < n +2 C b,k n − C b,k and t ≥ p − q . Denote the Bergmankernel on the the product domain Π n { k j } by ˆ K . Then we consider the following test functions f ( z ) := f ( G ( η )) := ρ − λ ( G ( η )) | det G ′ ( η ) | − /p ∗ ; h ( w ) := h ( G ( ζ )) := ρ − λ ( G ( ζ )) Q nj = k +1 | ζ j | m j ; h ( w ) := h ( G ( ζ )) := ρ − λ ( G ( ζ )) ˆ K /p − /q ( ζ, ζ ) | det G ′ ( ζ ) | − /p , (3.9)where ρ ( G ( η )) := Q lj =1 (1 − k ˜ η j k ) Q nj = k +1 (1 − | η j | ), p ∗ is the conjugate exponent of p , and theparameters λ and { m j } nj = k +1 satisfy ( < λ < min { q , p ∗ } , − j +1+ C b,k p ∗ < m j ≤ ( j − C b,k )( q − p ) pq . (3.10)The existence of m j follows that − j + 1 + C b,k p ∗ < ( j − C b,k )( q − p ) pq oeplitz Operator 9 ⇔ q > j −
1) + 2 C b,k j + 1 + C b,k − /p ⇐ q > n −
1) + 2 C b,k n + 1 + C b,k − /p . (3.11)The last inequality is obvious right according to the sufficient condition of Theorem 1.3 (2).Let r = ( p ∗ ) − in Lemma 2.3. Substituting (3.9) into (2.7) and (2.8) respectively, and making thechange of variables z = G ( η ) and w = G ( ζ ), we obtain Z H n { kj,b } | K ( z, w ) | rp ∗ h p ∗ ( w ) dv ( w )= 1 | det G ′ ( η ) | l Y j =1 Z B kj (1 − k ˜ ζ j k ) − λp ∗ | − h ˜ η j , ˜ ζ j i| k j +1 dv (˜ ζ j ) · n Y j = k +1 Z D ∗ (1 − | ζ j | ) − λp ∗ | ζ j | m j p ∗ + j − C b,k | − η j ¯ ζ j | dv ( ζ j ) . ρ − λp ∗ ( G ( η )) | det G ′ ( η ) | − = f p ∗ ( z ) (3.12)and Z H n { kj,b } | K ( z, w ) | (1 − r ) q f q ( z ) dv ( z )= 1 | det G ′ ( ζ ) | q/p l Y j =1 Z B kj (1 − k ˜ η j k ) − λq | − h ˜ η j , ˜ ζ j i| ( k j +1) q/p dv (˜ η j ) · n Y j = k +1 Z D ∗ (1 − | η j | ) − λq | η j | (2 − q )( j − C b,k ) | − η j ¯ ζ j | q/p dv ( η j ) . ˆ K q/p − ( ζ, ζ ) ρ − λq ( G ( ζ )) | det G ′ ( ζ ) | − q/p = h q ( w ) . (3.13)Here, for the inequalities in formulas (3.12) and (3.13), we employ the range of the parameters (3.10)and integral estimates in Lemma 2.6 and Lemma 2.7.On the other hand, put w = G ( ζ ), by the relationship ˆ K ( ζ, ζ ) = | det G ′ ( ζ ) | K ( G ( ζ ) , G ( ζ )), wesee that h − ( w ) h ( w ) K − t ( w, w )= K p − q − t ( G ( ζ ) , G ( ζ )) Y nj = k +1 | ζ j | ( j − C b,k )( p − q ) − m j . (3.14)By (3.10), we have ( j − C b,k )( p − q ) − m j ≥
0. Again since t ≥ p − q , it follows from formula(3.14) that h − h K − t ∈ L ∞ ( H { k j ,b } ). Thus, by Lemma 2.3, we obtain that T K − t is bounded from L p ( H n { k j ,b } ) to L q ( H n { k j ,b } ). This completes the proof of the sufficiency of Theorem 1.3 (2). Necessity.
Suppose now that n − C b,k n +1+ C b,k − /p < q < n +2 C b,k n − C b,k and T K − t is bounded from L p ( H n { k j ,b } )to L q ( H n { k j ,b } ). Let H := H n { k j ,b } . We set g w ( z ) := K ( z,w )det Ψ ′ ( z ) for z, w ∈ H . Then, similar computationas (3.12), we have k g w k L p ( H ) . ( K ( w, w )) − p | det Ψ ′ ( w ) | p − . (3.15)Assume that s ∈ R + . Then, by Lemma 2.5 and Lemma 2.4, we have Z H K − t ( z, z ) | g w ( z ) | dv ( z )0 Y. Tang & Z. Tu & Z { z ∈H ,G H ( · ,w ) < − s } (cid:12)(cid:12)(cid:12)(cid:12) K ( z, w )(det Ψ ′ ( z )) t (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) K ( z, z )(det Ψ ′ ( z )) (cid:12)(cid:12)(cid:12)(cid:12) − t dv ( z ) & (cid:12)(cid:12)(cid:12)(cid:12) K ( w, w )(det Ψ ′ ( w )) (cid:12)(cid:12)(cid:12)(cid:12) − t Z { z ∈H ,G H ( · ,w ) < − s } (cid:12)(cid:12)(cid:12)(cid:12) K ( z, w )(det Ψ ′ ( z )) t (cid:12)(cid:12)(cid:12)(cid:12) dv ( z ) & ( K ( w, w )) − t | det Ψ ′ ( w ) | − . On the other hand, we also have Z H K − t ( z, z ) | g w ( z ) | dv ( z )= Z H K − t ( z, z ) K ( w, z )det Ψ ′ ( z ) K ( z, w )det Ψ ′ ( z ) dv ( z )= Z H K − t ( z, z ) K ( w, z )det Ψ ′ ( z ) (cid:16) Z H K ( z, η ) K ( η, w )det Ψ ′ ( η ) dv ( η ) (cid:17) dv ( z )= Z H K ( η, w )det Ψ ′ ( η ) (cid:16) Z H K ( z, η ) K − t ( z, z ) K ( w, z )det Ψ ′ ( z ) dv ( z ) (cid:17) dv ( η )= Z H g w ( η ) T K − t ( g w )( η ) dv ( η ) . k g w k L q ∗ ( H ) k T K − t ( g w ) k L q ( H ) . k g w k L q ∗ ( H ) k g w k L p ( H ) . ( K ( w, w )) − p + q | det Ψ ′ ( w ) | p − q . For the last two lines, we apply the H¨older inequality, the L p − L q boundedness of the Toeplitz operator T K − t , and the estimate (3.15). Thus, comparing the above two formulas, we obtain( K ( w, w )) − t + p − q | det Ψ ′ ( w ) | − − p + q . constant . (3.16)Since | det Ψ ′ ( w ) | > w ∈ H and q ≥ p >
1, the second term is bounded. In addition, K ( w, w ) →∞ as w → ∂ H . Then it follows from (3.16) that p − q ≤ t . This proves the necessity of Theorem 1.3(2).3.3. Proof of Theorem 1.3 (3) Sufficiency.
Suppose that 1 < p ≤ q ≤ n − C b,k n +1+ C b,k − /p and t > p + (1 − p )2 p n +1+ C b,k n − C b,k . In order to provethe sufficiency of Theorem 1.3 (3), we only need to proceed as the proof of the sufficiency of Theorem1.3 (2). However, we should reset the value of parameters { m j } nj = k +1 by ( j + 1 + C b,k )(1 /p −
Suppose that 1 < p ≤ q ≤ n − C b,k n +1+ C b,k − /p and T K − t is bounded from L p ( H n { k j ,b } ) to L q ( H n { k j ,b } ). We argue by contradiction. Namely, we assume that t ≤ p + (1 − p )2 p n +1+ C b,k n − C b,k . We adoptoeplitz Operator 11a family of functions used in Chen [5], which was also applied by Khanh-Liu-Thuc [12]. Here, inorder to hold in our case, we make a little modification of the power of the functions. Next, for thecompleteness, we give the details. Define a sequence { f j } ∞ j =1 by f j ( z ) := ( h ( | z n | )¯ z n − C b,k n ; | z n | ∈ ( a j +1 , , | z n | ∈ (0 , a j +1 ] , where a j := j − j and the function h : (0 , → (0 , ∞ ) is defined by h ( r ) := r x , r ∈ ( a l +1 , a l ]; l = 1 , , · · · , where x = l − p ( n + C b,k ) − ( n − C b,k ). A simple calculation shows that k f j k pL p ( H n { kj,b } ) is controlledby P ∞ l =1 l − p . Then f j ∈ L p ( H n { k j ,b } ) for all p > T K − t ( f j )( G ( η )) = Z H n { kj,b } K ( G ( η ) , w ) K − t ( w, w ) f j ( w ) dv ( w )= Z Π { kj } K ( G ( η ) , G ( ζ )) K − t ( G ( ζ ) , G ( ζ )) f j ( G ( ζ )) | det G ′ ( ζ ) | dv ( ζ )= 1det G ′ ( η ) Z Π { kj } ˆ K ( η, ζ )det G ′ ( ζ ) ˆ K − t ( ζ, ζ ) f j ( G ( ζ )) | det G ′ ( ζ ) | t dv ( ζ ) . (3.18)Since by the proof of Lemma 2.2, ˆ K ( η, ζ ) could be written as P β ∈ N n | c β | e β ( η ) e β ( ζ ). Note that f j is only dependent on the last variables. Substituting the power series of ˆ K ( η, ζ ) into (3.18). Afterchanging the order of integral and summation, it is easy to obtain that the summation is only workon the index β = (0 , · · · , , k + C b,k , · · · , n − C b,k , | T K − t ( f j )( G ( η )) |≈ | η n | − n − C b,k | Z Π { kj } ζ − n − C b,k n ˆ K − t ( ζ, ζ ) f j ( G ( ζ )) | det G ′ ( ζ ) | t dv ( ζ ) |≈ | η n | − n − C b,k j X l =1 Z a l a l +1 (1 − r ) t r x +(2+2 t )( n − C b,k )+1 dr & j X l =1 Z a l a l +1 (1 − r ) t r x +(2+2 t )( n − C b,k )+1 dr. Since t ≤ p + (1 − p )2 p n +1+ C b,k n − C b,k , we have r x +(2+2 t )( n − C b,k )+1 ≥ r /l − and 2 t <
1. Therefore, we alsohave (1 − r ) t > − r for any r ∈ (0 , k T K − t ( f j ) k L q ( H n { kj,b } ) & j X l =1 Z a l a l +1 r /l − dr − j X l =1 Z a l a l +1 r /l dr. Since the first term goes to infinity and the second term converges as j → ∞ . So T K − t is not bounded,a contradiction. This proves the necessity condition of Theorem 1.3 (3). Acknowledgments
The project is supported by the National Natural Science Foundation of China(No. 11671306).2 Y. Tang & Z. Tu
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