Remark on the tautness modulo an analytic hypersurface of hartogs type domains
We present sufficient conditions for the tautness modulo an analytic hypersurface of Hartogs-type domains $\Omega_H(X)$ and Hartogs–Laurent-type domains $\Sigma_{u, v}(X).$ We also propose a version of Eastwood's theorem for the tautness modulo an analytic hypersurface.  
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| Дата: | 2020 |
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Institute of Mathematics, NAS of Ukraine
2020
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Репозитарії
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860506981193744384 |
|---|---|
| author | Pham, Duc Thoan Pham, Duc Thoan Pham, Duc Thoan |
| author_facet | Pham, Duc Thoan Pham, Duc Thoan Pham, Duc Thoan |
| author_sort | Pham, Duc Thoan |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-01-29T12:45:42Z |
| description | We present sufficient conditions for the tautness modulo an analytic hypersurface of Hartogs-type domains $\Omega_H(X)$ and Hartogs–Laurent-type domains $\Sigma_{u, v}(X).$ We also propose a version of Eastwood's theorem for the tautness modulo an analytic hypersurface.
  |
| first_indexed | 2026-03-24T02:02:03Z |
| format | Article |
| fulltext |
UDC 517.5
Pham Duc Thoan (Nat. Univ. Civil Engineering, Hanoi, Vietnam)
REMARK ON THE TAUTNESS MODULO AN ANALYTIC HYPERSURFACE
OF HARTOGS-TYPE DOMAINS
ЗАУВАЖЕННЯ ЩОДО НАТЯГУ ОБЛАСТЕЙ ТИПУ ХАРТОГСА
ЗА МОДУЛЕМ АНАЛIТИЧНОЇ ГIПЕРПОВЕРХНI
We present sufficient conditions for the tautness modulo an analytic hypersurface of Hartogs-type domains \Omega H(X) and
Hartogs – Laurent-type domains \Sigma u,v(X). We also propose a version of Eastwood’s theorem for the tautness modulo an
analytic hypersurface.
Наведено достатнi умови натягу областей типу Хартогса \Omega H(X) та Хартогса – Лорана \Sigma u,v(X) за модулем ана-
лiтичної гiперповерхнi. Сформульовано версiю теореми Iствуда для натягу за модулем аналiтичної гiперповерхнi.
1. Introduction. Let X be a complex space and let H : X \times \BbbC m \rightarrow [ - \infty ; +\infty ) be an upper
semicontinuous function such that H(z, w) \geq 0 and H(z, \lambda w) = | \lambda | H(z, w) with \lambda \in \BbbC , z \in X,
w \in \BbbC m. We put
\Omega H(X) := \{ (z, w) \in X \times \BbbC m : H(z, w) < 1\} ,
and call it a Hartogs-type domain. For each z \in X, we denote by \Omega H(z) := \{ w \in \BbbC m : H(z, w) <
< 1\} the fiber of \Omega H(X) at z. Here, if H(z, w) = h(w)eu(z) for z \in X,w \in \BbbC m, where h, u are
upper semicontinuous, h \not \equiv 0 and h(\lambda w) = | \lambda | h(w) with \lambda \in \BbbC , we denote \Omega H(X) by \Omega u,h(X)
and the fiber by \Omega h := \{ w \in \BbbC m : h(w) < 1\} .
The following properties are known (see [1]):
\Omega h \Subset \BbbC m if and only if there exists a positive constant C such that h(w) \geq C\| w\| for all
w \in \BbbC m;
h is plurisubharmonic on \BbbC m if and only if \mathrm{l}\mathrm{o}\mathrm{g} h is plurisubharmonic on \BbbC m;
\Omega h is taut if and only if \Omega h \Subset \BbbC m and h is continuous plurisubharmonic on \BbbC m.
For u, v are upper semicontinuous functions on X with u+ v < 0 on X, we put
\Sigma u,v(X) := \{ (z, \lambda ) \in X \times \BbbC : ev(z) < | \lambda | < e - u(z)\} ,
and call it a Hartogs – Laurent-type domain.
In the past ten years, much attention has been given to the properties of Hartogs-type domains
from the viewpoint of hyperbolicity and tautness complex analysis (see [2, 3, 5, 7, 9, 11 – 14]). In
[7], S. H. Park obtained necessary and sufficient conditions for the hyperbolicity and tautness of
certain Hartogs-type domains and Hartogs – Laurent-type domains. In particular, in [13], D. D. Thai,
M. A. Duc, P. J. Thomas and N. V. Trao considered the tautness modulo an analytic subset S of
Hartogs-type domains in general situation. However, the original proof in [13] is based on Zorn’s
lemma, and it is not elementary. Notice that the results in [13] were proved for any analytic subset.
When the analytic subset S = \varnothing , those results (see [13], Theorem 2.3) seem to be very different
from previous results (see [7], Theorem 5.2, and [14], Theorem 1.2) by observing that given results
in [13] do not need the tautness of the fibers \Omega H(z), but they do need in [7] and [14]. Moreover,
c\bigcirc PHAM DUC THOAN, 2020
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1 119
120 PHAM DUC THOAN
in the case of an analytic hypersurface, we can show a contradiction. For instance, we consider the
following example. Let
h(w) := h(w1, w2) = | w2
1 + w2
2|
1
2 , (w1, w2) \in \BbbC 2.
It is easy to check that h(\lambda w) = | \lambda | h(w) \geq 0, \lambda \in \BbbC , w = (w1, w2) \in \BbbC 2 and h is continuous on
\BbbC 2 and \mathrm{l}\mathrm{o}\mathrm{g} h is plurisubharmonic on \BbbC 2, but the fiber \Omega h is unbounded. Hence, \Omega h is not taut. Let
u(z) := \mathrm{l}\mathrm{o}\mathrm{g} | z| , it is a continuous subharmonic function on X := \BbbC \setminus \{ 0, 1\} . Put S = \{ - 1\} . It is
clear to see that X is taut, hence X is taut modulo S (see Definition 2.2). Consider
H : X \times \BbbC 2 \rightarrow [0; +\infty ), H(z, w) = h(w)eu(z),
which is continuous and log-plurisubharmonic on X \times \BbbC 2. Thank to [7] (Theorem 5.2), we deduce
that \Omega u,h(X \setminus S) is not taut. By Theorem 2.3 (iii) in [13] and Remark 2.1 below, we can see that
\Omega u,h(X) \setminus \widetilde S = \Omega u,h(X \setminus S) is taut, where \widetilde S := S \times \BbbC 2. This is a contradiction.
Hence, in our opinion, for the tautness modulo an analytic subset \widetilde S of certain Hartogs-type
domains \Omega H(X), the tautness of the fibers \Omega H(z) can not be dropped.
The first purpose of this paper is to give some general versions for the tautness modulo an
analytic hypersurface of Hartogs-type domains and Hartogs – Laurent-type domains. Finally, we give
a version of Eastwood’s theorem for the tautness modulo an analytic hypersurface of a complex
space. To finish the proofs, we use the ideas and arguments in [7] to avoid using Zorn’s lemma.
2. Preliminaries. Let \Delta be the open unit disk in the complex plane. For a complex space X,
we denote by \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) the set of all holomorphic maps from \Delta to X and by Bn(z, r) the n-
dimensional Euclidean open ball with center z and radius r > 0 and by \rho (a, b) := \mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h} - 1 | a - b|
| 1 - ab|
the Poincaré distance on the open unit disk \Delta .
Definition 2.1 [6, p. 68]. Let X be a complex space and let S be an analytic subset in X. We
say that X is hyperbolic modulo S if for every pair of distinct points p, q of X we have dX(p, q) > 0
unless both are contained in S, where dX is the Kobayashi pseudodistance of X.
If S = \varnothing , then X is said to be hyperbolic.
Definition 2.2 [6, p. 240]. Let X be a complex space and let S be an analytic subset in X.
We say that X is taut modulo S if \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) is normal modulo S, i.e., for every sequence \{ fn\} in
\mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) one of the following holds:
(i) there exists a subsequence of \{ fn\} which converges uniformly on every compact subset to
f \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X);
(ii) the sequence \{ fn\} is compactly divergent modulo S in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X), i.e., for each compact
set K \subset \Delta and each compact set L \subset X \setminus S, there exists an integer N such that fn(K) \cap L = \varnothing
for all n \geq N.
If S = \varnothing , then X is said to be taut. It is immediately from the definition that if S \subset S\prime \subset X
and X is taut modulo S, then it is taut modulo S\prime , so in particular if X is taut, it is taut modulo S
for any analytic subset S.
For z\prime , z\prime \prime in X, we put\widetilde kX(z\prime , z\prime \prime ) = \mathrm{i}\mathrm{n}\mathrm{f}\{ \rho (a, b) : a, b \in \Delta ,\exists \varphi \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X), \varphi (a) = z\prime , \varphi (b) = z\prime \prime \} =
= \mathrm{i}\mathrm{n}\mathrm{f}\{ \rho (0, a) : a \in \Delta ,\exists \varphi \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X), \varphi (0) = z\prime , \varphi (a) = z\prime \prime \} ,
which is called the Lempert function on X.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
REMARK ON THE TAUTNESS MODULO AN ANALYTIC HYPERSURFACE OF HARTOGS-TYPE DOMAINS 121
Related to the tautness of a complex space, there is a function k(2)X defined as follows.
Definition 2.3 [5]. Let X be a complex space. We define
k
(2)
X (z\prime , z\prime \prime ) = \mathrm{i}\mathrm{n}\mathrm{f}\{ \widetilde kX(z\prime , z1) + \widetilde kX(z1, z
\prime \prime ) : z1 \in X\} =
= \mathrm{i}\mathrm{n}\mathrm{f}\{ \rho (0, a) + \rho (0, b) : a, b \in \Delta ,\exists \varphi 1, \varphi 2 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X),
\varphi 1(0) = z\prime , \varphi 1(a) = \varphi 2(0), \varphi 2(b) = z\prime \prime \} , z\prime , z\prime \prime \in X.
Similar to the above definition, for the tautness modulo an analytic subset S of X, there is a
function \widetilde k(2)X defined as follows.
Definition 2.4 [3]. Let X be a complex space and let S be an analytic subset of X. We define
\widetilde k(2)X (z\prime , z\prime \prime ) = \mathrm{i}\mathrm{n}\mathrm{f}\{ \widetilde kX\setminus S(z
\prime , z1) + \widetilde kX(z1, z
\prime \prime ) : z1 \in X \setminus S\} =
= \mathrm{i}\mathrm{n}\mathrm{f}\{ \rho (0, a) + \rho (0, b) : a, b \in \Delta , \exists \varphi 1 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S),
\varphi 2 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X), \varphi 1(0) = z\prime , \varphi 1(a) = \varphi 2(0), \varphi 2(b) = z\prime \prime \} ,
where z\prime \in X \setminus S and z\prime \prime \in X.
Obviously, k(2)X \leq \widetilde k(2)X \leq \widetilde kX .
We recall the following result, which is similar to Royden’s criterion for the taut domains [8] .
Proposition 2.1 [3]. Let X be a complex space and let S be an analytic hypersurface in X.
Then X is taut modulo S if and only if
B\widetilde k(2)G
(z0, R) := \{ z \in X : \widetilde k(2)X (z0, z) < R\} \Subset X
for any R > 0 and z0 \in X \setminus S.
Proposition 2.2 [3]. Let X be a complex space, S be an analytic subset in X and X =
\bigcup
i\in I
Xi
be the irreducible decomposition of X. Then X is taut modulo S if and only if Xi is taut modulo
Si := Xi \cap S for all i \in I.
Lemma 2.1 [10]. Let Z be a complex manifold. Let S be a hypersurface of a complex space
X. If \{ \varphi n\} n\geq 1 \subset \mathrm{H}\mathrm{o}\mathrm{l}(Z,X \setminus S) converging uniformly on every compact subsets of Z to a mapping
\varphi \in \mathrm{H}\mathrm{o}\mathrm{l}(Z,X), then either \varphi (Z) \subset X \setminus S or \varphi (Z) \subset S.
The following statement is an immediate consequence of the criterion for the tautness modulo an
analytic hypersurface.
Corollary 2.1 [3]. Let X be a complex space and let S be an analytic hypersurface of X. If
X is not taut modulo S, then there exist a number R > 0, sequences \{ zn\} n\geq 0 \subset X, \{ fn\} n\geq 1 \subset
\subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S), \{ gn\} n\geq 1 \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) and \{ \alpha n\} n\geq 1, \{ \beta n\} n\geq 1 \subset [0; 1) such that for n \geq 1, we
have
(i) \widetilde k(2)X (z0, zn) < R,
(ii) fn(0) = z0 \in X \setminus S,
(iii) fn(\alpha n) = gn(0),
(iv) gn(\beta n) = zn, zn \rightarrow w \in \partial X or | zn| \rightarrow \infty ,
(v) \alpha n \rightarrow \alpha 0, \beta n \rightarrow \beta 0.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
122 PHAM DUC THOAN
Remark 2.1. We could have X \setminus S being taut without X being taut modulo S. For instance,
\BbbC \setminus \{ 0, 1\} is taut, but \BbbC is not taut modulo \{ 0, 1\} . On the other hand, there are examples of domains
taut modulo S such that X\setminus S is not taut. Just take X a taut domain and S such that the codimension
of S is at least 2. Then X \setminus S is not pseudoconvex, therefore not taut.
However, when S is an analytic hypersurface, using Lemma 2.1, we can show that if X is taut
modulo S then X\setminus S is also taut. Indeed, we take any sequence \{ fn\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X\setminus S). Suppose that
\{ fn\} is not compactly divergent in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S), we deduce that \{ fn\} is not compactly divergent
modulo S in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) either. By the tautness modulo hypersurface S of X, it implies that \{ fn\}
converges uniformly on every compact subset of \Delta to a mapping f \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X). By Lemma 2.1,
we have either f \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S) or f(\Delta ) \subset S. By the assumption, \{ fn\} is normally convergent
in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S).
3. The tautness of Hartogs-type domains and of Hartogs – Laurent-type domains. Firstly,
we give a theorem for the tautness modulo an analytic hypersurface of Hartogs-type domains as
follows.
Theorem 3.1. Let X be a complex space and let S be an analytic hypersurface in X. If X is
taut modulo S, the fiber \Omega H(z) is taut for each z \in X, H is continuous on X \times \BbbC m and H is
plurisubharmonic on (X \setminus S)\times \BbbC m, then \Omega HX is taut modulo \widetilde S := S \times Cm.
Proof. Suppose that \Omega H(X) is not taut modulo \widetilde S. By Proposition 2.2, we can assume that
X is an irreducible complex space. By Corollary 2.1, we can choose a number R > 0, sequences
\{ zn\} n\geq 0 \subset \Omega H(X) and
\{ fn\} n\geq 1 \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Omega H(X) \setminus \widetilde S), \{ gn\} n\geq 1 \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Omega H(X))
and \{ \alpha n\} n\geq 1, \{ \beta n\} n\geq 1 \subset [0; 1) satisfying the properties (i) to (v). By the definition of \widetilde k(2), we have
\widetilde k(2)\Omega H(X)(z0, zn) \geq \widetilde k(2)X (z10 , z
1
n),
where zn = (z1n, z
2
n) \in X \times \BbbC m. Since the property (i), it implies that \{ z1n\} n\geq 1 \subset B\widetilde k(2)X
(z10 , R). By
Proposition 2.1, we may see that \{ z1n\} n\geq 1 \rightarrow a10 \in X as n\rightarrow \infty . For each n \geq 1, we denote
fn := (f1n, f
2
n) \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S)\times \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\BbbC m)
and
gn := (g1n, g
2
n) \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X)\times \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\BbbC m).
By the property (ii), we have f1nk
(0) = z10 \in X \setminus S. Then, since the tautness modulo S of X,
we may choose a subsequence \{ f 1
nk
\} \subset \{ f1n\} such that
f1nk
K\Rightarrow f10 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X),
i.e., \{ f1nk
\} is converging uniformly on every compact subset of \Delta to f10 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X). It is clear
to see that f10 (0) = z10 \in X \setminus S. Since f1n \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S), applying Lemma 2.1, we have
f10 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S). Then the properties (iii) and (v) yield that
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
g1nk
(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
f1nk
(\alpha nk
) = f10 (\alpha 0) \in X \setminus S.
Hence, there exists a subsequence of \{ g1nk
\} , without loss of generality, we assume that g1nk
K\Rightarrow g10 \in
\in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X). Then we have
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
REMARK ON THE TAUTNESS MODULO AN ANALYTIC HYPERSURFACE OF HARTOGS-TYPE DOMAINS 123
g10(0) = f10 (\alpha 0) \in X \setminus S. (1)
In particular,
g10(\beta 0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
g1nk
(\beta nk
) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
z1nk
:= a10 \in X. (2)
Assume that \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty \| z2n\| = \infty . Put z2n = rnwn with \| wn\| = 1 and rn \in \BbbR , n \geq 1. It implies
that \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty rn = \infty and \mathrm{l}\mathrm{i}\mathrm{m}k\rightarrow \infty wnk
= w0 \not = 0 with some subsequence \{ wnk
\} of \{ wn\} . By
H(z1n, rnwn) = rnH(z1n, wn) < 1 and since continuity of H on X \times \BbbC m, we have
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
H
\bigl(
z1nk, wnk
\bigr)
= H
\bigl(
a10, w0
\bigr)
= 0.
Thus, H(a10, L) = 0 < 1 with a complex line L = tw0 in \BbbC m (t \in \BbbC ). It implies that the fiber
\Omega H(a10) = L. But L is not taut, so we have a contradiction to the assumption. Therefore, since the
property (iv), we obtain
\mathrm{l}\mathrm{i}\mathrm{m}
n\rightarrow \infty
zn = (a10, a
2
0) = a0 \in \partial \Omega H(X). (3)
Step 1: Choose c2 \in (0, 1) such that \beta n \in c22\Delta , n \geq 1. Since (1), we have (g10)(\Delta ) \not \subset S. It
implies that (g10)
- 1(S) is an analytic subset in the open unit disc \Delta , so it is a discrete set. Then
(g10)
- 1(S) does not have any accumulation point in \Delta . Therefore, we can assume that
c2\Delta \cap (g10)
- 1(S) = \varnothing . (4)
We put E2 := c2
- 1\Delta . For each n \geq 1, we define a map \widetilde gn : E2 \rightarrow X \times Cm by
\widetilde gn(\lambda ) = \bigl( \widetilde g1n(\lambda ), \widetilde g2n(\lambda )\bigr) := gn(\beta n\lambda ).
Clearly,
\{ \widetilde gn\} n\geq 1 \subset \mathrm{H}\mathrm{o}\mathrm{l}(E2,\Omega H(X)). (5)
Put F2 :=
\bigcup
n\geq 1(\beta nE2). Using (v), it is easy to check that F2 \Subset \Delta . Let M := g10(F2). Since
\beta n \in c22\Delta , n \geq 1, we have \beta n < c22. It implies that
\beta nE2 = \beta nc2
- 1\Delta \subset c2\Delta , n \geq 1.
Then g10(F2) \subset g10(c2\Delta ). This and (4) imply that M \Subset X \setminus S. Notice that X is hyperbolic modulo
S and d = dX is the Kobayashi pseudodistance, then \delta := \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(M,\partial (X \setminus S))/3 > 0. Since \{ g1nk
\}
converges uniformly on F 2, we may take n0 \in \BbbN such that d
\bigl(
g1nk
(\lambda ), g10(\lambda )
\bigr)
< \delta , \lambda \in F 2 and
nk > n0. Hence, for v0 \in \partial (X \setminus S), \lambda \in F 2, we obtain
d
\bigl(
g1nk
(\lambda ), v0
\bigr)
\geq d
\bigl(
g10(\lambda ), v0
\bigr)
- d
\bigl(
g1nk
(\lambda ), g10(\lambda )
\bigr)
\geq \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(M,\partial X) - \delta \geq 2\delta .
Then we get d
\bigl(
g1nk
(F 2), \partial (X \setminus S)
\bigr)
\geq 2\delta > 0, which implies that
K := g10(F 2) \cup
\left( \bigcup
nk\geq n0
g1nk
(F 2)
\right) \Subset X \setminus S. (6)
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
124 PHAM DUC THOAN
Particularly,
K \prime :=
\bigl\{
g1nk
(\beta nk
\lambda ), g10(\beta 0\lambda ) : \lambda \in E2, nk \geq n0
\bigr\}
\subset K. (7)
We now assume that the family \{ \widetilde g2n\} is not uniformly bounded in E2. Then there exist a subse-
quence \{ \widetilde g2nk
\} and a sequence \{ \lambda k\} \subset E2 such that
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\| \widetilde g2nk
(\lambda k)\| = \infty .
Put \widetilde g2nk
(\lambda k) = rkwk with \| wk\| = 1 and rk \in \BbbR . Similar to the above argument, we have
\mathrm{l}\mathrm{i}\mathrm{m}k\rightarrow \infty rk = \infty and \mathrm{l}\mathrm{i}\mathrm{m}k\rightarrow \infty wk = w0 \not = 0. Since (6) and (7), we deduce that
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde g1nk
(\lambda k) = b0 \in K \subset X \setminus S.
Since (5), we obtain
rkH
\bigl( \widetilde g1nk
(\lambda k), wk
\bigr)
= H
\bigl( \widetilde g1nk
(\lambda k), \widetilde g2nk
(\lambda k)
\bigr)
< 1.
By continuity of H, we have
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
H
\bigl( \widetilde g1nk
(\lambda k), wk
\bigr)
= H(b0, w0) = 0.
Repeat the same argument in the above, we will get a contradiction to the tautness of the fiber \Omega H(z)
again. Therefore, \widetilde g2nk
is uniformly bounded in E2. Applying to Montel’s theorem, without loss of
generality, we can assume \widetilde g2nk
K\Rightarrow \widetilde g20 \in \mathrm{H}\mathrm{o}\mathrm{l}(E2,\BbbC m)
as k \rightarrow \infty . In particular,
\widetilde g20(1) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde g2nk
(1) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
g2nk
(\beta nk
) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
z2nk
= a20. (8)
Put \varphi nk
:= H \circ \widetilde gnk
on E2. Since \varphi nk
< 1 on E2 for any nk \geq n0, we have \varphi 0 := H \circ \widetilde g0 \leq 1 on
E2, where \widetilde g0 := \bigl( \widetilde g10, \widetilde g20\bigr) and \widetilde g10(\lambda ) := g10(\beta 0\lambda ), \lambda \in E2. It follows from (2) that
\widetilde g10(1) = g10(\beta 0) = a10. (9)
Since (3), (8) and (9), we get \varphi 0(1) = H(a0) = 1. By H is plurisubharmonic on (X \setminus S)\times \BbbC m, \varphi 0
is subharmonic on E2. Thus, the maximum principle for subharmonic functions implies that \varphi 0 \equiv 1
on E2, and hence
\widetilde g0(0) = \bigl( \widetilde g10(0), \widetilde g20(0)\bigr) \in \partial \Omega H(X). (10)
Step 2: We are going to apply the same argument as in Step 1 to \{ fn\} n\geq 1 and \{ \alpha n\} n\geq 0. Choose
c1 \in (0, 1) such that \alpha n \in E1 := c1
- 1\Delta for n \geq 1. We define a holomorphic function \widetilde fn :
E1 \rightarrow \Omega H(X) by \widetilde fn(\lambda ) = ( \widetilde f1n(\lambda ), \widetilde f2n(\lambda )) := fn(\alpha n\lambda ), \lambda \in E1.
Then we also have \widetilde f2nk
K\Rightarrow \widetilde f20 \in \mathrm{H}\mathrm{o}\mathrm{l}(E1,\BbbC m)
as k \rightarrow \infty . Since condition (iii), we observe that
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
REMARK ON THE TAUTNESS MODULO AN ANALYTIC HYPERSURFACE OF HARTOGS-TYPE DOMAINS 125
\widetilde g0(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde gnk
(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
gnk
(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
fnk
(\alpha nk
) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde fnk
(1) = \widetilde f0(1), (11)
where \widetilde f0 := ( \widetilde f10 , \widetilde f20 ) and \widetilde f10 (\lambda ) := f10 (\alpha 0\lambda ), \lambda \in E1. Put
\psi 0(\lambda ) = H \circ ( \widetilde f0)(\lambda ) \leq 1, \lambda \in E1.
Obviously, since (10) and (11), we obtain \psi 0(1) = H( \widetilde f0(1)) = 1. It follows from the maximum
principle for \psi 0 that \psi 0 \equiv 1 on E1. This implies that
\widetilde f0(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde fnk
(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
fnk
(0) = z0 \in \partial \Omega H(X),
which contradicts the condition (ii).
Theorem 3.1 is proved.
We know that the tautness of X \setminus S does not imply the tautness modulo S of X in general.
But, using Theorem 1.2 in [14], we have the following assertion in special situations of Hartogs-type
domains.
Corollary 3.1. Let X be a complex space being taut modulo an analytic hypersurface S. Assume
that H is continuous on \widetilde S := S \times \BbbC m and the fiber \Omega H(z) is taut for each z \in S. If \Omega H(X) \setminus \widetilde S is
taut, then \Omega H(X) is taut modulo \widetilde S.
Due to Barth [1], \Omega h is taut if and only if \Omega h \Subset \BbbC m and h is continuous plurisubharmonic on
\BbbC m. In addition, if u is plurisubharmonic then eu is. We immediately have the following corollary.
Corollary 3.2. Let X be a complex space and S be an analytic hypersurface in X. If X is taut
modulo S, the fiber \Omega h is taut, u is continuous on X and u is plurisubharmonic on X \setminus S, then
\Omega u,h(X) is taut modulo \widetilde S := S \times Cm.
We recall the Example 2.4 in [13]. Let X = \{ (z1, z2) \in \BbbC 2 : z1z2 = 0\} and S = \{ (z1, z2) \in \BbbC 2 :
z2 = 0\} . We can check that X is taut modulo S. We put u(z) = u(z1, z2) := \mathrm{l}\mathrm{o}\mathrm{g} | z2| and h(w) = | w|
with w \in \BbbC . Obviously, u is plurisubharmonic on X \setminus S and continuous on X. It is easy to see that
the fiber \Omega h is taut, H(z, w) := h(w)eu(z) is continuous on X \times \BbbC and \mathrm{l}\mathrm{o}\mathrm{g}H is plurisubharmonic
on (X \setminus S)\times \BbbC . Applying Corollary 3.2, we deduce that \Omega u,h is taut modulo \widetilde S := S \times \BbbC .
Notice that we also obtain this conclusion by direct proof as in [13]. However, we can not get
one from Theorem 2.3 (iii) in [13]. Because \mathrm{l}\mathrm{o}\mathrm{g}H is not plurisubharmonic on X\times \BbbC , since u is not
plurisubharmonic on X.
Now, we give a necessary condition for the tautness modulo of Hartogs – Laurent-type domains.
Proposition 3.1. If \Sigma u,v(X) is taut modulo \widetilde S := S\times \BbbC , then u and v are continuous on X \setminus S,
where S is an analytic subset of X.
Proof. Suppose the contrary. Without loss of generality, we can assume that u is not continuous
at z0 \in X \setminus S. By upper semicontinuity of u, we can choose a number R \in \BbbR and a sequence
\{ zn\} \subset X \setminus S such that zn \rightarrow z0 as n \rightarrow \infty and - u(z0) < - R < - u(zn) for any n \in \BbbN . Since
u(z0) \not = - \infty and u(z0) + v(z0) < 0, we may take an \alpha \in \BbbR such that v(z0) < - \alpha < - u(z0).
Since upper semicontinuity of v, we can assume that v(zn) < - \alpha for n > 1. Put
C :=
1
2
\mathrm{m}\mathrm{i}\mathrm{n}
\bigl\{
- u(z0) + \alpha , - R+ u(z0)
\bigr\}
> 0
and
\^u := u - u(z0) -
C
2
, \^v := v + u(z0) +
C
2
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
126 PHAM DUC THOAN
Obviously, the mapping
(z, w) \in \Sigma u,v(X) \mapsto \rightarrow
\Bigl(
z, weu(z0)+
C
2
\Bigr)
\in \Sigma \^u,\^v(X)
is biholomorphic, so \Sigma \^u,\^v(X) is taut modulo \widetilde S. We put
\^R := - u(z0) +R - C
2
, \^\alpha := - u(z0) + \alpha - C
2
.
It is easy to show that \^v(zn) < - \^\alpha for any n \in \BbbN . Hence, for any n \geq 1,
\mathrm{m}\mathrm{a}\mathrm{x}\{ \^v(z0), \^v(zn)\} < - \^\alpha < - C < 0 < - \^u(z0) < C < - \^R < - \^u(zn). (12)
We define fn(\lambda ) := (zn, e
C\lambda ), \lambda \in \Delta for n \geq 1. Observe that
e\^v(zn) < e - C < | eC\lambda | < eC < e - \^u(zn), n \geq 1, \lambda \in \Delta .
It implies that \{ fn\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma \^u,\^v \setminus \widetilde S) by zn \in X \setminus S, n \geq 1. Because, e\^v(z0) < e - C < e0 <
< e - \^u(z0) and z0 \in X \setminus S, we have
fn(0) = (zn, 1) \rightarrow (z0, 1) \in \Sigma \^u,\^v \setminus \widetilde S.
By the tautness modulo \widetilde S of \Sigma \^u,\^v, we get
fn(\lambda )
K\Rightarrow f(\lambda ) = (z0, e
C\lambda ) \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma \^u,\^v)
as n \rightarrow \infty . It implies that e\^v(z0) < eCRe\lambda < e - \^u(z0) for any \lambda \in \Delta . By letting \lambda \rightarrow 1, we have a
contradiction to (12).
Hence, Proposition 3.1 is proved.
The following proposition gives a sufficient condition for the tautness modulo of Hartogs –
Laurent-type domains.
Proposition 3.2. If X is taut modulo an analytic hypersurface S, u is continuous on X,
plurisubharmonic on X \setminus S and v is continuous plurisubharmonic on X, then \Sigma u,v(X) is taut
modulo \widetilde S := S \times \BbbC .
Proof. Let a sequence \{ \varphi n\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma u,v(X)). We have \Sigma u,v(X) \subset \Omega u,| .| (X), where | .|
is the norm on \BbbC . By Corollary 3.2, \Omega u,| .| (X) is taut modulo \widetilde S. It implies that there exists a sub-
sequence \{ \varphi nk
\} \subset \{ \varphi n\} which is either normally convergent or compactly divergent modulo \widetilde S in
\mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Omega u,| .| (X)). In the latter case, the sequence \{ \varphi nk
\} as a subfamily of \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma u,v(X)), di-
verges compactly modulo \widetilde S. Then, we only suppose that \{ \varphi nk
\} is normally convergent in
\mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Omega u,| .| (X)). Put \varphi nk
:= (fnk
, gnk
), where \{ fnk
\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) and \{ gnk
\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\BbbC ).
We denote
\varphi := (f, g) \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Omega u,| .| (X)),
where f \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) and g \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\BbbC ), such that fnk
K\Rightarrow f and gnk
K\Rightarrow g as k \rightarrow \infty . We have
e(v\circ fnk
)(\lambda ) < | gnk
(\lambda )| < e - (u\circ fnk
)(\lambda )
and
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
REMARK ON THE TAUTNESS MODULO AN ANALYTIC HYPERSURFACE OF HARTOGS-TYPE DOMAINS 127
| g(\lambda )| < e - (u\circ f)(\lambda ), \lambda \in \Delta .
Since (gnk
) - 1(0) = \varnothing for any k \geq 1, it follows from Hurwitz’s theorem that either g \equiv 0 or g
never vanishes. If g \equiv 0 then \varphi (\Delta ) \subset \partial \Sigma u,v(X), which implies that \{ \varphi nk
\} as a subfamily of
\mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma u,v(X)) diverges compactly. Now, we suppose that g \not \equiv 0 and define
\^v :=
1
| g(\lambda )|
e(v\circ f)(\lambda ), \lambda \in \Delta .
It implies that \^v is continuous subharmonic on \Delta . By continuity of v, we have \^v(\lambda ) \leq 1 for any
\lambda \in \Delta . It follows from the maximum principle for subharmonic that either \^v \equiv 1 on \Delta or \^v < 1
on \Delta . Therefore, it is either \varphi (\Delta ) \subset \partial \Sigma u,v(X) or \varphi (\Delta ) \subset \Sigma u,v(X). Then \{ \varphi n\} is either normally
convergent in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma u,v(X)) or compactly divergent. Thus, since the above arguments, \{ fn\} is
either compactly divergent modulo \widetilde S or normally convergent in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta ,\Sigma u,v(X)). It implies that
\Sigma u,v(X) is taut modulo \widetilde S.
4. Eastwood’s theorem for the tautness modulo. Similar to Eastwood’s theorem for the
hyperbolicity and tautness of a complex space (see [4, 9, 11]), we give a version of Eastwood’s
theorem for the tautness modulo an analytic hypersurface of a complex space.
Theorem 4.1. Let \widetilde X and X be two complex spaces. Let \pi : \widetilde X \rightarrow X be a holomorphic
mapping and let S be an analytic hypersurface in X. Suppose that for each p \in X, there exists an
open neighborhood U := U(p) in X such that \pi - 1(U) is taut modulo \widetilde S := \pi - 1(S). If X is taut
modulo S, then \widetilde X is also taut modulo \widetilde S.
Proof. As in the proof of Theorem 3.1, we can consider X as an irreducible complex space.
Suppose that \widetilde X is not taut modulo \widetilde S. Then by Corollary 2.1, we can take sequences \{ zn\} n\geq 0 \subset
\subset \Omega H(X), \{ fn\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , \widetilde X \setminus \widetilde S), \{ gn\} \subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , \widetilde X) and sequences \{ \alpha n\} n\geq 1, \{ \beta n\} n\geq 1 \subset [0; 1)
satisfying the properties (ii) to (v). We put
\widetilde fn = \pi \circ fn \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S) (13)
and \widetilde gn = \pi \circ gn \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X), n \geq 1.
By the property (ii), \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty \widetilde fn(0) = \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty \pi (z0) \in X \setminus S. Since X is taut modulo S, there
exists a sequence \{ \widetilde fnk
\} \subset \{ \widetilde fn\} such that
\widetilde fnk
K\Rightarrow \varphi 1 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X)
as k \rightarrow \infty . By (13) and applying Lemma 2.1, it implies that \varphi 1 \in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X \setminus S). By the property
(iii), we get
\mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde gnk
(0) = \mathrm{l}\mathrm{i}\mathrm{m}
k\rightarrow \infty
\widetilde fnk
(\alpha nk
) = \varphi 1(\alpha 0) \in X \setminus S.
Then \{ \widetilde gn\} contains a subsequence \{ \widetilde gnk
\} converging uniformly on compact subsets to a map \varphi 2 \in
\in \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , X) as k \rightarrow \infty . Therefore, for any \lambda \in \Delta , there exists an open neighborhood V\lambda \Subset \Delta of
\lambda , U\varphi 2(\lambda ) and k\lambda \in \BbbN , such that \pi - 1(U\varphi 2(\lambda )) is taut modulo \widetilde S and gnk
(V\lambda ) \Subset \pi - 1(U\varphi 2(\lambda )) \subset \widetilde X,
for any k \geq k\lambda .
Now, we take a point 0 < s < 1. By the compactness of [ - s, \beta 0] \subset \Delta , we can choose a finite
set \{ x\mu : \mu = 1, . . . , q\} \subset [ - s, \beta 0] such that [ - s, \beta 0] \subset
\bigcup q
\mu =1 Vx\mu and for all \mu \in \{ 1, . . . , q\} exists
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
128 PHAM DUC THOAN
\nu \in \{ 1, . . . , q\} \setminus \{ \mu \} such that Vx\mu \cap Vx\nu \not = \varnothing . After a rearrangement, we can assume that \beta 0 \in Vq
and Vx\mu \cap Vx\mu +1 \not = \varnothing , \mu \in \{ 1, . . . , q - 1\} . For \lambda = \beta 0, we consider gnk
\in \mathrm{H}\mathrm{o}\mathrm{l}(V\beta 0 , \pi
- 1(U\varphi 2(\beta 0))).
Assume that there exists a subsequence \{ gnk1
\} \subset \{ gnk
\} converging uniformly on compact subset
to map g\beta 0 \in \mathrm{H}\mathrm{o}\mathrm{l}(V\beta 0 , \pi
- 1(U\varphi 2(\beta 0))) as k1 \rightarrow \infty . By the property (iv), we have
\mathrm{l}\mathrm{i}\mathrm{m}
k1\rightarrow \infty
znk1
= \mathrm{l}\mathrm{i}\mathrm{m}
k1\rightarrow \infty
gnk1
(\beta nk1
) = g\beta 0(\beta 0) \in \widetilde X.
That is a contradiction to the condition (iv). Hence, by the tautness modulo \widetilde S of \pi - 1(U\varphi 2(\beta 0)), it
implies that gnk
diverges compactly modulo on V\beta 0 . But, since \beta 0 \in Vxq \cap V\beta 0 \not = \varnothing , we can choose
a sequence \{ gnk2
\} \subset \{ gnk1
\} which diverges compactly modulo on Vxq . Because Vxq \cap Vxq - 1 \not = \varnothing ,
we also choose a subsequence \{ gnk3
\} \subset \{ gnk2
\} which diverges compactly modulo on Vxq - 1 . And,
we can proceed to q - 2, in this manner, we can choose \mu 0 \in \{ 1, . . . , q\} with 0 \in Vx\mu 0
and a
subsequence \{ gnk4
\} \subset \{ gnk3
\} diverges compactly modulo on Vx\mu 0
. Thus, in view of (iii), we have
either
\mathrm{l}\mathrm{i}\mathrm{m}
k4\rightarrow \infty
fnk4
(\alpha nk4
) = \mathrm{l}\mathrm{i}\mathrm{m}
k4\rightarrow \infty
gnk4
(0) = \^a0 \in \partial \widetilde X, (14)
or
\mathrm{l}\mathrm{i}\mathrm{m}
k4\rightarrow \infty
fnk4
(\alpha nk4
) \in \widetilde S. (15)
Applying the above argument for the sequence fnk4
\subset \mathrm{H}\mathrm{o}\mathrm{l}(\Delta , \widetilde X \setminus \widetilde S), we can choose a sub-
sequence \{ fnk5
\} of \{ fnk4
\} that diverges compactly modulo on V\alpha 0 . Because if fnk5
K\Rightarrow f\alpha 0 \in
\in \mathrm{H}\mathrm{o}\mathrm{l}(V\alpha 0 , \pi
- 1(U\varphi 1(\alpha 0))), by Lemma 2.1, we have f\alpha 0 \in \mathrm{H}\mathrm{o}\mathrm{l}(V\alpha 0 , \pi
- 1(U\varphi 1(\alpha 0)) \setminus \widetilde S). It implies
that
\mathrm{l}\mathrm{i}\mathrm{m}
k5\rightarrow \infty
fnk5
(\alpha nk5
) = f\alpha 0(\alpha 0) \in \pi - 1(U\varphi 1(\alpha 0)) \setminus \widetilde S \subset \widetilde X \setminus \widetilde S.
This is a contradiction to (14) and (15). So, we can take a subsequence \{ fnk6
\} \subset \{ fnk5
\} such that
\{ fnk6
(0)\} converges to a point in \partial \widetilde X or in \widetilde S. Obviously, this is a contradiction to the property (ii).
Therefore, \widetilde X is taut modulo \widetilde S.
Immediately, we get the following corollary.
Corollary 4.1. If \pi : \widetilde X \rightarrow X is a holomorphic covering between complex spaces, then \widetilde X is taut
modulo \widetilde S if and only if X is taut modulo an analytic hypersurface S in X, where \widetilde S := \pi - 1(S).
Acknowledgements. The author would like to thank Professor Do Duc Thai for suggesting the
problem and helpful advices during the preparation of this work.
References
1. T. J. Barth, The Kobayashi indicatrix at the center of a circular domain, Proc. Amer. Math. Soc., 88, 527 – 530
(1983).
2. N. Q. Dieu, D. D. Thai, Complete hyperbolicity of Hartogs domain, Manuscripta Math., 112, 171 – 181 (2003).
3. P. V. Duc, P. N. T. Trang, M. A. Duc, On tautness modulo an analytic subset of complex spaces, Acta Math. Vietnam,
42, 717 – 726 (2017).
4. A. Eastwood, À propos des variétés hyperboliques complètes, C. R. Acad. Sci. Paris, 280, 1071 – 1075 (1975).
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(1993).
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6. S. Kobayashi, Hyperbolic complex spaces, Springer-Verlag, Berlin (1998).
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Received 24.12.15,
after revision — 08.01.19
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 1
|
| id | umjimathkievua-article-182 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:02:03Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/2b/b9954c8467dbdb1c6b5d6905128b3c2b.pdf |
| spelling | umjimathkievua-article-1822020-01-29T12:45:42Z Remark on the tautness modulo an analytic hypersurface of hartogs type domains Зауваження щодо натягу областей типу Хартогса за модулем аналiтичної гiперповерхнi Зауваження щодо натягу областей типу Хартогса за модулем аналiтичної гiперповерхнi Pham, Duc Thoan Pham, Duc Thoan Pham, Duc Thoan Хартові типи доменів Домени типу Hartogs-Laurent Tautness modulo an analytic hypersurface Hartogs type domains Hartogs-Laurent type domains We present sufficient conditions for the tautness modulo an analytic hypersurface of Hartogs-type domains $\Omega_H(X)$ and Hartogs–Laurent-type domains $\Sigma_{u, v}(X).$ We also propose a version of Eastwood's theorem for the tautness modulo an analytic hypersurface. &nbsp; Наведено достатні умови натягу областей типу Хартогса $\Omega_H(X)$ та Хартогса–Лорана $\Sigma_{u, v}(X)$ за модулем аналітичної гіперповерхні. Сформульовано версію теореми Іствуда для натягу за модулем аналітичної гіперповерхні. Institute of Mathematics, NAS of Ukraine 2020-01-15 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/182 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 1 (2020); 119-129 Український математичний журнал; Том 72 № 1 (2020); 119-129 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/182/1550 |
| spellingShingle | Pham, Duc Thoan Pham, Duc Thoan Pham, Duc Thoan Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title | Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title_alt | Зауваження щодо натягу областей типу Хартогса за модулем аналiтичної гiперповерхнi Зауваження щодо натягу областей типу Хартогса за модулем аналiтичної гiперповерхнi |
| title_full | Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title_fullStr | Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title_full_unstemmed | Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title_short | Remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| title_sort | remark on the tautness modulo an analytic hypersurface of hartogs type domains |
| topic_facet | Хартові типи доменів Домени типу Hartogs-Laurent Tautness modulo an analytic hypersurface Hartogs type domains Hartogs-Laurent type domains |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/182 |
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