Issn 2091-5446 ilmiy axborotnoma научный вестник scientific journal

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ILMIY AXBOROTNOMA
MATEMATIKA
  2016-yil, 1-son
( )
(
)
(
)
0
cos
cos
cos
,
,...,
,
,
,
1
sin
1
2
2
3
4
3
2
1
3
2
=
+
⋅
⋅
⋅
+
−
θ
θ
θ
ε
θ
θ
θ
εφ
θ
x
x
x
n
n
n
(10)
with respect to
2
x
, where
i
θ
  are small numbers,  i = 2,3,…,n  and
(
)
ε
θ
θ
θ
φ
,
,...,
,
,
,
4
3
2
1
3
n
x
x
  is
defined by the following recurrent formula
(
)
(
)
(
)
ε
θ
θ
ε
θ
θ
θ
φ
ε
θ
θ
θ
φ
,
,...,
,
,
,...,
,
,
,...,
,
,...,
,
,...,
,
,
,...,
1
1
1
1
1
1
1
1
1
1
n
j
n
j
j
j
j
j
j
n
j
j
j
j
x
x
x
x
x
x
x
+
+
−
−
+
+
−
=
,
j= 3,4,…,n-1.  Moreover, we have the following iterated integral
( )
( )
∫
−
−
=
2
2
2
3
3
3
2
b
b
t
t
d
y
f
A
y
f
R
A
R
θ
θ
θ
θ
θ
,
  (11)
where b
2
> 0
and
2
θ
t
A
denotes the following averaging operator
( )
( )
(
)
(
)
(
)
∫








⋅⋅
⋅
+
−
+
−
=
+
R
n
n
n
n
n
t
dx
x
y
y
x
t
y
tx
y
f
y
f
A
1
3
2
1
2
1
3
3
3
2
1
2
2
1
1
,
,...,
,
,
,...,
,
cos
cos
,
,...,
,
,
1
1
,
:
2
ε
θ
θ
θ
ψ
θ
θ
ε
θ
θ
θ
φ
ε
θ
,
where
(
)
(
)
(
)
ε
θ
θ
ε
θ
θ
θ
φ
ε
θ
θ
θ
φ
,
,...,
,
,
,...,
,
,
,
,
,...,
,
,
3
3
2
1
2
1
3
3
2
1
2
n
n
n
x
x
x
x
=
,
(
)
ε
θ
θ
θ
ψ
,
,...,
,
,
3
2
1
2
n
x
is defined by following recurrent formula
(
)
(
) (
)
ε
θ
θ
θ
ε
θ
θ
ε
θ
θ
θ
ψ
ε
θ
θ
θ
ψ
,
,...,
,
,
,...,
)
,
,...,
),
,
,...,
,
,
,...,
(
,
,...,
:
,
,...,
,
,
,...,
1
1
1
1
1
1
1
1
1
1
1
1
1
n
k
k
k
n
k
n
k
k
n
k
k
k
n
k
k
k
k
x
x
J
x
x
x
x
x
x
x
+
−
+
+
−
−
+
+
−
=
,
(
)
ε
θ
θ
θ
,
,...,
,
,
,...,
1
1
1
n
k
k
k
x
x
J
+
−
denotes the Jacobian, k = 2,3,…,n-1.
Then we obtain
( )
( )
(
)
(
)
(
)
.
,
,...,
,
,
,...,
,
cos
cos
,
,...,
,
,
1
1
,
1
3
2
1
2
1
3
2
3
2
1
2
2
1
1
2
2
2
∫








⋅⋅
⋅
+
−
+
−
=
+
−
R
n
n
n
n
n
t
dx
x
y
y
x
t
y
tx
y
f
y
f
R
A
R
ε
θ
θ
θ
ψ
θ
θ
ε
θ
θ
θ
φ
ε
θ
θ
θ
(12)
In order to prove  formulas (10), (11), (12) we use the induction method.
Now we write the averaging operator in the form
( )
(
)
( )
∫ ∫ ∫
− −
−
−
−
−
=
n
n
n
n
b
b
b
b
b
b
t
t
d
y
f
R
A
R
y
f
A
1
1
2
2
...
θ
θ
θ
θ
,
where
θ = (θ
2

,θ
3

,…,θ
n
),
n
R
R
R
R
θ
θ
θ
θ
...
:
3
2
=
,
2
1
...
:
θ
θ
θ
θ
−
−
−
−
−
=
R
R
R
R
n
n
,
( )
( )
(
)
(
)
(
)
.
,
,...,
,
,
,...,
,
cos
cos
,
,...,
,
,
1
1
,
1
3
2
1
2
1
3
2
3
2
1
2
2
1
1
∫








⋅
⋅
⋅
+
−
+
−
=
+
−
R
n
n
n
n
n
t
dx
x
y
y
x
t
y
tx
y
f
y
f
R
A
R
ε
θ
θ
θ
ψ
θ
θ
ε
θ
θ
θ
φ
ε
θ
θ
θ

Thus, we apply to the last integral the Theorem 4.2 (in the first two variables) of the paper [5]
and obtain  that for  p > 2
( )
p
p
L
p
p
L
t
o
t
f
C
y
f
R
A
R
1
sup
−
−
>
≤
ε
θ
θ
θ
,
hence for any p >2 we have the following estimate
p
p
L
p
p
L
f
C
f
M
1
−
≤
ε
θ
,
where
( )
( )
: sup
.
t
t o
M f y
R A R f y
θ
θ
θ
θ
−
>
=
Moreover, the estimate is uniformly with respect to
θ. Then integrating over the set
{
}
2
2
,...,
b
b
n
n
<
<
θ
θ
we get a required bound.
Proof of Theorem 2. If
2
≥
m
is the minimal numberfor which
( )
0
0
≠
φ
m
d
, then there exists an unit
vektor ξ  such that
( )
0
0 =
∂
∂
k
l
ξ
φ
  for l=1,2,…, m-1  and
( )
0
0 ≠
∂
∂
m
m
ξ
φ
. Then by using rotation we may
assume
ξ
=
1
x
. Thus, theorem 2 can be proved by using the same arguments in the proof of Theorem 1.

13

ILMIY AXBOROTNOMA
MATEMATIKA
  2016-yil, 1-son
References
1. E.M.Stein.  Harmonic analysis: real-variable methods, orthogonality, and oscillatory
integrals, volume 43 of Princeton Mathematical Series. Princeton University Press,
Princeton, NJ, 1993.
2. J.Bourgain. Averages in the plane convex curves and maximal operators. J.Anal. Math.
1986. V. 47. P. 69-85.
3. A.Greenleaf. Principal curvature and harmonic analysis. Indiana Univ. Math. J., 30(4): 519-
537, 1981.
4. C.D.Sogge. Maximal operators associated to hypersurfaces with one nonvanishing principal
curvature. In Fourier analysis and partial differential equations, Stud. Adv. Math., pages
317 323.CRC, Boca Raton, FL, 1995.
5. I.A.Ikromov, Kempe, M., Müller, D., Estimates for maximal functions associated to
hypersurfaces in R
3
and related problems of harmonic analysis. Acta Math. 204 (2010), 151–
271.
6. A.N.Varchenko. Newton polyhedrons and the estimates of oscillatory integrals. 1976, 13-
38.
7. S.E.Usmanov. Adapted coordinates system for functions of several variables. Uzbek
Mathematical Journal
, 2013, № 4, pp. 128-139.
S.E. Usmanov

R
n+1
FAZODAGI EGRILIGI KICHIK
B
OʻLGAN GIPERSIRTLAR BILAN
BO
GʻLANGAN MAKSIMAL
OPERATORLARNING
CHEGARALANGANLIK MUAMMOSI

Mazkur maqolada egriligi kichik  b
oʻlgan
gipersirtlar bilan bo
gʻlangan
maksimal
operatorlar qaralgan. Gipersirt silliq funksiyaning
grafigi k
oʻrinishida berilganda maksimal
operatorning chegaralanganligi isbotlangan.

Kalit s
oʻzlar: maximal operator, oʻrtacha
operator, bogʻliqsizlik, tekis funskiya, giperyuza.
С.Э.Усманов

ПРОБЛЕМА ОГРАНИЧЕННОСТИ
МАКСИМАЛЬНЫХ ОПЕРАТОРОВ,
СВЯЗАННЫХ С ГИПЕРПОВЕРХНОСТЯМИ
НА R
n+1

С МАЛЫМИ КРИВИЗНАМИ

В
данной
работе
рассмотрены
максимальные
операторы,
связанные
с
гиперповерхностями
малыми
кривизнами.
Доказана    ограниченность  максимального
оператора,  когда  гиперповерхность  задана  в
виде графика гладких функций.

Ключевые
слова:
максимальный
оператор,  средний  оператор,  независимость,
ровная функция, гиперповерхность.

УДК: 517.3
НЕКОТОРЫЕ СВОЙСТВА БИСИНГУЛЯРНОГО ИНТЕГРАЛА И ЕГО ПРИЛОЖЕНИЯ
Т.Абсаламов, А.Абсаламов
Самаркандский государственный университет
Аннотация. Получены оценки типа оценки Зигмунда для бисингулярного интеграла.  На
основе  полученных  оценок  строится  класс  функций  инвариантного  относительно
бисингулярного  оператора.  Полученные  результаты  применены  для  нелинейного
бисингулярного  интегрального  уравнения.
Ключевые  слова:    бисингулярный  интеграл,  оценка  Зигмунда,  инвариантное
пространство.
Рассмотрим бисингулярный интеграл вида:
??????�(??????
1
, ??????
2
) = � �
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
??????
2
????????????
1
????????????
2
  ,
??????
1
??????
1

где функция
?????? ?????? ??????
??????
(∆), ?????? > 1   ∆= (??????
1
, ??????
1
; ??????
2
, ??????
2
).
Введем характеристики:

14

ILMIY AXBOROTNOMA
MATEMATIKA
  2016-yil, 1-son
Ω
??????,1
(??????, ??????
1
, ??????
2
) = � � � |??????(??????
1
, ??????
2
)|
??????
??????
2
+??????
2
??????
2
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
,
Ω
??????,2
(??????, ??????
1
, ??????
2
) = � �
� |??????(??????
1
, ??????
2
)|
??????
??????
2
??????
2
−??????
2
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
,
Ω
??????,3
(??????, ??????
1
, ??????
2
) = � � � |??????(??????
1
, ??????
2
)|
??????
??????
2
+??????
2
??????
2
????????????
1
????????????
2
??????
1
??????
1
−??????
1
�
1
??????
,
Ω
??????,4
(??????, ??????
1
, ??????
2
) = � � � |??????(??????
1
, ??????
2
)|
??????
??????
2
??????
2
−??????
2
????????????
1
????????????
2
??????
1
??????
1
−??????
1
�
1
??????
,

??????
??????
??????(0, ??????
??????
], ??????
??????
= ??????
??????
− ??????
??????
, ?????? = 1,2
��.
Пользуясь
[1 − 6] доказана следующая
Теорема  1.  Пусть  ?????? ?????? ??????
??????
(∆).  Тогда  при  сходимости  соответствующих  интегралов
справедливо неравенство
??????
??????,??????
(??????�, ??????
1
, ??????
2
) ≤ ??????
??????
1
((??????
1
??????
2
)
1
??????
� �
??????
??????,??????
(??????, ??????
1
, ??????
2
)
(??????
1
??????
2
)
1+1??????
??????
2
??????
2
????????????
1
????????????
2
??????
1
??????
1
+
+(??????
1
??????
2
)
1
??????
�
??????
??????
(??????, ??????
1
, ??????
2
)
(??????
2
)
1+1??????
????????????
2
??????
1
??????
2
+ (??????
1
??????
2
)
1
??????
�
??????
??????
(??????, ??????
1
, ??????
2
)
(??????
1
)
1+1??????
????????????
1
??????
1
??????
1
+
+(??????
1
??????
2
)
1
??????
??????
??????
(??????, ??????
1
, ??????
2
)),
при
??????
??????
??????  �0,
??????
??????
4
� , ?????? = 1,2 ,
??????
??????,??????
(??????�, ??????
1
, ??????
2
) ≤ ??????
??????
2
(??????
1
1
??????
�
??????
??????
(??????, ??????
1
, ??????
2
)
(??????
1
)
1+1??????
????????????
1
??????
1
??????
1
+
+(??????
1
)
1
??????
??????
??????,??????
(??????, ??????
1
, ??????
2
)),
??????
1
??????  �0,
??????
1
4� , ??????
2
??????  �
??????
2
4 , ??????
2
�,
??????
??????,??????
(??????�, ??????
1
, ??????
2
) ≤ ??????
??????
3
(??????
2
1
??????
�
??????
??????
(??????, ??????
1
, ??????
2
)
(??????
2
)
1+1??????
????????????
2
??????
2
??????
2
+
+(??????
2
)
1
??????
??????
??????,??????
(??????, ??????
1
, ??????
2
)),
??????
1
??????  �
??????
1
4 , ??????
1
� , ??????
2
??????  �0,
??????
2
4�,

Ω
??????,??????
(??????�, ??????
1
, ??????
2
) ≤ ??????
??????
4
Ω
??????,??????
(??????, ??????
1
, ??????
2
),       ??????
??????
??????  �
??????
??????
4 , ??????
??????
� , ?????? = 1,2 ,
где постоянные
??????
??????
??????
   (??????, ?????? = 1,4
��) зависят лишь от  ??????,    ??????
??????
(?????? = 1,2).
Доказательство. Пусть ??????
??????
??????  �0,
??????
??????
4
� , ?????? = 1,2.   ??????�(??????
1
, ??????
2
) представим в следующим виде:
??????�(??????
1
, ??????
2
) = �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
+2??????
2
??????
2
????????????
1
????????????
2
??????
1
+2??????
1
??????
1
+

15

ILMIY AXBOROTNOMA
MATEMATIKA
  2016-yil, 1-son
+ �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
??????
2
+2??????
2
????????????
1
????????????
2
??????
1
+2??????
1
??????
1
+
+ �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
+2??????
2
??????
2
????????????
1
????????????
2
??????
1
??????
1
+2??????
1
+
+ �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
??????
2
+2??????
2
????????????
1
????????????
2
??????
1
??????
1
+2??????
1
= � ??????
??????
4
??????=1
(??????
1
, ??????
2
)
Следовательно,
Ω
??????,1
(??????�, ??????
1
, ??????
2
) ≤ � � � �� ??????
??????
(??????
1
, ??????
2
)
4
??????=1
�
??????
??????
2
+??????
2
??????
2
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
≤
≤ � � �
� |??????
??????
(??????
1
, ??????
2
)|
??????
??????
2
+??????
2
??????
2
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
4
??????=1
= � ??????
??????
4
1

В силу теоремы М.Рисса
[6]
??????
1
= � �
� |??????
??????
(??????
1
, ??????
2
)|
??????
??????
2
+??????
2
??????
2
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
=
= � �
� � �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
+2??????
2
??????
2
????????????
1
????????????
2
??????
1
+2??????
1
??????
1
�
??????
2
+??????
2
??????
2
??????
????????????
1
????????????
2
??????
1
+??????
1
??????
1
�
1
??????
≤
≤ � �
� � �
�
??????(??????
1
, ??????
2
)
(??????
1
− ??????
1
)(??????
2
− ??????
2
)
??????
2
+2??????
2
??????
2
????????????
1
????????????
2
??????
1
+2??????
1
??????
1
�
??????
??????
2
+2??????
2
??????
2
????????????
1
????????????
2
??????
1
+2??????
1
??????
1
�
1
??????
≤
≤ ??????
??????
‖??????‖
??????
??????
[??????
1
,??????
1
+2??????
1
;??????
2
,??????
2
+2??????
2
]
= ??????
??????
Ω
??????,1
(??????, 2??????
1
, 2??????
2
)
так как
Ω
??????,1
(??????, ??????
1
, ??????
2
) неубывающие функции по ??????
1
, ??????
2
, то при
??????
??????
??????  �0,
??????
??????
4
� , ?????? = 1,2,
(??????
1
??????
2
)
1
??????
� �
Ω
??????,1
(??????, ??????
1
, ??????
2
)
(??????
1
??????
2
)
1+1??????
????????????
1
????????????
2
≥
??????
2
2ξ
2
??????
1
2ξ
1

≥ (ξ
1
ξ
2
)
1
??????
Ω
??????,1
(??????, 2ξ
1
, 2ξ
2
) ∙ ?????? ∙
??????
1
1
??????
− (2??????
1
)
1
??????
(2??????
1
)
1
??????
??????
1
1
??????
∙ ?????? ∙
??????
2
1
??????
− (2??????
2
)
1
??????
(2??????
2
)
1
??????
??????
2
1
??????
≥
≥
??????
2
�2
1
??????
− 1�
2
16
Ω
??????,1
(??????, 2??????
1
, 2??????
2
).
Отсюда,
Ω
??????,1
(??????, 2??????
1
, 2??????
2
) ≤ ??????
??????
(??????
1
??????
2
)
1
??????
∫ ∫
Ω
??????,1
(??????,??????
1
,??????
2
)
(??????
1
??????
2
)
1
??????+1
????????????
1
????????????
2
??????
2
2??????
2
??????
1
2??????
1
.
Теперь, оценим
??????
2
. Учитывая, что
??????
2
− ??????
2
− ??????
2
≤ ??????
2
− ??????
2

при

16

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