Armenian Journal of Mathematics
Volume 2, Number 3, 2009, 77–93
Reverse weighted 𝑙𝑝− norm inequalities for convolution
type integrals
Nguyen Du Vi Nhan*, Dinh Thanh Duc** and Vu Kim Tuan***
* Department of Mathematics,
Quy Nhon University, Binh Dinh, Vietnam.
[email protected]
** Department of Mathematics,
Quy Nhon University, Binh Dinh, Vietnam.
[email protected]
*** Department of Mathematics,
University of West Georgia, Carrollton, GA 30118, USA.
[email protected]
Abstract
By using reverse Hölder’s inequalities we obtain reverse weighted 𝐿𝑝 norm inequalities
for various iterated convolutions.
Key Words: Convolution, integral transform, inverse problem, reverse Hölder inequality,
stability, weighted 𝐿𝑝 inequality, Bernoulli-Euler beam equation, Helmholtz equation, heat
equation.
Mathematics Subject Classification 2000: Primary 44A35, 35A22; Secondary 26D20
1
Introduction
In order to study stability of some inverse problems, in a series of papers, S. Saitoh, M.
Yamamoto and the last author ([10, 11, 12]) derived reverse inequalities for convolution in
some weighted 𝐿𝑝 spaces by using the following reverse Hölder’s inequality:
Proposition 1.1 ([14]; see also [4,
satisfying
pp. 125-126]) For two positive functions 𝑓 and 𝑔
0<𝑚≤
𝑓
≤𝑀 <∞
𝑔
77
(1.1)
on the set 𝑋, and for 𝑝, 𝑞 > 1, 𝑝−1 + 𝑞 −1 = 1,
(∫
) 𝑝1 (∫
) 1𝑞
(𝑚)∫
1
1
𝑓 𝑑𝜇
𝑔𝑑𝜇
≤ 𝐴𝑝,𝑞
𝑓 𝑝 𝑔 𝑞 𝑑𝜇,
𝑀
𝑋
𝑋
𝑋
(1.2)
if the right hand side integral converges. Here
1
𝐴𝑝,𝑞 (𝑡) = 𝑝
− 𝑝1 − 1𝑞
𝑞
𝑡− 𝑝𝑞 (1 − 𝑡)
(
) 𝑝1 (
) 1𝑞 .
1
1
𝑝
𝑞
1−𝑡
1−𝑡
Then, by using Proposition 1.1 the first two authors obtained the following
Proposition 1.2 ([3]) Let 𝐹1 and 𝐹2 be positive functions satisfying
1
1
1
1
0 < 𝑚1𝑝 ≤ 𝐹1 (𝒛) ≤ 𝑀1𝑝 < ∞, 0 < 𝑚2𝑝 ≤ 𝐹2 (𝒛) ≤ 𝑀2𝑝 < ∞, 𝒛 ∈ ℝ𝑛 .
(1.3)
Then for any positive continuous functions 𝜌1 and 𝜌2 on ℝ𝑛 , we have the following reverse
weighted 𝐿𝑝 (𝑝 > 1)−norm inequality for convolution
1
−1 ((𝐹1 𝜌1 ) ∗ (𝐹2 𝜌2 ))(𝜌1 ∗ 𝜌2 ) 𝑝 𝐿𝑝 (ℝ𝑛 )
)
(
(1.4)
𝑚1 𝑚2
−𝑛
≥ 𝐴𝑝,𝑞
∥𝐹1 ∥𝐿𝑝 (ℝ𝑛 ,𝜌1 ) ∥𝐹2 ∥𝐿𝑝 (ℝ𝑛 ,𝜌2 ).
𝑀1 𝑀2
In [8, 3] we gave various applications of (1.4) from the viewpoint of stability in inverse
problems.
Later in [7], we introduced several iterated convolution type transformations. Using the
Hölder’s inequalities we established weighted 𝐿𝑝 , 𝑝 > 1, norm inequalities for these iterated
convolutions. In this paper, by using the reverse Hölder’s inequalities we will derive reverse
type inequalities for the iterated convolutions.
2
Preliminaries
Throughout this paper, by 𝒙 = (𝑥1 , ..., 𝑥𝑛 ), 𝑥𝑗 ∈ ℝ, 𝑗 = 1, 2, ..., 𝑛, we denote a vector in ℝ𝑛 .
In particular,
1 = (1, 1, ..., 1), 2 = (2, 2, ..., 2), ...
(2.1)
We shall write 𝒙 > 𝒚 to denote 𝑥𝑗 > 𝑦𝑗 , 𝑗 = 1, 2, ..., 𝑛. Anologously one has to understand
𝒙 ≥ 𝒚, 𝒙 < 𝒚, 𝒙 ≤ 𝒚.
We always assume that 𝜌, 𝜌𝑗 and 𝜌𝑗,𝑟 (𝑗 = 1, 2, ..., 𝑚; 𝑟 = 1, 2, ..., 𝑠)− the weight functions,
to be nonnegative, and 1 < 𝑝 < ∞. When we write 𝐴 ≤ 𝐵, we understand that if 𝐵 is
finite, then 𝐴 is also finite, and bounded above by 𝐵.
We shall denote some subsets of ℝ𝑛
ℝ𝑛+ (𝒕) = {𝒙 : 𝒙 ∈ ℝ𝑛 , 0 < 𝒙 < 𝒕}
(2.2)
ℝ𝑛𝒕 = {𝒙 : 𝒙 ∈ ℝ𝑛 , 𝒕 < 𝒙 < ∞} .
(2.3)
Then, by using Proposition 1.1 we obtain the following lemma
78
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
79
Lemma 2.1 Let Ω be a measurable set in ℝ𝑛 and 𝜌 be a positive function belonging to 𝐿1 (Ω).
For a positive function 𝑓 satisfying
0 < 𝑚1/𝑝 ≤ 𝑓 (𝒙) ≤ 𝑀 1/𝑝 < ∞
(2.4)
on the set Ω, and for 𝑝, 𝑞 > 1, 𝑝−1 + 𝑞 −1 = 1,
) 𝑝1 (∫
) 1𝑞
(𝑚)∫
𝑛
[𝑓 (𝒙)] 𝜌(𝒙)𝑑𝒙
𝜌(𝒙)𝑑𝒙
≤ 𝐴𝑝,𝑞
𝑓 (𝒙)𝜌(𝒙)𝑑𝒙.
𝑀
Ω
Ω
Ω
(∫
𝑝
(2.5)
Let 𝐷 be a domain in ℝ𝑛 , 𝐹𝑖 (.) : 𝐷 → ℝ, 𝑖 = 1, ⋅ ⋅ ⋅ , 𝑞, and 𝜑(., .), 𝜓(., .) : 𝐷 × 𝐷 → 𝐷.
Then, we introduce a convolution type integral, called the 𝜑− convolution
Definition 2.2 ([7]) The 𝜑− convolution of 𝐹1 and 𝐹2 , denoted by 𝐹1 ∗𝜑 𝐹2 , is defined by
∫
(𝐹1 ∗𝜑 𝐹2 )(𝜼) =
𝐹1 (𝝃)𝐹2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣𝑑𝝃,
(2.6)
𝐷
when this integral exists. Here,
(
∣𝜑𝜼 (𝝃, 𝜼)∣ := det
)
∂
𝜑(𝝃, 𝜼)
∂𝜼
is the Jacobian of the transformation 𝜼 → 𝜑(., 𝜼).
Example 2.3 The following convolutions are particular cases of the 𝜑−convolution ([1],
[2]):
∙ The Fourier convolution
∫
𝐹1 (𝒚)𝐹2 (𝒙 − 𝒚)𝑑𝒚,
𝐹1 ∗𝔉 𝐹2 :=
𝒙 ∈ ℝ𝑛 .
(2.7)
ℝ𝑛
∙ The Laplace convolution
∫
𝐹1 (𝝉 )𝐹2 (𝒕 − 𝝉 )𝑑𝝉 ,
𝒕 ∈ ℝ𝑛+ .
(2.8)
𝐹1 (𝒙)𝐹2 (𝒕/𝒙)𝒙−1 𝑑𝒙,
𝒕 ∈ ℝ𝑛+ .
(2.9)
𝐹1 ∗𝔏 𝐹2 :=
ℝ𝑛
+ (𝒕)
∙ The Mellin convolution
∫
𝐹1 ∗𝔐 𝐹2 :=
ℝ𝑛
+
Definition 2.4 ([7]) Let 𝐺𝑗 (., .) : ℝ𝑛 × ℝ𝑚
+ → ℝ. The Fourier-Laplace convolution of
𝐺1 (𝝉 , 𝜻) and 𝐺2 (𝝉 , 𝜻) is defined by
∫
∫
𝑑𝝉
(𝐺1 ∗𝔉,𝔏 𝐺2 )(𝝃, 𝜼) :=
𝐺1 (𝝉 , 𝜻)𝐺2 (𝝃 − 𝝉 , 𝜼 − 𝜻)𝑑𝜻.
(2.10)
ℝ𝑛
+ (𝝃)
ℝ𝑛
79
Remark 2.5 We observe that the Fourier-Laplace convolution is a special case of the 𝜑 convolution.
By Lemma 2.1, we have
Lemma 2.6 Let 𝜌1 and 𝜌2 be two positive functions on 𝐷 such that the convolution 𝜌1 ∗𝜑 𝜌2
exists. For two positive functions 𝐹1 and 𝐹2 satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞,
𝑗 = 1, 2,
(2.11)
on the set 𝐷, and for 𝑝 > 1,
𝐴𝑛𝑝
𝑝,𝑞
(
𝑚1 𝑚2
𝑀1 𝑀2
)
[((𝐹1 𝜌1 ) ∗𝜑 (𝐹2 𝜌2 ))(𝜼)]𝑝
𝑝−1
≥ [(𝜌1 ∗𝜑 𝜌2 )(𝜼)]
((𝐹1𝑝 𝜌1 )
(2.12)
∗𝜑 (𝐹2𝑝 𝜌2 ))(𝜼)
for all 𝜼 ∈ 𝐷.
¿From the 𝜑− convolution and the 𝜓− convolution, we get the following definition
Definition 2.7 ([7]) Under suitable hypotheses for the 𝜑− convolution 𝐹1 ∗𝜑 𝐹2 and the 𝜓−
convolution 𝐹1 ∗𝜓 𝐹2 , the (𝜑 + 𝜓)− convolution, denoted by 𝐹1 ∗𝜑+𝜓 𝐹2 , is defined by
∫
𝐹1 (𝝉 ) [𝐹2 (𝜑(𝝉 , 𝝃))∣𝜑𝝃 (𝝉 , 𝝃)∣ + 𝐹2 (𝜓(𝝉 , 𝝃))∣𝜓𝝃 (𝝉 , 𝝃)∣] 𝑑𝝉 .
(2.13)
(𝐹1 ∗𝜑+𝜓 𝐹2 )(𝝃) :=
𝐷
Example 2.8 The Fourier cosine convolution (see [13])
∫
1
𝐹1 ∗𝔉𝑐 𝐹2 (𝒙) := √ 𝑛
𝐹1 (𝒚)[𝐹2 (𝒙 + 𝒚) + 𝐹2 (∣𝒙 − 𝒚∣)]𝑑𝒚
2𝜋 ℝ𝑛+
(2.14)
is a special case of the (𝜑 + 𝜓)−convolution.
Then, we have
Lemma 2.9 Let 𝜌1 and 𝜌2 be two positive functions on 𝐷 such that the convolution 𝜌1 ∗𝜑+𝜓 𝜌2
exists. For two positive functions 𝐹1 and 𝐹2 satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞
on the set 𝐷, and for 𝑝 > 1,
(
)
(
)
𝑚1 𝑚2
𝑚2
𝑛𝑝
𝑝
𝐴𝑝,𝑞
𝐴𝑝,𝑞
[((𝐹1 𝜌1 ) ∗𝜑+𝜓 (𝐹2 𝜌2 ))(𝜼)]𝑝
𝑀1 𝑀2
𝑀2
𝑝−1
≥ [(𝜌1 ∗𝜑+𝜓 𝜌2 )(𝜼)]
((𝐹1𝑝 𝜌1 )
for all 𝜼 ∈ 𝐷.
80
∗𝜑+𝜓 (𝐹2𝑝 𝜌2 ))(𝜼)
(2.15)
(2.16)
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
81
Proof. Application of the reverse Hölder’s inequality gives
(
)
𝑚2
𝐴𝑝,𝑞
[𝐹2 (𝜑(𝝃, 𝜼))𝜌2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣ + 𝐹2 (𝜓(𝝃, 𝜼))𝜌2 (𝜓(𝝃, 𝜼))∣𝜓𝜼 (𝝃, 𝜼)∣]
𝑀2
1
≥ [𝐹2𝑝 (𝜑(𝝃, 𝜼))𝜌2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣ + 𝐹2𝑝 (𝜓(𝝃, 𝜼))𝜌2 (𝜓(𝝃, 𝜼))∣𝜓𝜼 (𝝃, 𝜼)∣] 𝑝
× [𝜌2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣ + 𝜌2 (𝜓(𝝃, 𝜼))∣𝜓𝜼 (𝝃, 𝜼)∣]
𝑝−1
𝑝
.
Therefore,
)
(
)
𝑚1 𝑚2
𝑚2
𝐴𝑝,𝑞
[((𝐹1 𝜌1 ) ∗𝜑+𝜓 (𝐹2 𝜌2 ))(𝜼)]
𝑀1 𝑀2
𝑀2
)∫
(
𝑚1 𝑚2
≥ 𝐴𝑛𝑝,𝑞
𝐹1 (𝝃)𝜌1 (𝝃)
𝑀1 𝑀2
𝐷
𝐴𝑛𝑝,𝑞
(
1
× [𝐹2𝑝 (𝜑(𝝃, 𝜼))𝜌2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣ + 𝐹2𝑝 (𝜓(𝝃, 𝜼))𝜌2 (𝜓(𝝃, 𝜼))∣𝜓𝜼 (𝝃, 𝜼)∣] 𝑝
× [𝜌2 (𝜑(𝝃, 𝜼))∣𝜑𝜼 (𝝃, 𝜼)∣ + 𝜌2 (𝜓(𝝃, 𝜼))∣𝜓𝜼 (𝝃, 𝜼)∣]
𝑝−1
𝑝
𝑑𝝃,
which is, by Lemma 2.1,
≥ [(𝜌1 ∗𝜑+𝜓 𝜌2 )(𝜼)]
𝑝−1
𝑝
1
[((𝐹1𝑝 𝜌1 ) ∗𝜑+𝜓 (𝐹2𝑝 𝜌2 ))(𝜼)] 𝑝 .
Thus, the proof is complete. □
Definition 2.10 ([7]) For the 𝜑− convolution, define the 𝜑− convolution product
by
[𝑚−1
]
𝑚
∏
∏
∗𝜑 𝐹𝑗 (𝝃 𝑚 ) =
∗𝜑 𝐹𝑗 ∗𝜑 𝐹𝑚 (𝝃 𝑚 ).
𝑗=1
3
∏𝑚
𝑗=1
∗𝜑 𝐹𝑗
(2.17)
𝑗=1
Reverse weighted 𝐿𝑝 inequalities in the iterated convolutions
Our main inequality is the following:
Theorem 3.1 Let functions 𝜌𝑗 , 𝑗 = 1, 2, ..., 𝑚, be positive on 𝐷 such that the convolution
∏𝑚
𝑗=1 ∗𝜑 𝜌𝑗 . For 𝑚 positive functions 𝐹𝑗 satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞
on the set 𝐷(𝑗 = 1, 2, ..., 𝑚), and for 𝑝 > 1,
(
)( 𝑚
) 𝑝1 −1 ∏
𝑚
∏
∗𝜑 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
∗𝜑 𝜌𝑗 (𝝃 𝑚 )
𝑗=1
𝑗=1
𝐿𝑝 (𝐷)
{
( 𝑖
)} 𝑚
𝑚
∏
∏ 𝑚𝑗
∏
≥
𝐴−𝑛
∥𝐹𝑗 ∥𝐿𝑝 (𝐷,𝜌𝑗 ) .
𝑝,𝑞
𝑀𝑗
𝑖=2
𝑗=1
𝑗=1
81
(3.1)
(3.2)
Inequality (3.2) and others should be understood in the sense that if the left hand side is
finite, then so is the right hand side, and in this case the inequality holds.
Corollary 3.2 For positive functions 𝜌𝑗,𝑟 (𝑗 = 1, 2, ..., 𝑚; 𝑟 = 1, 2, ..., 𝑠) on 𝐷 such that the
∏
convolution 𝑚
𝑗=1 ∗𝔉 𝜌𝑗,𝑟 exists and for positive functions 𝐹𝑗,𝑟 satisfying
1
1
𝑝
𝑝
0 < 𝑚𝑗,𝑟
≤ 𝐹𝑗,𝑟 ≤ 𝑀𝑗,𝑟
<∞
(3.3)
on the set 𝐷(𝑗 = 1, 2, ..., 𝑚; 𝑟 = 1, 2, ..., 𝑠), and for 𝑝 > 1, we have
(
)( 𝑠 𝑚
) 𝑝1 −1 ∑
𝑝
𝑚
∑∏
𝑠 ∏
∗
(𝐹
𝜌
)
∗
𝜌
𝜑
𝑗,𝑟
𝑗,𝑟
𝜑
𝑗,𝑟
𝑟=1 𝑗=1
𝑟=1 𝑗=1
𝐿𝑝 (𝐷)
{𝑚
) 𝑚
}
( 𝑖
)∑
(
𝑠
∏
∏ 𝑚𝑗,𝑟 ∏
𝑚0
𝑝
−𝑝
−𝑛𝑝
≥ 𝐴𝑝,𝑞
𝐴
∥𝐹𝑗,𝑟 ∥𝐿𝑝 (𝐷,𝜌𝑗,𝑟 ) ,
𝑀0 𝑟=1 𝑖=2 𝑝,𝑞 𝑗=1 𝑀𝑗,𝑟 𝑗=1
(3.4)
where
𝑚0 = min
𝑟
{𝑚
∏
}
𝑚𝑗,𝑟
,
𝑀0 = max
{𝑚
∏
𝑟
𝑗=1
}
𝑀𝑗,𝑟
.
𝑗=1
Proof of Theorem 3.1 We first proof the following inequality
(𝑚
∏
)𝑝 ( 𝑚
)1−𝑝
∏
∗𝜑 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
∗𝜑 𝜌𝑗 (𝝃 𝑚 )
𝑗=1
≥
𝑗=1
𝑚
∏
{
(
𝐴−𝑛𝑝
𝑝,𝑞
𝑖=2
𝑖
∏
𝑚𝑗
𝑀𝑗
𝑗=1
)}
𝑚
∏
(3.5)
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 ),
𝝃 𝑚 ∈ 𝐷.
𝑗=1
We use induction on 𝑚. When 𝑚 = 2, the inequality (3.5) is reduced to Lemma 2.1. Now
suppose (3.5) holds for some integer 𝑚 ≥ 2. We claim that it also holds for 𝑚 + 1. For all
𝝃 𝑚+1 , put
𝑓𝝃𝑚+1 (𝝃 𝑚 ) =
{𝑚
∏
} 𝑝−1
𝑝
∗𝜑 𝜌𝑗 (𝝃 𝑚 )𝜌𝑚+1 (𝜑(𝝃 𝑚 , 𝝃 𝑚+1 ))𝜑𝝃𝑚+1 (𝝃 𝑚 , 𝝃 𝑚+1 )
𝑗=1
and
𝑔𝝃𝑚+1 (𝝃 𝑚 ) =
{𝑚
∏
} 𝑝1
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 )𝜌𝑚+1 (𝜑(𝝃 𝑚 , 𝝃 𝑚+1 ))𝜑𝝃𝑚+1 (𝝃 𝑚 , 𝝃 𝑚+1 )
𝑗=1
× 𝐹𝑚+1 (𝜑(𝝃 𝑚 , 𝝃 𝑚+1 )).
82
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
By induction hypothesis, we arrive at
( 𝑖
)} 𝑚+1
{
𝑚
∏
∏ 𝑚𝑗
∏
𝐴𝑛𝑝,𝑞
∗𝜑 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚+1 )
𝑀𝑗
𝑖=2
𝑗=1
𝑗=1
)} ( 𝑚
)
{
( 𝑖
𝑚
∏
∏
∏ 𝑚𝑗
∗𝜑 (𝐹𝑗 𝜌𝑗 ) ∗𝜑 (𝐹𝑚+1 𝜌𝑚+1 ) (𝝃 𝑚+1 )
=
𝐴𝑛𝑝,𝑞
𝑀𝑗
𝑗=1
𝑖=2
𝑗=1
∫
𝑓𝝃𝑚+1 (𝝃 𝑚 )𝑔𝝃𝑚+1 (𝝃 𝑚 )𝑑𝝃 𝑚 .
≥
𝐷
The condition (3.1) implies
𝑚+1
∏
𝑚𝑗 ≤
𝑗=1
𝑔𝝃𝑝𝑚+1 (𝝃 𝑚 )
𝑓𝝃𝑞𝑚+1 (𝝃 𝑚 )
≤
𝑚+1
∏
𝑀𝑗 ,
𝝃 𝑚 ∈ 𝐷.
𝑗=1
Hence, one can apply the reverse Hölder inequality to 𝑓𝝃𝑚+1 (𝝃 𝑚 ) and 𝑔𝝃𝑚+1 (𝝃 𝑚 ) to obtain
}𝑝
)
{
(𝑚+1
∏ 𝑚𝑗 ∫
𝑓𝝃𝑚+1 (𝝃 𝑚 )𝑔𝝃𝑚+1 (𝝃 𝑚 )𝑑𝝃 𝑚
𝐴𝑛𝑝,𝑞
𝑀
𝑗
𝐷
𝑗=1
}𝑝−1 ∫
{∫
𝑞
[𝑔𝝃𝑚+1 (𝝃 𝑚 )]𝑝 𝑑𝝃 𝑚
[𝑓𝝃𝑚+1 (𝝃 𝑚 )] 𝑑𝝃 𝑚
≥
𝐷
𝐷
=
{𝑚+1
∏
∗𝜑 𝜌𝑗 (𝝃 𝑚+1 )
}𝑝−1 𝑚+1
∏
𝑗=1
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚+1 )
𝑗=1
and so the assertion follows.
Now, by taking integration of both sides of (3.5) with respect to 𝝃 𝑚 on 𝐷 we obtain the
inequality
)𝑝 ( 𝑚
)1−𝑝
∫ (∏
𝑚
∏
∗𝜑 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
∗𝜑 𝜌𝑗 (𝝃 𝑚 )
𝑑𝝃 𝑚
𝐷
≥
𝑗=1
𝑚
∏
𝑖=2
𝑗=1
{
( 𝑖
)} ∫ 𝑚
∏ 𝑚𝑗
∏
𝐴−𝑛𝑝
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 )𝑑𝝃 𝑚 .
𝑝,𝑞
𝑀𝑗
𝐷 𝑗=1
𝑗=1
Here we have, by definition,
∫ ∏
𝑚
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 )𝑑𝝃 𝑚
𝐷 𝑗=1
∫
=
∫
𝑑𝝃 𝑚
𝐷
×
𝐷𝝃𝜑
𝑚−1
[𝑚−1
∏
]
∗𝜑 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚−1 )
𝑗=1
𝐹𝑚𝑝 (𝜑(𝝃 𝑚−1 , 𝝃 𝑚 ))𝜌𝑚 (𝜑(𝝃 𝑚−1 , 𝝃 𝑚 ))∣𝜑𝝃𝑚 (𝝃 𝑚−1 , 𝝃 𝑚 )∣𝑑𝝃 𝑚−1 ,
which is, by the Fubini’s theorem and the change of variables 𝒙𝑚 = 𝜑(𝝃 𝑚−1 , 𝝃 𝑚 ),
]
∫ [𝑚−1
∫
∏
𝑝
=
∗𝜑 (∣𝐹𝑗 ∣ 𝜌𝑗 )(𝝃 𝑚−1 ) 𝑑𝝃 𝑚−1
∣𝐹𝑚 (𝒙𝑚 )∣𝑝 𝜌𝑚 (𝒙𝑚 )𝑑𝒙𝑚 .
𝐷
𝐷
𝑗=1
83
83
Therefore,
∫ (∏
𝑚
𝐷
≥
)𝑝 (
∗𝜑 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
𝑗=1
𝑚
∏
𝑚
∏
)1−𝑝
∗𝜑 𝜌𝑗 (𝝃 𝑚 )
𝑑𝝃 𝑚
𝑗=1
(
{
𝐴−𝑛𝑝
𝑝,𝑞
𝑖=2
𝑖
∏
𝑚𝑗
𝑀𝑗
𝑗=1
)}
(3.6)
𝑚 ∫
∏
𝐹𝑗𝑝 (𝒙)𝜌𝑗 (𝒙)𝑑𝒙.
𝐷
𝑗=1
Raising both sides of the inequality (3.6) to power 1/𝑝 yields the inequality (3.1). □
Proof of Corollary 3.2 For every 𝝃 𝑚 ∈ 𝐷, take
𝑎𝑟 =
(𝑚
∏
)( 𝑚
)−1/𝑞
∏
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑗=1
𝑗=1
and
𝑏𝑟 =
(𝑚
∏
)1/𝑞
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
,
𝑟 = 1, 2, ..., 𝑠.
𝑗=1
The condition (3.3) implies
𝑚0 ≤
𝑚
∏
𝑚
𝑚𝑗,𝑟
𝑗=1
𝑎𝑝 ∏
𝑀𝑗,𝑟 ≤ 𝑀0 ,
≤ 𝑞𝑟 ≤
𝑏𝑟
𝑗=1
𝑟 = 1, 2, ..., 𝑠.
Hence, we obtain
(
)( 𝑠
)𝑝−1
)𝑝
(
) (∑
𝑠
∑
𝑚
0
𝑎𝑝𝑟
𝑏𝑞𝑟
≤ 𝐴𝑝𝑝,𝑞
𝑎𝑟 𝑏 𝑟
𝑀0
𝑟=1
𝑟=1
𝑟=1
𝑠
∑
or, equivalent
(𝑚
𝑠
∑
∏
𝑟=1
)𝑝 (
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
𝑗=1
≤ 𝐴𝑝𝑝,𝑞
(
𝑚
∏
)1−𝑝 { 𝑠 𝑚
}𝑝−1
∑∏
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑟=1 𝑗=1
𝑗=1
𝑚0
𝑀0
) (∑
𝑠 ∏
𝑚
)𝑝
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
.
𝑟=1 𝑗=1
Therefore,
(
𝑠 ∏
𝑚
∑
)𝑝 (
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
𝑟=1 𝑗=1
≥ 𝐴−𝑝
𝑝,𝑞
(
𝑠 ∏
𝑚
∑
)1−𝑝
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑟=1 𝑗=1
𝑚0
𝑀0
(𝑚
)∑
𝑠
∏
𝑟=1
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
𝑗=1
)𝑝 ( 𝑚
∏
𝑗=1
84
(3.7)
)1−𝑝
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
.
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
85
Now, by taking integration of both sides of (3.7) with respect to 𝝃 𝑚 on 𝐷 we obtain
)𝑝 ( 𝑠 𝑚
)1−𝑝
∫ (∑
𝑠 ∏
𝑚
∑∏
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑑𝝃 𝑚
𝐷
𝑟=1 𝑗=1
≥ 𝐴−𝑝
𝑝,𝑞
= 𝐴−𝑝
𝑝,𝑞
(
𝑀0
(
𝑟=1 𝑗=1
(𝑚
)∫ ∑
𝑠
∏
𝑚0
𝐷 𝑟=1
)𝑝 (
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
𝑚
∏
𝑗=1
𝑗=1
) 𝑠 ∫ (∏
𝑚
𝑚0 ∑
)𝑝 ( 𝑚
∏
𝑀0
𝑟=1
𝐷
∗𝜑 (𝐹𝑗,𝑟 𝜌𝑗,𝑟 )(𝝃 𝑚 )
𝑗=1
)1−𝑝
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑑𝝃 𝑚
)1−𝑝
∗𝜑 𝜌𝑗,𝑟 (𝝃 𝑚 )
𝑑𝝃 𝑚 ,
𝑗=1
which is, by Theorem 3.1,
{𝑚
( 𝑖
) 𝑚
}
)∑
(
𝑠
∏
∏
∏
𝑚
𝑚
0
𝑗,𝑟
𝐴−𝑝𝑛
∥𝐹𝑗,𝑟 ∥𝑝𝐿𝑝 (𝐷,𝜌𝑗,𝑟 ) ,
≥ 𝐴−𝑝
𝑝,𝑞
𝑀0 𝑟=1 𝑖=2 𝑝,𝑞 𝑗=1 𝑀𝑗,𝑟 𝑗=1
that completes the proof of (3.4). □
¿From Theorem 3.1, we obtain an inequality for the iterated Fourier-Laplace convolution
here.
Corollary 3.3 Let 𝐹𝑗 (𝒙, 𝑡), 𝑗 = 1, 2, ..., 𝑚, be positive functions satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞
(3.8)
on the set ℝ𝑛 × ℝ+ . Then for any positive functions 𝜌𝑗 on ℝ𝑛 × ℝ+ such that there exists
∏𝑚
∏𝑚
𝑗=1 ∗𝔉,𝔏 , we have the inequality
𝑗=1 ∗𝔉,𝔏 , and for the iterated Fourier Laplace convolution
(
)( 𝑚
) 𝑝1 −1 𝑚
∏
∏
∗
(𝐹
𝜌
)
∗
𝜌
𝔉,𝔏
𝑗
𝑗
𝔉,𝔏
𝑗
𝑗=1
𝑗=1
𝐿𝑝 (ℝ𝑛 ×ℝ+ )
(3.9)
{
(
)}
−(𝑛+1) 𝑚
𝑚
𝑖
∏
∏
∏
𝑚𝑗
≥
𝐴𝑝,𝑞
∥𝐹𝑗 ∥𝐿𝑝 (ℝ𝑛 ×ℝ+ ,𝜌𝑗 ) .
𝑀
𝑗
𝑖=2
𝑗=1
𝑗=1
Remark 3.4 In formula 3.9, when 𝑚 = 2 replacing 𝜌2 by 1, and 𝐹2 (𝒙 − 𝝃, 𝑡 − 𝜏 ) by 𝐺(𝒙 −
𝝃, 𝑡 − 𝜏 ), and integrating with respect to 𝒙 from 𝒂 to 𝒃 and respect to 𝑡 from 0 to 𝑇 (> 0), we
arrive at the following inequality
(
)𝑝
∫ 𝑇 ∫ 𝒃 ∫ 𝑡 ∫ 𝑛 𝐹 (𝝃, 𝜏 )𝜌(𝝃, 𝜏 )𝐺(𝒙 − 𝝃, 𝑡 − 𝜏 )𝑑𝝃𝑑𝜏
0 ℝ
𝑑𝒙
𝑑𝑡
(∫ ∫
)𝑝−1
𝑡
0
𝒂
𝜌(𝝃, 𝜏 )𝑑𝝃𝑑𝜏
(3.10)
0 ℝ𝑛
∫
∫ 𝑇 −𝜏 ∫ 𝒃−𝝃
{
( 𝑚 )}−(𝑛+1)𝑝 ∫ 𝑇
≥ 𝐴𝑝,𝑞
𝑑𝜏
𝐹 𝑝 (𝝃, 𝜏 )𝜌(𝝃, 𝜏 )𝑑𝝃
𝑑𝑡
𝐺𝑝 (𝒙, 𝑡)𝑑𝒙,
𝑀
0
ℝ𝑛
0
𝒂−𝝃
if positive functions 𝜌, 𝐹 and 𝐺 satisfy
1
1
0 < 𝑚 𝑝 ≤ 𝐹 (𝝃, 𝑡)𝐺(𝒙 − 𝝃, 𝑡 − 𝜏 ) < 𝑀 𝑝 < ∞
for all (𝒙, 𝑡) ∈ [𝒂, 𝒃] × [0, 𝑇 ], (𝝃, 𝜏 ) ∈ ℝ𝑛 × [0, 𝑇 ].
85
(3.11)
Theorem 3.5 Let 𝑝 ≥ 1, 0 < 𝜶 < 𝑻 1 , 0 < 𝑻 𝑖 (𝑖 = 2, ..., 𝑚), and 𝐹𝑗 ∈ 𝐿∞ (0, 𝑻 1 + ⋅ ⋅ ⋅ +
𝑻 𝑚 )(𝑗 = 1, 2, ..., 𝑚), satisfy
0 ≤ 𝐹𝑗 ≤ 𝑀 < ∞,
0<𝒕<
𝑚
∑
𝑻 𝑗.
(3.12)
𝑗=1
Then
𝑚(1−𝑝)
𝑝
𝑀
𝑚
∏
∥𝐹1 ∥𝐿𝑝 (𝜶,𝑻 1 )
∥𝐹𝑗 ∥𝐿𝑝 (0,𝑻 𝑗 )
𝑗=2
(∫
𝑻 1 +⋅⋅⋅+𝑻 𝑚
𝒕𝑚
∫
≤
𝒕2
∫
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑑𝒕𝑚
𝐹1 (𝒕1 )
𝜶
𝜶
𝜶
𝑚
∏
(3.13)
)1/𝑝
𝐹𝑗 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
.
𝑗=2
In particular; for 𝜶 = 0, we have
𝑀
𝑚(1−𝑝)
𝑝
𝑚
∏
(∫
∥𝐹𝑗 ∥𝐿𝑝 (0,𝑻 𝑗 ) ≤
𝑚
𝑻 1 +⋅⋅⋅+𝑻 𝑚 ∏
0
𝑗=1
𝒕𝑚
𝒕2
∫
𝐹1𝑝 (𝒕1 )
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝜶
𝜶
∫
𝒕𝑚
𝜶
≤𝑀
∫
(3.14)
∑𝑚
𝑗=1
𝑻 𝑗 , we have
𝐹𝑗𝑝 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
𝒕2
𝐹1𝑝−1 (𝒕1 )
𝜶
𝑚(𝑝−1)
.
𝑗=2
∫
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
=
𝑚
∏
∗𝔏 𝐹𝑗 (𝒕)𝑑𝒕
𝑗=1
Proof. Since 0 ≤ 𝐹𝑗 (𝑗 = 1, 2, ..., 𝑚) ≤ 𝑀 for 0 < 𝒕 <
∫
)1/𝑝
𝑚
∏
𝐹𝑗𝑝−1 (𝒕𝑗
− 𝒕𝑗−1 )𝐹1 (𝒕1 )
𝑗=2
𝒕𝑚
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑚
∏
𝐹1 (𝒕1 )
𝜶
𝜶
𝐹𝑗 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
𝑗=2
𝒕2
∫
𝑚
∏
𝐹𝑗 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1 .
𝑗=2
Hence
∫
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝒕𝑚
≤𝑀
𝜶
𝑚(𝑝−1)
∫
𝒕2
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑑𝒕𝑚
𝜶
∫
𝑚
∏
𝐹1𝑝 (𝒕1 )
𝜶
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝑗=2
𝒕𝑚
∫
𝒕2
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑑𝒕𝑚
𝜶
𝐹𝑗𝑝 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
𝜶
𝐹1 (𝒕1 )
𝜶
𝑚
∏
(3.15)
𝐹𝑗 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1 .
𝑗=2
On the other hand, let
⎧
⎨1 for 𝑥 ≥ 0
𝜃(𝑥) =
⎩0 for 𝑥 < 0,
and let
∫
𝒕𝑚−1
∫
𝒕2
𝑑𝒕𝑚−2 ⋅ ⋅ ⋅
𝐺(𝒕𝑚−1 ) =
𝜶
𝜶
86
𝐹1𝑝 (𝒕1 )
𝑚−1
∏
𝑗=2
𝐹𝑗𝑝 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1 ,
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
87
we have
∫
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝒕𝑚
∫
𝜶
𝑻 1 +⋅⋅⋅+𝑻 𝑚 ∫
𝜶
𝑻 1 +⋅⋅⋅+𝑻 𝑚
𝑚
∏
𝐹𝑗𝑝 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
𝑗=2
𝒕𝑚
=
∫
𝐹1𝑝 (𝒕1 )
𝜶
𝜶
∫
𝒕2
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑑𝒕𝑚
𝐺(𝒕𝑚−1 )𝐹𝑚𝑝 (𝒕𝑚 − 𝒕𝑚−1 )𝑑𝒕𝑚−1 𝑑𝒕𝑚
𝜶
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝐺(𝒕𝑚−1 )𝐹𝑚𝑝 (𝒕𝑚 − 𝒕𝑚−1 )𝜃(𝒕𝑚 − 𝒕𝑚−1 )𝑑𝒕𝑚−1 𝑑𝒕𝑚 ,
=
𝜶
𝜶
which is, by Fubini’s theorem and the change of variables 𝒙 = 𝒕𝑚 − 𝒕𝑚−1 ,
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝑻 1 +⋅⋅⋅+𝑻 𝑚 −𝒕𝑚−1
(∫
𝐹𝑚𝑝 (𝒙)𝑑𝒙
)
𝐺(𝒕𝑚−1 )𝑑𝒕𝑚−1
)
(∫ 𝑻 1 +⋅⋅⋅+𝑻 𝑚 −𝒕𝑚−1
𝑝
𝐹𝑚 (𝒙)𝑑𝒙 𝐺(𝒕𝑚−1 )𝑑𝒕𝑚−1
≥
0
𝜶
)
∫ 𝑻 1 +⋅⋅⋅+𝑻 𝑚−1 (∫ 𝑻 𝑚
𝑝
𝐹𝑚 (𝒙)𝑑𝒙 𝐺(𝒕𝑚−1 )𝑑𝒕𝑚−1
≥
0
𝜶
∫ 𝑻 1 +⋅⋅⋅+𝑻 𝑚−1
∫ 𝑻𝑚
=
𝐺(𝒕𝑚−1 )𝑑𝒕𝑚−1
𝐹𝑚𝑝 (𝒙)𝑑𝒙,
=
𝜶
∫ 𝑻 1 +⋅⋅⋅+𝑻 𝑚−1
0
𝜶
0
and so,
∫
𝑻 1 +⋅⋅⋅+𝑻 𝑚
∫
𝒕𝑚
∫
𝑑𝒕𝑚−1 ⋅ ⋅ ⋅
𝑑𝒕𝑚
𝜶
𝜶
∫
𝑻1
≥
𝜶
𝐹1𝑝 (𝒙)𝑑𝒙
𝒕2
𝜶
𝑚 ∫ 𝑻𝑗
∏
𝑗=2
𝐹1𝑝 (𝒕1 )
𝑚
∏
𝐹𝑗𝑝 (𝒕𝑗 − 𝒕𝑗−1 )𝑑𝒕1
𝑗=2
𝐹𝑗𝑝 (𝒙)𝑑𝒙.
0
Combining with (3.15), we have the desired inequality (3.13). □
We next obtain the inequalities for the (𝜑 + 𝜓)−convolution.
Theorem 3.6 Let functions 𝜌𝑗 , 𝑗 = 1, 2, ..., 𝑚, be positive on 𝐷 such that the convolution
∏𝑚
𝑗=1 ∗𝜑+𝜓 𝜌𝑗 exists. For positive functions 𝐹𝑗 satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞
on the set 𝐷, 𝑗 = 1, 2, ..., 𝑚, and for 𝑝 > 1,
(
)( 𝑚
) 𝑝1 −1 ∏
𝑚
∏
∗𝜑+𝜓 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
∗𝜑+𝜓 𝜌𝑗 (𝝃 𝑚 )
𝑗=1
𝑗=1
𝐿𝑝 (𝐷)
{
( 𝑖
)} 𝑚
(
)
𝑚
𝑚
∏ 𝑚𝑗
∏
𝑚−1 ∏
𝑚𝑗 ∏
≥2 𝑝
𝐴−𝑛
𝐴
∥𝐹𝑗 ∥𝐿𝑝 (𝐷,𝜌𝑗 ) .
𝑝,𝑞
𝑝,𝑞
𝑀𝑗
𝑀𝑗 𝑗=1
𝑖=2
𝑗=1
𝑗=2
87
(3.16)
(3.17)
Proof. From Lemma 2.9 and by induction on 𝑚, we have
(𝑚
)𝑝 ( 𝑚
)1−𝑝
∏
∏
∗𝜑+𝜓 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
∗𝜑+𝜓 𝜌𝑗 (𝝃 𝑚 )
𝑗=1
≥
𝑗=1
𝑚
∏
{
(
𝐴−𝑛𝑝
𝑝,𝑞
𝑖=2
𝑖
∏
𝑚𝑗
𝑀𝑗
𝑗=1
)}
𝑚
∏
𝐴𝑝𝑝,𝑞
𝑗=2
(
𝑚𝑗
𝑀𝑗
)∏
𝑚
∗𝜑+𝜓 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 ).
𝑗=1
Therefore,
∫ (∏
𝑚
𝐷
)𝑝 (
∗𝜑+𝜓 (𝐹𝑗 𝜌𝑗 )(𝝃 𝑚 )
𝑗=1
𝑚
∏
)1−𝑝
∗𝜑+𝜓 𝜌𝑗 (𝝃 𝑚 )
𝑑𝝃 𝑚
𝑗=1
)} 𝑚
(
)∫ ∏
𝑚
𝑖
∏
∏
𝑚𝑗
𝑚𝑗
𝑝
−𝑛𝑝
∗𝜑+𝜓 (𝐹𝑗𝑝 𝜌𝑗 )(𝝃 𝑚 )𝑑𝝃 𝑚
𝐴𝑝,𝑞
≥
𝐴𝑝,𝑞
𝑀𝑗
𝑀𝑗
𝐷 𝑗=1
𝑗=2
𝑖=2
𝑗=1
{
(
)}
) 𝑚 ∫
(
𝑚
𝑖
𝑚
∏
∏
∏
𝑚𝑗
𝑚𝑗 ∏
𝑚−1
−𝑛𝑝
𝑝
=2
𝐴𝑝,𝑞
𝐹𝑗𝑝 (𝒙)𝜌𝑗 (𝒙)𝑑𝒙.
𝐴𝑝,𝑞
𝑀
𝑀
𝑗
𝑗
𝑖=2
𝑗=1
𝑗=1 𝐷
𝑗=2
𝑚
∏
{
(
The Theorem is thus proved. □
¿From Theorem 3.6, we have the following inequalities:
Corollary 3.7 Let 𝐹𝑗 (𝑗 = 1, 2, ..., 𝑚) be positive functions satisfying
1
1
0 < 𝑚𝑗𝑝 ≤ 𝐹𝑗 ≤ 𝑀𝑗𝑝 < ∞
(3.18)
on the set ℝ𝑛+ . Then for any positive continuous functions 𝜌𝑗 on ℝ𝑛+ , and for the iterated
∏
Fourier cosin convolution 𝑚
𝑗=1 ∗𝔉𝑐 , we have the inequality
(
)( 𝑚
) 𝑝1 −1 𝑚
∏
∏
∗𝔉𝑐 (𝐹𝑗 𝜌𝑗 )
∗𝔉𝑐 𝜌𝑗
𝑗=1
𝑗=1
𝐿𝑝 (ℝ𝑛
+)
(3.19)
{
)}
(
] 𝑚−1
)∏
[
(
𝑚
𝑚
𝑚
𝑖
∏
∏
∏
𝑝
2
𝑚𝑗
𝑚𝑗
≥ √ 𝑛
𝐴−𝑛
𝐴𝑝,𝑞
∥𝐹𝑗 ∥𝐿𝑝 (ℝ𝑛+ ,𝜌𝑗 ) .
𝑝,𝑞
𝑀
𝑀
2𝜋
𝑗
𝑗
𝑖=2
𝑗=1
𝑗=2
𝑗=1
Corollary 3.8 Let functions 𝜌𝑗,𝑟 (𝑗 = 1, 2, ..., 𝑚; 𝑟 = 1, 2, ..., 𝑠) be positive on 𝐷 such that
∏
the convolution 𝑚
𝑗=1 ∗𝜑,𝜓 𝜌𝑗,𝑟 exists. For some positive functions 𝐹𝑗,𝑟 satisfying
1
1
𝑝
𝑝
≤ 𝐹𝑗,𝑟 ≤ 𝑀𝑗,𝑟
<∞
0 < 𝑚𝑗,𝑟
on the set 𝐷 (𝑗 = 1, 2, ..., 𝑚; 𝑟 = 1, 2, ..., 𝑠), and for 𝑝 > 1,
(
)( 𝑠 𝑚
) 𝑝1 −1 𝑝
∑
𝑠 ∏
𝑚
∑∏
∗
(𝐹
𝜌
)
∗
𝜌
𝜑,𝜓
𝑗,𝑟
𝑗,𝑟
𝜑,𝜓
𝑗,𝑟
𝑟=1 𝑗=1
𝑟=1 𝑗=1
𝐿𝑝 (𝐷)
(
)
𝑚0
≥ 2𝑚−1 𝐴−𝑝
𝑝,𝑞
𝑀0
{
( 𝑖
) 𝑚
}
(
) 𝑚
𝑠
𝑚
∑
∏
𝑚𝑗,𝑟 ∏ −𝑛𝑝 ∏ 𝑚𝑗,𝑟 ∏
×
𝐴𝑝,𝑞
𝐴𝑝,𝑞
∥𝐹𝑗,𝑟 ∥𝑝𝐿𝑝 (𝐷,𝜌𝑗,𝑟 ) ,
𝑀
𝑀
𝑗,𝑟
𝑗,𝑟
𝑟=1
𝑗=2
𝑖=2
𝑗=1
𝑗=1
88
(3.20)
(3.21)
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
where
𝑚0 = min
{𝑚
∏
𝑟
4
}
𝑚𝑗,𝑟
,
𝑀0 = max
𝑟
𝑗=1
{𝑚
∏
89
}
𝑀𝑗,𝑟
.
𝑗=1
Applications
4.1
The Bernoulli-Euler Beam Equation
We consider the vertical deflection 𝑢(𝑥) of an infinite beam on an elastic foundation under
the action of a prescribed vertical load 𝑊 (𝑥). The deflection 𝑢(𝑥) satisfies the ordinary
differential equation
𝑑4 𝑢
𝐸𝐼 4 + 𝜅𝑢 = 𝑊 (𝑥), −∞ < 𝑥 < ∞,
(4.1)
𝑑𝑥
where 𝐸𝐼 is the flexural rigidity and 𝜅 is the foundation modulus of the beam. We find
the solution assuming that 𝑊 (𝑥) has a compact support and 𝑢, 𝑢′ , 𝑢′′ , 𝑢′′′ all tend to zero as
∣𝑥∣ → ∞. Put
𝑊 (𝑥)
𝜅
, 𝐹 (𝑥)𝜌(𝑥) =
.
𝑎4 =
𝐸𝐼
𝐸𝐼
By using the Fourier transform, we obtain the solution ([2, pp. 63-64])
∫ ∞
1
𝐹 (𝜉)𝜌(𝜉)𝐺(𝑥 − 𝜉)𝑑𝜉,
(4.2)
𝑢(𝑥) = 3
2𝑎 −∞
where
(
)
(
)
𝑎
𝑎𝜉
𝜋
𝐺(𝜉) = exp − √ ∣𝜉∣ sin √ +
.
2
2 4
Let 𝑏, 𝑐 ∈ ℝ and
𝑥 ∈ [−𝑏, 𝑏],
(4.3)
√
𝜉 ∈ [−𝑐, 𝑐],
√
2
3 2
−
𝜋 < −𝑏 − 𝑐 < 𝑏 + 𝑐 < 𝜋
.
4𝑎
4𝑎
Clearly,
(
)
(
)
𝜋 𝑎(𝑏 + 𝑐)
𝑎(𝑥 − 𝜉) 𝜋
√
0 < 𝛼 := sin
− √
+
≤ sin
.
4
4
2
2
Since
}
{
}
{
𝑎∣𝑏 − 𝑐∣
𝑎(𝑏 + 𝑐)
≤ 𝐺(𝑥 − 𝜉) ≤ exp − √
𝛼 exp − √
2
2
we see that the condition
1
1
0 < 𝑚 𝑝 ≤ 𝐹 (𝜉)𝐺(𝑥 − 𝜉) ≤ 𝑀 𝑝 ,
(4.4)
holds if
{
}
{
}
1
1
1
𝑎(𝑏 + 𝑐)
𝑎∣𝑏
−
𝑐∣
√
√
0 < exp
𝑚 𝑝 ≤ 𝐹 (𝜉) ≤ 𝑀 𝑝 exp
.
𝛼
2
2
Thus, for −𝑏 ≤ 𝑑 < 𝑒 ≤ 𝑏, the formal solution 𝑢(𝑥) satisfies the inequality
{
}
∫ 𝑒
( 𝑚 )}−𝑝
𝑝𝑎(𝑏 + 𝑐) {
𝑝
𝑝
√
𝑢 (𝑥)𝑑𝑥 ≥ (𝑒 − 𝑑)𝛼 exp −
𝐴𝑝,𝑞
𝑀
2
𝑑
(∫ 𝑒
)𝑝−1 ∫ 𝑒
×
𝜌(𝑥)𝑑𝑥
𝐹 𝑝 (𝑥)𝜌(𝑥)𝑑𝑥.
𝑑
𝑑
89
(4.5)
(4.6)
4.2
The Helmholtz Equation
We consider the Dirichlet problem for the Helmholtz equation in a half space of ℝ3 , i.e. the
determination of the bounded solution of
(𝑥, 𝑦) ∈ ℝ2 ,
Δ3 𝑢(𝑥, 𝑦, 𝑡) + 𝑘𝑢(𝑥, 𝑦, 𝑡) = 0,
𝑡 ∈ ℝ+ ,
𝑘 ∈ ℝ+ ,
(4.7)
under the boundary value condition
𝐹 (𝑥, 𝑦)𝜌(𝑥, 𝑦) ∈ 𝐿1 (ℝ2 ).
𝑢(𝑥, 𝑦, 0) = 𝐹 (𝑥, 𝑦)𝜌(𝑥, 𝑦),
(4.8)
Its solution has the form ([1, pp. 75-76])
(
𝑢(𝑥, 𝑦, 𝑡) = 2𝑡
𝑘
2𝜋
) 32 ∫
∞
∫
∞
𝐹 (𝜉, 𝜏 )𝜌(𝜉, 𝜏 )
3
−∞ (𝑡2
+ ∣𝜉 − 𝑥∣2 + ∣𝜏 − 𝑦∣2 ) 2
)
( √
2
2
2
× 𝐾 3 𝑘 𝑡 + ∣𝜉 − 𝑥∣ + ∣𝜏 − 𝑦∣ 𝑑𝜉𝑑𝜏,
−∞
(4.9)
2
where 𝐾𝜈 (𝑥) denotes the McDonald function and for 𝜈 =
√
𝐾 3 (𝑥) =
2
𝜋
2𝑥
(
1
1+
𝑥
)
3
2
(see [9, Supp. 10]):
𝑒−𝑥 .
We rewrite (4.9) as
𝑡
𝑢(𝑥, 𝑦, 𝑡) =
2𝜋
∫
∞
∫
∞
𝐹 (𝜉, 𝜏 )𝜌(𝜉, 𝜏 )𝐺(𝜉 − 𝑥, 𝑦 − 𝜏 )𝑑𝜉𝑑𝜏,
−∞
where
𝐺(𝜉, 𝜏 ) =
(4.10)
−∞
1+𝑘
(𝑡2
√
𝑡2 + 𝜉 2 + 𝜏 2
+
𝜉2
+
9
𝑒−𝑘
√
𝑡2 +𝜉 2 +𝜏 2
.
𝜏 2) 4
Let 𝑎1 , 𝑎2 , 𝑏1 , 𝑏2 , 𝑐1 , 𝑐2 , 𝑑1 , 𝑑2 be some real numbers and
𝑥 ∈ [𝑎1 , 𝑎2 ],
𝜉 ∈ [𝑏1 , 𝑏2 ],
𝑦 ∈ [𝑐1 , 𝑐2 ],
𝜏 ∈ [𝑑1 , 𝑑2 ].
Denote
𝛼 = max{∣𝑎1 − 𝑏1 ∣, ∣𝑎1 − 𝑏2 ∣, ∣𝑎2 − 𝑏1 ∣, ∣𝑎2 − 𝑏2 ∣},
𝛽 = max{∣𝑐1 − 𝑑1 ∣, ∣𝑐1 − 𝑑2 ∣, ∣𝑐2 − 𝑑1 ∣, ∣𝑐2 − 𝑑2 ∣}.
We have
1 + 𝑘𝑡
0<
Thus
∫
𝑐2 −𝜏
√
𝑡2 +𝛼2 +𝛽 2
(𝑡 + 𝛼2 + 𝛽 2 ) 4
∫
𝑎2 −𝜉
𝑑𝑦
𝑐1 −𝜏
−𝑘
9 𝑒
𝑎1 −𝜉
𝐺𝑝 (𝑥, 𝑦)𝑑𝑥 ≥
≤ 𝐺(𝜉 − 𝑥, 𝑦 − 𝜏 ) ≤
(𝑐2 − 𝑐1 )(𝑎2 − 𝑎1 )(1 + 𝑘𝑡)𝑝
(𝑡 + 𝛼2 + 𝛽 2 )
90
9𝑝
4
𝑒−𝑝𝑘
1
9
𝑡2
.
√
𝑡2 +𝛼2 +𝛽 2
.
REVERSE CONVOLUTION INEQUALITIES AND THEIR APPLICATIONS
91
Hence, for a function 𝐹 satisfying
1 9
(𝑡 + 𝛼2 + 𝛽 2 ) 4 𝑘√𝑡2 +𝛼2 +𝛽 2 𝑝1
0<
𝑒
𝑚 ≤ 𝐹 (𝜉, 𝜏 ) ≤ 𝑀 𝑝 𝑡 2
1 + 𝑘𝑡
9
(4.11)
and for a positive continuous function 𝜌 on [𝑏1 , 𝑏2 ] × [𝑑1 , 𝑑2 ], we obtain
∫ 𝑐2 ∫ 𝑎2
𝑢𝑝 (𝑥, 𝑦, 𝑡)𝑑𝑥
𝑑𝑦
𝑐1
𝑎1
𝑡𝑝 (1 + 𝑘𝑡)𝑝
≥ (𝑐2 − 𝑐1 )(𝑎2 − 𝑎1 )
−𝑝𝑘
9𝑝 𝑒
(2𝜋)𝑝 (𝑡 + 𝛼2 + 𝛽 2 ) 4
)𝑝−1 ∫ 𝑑2 ∫
(∫ 𝑑2 ∫ 𝑏2
𝜌(𝜉, 𝜏 )𝑑𝜉
𝑑𝜏
×
𝑑𝜏
𝑑1
4.3
𝑏1
√
𝑡2 +𝛼2 +𝛽 2
{
( 𝑚 )}−2𝑝
𝐴𝑝,𝑞
𝑀
(4.12)
𝑏2
𝐹 𝑝 (𝜉, 𝜏 )𝜌(𝜉, 𝜏 )𝑑𝜉.
𝑏1
𝑑1
The Cauchy Problem for the Inhomogeneous Heat Equation
The equation of heat conduction with sources is given by
𝑢𝑡 (𝑡, 𝒙) − 𝑐2 Δ𝑛 𝑢(𝑡, 𝒙) = 𝐹 (𝑡, 𝒙)𝜌(𝑡, 𝒙),
(𝑡, 𝒙) ∈ ℝ+ × ℝ𝑛 ,
(4.13)
where 𝐹 𝜌 ∈ 𝐿1 for every 𝑡 ∈ ℝ+ , under the initial value condition
𝑢(0, 𝒙) = 0.
(4.14)
Its solution has the form (see [1, pp. 58-59])
1
𝑢(𝑡, 𝒙) =
(4𝜋𝑐2 )𝑛/2
𝑡
∫
∫
𝑑𝜏
0
ℝ𝑛
{
}
𝐹 (𝜏, 𝝃)𝜌(𝜏, 𝝃)
∣𝝃 − 𝒙∣2
exp − 2
𝑑𝝃.
(𝑡 − 𝜏 )𝑛/2
4𝑐 (𝑡 − 𝜏 )
Take
𝐺(𝝃, 𝜏 ) =
1
𝜏 𝑛/2
(4.15)
{
}
∣𝝃∣2
exp − 2
.
4𝑐 𝜏
Let
𝝃 ∈ [−𝒂, 𝒂],
Since
1
𝑛/2
𝑇2
𝝃 ∈ [−𝒃, 𝒃],
𝑡 ∈ [𝑇1 , 𝑇2 ],
𝑇1 > 0.
{
}
{
}
∣𝒂 + 𝒃∣2
1
∣𝒙 − 𝝃∣2
1
exp −
<
exp
−
<
,
𝑛/2
4𝑐2 𝑇1
(𝑡 − 𝜏 )𝑛/2
4𝑐2 𝜏
𝑇1
we see that the condition
1
1
0 < 𝑚 𝑝 < 𝐹 (𝝃, 𝜏 )𝐺(𝒙 − 𝝃, 𝑡 − 𝜏 ) < 𝑀 𝑝
holds if
0<
𝑛/2
𝑇2
{
exp
∣𝒂 + 𝒃∣2
4𝑐2 𝑇1
}
1
1
(4.16)
𝑛/2
𝑚 𝑝 < 𝐹 (𝝃, 𝜏 ) < 𝑀 𝑝 𝑇1 .
91
(4.17)
Thus, for −𝒂 < 𝒅, 𝒆 < 𝒂, the inequality (3.10) yields
∫ 𝑇2 ∫ 𝒆
𝑢𝑝 (𝒙, 𝑡)𝑑𝒙
𝑑𝑡
𝒅
𝑇1
}
{
( 𝑚 )}−(𝑛+1)𝑝
1
(𝒆 − 𝒅)
∣𝒂 + 𝒃∣2 {
≥
𝐴
exp
−𝑝
𝑝,𝑞
(4𝜋𝑐2 )𝑛𝑝/2 𝑇2𝑛𝑝/2
4𝑐2 𝑇1
𝑀
)𝑝−1 ∫ 𝑇2
(∫ 𝑇2 ∫ 𝒆
∫ 𝒆
𝐹 𝑝 (𝒙, 𝑡)𝜌(𝒙, 𝑡)𝑑𝒙,
(𝑇2 − 𝑡)𝑑𝑡
𝜌(𝒙, 𝑡)𝑑𝒙
×
𝑑𝑡
𝑇1
𝒅
𝑇1
(4.18)
𝒅
where 𝜌 is a positive continuous function on [𝒅, 𝒆] × [𝑇1 , 𝑇2 ], and 𝐹 satisfies (4.17).
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