The envelope of holomorphy of a classical truncated tube domain
aa r X i v : . [ m a t h . C V ] F e b THE ENVELOPE OF HOLOMORPHY OF A CLASSICALTRUNCATED TUBE DOMAIN
MAREK JARNICKI AND PETER PFLUG
Abstract.
We present the envelope of holomorphy of a classical truncatedtube domain.
Recall that the envelope of holomorphy of a tube domain G = D + i R n ⊂ C n isgiven by the convex hull of G . Discussions of so called generalized tube domains G = D + iD (also called truncated tube domains) have shown that Bochner’s resultbecomes false, i.e. the envelope of holomorphy of G is, in general, a strict subdomainof conv( D ) + i conv( D ) (see [Jar-Pfl 2020]). The simplest known example of thistype is given by the following domain G := { x + iy ∈ C n : R ′ < k x k < R ′′ , k y k < R } , where k k stands for the Euclidean norm in R n , see [Jar-Pfl 2020].Recently, discussing the tube theorem, (see [Nog 2020]) J. Noguchi were alsoasking for the envelope of holomorphy E ( G ) of G . Here we present the answer. Theorem 1. E ( G ) = { x + iy ∈ b G : k y k < k x k − ( R ′ − R ) } , where b G is theconvex hull of G . In particular, the envelope is univalent. Remark 2.
This result shows that the envelope of our truncated tube domainis never the convex hull of G except when R = 0 or R = ∞ (see the paper ofJ. Kajiwara [Kaj 1963]); the last case is just Bochner’s tube theorem while theother case can be also handled using [Shi 1968] if n > n = 2. Moreover, it is univalent; but we don’t know whether the envelope of anygeneralized tube domain is univalent. Proof.
Note that without loss of generality we may assume that R ′′ = 1; then weset R := R ′ to simplify the notation, i.e. the domain we discuss is given as G := { x + iy ∈ C n : R < k x k < , k y k < R } . The proof is done via induction. Let n = 2. Then it is based on the work ofIvashkovich ([Iva 1982]), in particular on Lemma 8 in this work.Let f ∈ O ( G ).Put ϕ ε ( z ) := z + z − ( R − R ) − ε , z = ( z , z ) ∈ C , where ε ∈ (0 , R ),Π ε := { z ∈ C : Re ϕ ε ( z ) = 0 } , and finally Π + ε := { z ∈ C : Re ϕ ε ( z ) > } . Notethat G ⊂ Π +0 .Put R ,ε := p R + ε , and fix R ∗ ,ε ∈ ( R , R ,ε ).Finally, define D ε := { z ∈ C : k x k < R ∗ ,ε , k y k < k x k − ( R − R ) − ε } ⊂ Π + ε ∩ G . Mathematics Subject Classification.
Key words and phrases. truncated tube domain; envelope of holomorphy. hen S ε := ∂D ε \ ( D ε ∩ Π ε ) is connected. Note that f is holomorphic in aneighborhood of S ε . Then f extends to a holomorphic functions f ε on D ε which isidentically to f near S . Using the identity theorem and ε −→ E ( G ).It remains to note that E ( G ) is pseudoconvex. Therefore we have got the envelopeof holomorphy of G .Now let n ≥ n −
1. Takean f ∈ O ( G ) and fix a point z n = x n + iy n with | x n | ≤ R , | y n | < R . Then G intersect the hyperplane C n − × { z n } along G ( z n ) := { e z ∈ C n − : R − | x n | < k e x k < − | x n | , k e y k < R − | y n | } . Observe that G ( z n ) is our domain, now in C n − .Therefore, by induction, there exists an e f ( · , z n ) ∈ O ( E ( G ( z n )) extending f ( · , z n ).In the remaining case where | x n | > R we have that e f ( · , z n ) := f ( · , z n ) is alreadyholomorphic on G ( z n ).Summarizing, we end up with a function e f on E ( G ) satisfying for all z n = x n + iy n with | x n | < | y n | < R , that f ( · , z n ) is holomorphic. Moreover, e f = f on G ,i.e. e f is holomorphic on an open set. Consequentially, using Hartogs theorem, e f is aholomorphic on E ( G ). Hence we have shown that E ( G ) is an holomorphic extensionof G . On the other hand, as above, it is easy to see that E ( G ) is pseudoconvex.Therefore, it is the envelope of holomorphy of G . (cid:3) Remark 3.
Observe that the generalized tube domain G = D + iD discussedabove has a special symmetry; namely, whenever x, x ′ ∈ R n , k x k = k x ′ k , and x ∈ D , then x ′ ∈ D (resp. whenever y, y ′ ∈ R n , k y k = k y ′ k , and y ∈ D , then y ′ ∈ D ). Moreover D is convex. So one may ask whether any generalized tubedomain with this kind of an additional geometric property (e.g. substitute in thedefinition of G the norm k · k by another R -norm) has an univalent envelope ofholomorphy and what is its desription.Let us end this short note with emphasizing some open general problems (see[Nog 2020]): Remark 4. a) Does every generalized tube domain have an univalent envelope ofholomorphy ?b) In case of univalence what is the description of the envelope ?c) What is the shape of the envelope of holomorphy in general ?
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Carl von Ossietzky Universit¨at Oldenburg, Institut f¨ur Mathematik, Postfach 2503,D-26111 Oldenburg, Germany
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