A note on the removability of totally disconnected sets for analytic functions
UDC 517.537.38 We prove that each totally disconnected closed subset $E$ of a domain $G$ in the complex plane is removable for analytic functions $f(z)$ defined in $G\setminus E$ and such that for any point $z_0\in E$ the real or imaginary part of $f(z)$ vanishes at $z_0$.  
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Institute of Mathematics, NAS of Ukraine
2020
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| author | Pokrovskii, A. V. Покровський, А.В. |
| author_facet | Pokrovskii, A. V. Покровський, А.В. |
| author_sort | Pokrovskii, A. V. |
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| description | UDC 517.537.38
We prove that each totally disconnected closed subset $E$ of a domain $G$ in the complex plane is removable for analytic functions $f(z)$ defined in $G\setminus E$ and such that for any point $z_0\in E$ the real or imaginary part of $f(z)$ vanishes at $z_0$.
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| doi_str_mv | 10.37863/umzh.v72i3.6046 |
| first_indexed | 2026-03-24T03:25:49Z |
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К О Р О Т К I П О В I Д О М Л Е Н Н Я
UDC 517.537.38
A. V. Pokrovskii (Inst. Math. Nat. Acad. Sci. Ukraine, Kyiv)
A NOTE ON THE REMOVABILITY OF TOTALLY DISCONNECTED SETS
FOR ANALYTIC FUNCTIONS
ЗАУВАЖEННЯ ПРО УСУВНIСТЬ СКРIЗЬ РОЗРИВНИХ МНОЖИН
ДЛЯ АНАЛIТИЧНИХ ФУНКЦIЙ
We prove that each totally disconnected closed subset E of a domain G in the complex plane is removable for analytic
functions f(z) defined in G \setminus E and such that for any point z0 \in E the real or imaginary part of f(z) vanishes at z0 .
Доведено, що будь-яка скрiзь розривна замкнена пiдмножина E областi G на комплекснiй площинi є усувною для
аналiтичних функцiй f(z), визначених у G \setminus E i таких, що для довiльної точки z0 \in E дiйсна або уявна частина
f(z) зникає в z0.
Let G be a domain in the complex plane \BbbC , E a totally disconnected closed subset of G, and
f(z) = u(z) + iv(z) an analytic function in G \setminus E (u(z) = \mathrm{R}\mathrm{e} f(z), v(z) = \mathrm{I}\mathrm{m} f(z)). Fedorov
[1] proved that, if f(z) is continuously extended from G \setminus E to G and u(z) vanishes on E , then
this extension is an analytic function in G. Ischanov [2] (see also [3, 4]) generalized this result as
follows: if u(z) vanishes on E , then f(z) is analytically extended from G \setminus E to G. The aim of
this paper is to prove the following generalization of the mentioned results.
Theorem 1. Let G be a domain in \BbbC , E a totally disconnected closed subset of G, and f(z) =
u(z) + iv(z) an analytic function in G \setminus E such that for any z0 \in E we have either u(z) \rightarrow 0 or
v(z) \rightarrow 0 as z \rightarrow z0 , z \in G \setminus E . Then the function f(z) can be analytically extended from G \setminus E
to G.
Proof. Let the conditions of Theorem 1 be satisfied and let z0 \in E . Then we have one of the
following cases:
(a) the function f(z) is bounded in the intersection of G \setminus E with some neighborhood of the
point z0;
(b) u(z) \rightarrow 0 as z \rightarrow z0 , z \in G \setminus E , and \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
z\rightarrow z0,z\in G\setminus E
| v(z)| = +\infty ;
(c) v(z) \rightarrow 0 as z \rightarrow z0 , z \in G \setminus E , and \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
z\rightarrow z0,z\in G\setminus E
| u(z)| = +\infty .
Consider the case (a). Then there is an r > 0 such that the disk D(z0, r) := \{ z \in \BbbC : | z - z0| < r\}
is contained in G and the function f(z) is bounded in D(z0, r) \setminus E . Define the function
f1(z) := - if2(z), z \in D(z0, r) \setminus E.
Then we have
u1(z) := \mathrm{R}\mathrm{e}f1(z) = 2u(z)v(z), v1(z) = \mathrm{I}\mathrm{m} f1(z) = v2(z) - u2(z).
c\bigcirc A. V. POKROVSKII, 2020
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 3 425
426 A. V. POKROVSKII
Since the functions u(z) and v(z) are bounded in D(z0, r) \setminus E , then for any \zeta \in D(z0, r) \cap E we
have u1(z) \rightarrow 0 as z \rightarrow \zeta , z \in D(z0, r) \setminus E . The Ischanov theorem implies the existence of an
analytic extension F (z) of the function f1(z) from D(z0, r) \setminus E to D(z0, r). Let \zeta \in E \cap D(z0, r).
Suppose that F (\zeta ) \not = 0 and take an \varepsilon \in (0, r) such that
D(\zeta , \varepsilon ) \subset D(z0, r) and | F (z) - F (\zeta )| < | F (\zeta )| /2 for all z \in D(\zeta , \varepsilon ).
Then
\sqrt{}
iF (z) is a univalent analytic function in D(\zeta , \varepsilon ), where the branch of the square root in
D(iF (\zeta ), | F (\zeta )| /2) is fixed by the condition
\sqrt{}
iF (z) = f(z) for all z \in D(\zeta , \varepsilon ) \setminus E .
Thus we justified the existence of an analytic continuation \=f(z) of the function f(z) from
D(z0, r) \setminus E to D(z0, r) \setminus (F - 1(0) \cap E), where the set F - 1(0) := \{ z \in D(z0, r) : F (z) = 0\}
contains only isolated points. Since the function \=f(z) is bounded in D(z0, r) \setminus (F - 1(0) \cap E), then
each point of the set F - 1(0) \cap E is a removable singular point for the function \=f(z).
The above arguments show that we can assume without loss of generality in the proof of Theorem
1 that for any z0 \in E we have either the case (b) or the case (c). Fix an arbitrary domain G0 \Subset G,
define E1 and E2 as the sets consisting of all points z0 \in E \cap G0 satisfying the conditions (b) and
(c), respectively, and denote
\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(E1, E2) := \mathrm{i}\mathrm{n}\mathrm{f}\{ | z1 - z2| : z1 \in E1, z2 \in E2\} .
Suppose that \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(E1, E2) = 0. Then there are sequences \{ z1n\} \infty n=1 \subset E1 and \{ z2n\} \infty n=1 \subset E2 such
that | z1n - z2n| \rightarrow 0 as n \rightarrow \infty whence the compactness of the set E \cap G0 , where G0 is the closure
of G0 , implies the existence of a point z0 \in E \cap G0 such that
\mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
\zeta \rightarrow z0,\zeta \in G\setminus E
| u(\zeta )| = \mathrm{l}\mathrm{i}\mathrm{m} \mathrm{s}\mathrm{u}\mathrm{p}
\zeta \rightarrow z0,\zeta \in G\setminus E
| v(z)| = +\infty .
Therefore, the case (b) or (c) is impossible. Hence, \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(E1, E2) > 0 and consequently E1 and
E2 are totally disconnected closed subsets of G0 such that for any z0 \in E1 we have u(z) \rightarrow 0 as
z \rightarrow z0 , z \in G0 \setminus (E1\cup E2), and for any z0 \in E2 we have v(z) \rightarrow 0 as z \rightarrow z0 , z \in G0 \setminus (E1\cup E2).
Since \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(E1, E2) > 0, then applying Ischanov’s theorem once again we conclude that the function
f(z) has an analytic continuation from G0 \setminus E to G0 . Taking into account the arbitrariness in the
selection of the domain G0 we complete the proof of Theorem 1.
References
1. W. Fédoroff, Sur la continuité des functions analytiques’, Math. Sb., 32, № 1, 115 – 121 (1924).
2. B. Zh. Ischanov, On one Fedorov’s theorem, Vestn. Moskov. Univ. Ser. Mat., № 3, 34 – 37 (1981).
3. B. Zh. Ischanov, Generalization of Fedorov’s theorem for harmonic functions of several variables, Vestn. Moskov.
Univ. Ser. Mat., № 1, 100 – 102 (1986).
4. B. Zh. Ischanov, Generalization of Fedorov’s theorem to M -harmonic functions, Math. Notes, 56, № 5-6, 1132 – 1136
(1994).
Received 19.04.17
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 3
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| id | umjimathkievua-article-6046 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:25:49Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/ed/526f591b1514fc39b39a8f025e0281ed.pdf |
| spelling | umjimathkievua-article-60462020-05-26T09:28:46Z A note on the removability of totally disconnected sets for analytic functions Зауважeння про усувнiсть скрiзь розривних множин для аналiтичних функцiй Pokrovskii, A. V. Покровський, А.В. UDC 517.537.38 We prove that each totally disconnected closed subset $E$ of a domain $G$ in the complex plane is removable for analytic functions $f(z)$ defined in $G\setminus E$ and such that for any point $z_0\in E$ the real or imaginary part of $f(z)$ vanishes at $z_0$. &nbsp; УДК 517.537.38 Доведено, що будь-яка скрізь розривна замкнена підмножина $E$ області $G$ на комплексній площині є усувною для аналітичних функцій $f(z),$ визначених у $G\setminus E$ і таких, що для довільної точки $z_0 \in E$ дійсна або уявна частина $f(z)$ зникає в $z_0.$ Institute of Mathematics, NAS of Ukraine 2020-03-28 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6046 10.37863/umzh.v72i3.6046 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 3 (2020); 425-426 Український математичний журнал; Том 72 № 3 (2020); 425-426 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6046/8672 |
| spellingShingle | Pokrovskii, A. V. Покровський, А.В. A note on the removability of totally disconnected sets for analytic functions |
| title | A note on the removability of totally disconnected sets for analytic functions |
| title_alt | Зауважeння про усувнiсть скрiзь розривних множин для аналiтичних функцiй |
| title_full | A note on the removability of totally disconnected sets for analytic functions |
| title_fullStr | A note on the removability of totally disconnected sets for analytic functions |
| title_full_unstemmed | A note on the removability of totally disconnected sets for analytic functions |
| title_short | A note on the removability of totally disconnected sets for analytic functions |
| title_sort | note on the removability of totally disconnected sets for analytic functions |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6046 |
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