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Geometric Properties of a General Kohn-Nirenberg Domain in $\mathbb{C}^{n}$

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31 December 2024

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31 December 2024

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Abstract
The Kohn-Nirenberg domains are unbounded domains in $\mathbb{C}^{n}$. In this article, we modify the Kohn-Nirenberg domain $\Omega_{K,L} =\left\{(z_{1},\ldots, z_{n}) \right.\in \mathbb{C}^{n} : Re z_{n} + g \mid z_{n}\mid^{2} + \sum_{j = 1}^{n-1} (\mid z_{j}\mid^{p} +K_{j} \mid z_{j}\mid^{p-q} Re z_{j}^{q} + L_{j} \mid z_{j}\mid^{p-2q} Im z_{j}^{2q}) < 0 \}$ and discuss the existence of supporting surface and peak function at the origin.
Keywords: 
;  ;  

1. Introduction

Consider a domain Ω C n with smooth boundary Ω . As the boundary point is a strongly pseudoconvex point p Ω , we can find a local system of holomorphic coordinates. Hakim [1] and Pflug[2] showed that every strongly pseudoconvex point of Ω is a peak point. But this property fails for weakly pseudoconvex boundary point in general. Kohn and Nirenberg have an example that is defined with the boundary point ( 0 , 0 ) Ω . The representing example [3] is Ω = { z = ( z 1 , z 2 ) T C 2 : r ( z ) : = R e z 2 + | z 1 | 8 + 15 7 | z 1 | 2 R e z 1 6 < 0 } , which is a pseudoconvex domain with point in the boundary that does not admit any peak function, supporting surface and the boundary can not be convexifiable by any local holomorphic coordinates[4,5,6]. The existence of supporting functions and smooth peak functions and the properties of convexifiability have been done by Pflug [2], by Kola r ˘ [7], by J.Han [8], D.Zhao [9] and by J. Byun and H. R. Cho [10]. In [11,12], Taeyong Ahn etc. provide a tool to construct global holomorphic peaks from local holomorphic supporting functions for a class of unbounded domains in C n . But it is still an open question whether any Kohn-Nirenberg domain is biholomorphic to a bounded domain. In order to better understand the properties of the Kohn-Nirenberg domain, in [13], Simone Calamai provides some new examples of Kohn-Nirenberg domain that develop some properties and theories about convexifiability in C 2 .
Let O α ( Ω ) ( 0 α ) denote the space of functions holomorphic on Ω and of class C α on Ω ¯ . Recall a point ξ Ω is a peak point to Ω if there is a function f O α ( Ω ) satisfying f ( ξ ) = 1 and | f ( z ) | < 1 for all z Ω ¯ { ξ } . We call f a peak function. A holomorphic supporting surface for Ω at ξ is a complex manifold M of co-dimension 1 with the property: there exists a neighborhood N ( ξ ) of ξ such that Ω ¯ N ( ξ ) M = { ξ } . In [9], D. Zhao etc. considered a general moditication of the Kohn-Nirenberg domain near the origin in C n , namely, the domain Ω K = { ( z 1 , , z n ) C n : R e z n + j = 1 n 1 ( | z j | p + K j | z j | p q R e ( z j q ) ) < 0 } , where K = ( K 1 , . . . , K n 1 ) R n 1 , p , q Z + and p q 2 0 . They proved the following sufficient condition.
Theorem 1 
([9]). Given the above domain Ω K with | K j | p 2 p 2 q 2 ( 1 j n 1 ) , then
1. 
Ω K is a pseudoconvex domain.
2. 
If q p or q p but | K j | < 1 ( 1 j n 1 ) , there exists a C -peak function and a supporting surface at the origin 0 Ω K .
3. 
If q p and j = 1 n 1 | K j | > n 1 , there does not exist any C -peak function and supporting surface at 0 Ω K .
In this paper, based on the domain Ω K , we define a general Kohn-Nirenberg type domain as follows.
Ω K , L = { ( z 1 , , z n ) C n : R e z n + g | z n | 2 + j = 1 n 1 ( | z j | p + K j | z j | p q R e ( z j q ) + L j | z j | p 2 q I m ( z j 2 q ) ) < 0 }
where K = ( K 1 , K 2 , , K n 1 ) R n 1 , L = ( L 1 , L 2 , , L n 1 ) R n 1 , g 0 , p Z + , q Z + and p 2 q 1 0 .
We see that Ω K , L is a modificationn of Ω K . If we do not consider the term g | z n | 2 , the general modified domain Ω K , L is a weighted-bumped domain [12], denoted by Ω W B . If we consider this term, the Theorem 4.6 [11] has a argument that there exists global holomorphic supporting function when g > 0 and a bound point of Ω W B admits a local holomorphic supporting function. Thus Ω K , L keeps the main features. It will be interesting to study whether the existence of supporting surface and peak function at the origin in [9] can be generalized to the domain Ω K , L .
For the domain Ω K , L , we study the existence of the holomorphic peak function, supporting surface at the boundary points. The main result of this article is the following theorem.
Theorem 2 
Let Ω K , L be the above domain with | K j | + 2 | L j | p 2 p 2 q 2 ( 1 j n 1 ) , we have
1. 
Ω K , L is a peseudoconvex domain.
2. 
If q p or q p and | K j | + | L j | < 1 ( 1 j n 1 ) , there exists holomorphic peak function and supporting surface at the origin 0 Ω K , L .
3. 
If q p and j = 1 n 1 | K j | > n 1 , there does not exist any supporting surface at 0 Ω K , L .
The structure of this article is as follows. In Sect. Section 2, we give some basic definitions for Kohn-Nirenberg domain. In Sect. Section 3, we provide the proof of theorem 2 about (1)-(3).

2. Basic Definitions and Lemmas

Let Ω = { ( z 1 , , z n ) C n : r ( z ) < 0 } be a domain in C n with smooth boundary, its defining function is r ( z ) . Let O α ( Ω ) ( 0 α ) be the space of holomorphic functions on Ω K , L and C α -continuous on Ω ¯ K , L . For a point ξ Ω and a vector t = ( t 1 , , t n ) C n , we write r ξ ( t ) = j = 1 n r z j ( ξ ) t j . The Levi form of r ( z ) at ξ applied to t C n is L ( ξ , t ) = j , k = 1 n 2 r z j z ¯ k ( ξ ) t j t ¯ k . ξ is called a pseudoconvex point if L ( ξ , t ) = j , k = 1 n 2 r z j z ¯ k ( ξ ) t j t ¯ k 0 for all t T ξ Ω = { t C n : r ξ ( t ) = j = 1 n r z j ( ξ ) t j = 0 } , where T ξ Ω = { t C n : r ξ ( t ) = 0 } is the corresponding complex tangent space. If the Levi form is positive at boundary point ξ , i.e. L ( ξ , t ) > 0 , we call ξ a strong pseudoconvex point. If all the boundary points are (strong) pseudoconvex points, the domain Ω is called a (strong) pseudoconvex domain.
Lemma 1 
([9]). For any real number K with | K | m 2 m 2 1 ( 1 < m Z + ) , there exist constants b R and C > 0 such that
1 + K c o s t + b c o s ( m t ) > C , for all t [ 0 , 2 π )
where b = b ( m ) and C = C ( m ) depend only on m .
Lemma 2 
If | K | + 2 | L | m 2 m 2 1 ( 1 < m Z + ) , there exist constants b R and C > 0 such that
1 + K c o s ( t ) + L s i n ( 2 t ) + b c o s ( m t ) > C , for t [ 0 , 2 π )
Proof. 
If | K | + 2 | L | m 2 m 2 1 , then | K + 2 L s i n ( t ) | | K | + 2 | L | m 2 m 2 1 . From Lemma 1 and K c o s ( t ) + L s i n ( 2 t ) = ( K + 2 L s i n ( t ) ) c o s ( t ) , we have constants b R and C > 0 such that
1 + K c o s ( t ) + L s i n ( 2 t ) + b c o s ( m t ) > C , for t [ 0 , 2 π )
where b = b ( m ) and C = C ( m ) depend only on m.
Set the domain Ω K , L = { ( z 1 , , z n ) C n : r ( z ) < 0 } , its defining function r ( z ) on C n is as follows
r ( z ) = R e z n + g | z n | 2 + j = 1 n 1 ( | z j | p + K j | z j | p q R e z j q + L j | z j | p 2 q I m z j 2 q )
where K = ( K 1 , . . . , K n 1 ) R n 1 , L = ( L 1 , . . . , L n 1 ) R n 1 , g 0 , p Z + , q Z + and p 2 q 1 0 . □
Lemma 3 
If ( p 2 q 2 ) | K j | + ( p 2 4 q 2 ) | L j | p 2 ( 1 j n 1 ) ( g 0 ) , then Ω K , L is a pseudoconvex domain.
Proof. 
For boundary point ξ = ( z 1 , , z n ) Ω and tangent vector t = ( t 1 , , t n ) , we compute the Levi form and get
r z j = 1 2 p z j 1 2 p 1 ( z ¯ j ) 1 2 p + K j ( p + q ) 4 z j 1 2 p + 1 2 q 1 ( z ¯ j ) 1 2 p 1 2 q + K j ( p q ) 4 z j 1 2 p 1 2 q 1 ( z ¯ j ) 1 2 p + 1 2 q + L j ( p + 2 q ) 4 i z j 1 2 p + q 1 ( z ¯ j ) 1 2 p q L j ( p 2 q ) 4 i z j 1 2 p q 1 ( z ¯ j ) 1 2 p + q
where i = 1 . r z j z ¯ k = 0 ( j k ) and r z n z ¯ n = g .
For 1 j n 1 , there is
r z j z ¯ j = 1 4 p 2 ( z j ) 1 2 p 1 + K j ( p + q ) ( p q ) 8 ( z j ) 1 2 p + 1 2 q 1 ( z ¯ j ) 1 2 p 1 2 q 1 + K j ( p q ) 8 ( p + q ) ( z j ) 1 2 p 1 2 q 1 ( z ¯ j ) 1 2 p + 1 2 q 1 + L j ( p + 2 q ) ( p 2 q ) 8 i ( z j ) 1 2 p + q 1 ( z ¯ j ) 1 2 p q 1 L j ( p 2 q ) ( p + 2 q ) 8 i ( z j ) 1 2 p q 1 ( z ¯ j ) 1 2 p + q 1 = 1 4 p 2 z j p 2 + K j ( p 2 q 2 ) ( z j ) 1 2 p 1 2 q 1 ( z ¯ j ) 1 2 p 1 2 q 1 ( z j ) q + ( z ¯ j ) q 2 + L j ( p 2 4 q 2 ) ( z j ) 1 2 p q 1 ( z ¯ j ) 1 2 p q 1 ( z j ) 2 q ( z ¯ j ) 2 q 2 i = 1 4 p 2 | z j | p 2 + ( p 2 q 2 ) K j | z j | p q 2 R e ( z j ) q + ( p 2 4 q 2 ) L j | z j | p 2 q 2 I m ( z j ) 2 q
If ( p 2 q 2 ) | K j | + ( p 2 4 q 2 ) | L j | p 2 and g 0 , we have the Levi form
L ( ξ , t ) = j , k = 1 n 2 r z j z ¯ k ( ξ ) t j t ¯ k = j = 1 n 1 r z j z ¯ j | t j | 2 + g | t n | 2 1 4 j = 1 n 1 ( p 2 ( p 2 q 2 ) ) | K j | ( p 2 4 q 2 ) | L j | ) | z j | p 2 | t j | 2 + g | t n | 2 0 .
Thus the Levi form L ( ξ , t ) is semi-positive definite, which proves that Ω K , L is pseudoconvex. □

3. Holomorphic Peak Function and Supporting Surface of the General Modified Domain Ω K , L

Here we prove the main Theorem 2 about (1)-(3).
Proof 
(Proof of Theorem 2 about (1)). Let Ω K , L = { z C n : R e z n + g | z n | 2 + j = 1 n 1 ( | z j | p + K j | z j | p q
R e z j q + L j | z j | p 2 q I m z j 2 q ) < 0 } . Suppose | K j | + 2 | L j | p 2 p 2 q 2 ( 1 j n 1 , g 0 ) , then ( p 2 q 2 ) | K j | + ( p 2 4 q 2 ) | L j | ( p 2 q 2 ) | K j | + 2 ( p 2 q 2 ) | L j | p 2 ( 1 j n 1 , g 0 ) . So Lemma 3 implies that Ω K , L is pseudoconvex. □
Remark 1. 
Let Ω K , L be the above domain. If | K j | + 2 | L j | p 2 p 2 q 2 ( 1 j n 1 , g 0 ) , it is easy to see the origin is a weakly pseudoconvex (not strong pseudoconvex ) boundary point of Ω K , L .
Proof 
(Proof of Theorem 2 about (2)). Let Ω K , L be the above domain with | K j | + 2 | L j | p 2 p 2 q 2 ( 1 j n 1 , g 0 ) .
For r ( z ) = R e z n + g | z n | 2 + j = 1 n 1 ( | z j | p + K j | z j | p q R e z j q + L j | z j | p 2 q I m z j 2 q ) , we consider two cases.
  • The case q p ( p = m q ) . Let z j = r j e i θ j ( 1 j n 1 ) and z n = u + i v . In polar coordinate system, we have r ( z 1 , z 2 , , z n ) = u + g u 2 + g v 2 + j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) ] r j p . Then we consider the coordinate transformation z 1 * = z 1 ; z 2 * = z 2 ; ;
    z n 1 * = z n 1 ; z n * = z n j = 1 n 1 b j z j p , and b j R . In the new coordinate system, after dropping the stars we have
    r ( z 1 , z 2 , , z n ) = u + g ( u + j = 1 n 1 b j c o s ( p θ j ) r j p ) 2 + g ( v + j = 1 n 1 b j s i n ( p θ j ) r j p ) 2 + j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) + b j c o s ( p θ j ) ] r j p
    Note that | K j | + 2 | L j | p 2 p 2 q 2 = m 2 m 2 1 ( p = m q ) . Lemma 2 implies that there exists C > 0 such that
    j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) + b j c o s ( p θ j ) ] r j p > C j = 1 n 1 r j p
    The point 0 belongs to the set { ( z 1 , z 2 , , 0 ) : z j C , 1 j n 1 } N ( 0 ) Ω ¯ K , L , where N ( 0 ) is a neighborhood of 0. For all 0 z { ( z 1 , z 2 , . . . , 0 ) : z j C , 1 j n 1 } N ( 0 ) Ω ¯ K , L , there exists j such that z j 0 . We have
    r ( z 1 , z 2 , , 0 ) = g ( j = 1 n 1 b j c o s ( p θ j ) r j p ) 2 + j = 1 n 1 b j s i n ( p θ j ) r j p ) 2 + 0 + j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) + b j c o s ( p θ j ) ] r j p > C j = 1 n 1 r j p > 0
    This is a contradiction with the definition r ( z 1 , z 2 , , z n ) 0 and implies that
    { ( z 1 , z 2 , . . . , 0 ) : z j C , 1 j n 1 } N ( 0 ) Ω ¯ K , L = { 0 } .
    Thus in the new coordinates, the complex manifold { ( z 1 , z 2 , , 0 ) : z j C , 1 j n 1 } is a holomorphic supporting surface at the origin 0 Ω K , L . The holomorphic supporting function is f ( z 1 , z 2 , , z n ) = z n at 0. And the corresponding holomorphic peak function is h ( z 1 , z 2 , , z n ) = e z n + z n 2 for the origin 0.
    In fact, it is obvious that h ( 0 ) = 1 . Put z n = u + i v . For z Ω ¯ K , L { 0 } , we have r ( z 1 , z 2 , , z n ) 0 , i.e.
    u + g ( u + j = 1 n 1 b j c o s ( p θ j ) r j p ) 2 + g ( v + j = 1 n 1 b j s i n ( p θ j ) r j p ) 2 + j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) + b j c o s ( p θ j ) ] r j p 0
    Thus
    u j = 1 n 1 [ 1 + K j c o s ( q θ j ) + L j s i n ( 2 q θ j ) + b j c o s ( p θ j ) ] r j p C j = 1 n 1 r j p 0 .
    If u = 0 , then v < 0 and | h ( z 1 , z 2 , , z n ) | = e u + u 2 v 2 = e v 2 < 1 .
    If u < 0 , then | h ( z 1 , z 2 , , z n ) | = e u + u 2 v 2 = e v 2 < 1 .
    So, the function h ( z 1 , z 2 , , z n ) is a local peak function at 0 Ω K , L . Further, Hakim and Sibony [1,2] show that there is a global peak function with the same regularity as h ( z 1 , z 2 , , z n ) .
  • The case q p and | K j | + | L j | < 1 ( 1 j n 1 ) .There exists holomorphic peak function and supporting surface at the origin. Note that j = 1 n 1 ( | z j | p + K j | z j | p q R e ( z j q ) + L j | z j | p 2 q I m ( z j 2 q ) ) j = 1 n 1 [ 1 ( | K j | + | L j | ) ] | z j | p , similar to case (I), we have the complex manifold { ( z 1 , z 2 , . . . , 0 ) : z j C , 1 j n 1 } , which is also a holomorphic supporting surface at the origin. At the same time, h ( z 1 , z 2 , . . . , z n ) = e z n + z n 2 is a local peak function at 0 Ω K , L . In [1,2], Hakim and Sibony show that there is a global peak function with the same regularity as h.
Proof 
(Proof of Theorem 2 about (3)). Assume that there exists supporting surface at the origin 0 Ω K , L . The support surface M as a complex manifold of co-dimension 1 implies that there are an open neighbourhood N ( 0 ) C n and holomorphic function f on N ( 0 ) such that
  • M N ( 0 ) = { z N ( 0 ) : f ( z ) = 0 } ;
  • r a n k ( f z 1 , . . . , f z n ) = 1 .
We shall study two different cases.
  • The case f z n = 0 , there is some j such that f z j 0 . The implicit function theorem implies that
    M = { ( z 1 , , z n ) C n : z j = ϕ ( z 1 , , z j 1 , z j , , z n ) }
    Now let z 1 = = z j 1 = z j + 1 = = z n = ε , then
    r ( z ) = ε + g ε 2 + ( n 2 ) ε p + ( 1 ) p l j K l | ϕ ( ε , , ε ) | p + K j | ϕ ( ε , , ε ) | p q R e [ ϕ ( ε , , ε ) ] q + L j | ϕ ( ε , , ε ) | p 2 q R e [ ϕ ( ε , , ε ) ] 2 q = ε + O ( ε 2 ) < 0 .
    If ε is small, then M Ω ¯ K , L { 0 } in every small neighborhood of 0. Therefore, we have a contradiction with M as a support surface.
  • The case f z n 0 , the implicit function theorem implies that
    M = { ( z 1 , . . . , z n ) C n : z n = ϕ ( z 1 , . . . , z n 1 ) }
We shall divide this into three different cases.
  • When ϕ ( z 1 , , z n 1 ) = s = t F s ( z 1 , , z n 1 ) , t p + 1 , where F s is the sum of those terms C α 1 , , α n 1 z 1 α 1 z n 1 α n 1 in the power series for which α 1 + + α n 1 = s .
    We let z j = ε e i ( χ ( K j ) + 1 ) q π , 1 j n 1 , where χ ( K j ) is defined by
    χ ( K j ) = 0 K j > 0 1 K j < 0 ( 1 j n 1 ) .
    And z n = ϕ ( z 1 , . . . , z n 1 ) . If ε is sufficiently small,
    r ( z ) = R e [ ϕ ( ε e i χ ( K 1 ) + 1 q π , , ε e i χ ( K n 1 ) + 1 q π ) ] + g | ϕ ( · ) | 2 + ( n 1 ) ε p ε p j = 1 n 1 | K j | = [ ( n 1 ) j = 1 n 1 | K j | ] ε p + O ( ε p + 1 ) < 0 .
    Hence M is not a supporting surface. It is a contradiction.
  • When ϕ ( z 1 , , z n 1 ) = s = t F s ( z 1 , , z n 1 ) , 1 t p 1 , F t ( z 1 , , z n 1 ) 0 . We can suppose F t ( t 1 , , t n 1 ) = λ 1 0 and choose θ such that λ 1 e i θ t < 0 .
    Let z 1 = t 1 ε e i θ , z 2 = t 2 ε e i θ , , z n 1 = t n 1 ε e i θ , it is easy to see that
    F t ( z 1 , , z n 1 ) = F t ( t 1 , , t n 1 ) ε t e i θ = λ 1 e i t θ ε t = | λ 1 | ε t .
    Then
    r ( z ) = R e [ ϕ ( z 1 , , z n 1 ) ] + g | ϕ ( z 1 , , z n 1 ) | 2 + ε p j = 1 n 1 | t j | p + ε p c o s q θ ( j = 1 n 1 K j | t j | p ) + ε p s i n 2 q θ ( j = 1 n 1 L j | t j | p ) = | λ 1 | ε t + O ( ε t + 1 ) < 0
    if ε is sufficiently small. Hence M is not a supporting surface. It is a contradiction.
  • Then the only remaining case is when
    ϕ ( z 1 , , z n 1 ) = s = p F s ( z 1 , , z n 1 ) .
    Let z j l = ε e i χ ( K j ) + l q π ( 1 j n 1 ) , where χ ( K j ) is defined by
    χ ( K j ) = 0 K j > 0 1 K j < 0
    and z n l = ϕ ( z 1 l , , z n 1 l ) . Then
    r ( z l ) = R e [ ϕ ( z 1 l , , z n 1 l ) ] + g | ϕ ( z 1 l , , z n 1 l ) | 2 + ( n 1 ) ε p + ε p c o s l π ( j = 1 n 1 | K j | ) = ε p [ R e ( λ e i p π q l ) + O ( ε ) + ( n 1 ) + c o s l π ( j = 1 n 1 | K j | ) ] ,
    where λ = F p ( e i χ ( K 1 ) q π , . . . , e i χ ( K n 1 ) q π ) . We take l = 1 , 3 , . . . , 2 q 1 , then
    l = 1 , 3 , , 2 q 1 r ( z l ) = ε p R e λ ( l = 1 , 3 , , 2 q 1 e i p π q l ) + O ( ε ) + q ( n 1 ) + l = 1 , 3 , , 2 q 1 c o s l π ( j = 1 n 1 | K j | )
    Since l takes odd integers, we obtain
    l = 1 , 3 , , 2 q 1 c o s l π = q .
    If q p , we have l = 1 , 3 , , 2 q 1 e i p π q l = 0 . Therefore, when ε is small,
    l = 1 , 3 , , 2 q 1 r ( z l ) = ε p O ( ε ) + q [ ( n 1 ) j = 1 n 1 | K j | ] < 0 .
    Since M Ω ¯ K , L = { 0 } and z l M , it follows that r ( z l ) > 0 for all l.
    Thus
    l = 1 , 3 , , 2 q 1 r ( z l ) > 0 .
    Hence we get a contradiction. This completes the proof.
Conjecture. According to the result of Theorem 2 about (3), we can conjecture that Ω K , L has no peak functions at 0, under the same condition.

Author Contributions

Conceptualization, K.H. and D.Z.; Methodology, D.Z. and K.H.; Formal analysis, D.Z. and K.H.; Writing—original draft, K.H.; Writing—review and editing, H.L. and D.Z.; Supervision, H.L. and Y.J.; Funding acquisition, K.H. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the China Postdoctoral Science Foundation certificate number: 2023M744095, National Natural Science Foundation of China grant number 61771001.

Conflicts of Interest

The authors declare no conflicts of interest.

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