Rigidity theorem of the Bergman kernel by analytic capacity
aa r X i v : . [ m a t h . C V ] J a n Rigidity theorem of the Bergman kernel byanalytic capacity
John Treuer
Abstract
In [7], Dong and I proved that the domains D ⊂ C of finite volume whose on-diagonalBergman kernels K ( · , · ) satisfy K ( z , z ) = V olume ( D ) − are disks minus closed polar sets.We utilized the solution of the Suita conjecture, a deep theorem of several complex variables.In this note, I present a significantly more elementary proof of this theorem that does not useseveral complex variables. As a corollary, a new lower bound for the on-diagonal Bergmankernel is given. Finally, I show that the only real ellipsoid in Webster normal form whichsatisfies K (0 ,
0) =
V olume ( D ) − is the unit ball. Let D be a domain in C n , and let A ( D ) be the closed holomorphic subspace of L ( D ). If { φ n } ∞ n =0 is an orthonormal basis of A ( D ), then the Bergman kernel K ( · , · ) : D × D → C is definedby K ( z, w ) = ∞ X n =0 φ n ( z ) φ n ( w ) . (1.1)This definition is independent of the choice of orthonormal basis. Let v denote the Euclideanvolume measure in R n and k · k the L ( D )-norm with respect to the measure v . If v ( D ) < ∞ ,then k v ( D ) − / k = 1. After completing { v ( D ) − / } to an orthonormal basis { v ( D ) − / }∪{ φ n } ∞ n =1 of A ( D ), K ( z, w ) = v ( D ) − + ∞ X n =1 φ n ( z ) φ n ( w ) . (1.2)When z = w , the on-diagonal Bergman kernel satisfies K ( z, z ) ≥ v ( D ) − . (1.3)In [7], Dong and I completely characterized the domains in C for which the on-diagonal Bergmankernel achieves the lower bound in (1.3). For finite-volume domains we proved the following: heorem 1. [7] Let D ⊂ C be a domain with v ( D ) < ∞ . Suppose there exists a z ∈ D suchthat K ( z , z ) = v ( D ) − . (1.4)Then D = D ( z , r ) \ P . Here D ( z , r ) denotes the disk centered at z of radius r and P is apossibly empty, polar set closed in the relative topology of D ( z , r ).The authors proved Theorem 1 by using the solution of the Suita conjecture. The Suitaconjecture was conjectured by Suita [15] in 1972. It was studied by Ohsawa [14] and proved byB locki [4] and Guan and Zhou [10] using the Ohsawa-Takegoshi L extension theorem. Dong [6]reproved parts of Guan and Zhou’s proof using Maitani and Yamaguchi’s variational formula forthe Bergman kernel. Recently, Dong and Zhang [8] gave generalizations of Theorem 1.Since Theorem 1 pertains to domains in C , it would be desirable to prove it using only complexanalysis of one variable and Riemann surface theory and not rely on results proved using severalcomplex variables. In this paper, I give such a proof. In the new proof, I use results on theanalytic capacity of domains in C due to Suita [15] and Ahlfors and Beurling [2]. As a corollary,I also improve the lowerbound in (1.3). Corollary 2.
Let f z ( z ) = z − z and D ⊂ C be a domain with v ( D ) < ∞ . Then K ( z , z ) ≥ v ( C ∞ \ f z ( D )) π ≥ v ( D ) , where C ∞ = C ∪ {∞} and the second inequality is strict if and only if D does not equal a diskminus a relatively closed set of measure 0. In [7] we also proved that when v ( D ) = ∞ , (1.4) holds if and only if D = C \ P where P isa closed polar set. We also used the Suita conjecture for this proof. B locki and Zwonek [5] alsoproved the v ( D ) = ∞ case, but without the Suita conjecture.When n >
1, a much wider class of domains have Bergman kernels which satisfy (1.4). Adomain D is said to be circular containing its center z if z ∈ D and { z } + e iθ ( D − { z } ) ⊂ D .The Bergman kernel of such a domain satisfies (1.4), see [3]. The final part of this paper investigateswhether (1.4) also holds for the real ellipsoids { ( x + ix , . . . , x n − + ix n ) ∈ C n : n X j =1 a j x j − + b j x j < } , a j ≥ b j > , j = 1 , . . . , n. (1.5)After a complex linear change of variables, the ellipsoids in (1.5) can be written in its Websternormal form [13] E A = { z ∈ C n : ρ ( z ) = | z | + n X j =1 A j Re ( z j ) < } , ≤ A j = a j − b j a j + b j < , j = 1 , . . . , n, where A = ( A , . . . , A n ). The Bergman kernels of the ellipsoids were studied by Hirachi, [13]. Theellipsoids have also been studied in pseudo-Hermitian CR geometry, see [11, 12]. The final resultof the paper characterizes the ellipsoids which satisfy (1.4). Theorem 3.
Let K A ( · , · ) denote the Bergman kernel of the normalized real ellipsoid E A . Then K A (0 ,
0) = v ( E A ) − if and only if A = (0 , . . . , . In Section 2, the relevant theory about the analytic capacity is recalled. In Section 3, Theorem1, Corollary 2, and Theorem 3 are proved. 2
The Analytic Capacity
The analytic capacity was introduced by Ahlfors [1, 2] to study the domains where the set ofthe bounded analytic functions on the domain consists of only constant functions.
Definition 4.
The analytic capacity of a domain D at z ∈ D is c D ( z ) = sup {| g ′ ( z ) | : g ∈ O ( D ) , g ( z ) = 0 , | g ( z ) | ≤ } , where O ( D ) denotes the holomorphic functions on D . The analytic capacity satisfies a transformation property under biholomorphic mappings. If h : D → h ( D ) is biholomorphic and z ∈ D , then c D ( z ) = | h ′ ( z ) | c h ( D ) ( h ( z )) . (2.1)In 1972, Suita [15] showed that the on-diagonal Bergman kernel and the analytic capacity arerelated as follows: Theorem 5. [15]
Suppose D is an open Riemann surface which admits a Green’s function. Then p πK ( z , z ) ≥ c D ( z ) and equality holds if and only if D is biholomorphically equivalent to theunit disk minus a (possibly empty) closed set of inner capacity zero. A closed set of inner capacity zero is a closed polar set. In the literature, the analytic capacityis often defined in terms of compact sets.
Definition 6. [9]
The analytic capacity γ ( E ) of a compact subset E ⊂ C is γ ( E ) = sup {| g ′ ( ∞ ) | : g ∈ O ( C ∞ \ E ) , g ( ∞ ) = 0 , | g ( z ) | ≤ } where g ′ ( ∞ ) = lim z →∞ z ( g ( z ) − g ( ∞ )) . A compact set E ⊂ C satisfies γ ( E ) = 0 if and only if every bounded holomorphic function on C \ E is constant. The two definitions of analytic capacity given are related as follows. Definition 7.
Let f z ( z ) = z − z . For ease of notation, when z = 0 , let f = f z . Lemma 8.
For any domain D ⊂ C , c D ( z ) = γ ( C ∞ \ f z ( D )) . Proof.
Since f ( D − { z } ) = f z ( D ) and by (2.1), c D ( z ) = c D −{ z } (0) , it suffices to prove thelemma when z = 0 ∈ D . Notice that for g holomorphic in a neighborhood of ∞ with g ( ∞ ) = 0,( g ◦ f ) ′ ( ∞ ) = lim w →∞ w (( g ◦ f )( w ) − ( g ◦ f )( ∞ ))= lim w → g ◦ f ( w ) − g (0) w −
0= lim w → g ( w ) − g (0) w − g ′ (0) . E = C ∞ \ f ( D ). Then c D (0) = sup {| ( g ◦ f ) ′ ( ∞ ) | : g ◦ f ∈ O ( f ( D )) , ( g ◦ f )( ∞ ) = 0 , | g ◦ f | ≤ }≤ γ ( E )= sup {| ( g ◦ f ) ′ (0) | : g ◦ f ∈ O ( D ) , ( g ◦ f )(0) = 0 , | g ◦ f | ≤ }≤ c D (0) . The analytic capacity satisfies the following lower bound due to Ahlfors and Beurling, see [2], [9,Theorem 4.6].
Theorem 9 (Ahlfors-Beurling Inequality) . For any compact set E ⊂ C , γ ( E ) ≥ v ( E ) π . Lemma 10.
Let D ⊂ C be a domain with z ∈ D and v ( D ) < ∞ . Then πv ( D ) ≤ v ( C ∞ \ f z ( D )) π (3.1) and equality holds if and only if D is a disk centered at z minus a relatively closed set of measure0.Proof. If M ⊂ C is a set with v ( M ) = 0, then v ( f z ( M )) = 0. With this fact it is straightforwardto verify that equality holds for a disk minus a relatively closed set of measure 0. Since v ( D ) = v ( D − { z } ) and f z ( D ) = f ( D − { z } ), without loss of generality we may suppose z = 0 ∈ D .Since f is an automorphism of C ∞ , f ( D ) ⊔ ( C ∞ \ f ( D )) = C ∞ = D ⊔ ( C ∞ \ D ) . Hence C ∞ \ f ( D ) = f ( C ∞ \ D ) . (3.2)By (3.2), (3.1) is equivalent to v ( f ( C ∞ \ D )) v ( D ) ≥ π . (3.3)Let ∆ = D (0 , a ) denote the disk centered at 0 of radius a where a is chosen so that v (∆) = v ( D ).Since π = v (∆) v ( f ( C ∞ \ ∆)) , (3.4)(3.3) is equivalent to v ( f ( C ∞ \ D )) ≥ v ( f ( C ∞ \ ∆)) . (3.5)Let S = ( C ∞ \ D ) ∩ ∆ , S = ( C ∞ \ ∆) ∩ D. C ∞ \ D = C ∞ \ ( D ∪ ∆) ⊔ ( C ∞ \ D ) ∩ ∆and f is an automorphism of C ∞ , v ( f ( C ∞ \ D )) = v ( f (( C ∞ \ ( D ∪ ∆)))) + v ( f ( S )) . Similarly, v ( f ( C ∞ \ ∆)) = v ( f (( C ∞ \ (∆ ∪ D )))) + v ( f ( S )) . Thus, (3.5) is equivalent to v ( f ( S )) ≥ v ( f ( S )) . (3.6)For all z ∈ S , | z | ≤ a and for all z ∈ S , | z | ≥ a . Hencemin z ∈ S | z | ≥ a ≥ max z ∈ S | z | . Notice also that v ( S ) = v (∆ \ D ) = v ( D \ ∆) = v ( S ) . Thus, (3.6) is equivalent to v ( f ( S )) = Z f ( S ) dv ( z )= Z S | z | dv ( z ) ≥ a v ( S )= 1 a v ( S ) ≥ Z S | z | dv ( z )= v ( f ( S )) . (3.7)This completes the proof of the inequality part. For the equality part, first replace the inequalitiesin (3.3)-(3.6) with equalities. By (3.6) and (3.7), Z C | z | ( χ S ( z ) − χ S ( z )) dv ( z ) = 0 . Since v ( S ) = v ( S ), α Z C ( χ S ( z ) − χ S ( z )) dv ( z ) = 0 , α ∈ C . Thus, Z C ( 1 | z | + α )( χ S ( z ) − χ S ( z )) dv ( z ) = 0 . (3.8)Without loss of generality, there exists r so that S ⊂ D (0 , a ) \ D (0 , r ) , S ⊂ { z ∈ C : | z | ≥ a } . r , z ∈ S ⇒ a < | z | < r z ∈ S ⇒ | z | ≤ a . Let α = − a . Then ( 1 | z | + α ) χ S ( z ) ≤ | z | + α )( − χ S ( z )) ≤ α that v ( S ) = v ( S ) = 0. Furthermore, since D is open, S = D \ ∆ = ∅ . Thus, D = ∆ \ S . Notice S = ∆ \ D is closed in ∆. Proof of Theorem 1.
By Theorems 5 and 9 and Lemmas 8 and 10, πK ( z , z ) ≥ c D ( z ) = γ ( C ∞ \ f z ( D )) ≥ v ( C ∞ \ f z ( D )) π ≥ πv ( D ) = πK ( z , z ) . (3.11)By Theorem 5, D is biholomorphic to a disk minus a closed polar set. Let F : D → D (0 ,
1) bebiholomorphic with F ′ ( z ) > F ( z ) = 0. By the transformation law of the Bergman kernel, F ′ ( z ) F ′ ( z ) K D (0 , ( F ( z ) ,
0) = K D ( z, z ) . (3.12)By (1.2), since K D ( z , z ) = v ( D ) − , K D ( · , z ) is a constant function. Since K D (0 , ( · ,
0) is alsoconstant, so is F ′ ( · ). Thus, F is affine and D is a disk minus a closed polar set. Proof of Corollary 2.
As with (3.11), this follows from Theorems 5 and 9 and Lemmas 8 and10.
Proof of Theorem 3. If A = (0 , . . . , E A is the unit ball and (1.4) holds. The otherdirection will be proved by contradiction. Suppose without loss of generality that A n = 0. If K A (0 ,
0) = v ( E A ) − , then by (1.2), K A (0 , z ) = K A (0 , v ( E A ) Z E A z n dv ( z ) . (3.13)Notice that Re ( Z E A z n dv ( z )) = Z E A x n − dv ( x , . . . , x n ) − Z E A x n dv ( x , . . . , x n ) =: I − II. X = ( X , . . . , X n ) where X i − = x i − p A i , X i = x i p − A i , i = 1 , . . . , n, and α = ( Q ni =1 (1 + A i )(1 − A i )) − . Using these coordinates, I = α Z { X + ··· + X n < } X n − (1 + A n ) dv ( X ) =: 1(1 + A n ) γ. Similarly, II = − A n ) γ . Notice I − II = 0, which implies that (3.13) does not hold. Thus, K A (0 , = v ( E A ) − . References [1] L. Ahlfors,
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