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Discrete iterated Hardy-type inequalities . . .

in this type of inequalities are caused by their applicability to spaces of the Morrey type ([16], [17]). Moreover, the characterizations of these inequalities can be applied to research weighted inequalities for Hardy’s bilinear inequalities ([18]-[20]). However, the discrete Hardytype inequality (1) is study very little. For example, see the papers [21] and [22], where in particular, in [22] a criterion for the fulﬁlment of inequality (1) was obtained for the case

0 < q < θ < p < ∞, p > 1.

3 Material and research methods

The research methods are as follows: in this paper a method of partition of the sequence of elements of the Hardy operator on the part in each point is developed, which allows us to e ectively estimate the sum on the parts. Note that such "blocking technic" was developed in [4]. During the estimate, various classical inequalities are used, such as Minkowski inequality, Holder inequality and the following elementary inequalities:

If ai > 0, i = 1, 2, ..., k, then

m=1 ai		≤ m=1 aiα,	0 < α ≤ 1,	(2)
k		k
	α
	α
and		≥ m=1 aiα,	α ≥ 1.
m=1 ai		≥ m=1 aiα,	α ≥ 1.	(3)
k		k
	α
	α

In the proofs of our main results we will need the following well-known version of the discrete Minkowski inequality:

Lemma. Let {ai,j }, i = 1, 2, ..., n ≤ +∞, j = 1, 2, ..., m, be a positive matrix. Then the inequalities

n	m		1	m	n		1
			σ				σ ,	(4)
i=1	j=1 ai,j	σ	σ	≤ j=1	i=1 \|ai,j \|σ		σ ,	(4)
and
and
n	i		1	n	n		1
			σ			\|ai,j \|σ	σ ,	(5)
i=1	j=1 ai,j	σ	σ	≤ j=1	i=j	\|ai,j \|σ	σ ,	(5)

hold, where σ ≥ 1.

Convention: The symbol M K means that M ≤ cK, where c > 0 is a constant depending only on unessential parameters. If M K M, then we write M ≈ K.

4 Results and discussion

4.1 Main result

Our main result reads as follows.

22 R. Oinarov et al.

Theorem. (i) If 1 < p ≤ min{q, θ} < ∞, then the inequality (1) holds, if and only if

A1 < ∞, where

∞

ϕq

u−p

p .

sup

n=1

ωθ

k=n

r≥1

i=r

Moreover, C ≈ A1, where C is the best constant in (1).

(ii) If p (0, 1] and p ≤ min{q, θ} < ∞, then the inequality (1) holds, if and only if A2 < ∞,

where			r		r		θ	1
A	2	:= sup	n=1	ωθ	s=n	ϕq	q	θ	sup u−1.
	2	r≥1	n=1	n	s=n	s			r≤k k

Moreover, C ≈ A2, where C is the best constant in (1).

Proof. Necessity: Suppose that the inequality (1) holds with best constant C > 0.

(i) Let us show that A1 < ∞. Let 1 ≤ r < N < ∞ and take a test sequence f = {fs}s∞=1

such that fs = 0 for 1 ≤ s < r and s > N and fs = us−p

for r ≤ s ≤ N < ∞.

Then

∞

f lp,u =

s=1 |fs · us|p p

s=r

|us−p · us|p

p =

s=r

us−p p .

(6)

By substituting f in the left hand side of inequality (1), we can deduce that

∞

ϕk

q θ

≥

n=1

ϕk i=k

q θ

≥

I(f) := n=1 ωn

k=n

i=k

ωn k=n

ϕk

θ 1

≥

(7)

≥ n=1 ωnθ k=n

i=r

ui−p q q

n=1

ωnθ k=n ϕkq q

i=r ui−p .

From (1), (6) and (7), it follows that

C ≥ n=1 ωnθ k=n

ϕkq q

i=r

ui−p

for

all

1 ≤ r < N.

Since r ≥ 1 is arbitrary, passing to the limit as N → ∞, we have that

∞

u−p

= sup

ωθ

≤

(8)

r≥1

n=1

n k=n ϕk

i=r

(ii) Let us show that A2 <

. Now for 1 < r

≤

k <

∞

we assume that g =

{

}

s∞=1, where

∞

gs = 0 for s = k and gs = u−

for s = k, where uk

= 0. Then

· uk = 1.

(9)

g lp,u = uk−

Substituting g in the left hand side of inequality (1), we ﬁnd that

∞

ϕs

∞

−1

q θ

k ≥ r,

I(g) := n=1

ωn s=n

i=s

≥ n=1 ωn s=n ϕs

Discrete iterated Hardy-type inequalities . . .

≥

n=1

s=n

ϕs

r≤k k

≥

I(g)

ωθ

sup u−1,

Therefore

sup

−1

I(g)

n=1

≥ r≥1

ωn s=n

ϕs

r≤k uk

= A2.

(10)

From (1), (9) and (10), we have that

θ sup u−1

sup

ωθ

ϕq

≤

2 = r≥1 n=1

s=n

r≤k k

(11)

Su ciency: Now, we prove that the inequality (1) holds. Let 0 ≤ f lp,u be such that

∞

(12)

fi < ∞.

Let

∞

k1 := inf{k Z :

fi ≤ 2k},

then

∞

2k1−1 ≤ fi

< 2k1 .

We consider the sequence {jk}, where jk are deﬁned by

∞

jk := min{j ≥ 1 :

fi ≤ 2k1−k+1}.

We note that

∞

fi ≤ 2k1 } = 1.

j1 = min{j ≥ 1 :

For all k ≥ 1 it yields that

∞

(13)

fi ≤ 2k1−k+1 <

fi.

i=jk

i=jk−1

Therefore the set of natural numbers N can be written

N = [jk, jk+1 − 1].

k≥1

Moreover,

	∞		jm+1−1		∞	jm+1−1
2k1−m+1 <	−	fi =	−	fi +		−	fi + 2k1	−(m+1)+1, m ≥ 2.
					fi <
	i=jm 1		i=jm 1		i=jm+1	i=jm 1

R. Oinarov et al.

Hence,

jm+1

−1

2k1−m <

−

fi, m ≥ 2.

i=jm 1

jm+2−1

2k1−m = 2k1−(m+1)+1 = 2 · 2k1−(m+1) < 2

−

fi,

m ≥ 2.

i=jm+1 1

We have that

j2+1

−1

2k1−1+1 = 2k1 = 4 · 2k1−2 < 4

fi.

i=j2−1

Then we obtain that

jm+2−1

2k1−m+1 < 4

−

fi, m

≥ 1.

(14)

i=jm+1 1

Therefore, in view of (13),

= k=1

ωnθ s=n ϕs i=s

Iθ(f) := n=1 ωnθ s=n ϕs i=s fi

n=jk

≤

∞

jk+1−1

∞

≤

ϕsq

≤

ωnθ

∞

jk+1−1

∞

jm+1−1

∞

k=1

n=j

m=k

=max(n,j

)

i=jm

∞

jk+1−1

∞

jm+1−1

≤ k=1

n=jk

ωnθ m=k s=max(n,jm) ϕsq 2k1−m+1 .

Hence, by applying (14) we have that

ϕsq

Iθ(f) ≤ 4θ

ωnθ

(15)

∞

jk+1−1

∞

jm+1−1

jm+2−1

−

k=1

n=j

m=k

=max(n,j

)

i=jm+1

We must now consider the cases θ ≤ q and θ > q separately.

4.2 The case θ ≤ q

(i) Let 1 < p ≤ θ ≤ q. By applying (2) in (15), we ﬁnd that

Iθ(f) ≤ 4θ

ωnθ

ϕsq

fi .

∞

jk+1−1

∞

jm+1−1

jm+2

−1

−

k=1

n=j

m=k

=max(n,j

)

i=jm+1

Discrete iterated Hardy-type inequalities . . .

Now, by changing the orders of sums, we get that

ωnθ

ϕsq

Iθ(f) ≤ 4θ

≤

∞

jm+2

−

jk+1−1

jm+1

−

m=1 i=jm+1

k=1 n=j

=max(n,j

)

≤

≤ 4θ

ωnθ

ϕsq

ωnθ

ϕsq

∞

−

jm+1

−

m jk+1

−1

jm+1

−

jm+2 1

−

m=1 i=jm+1

n=j1

s=n

ϕsq

k=2 n=j

s=jm

≤

≤ 4θ

ωnθ

ϕsq

∞

−1

j2−1

jm+1−1

−1

jm+2

q jm+1

jm+1

−

m=1 i=jm+1

n=j

s=n

ϕsq

n=j

ωnθ

s=jm

ϕsq

≤

≤ 4θ

ωnθ

∞

−1

j2−1

jm+1−1

−1

jm+2

q jm+1

jm+1

−

m=1 i=jm+1

n=j

s=n

ωnθ

n=j

s=n

≤ 4θ

∞

jm+2−1

θ jm+1−1

jm+1

ϕsq

−1

−

m=1 i=jm+1

n=j

s=n

so that

ϕsq .

Iθ(f)

∞

θ jm+1−1

ωnθ

jm+1−1

(16)

jm+2−1

−

m=1 i=jm+1

n=j

s=n

Therefore, by using Holder’s inequality and (3) in (16), we obtain that

Iθ(f) ≤

|fi · ui|p

ωnθ

ϕsq

ui−p

≤

∞

jm+2−1

p jm+1

−1

jm+1

−1

jm+2−1

−

m=1 i=jm+1

n=j1

s=n

i=jm+1

≤

| ·

m 1

∞ jm+2−1

jm+1

−1

jm+1

−1

≥

−

ui p

ωnθ

ϕsq

sup

m=1 i=jm+1 1

r 1 n=1

n=j

s=n

≤ =1 |

s=n

i=r

∞

≥

fiui

ωnθ

ϕsq

sup

Hence,

I(f) A1 f p,u,

if 1 < p ≤ θ ≤ q.

q	θ			∞		ui−p p			≤
θ	1						1		θ
				−

			i=jm+1		1
		1		θ
		1
ui−p		p		=	A1 f		p,u	θ.

(17)

R. Oinarov et al.

(ii) Let 0 < p ≤ 1. We start with the inequality (16):

ωnθ

ϕsq .

Iθ(f)

fi · ui · ui−1

∞

jm+2−1

p· p jm+1−1

jm+1−1

−

m=1 i=jm+1

n=1

s=n

By applying (2) with 0 < p ≤ 1, we obtain that

1 ui−1

ωnθ

ϕsq ≤

Iθ(f) ≤

|fi · ui|p

jm+1

1maxi jm+2

∞

jm+2−1

jm+1−1

jm+1

−1

−

− ≤ ≤ −

m=1 i=jm+1

| · |

n=j1

s=n

≤

jm+1 1 k

∞

jm+2

−1

θ jm+1−1

jm+1

−1

−

− ≤

ui p

sup

uk−1

ωnθ

ϕsq .

m=1 i=jm+1 1

n=1

s=n

By using (3), we get

≤

| · |

m 1

jm+1 1 k

≤

∞

jm+2−1

jm+1−1

Iθ(f)

−

fi ui

≥

ωnθ

ϕsq

− ≤

uk−1

sup

≤

m=1 i=jm+1 1

r 1

n=j

s=n

| · |

r k

∞

sup uk−1

ui p

ωnθ

ϕsq

A2 f

p,u

sup

i=1

≥

n=1

s=n

≤

so that

I(f) A2 f p,u, if p ≤ θ ≤ q, (18) where p (0, 1].

4.3 The case θ > q

(i) Let 1 < p ≤ q < θ. We start with the inequality (15). First we raise both sides in (15) to the power θq ≤ 1, i.e.,

	−1			jm+1−1
∞ jk+1		ωnθ	∞
Iq(f) ≤ 4q

ϕsq		fi		.
			θ	q
jm+2	−1	q	q	θ

k=1 n=jk

m=k s=max(n,jm)

i=jm+1−1

Next we apply (4) in the inner sum with σ = θ and obtain that
					q
≤						θ
Iq(f) 4q	∞	∞ jk+1−1		ωnθ	jm+1−1	ϕsq q	jm+2−1
								−




			n=jk		s=max(n,jm)		i=jm+1 1
	k=1 m=k		n=jk		s=max(n,jm)		i=jm+1 1

fi θ θq q		θ	,
	θ	q

Discrete iterated Hardy-type inequalities . . .

Using (5), we get that

ωnθ

ϕsq

≤

Iq(f) ≤ 4q

∞

jk+1−1

jm+1

−1

jm+2−1

−

m=1

k=1

n=j

=max(n,j )

i=jm+1

≤ 4q

ωnθ

ϕsq

≤

∞

jm+2−1

jk+1

−1

jm+1−1

−

=max(n,j

)

m=1 i=jm+1

k=1 n=j

≤ 4q

ωnθ

ϕsq

∞

jm+2−1

jm+1

−1

jm+1−1

−

m=1 i=jm+1

n=j

s=n

so that

ωnθ

ϕsq

(19)

Iq(f)

∞

jm+2

−1

jm+1

−1

jm+1

−1

−

m=1 i=jm+1

n=j

s=n

Hence, by using Holder’s inequality,

ui−p

ωnθ

Iq(f) ≤

|fi · ui|p

ϕsq

∞

jm+2

−1

jm+2

−1

jm+1−1

jm+1

−1

−

m=1 i=jm+1

i=jm+1

n=j

s=n

Therefore, by applying (3) with α = q

, we obtain that

ui p

ωnθ

ϕsq

ui−p

Iq(f)

sup

≤

| · |

m 1

∞

jm+2

−1

jm+1−1

∞

−

≥

−

m=1 i=jm+1 1

s=n

≤

r 1

n=j

i=jm+1 1

∞

fiui

ϕsq

ui−p

= A1

p,u

sup

ωnθ

i=1

≥

n=1

s=n

i=r

so that

I(f) A1 f p,u,

when 1 < p ≤ q < θ.

(20)

(ii) Let 0 < p ≤ 1. We start with the inequality (19):

ωnθ

ϕsq

Iq(f)

fi · ui · ui−1

∞

jm+2−1

jm+1−1

p· p

−

m=1 i=jm+1

n=1

s=n

R. Oinarov et al.

By applying (2) with 0 < p ≤ 1, we obtain that

Iq(f)

fi ui p

sup

uk−1

ωnθ

ϕsq

≤

| · |

jm+1 1 k

∞

jm+2−1

jm+1

−1

jm+1−1

−

− ≤

m=1 i=jm+1 1

n=1

s=n

By using (3), we get that

≤

| · |

m 1

jm+1 1 k

≤

∞

jm+2−1

jm+1

−1

jm+1

−1

Iq(f)

−

fi ui p

≥

ϕsq

− ≤

sup

ωnθ

sup

uk−1

m=1 i=jm+1 1

n=j

s=n

≤

| · |

r 1

r k

∞

sup uk−1

ui p

ωnθ

ϕsq

A2 f

p,u

sup

i=1

≥

n=1

s=n

≤

so that

I(f) A2 f p,u, if

p ≤ q < θ,

(21)

where p (0, 1].

The estimates (17), (20) and (18), (21) were obtained under the condition (12). Let M = {f lp,u : N = N(f) > 0 and fi = 0, i ≥ N}. Since f from M satisfy the condition (12) and the set M is everywhere dense in lp,u, then the estimates (17), (20) and (18), (21) are satisﬁed for all f lp,u. Therefore from the inequalities (8), (17) and (20), we get C ≈ A1 and from the inequalities (8) and (18), (21), we get C ≈ A2, where C is the best constant in

(1). The proof of Theorem is complete.

5 Conclusion

In conclusion, we have established necessary and su cient conditions on functions u and ω are ensuring boundedness of a discrete Hardy-type operator from a weighted sequence space lp,u to a weighted sequence space for a wide range of the numerical parameters p, u and θ.

6 Acknowledgement

We thank Professor Lars-Erik Persson for several suggestions, which have improved ﬁnal version of this paper.

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30 ISSN 1563-0277, eISSN 2617-4871

Journal of Mathematics, Mechanics, Computer Science. № 1(105). 2020

IRSTI 27.33.19

https://doi.org/10.26577/JMMCS.2020.v105.i1.04

1A.N. Stanzhitskii , 2S.G. Karakenova , 3S.S. Zhumatov

1Doctor of Physical and Mathematical Sciences, E-mail: ostanzh@gmail.com, Taras Shevchenko National University of Kyiv, Kyiv, Ukraine

2PhD student, E-mail: sayakhat.karakenova05@gmail.com, Al-Farabi Kazakh National University, Almaty, Kazakhstan

3Doctor of Physical and Mathematical Sciences, E-mail: sailau.math@gmail.com, Institute of Mathematics and Mathematical Modeling, Almaty, Kazakhstan

ON A COMPARISON THEOREM FOR STOCHASTIC INTEGRO-FUNCTIONAL EQUATIONS OF NEUTRAL TYPE

In this paper, we will discuss a comparison result for solutions to the Cauchy problems for two stochastic di erential equations with delay. On this subject number of authors have obtained their comparison results. We deal with the Cauchy problems for two integro-di erential equations. Except transient- (or drift-) and di usion coe cients our equations include also one integrodi erential term. Basic di erence of our case from the case of all earlier investigated problems is presence of this term. We introduce a concept of solutions to our problems and prove the comparison theorem for them. According to our result under certain assumptions on coe cients of equations under consideration, their solutions depend on the transient-coe cients in a monotone way.

Key words: stochastic di erential equation, comparison theorem, Hilbert space.

1А.Н. Станжицкий , 2С.Г. Каракенова , 3С.С. Жуматов

1ф.-м.ғ.д., Т. Шевченко атындағы Киев ұлттық университетi, Киев қ., Украина, E-mail: ostanzh@gmail.com

2докторант, әл-Фараби атындағы Қазақ ұлттық университетi, Алматы қ., Қазақстан, E-mail: sayakhat.karakenova05@gmail.com 3ф.-м.ғ.д., Математика және математикалық моделдеу институты Алматы қ., Қазақстан, E-mail: sailau.math@gmail.com

Бейтарап типтегi стохастикалық интегралдық-функционалдық теңдеулер үшiн салыстыру теоремасы

Бұл мақалада кiдiрiс әсерлi екi стохастикалық дифференциалдық теңдеулер үшiн Коши есебi шешiмдерiн салыстыру есебi қарастырылған. Осы сияқты есептердiң шешiмдерiн салыстыруға қатысты көптеген авторлар өз нәтижелерiн алды. Мақалада бейтарап типтi екi стохастикалық интегралдық-дифференциалдық теңдеулер үшiн Коши есебi қарастырылған. Қарастырылатын есептiң тасымалдау және диффузия коэффициенттерiнен басқа бiр интегралдық-дифференциалдық бөлiктен тұрады. Бұрын зерттелген жұмыстардан негiзгi айырмашылығы интегралдық бөлiктiң бар болуы болып табылады. Алынған нәтижелерге сәйкес, олардың шешiмi тасымалдау коэффициентiне монотонды тәуелдi болады.

Түйiн сөздер: стохастикалық дифференциалдық теңдеу, салыстыру теоремасы, гильберт кеңiстiгi.

1А.Н. Станжицкий , 2С.Г. Каракенова , 3С.С. Жуматов

1д.ф.-м.н., Киевский национальный университет имени Т. Шевченко,

г.Киев, Украина, E-mail: ostanzh@gmail.com

2докторант, Казахский национальный университет имени аль-Фараби, г. Алматы, Казахстан, E-mail: sayakhat.karakenova05@gmail.com

3д.ф.-м.н., Институт математики и математического моделирования,

г.Алматы, Казахстан, E-mail: sailau.math@gmail.com

Об одной теореме сравнения для стохастических интегро-функциональных уравнений нейтрального типа

c 2020 Al-Farabi Kazakh National University

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