Volume 2009, Article ID 192197,13pages doi:10.1155/2009/192197
Research Article
On a Multiple Hilbert-Type Integral
Operator and Applications
Qiliang Huang and Bicheng Yang
Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China
Correspondence should be addressed to Qiliang Huang,qlhuang@yeah.net
Received 19 August 2009; Accepted 9 December 2009
Academic Editor: Wing-Sum Cheung
Copyrightq2009 Q. Huang and B. Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
By using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral operator with the homogeneous kernel of−λ-degreeλ∈Rand its norm are considered. As for applications, two equivalent inequalities with the best constant factors, the reverses, and some particular norms are obtained.
1. Introduction
Ifp > 1,1/p1/q 1,f≥ 0 ∈ Lp0,∞, g≥ 0 ∈ Lq0,∞,f p
∞
0 fpxdx
1/p
>
0,gq > 0,then we have the following famous Hardy-Hilbert’s integral inequality and its equivalent formcf.1:
∞
0
∞
0
fxgy
xy dx dy < π
sinπ/pfpgq, 1.1
∞
0
∞
0 fx xydx
p
dy
1/p
< π
sinπ/pfp, 1.2
where the constant factorπ/sinπ/pis the best possible. Define the Hardy-Hilbert’s integral operator T : Lp0,∞ → Lp0,∞ as follows: for f ∈ Lp0,∞, Tfy : ∞
01/x yfxdxy ∈0,∞.Then in view of1.2, it follows thatTfp < π/sinπ/pfpand
T ≤π/sinπ/p.Since the constant factor in1.2is the best possible, we find thatcf.2
Inequalities1.1and1.2are important in analysis and its applicationscf.3. In 2002, reference4considered the property of Hardy-Hilbert’s integral operator and gave an improvement of 1.1 for p q 2. In 2004-2005, introducing another pair of conjugate exponentsr, sr >1,1/r1/s 1and an independent parameterλ >0,5,6gave two best extensions of1.1as follows:
∞
0
∞
0
fxgy
xλyλ dx dy <
π
λsinπ/rfp,φgq,ψ, 1.3
∞
0
∞
0
fxgy
xyλ dx dy < B
λ r,
λ s
fp,φgq,ψ, 1.4
where Bu, v is the Beta function φx xp1−λ/r−1, ψx xq1−λ/s−1,f
p,φ :
∞
0 φxfpxdx
1/p
> 0,gq,ψ > 0. In 2009, 7, Theorem 9.1.1 gave the following multiple Hilbert-type integral inequality: suppose that n ∈ N\ {1}, pi > 1,
n
i11/pi
1, λ > 0,thenkλx1, . . . , xn ≥ 0 is a measurable function of−λ-degree in Rn and for any
r1, . . . , rnri>1satisfiesni11/ri 1 and
kλ
Rn−1
kλu1, . . . , un−1,1
n−1
j1
ujλ/rj−1du1· · ·dun−1>0. 1.5
Ifφix xpi1−λ/ri−1, fi≥0∈Lφpii0,∞,fpi,φi >0i1, . . . , n,then we have the following
inequality:
Rn
kλx1, . . . , xn n
i1
fixidx1· · ·dxn< kλ n
i1
fi
pi,φi, 1.6
where the constant factor kλ is the best possible. For n 2, kλx, y 1/xλ yλ, and
1/xyλin1.6, we obtain1.3and1.4. Inequality1.6is some extensions of the results in6,8–11. In 2006, reference12also considered a multiple Hilbert-type integral operator with the homogeneous kernel of−n1-degree and its inequality with the norm, which is the best extension of1.2.
In this paper, by using the way of weight functions and the technic of real analysis, a new multiple Hilbert-type integral operator with the norm is considered, which is an extension of the result in12. As for applications, an extended multiple Hilbert-type integral inequality and the equivalent form, the reverses, and some particular norms are obtained.
2. Some Lemmas
Lemma 2.1. Ifn∈N\ {1}, pi∈R\ {0,1}, λi∈Ri1, . . . , n,ni11/pi1,then
A:
n
i1
⎡
⎣xλi−11−pi
i
n
j1j /i
xλj−1
j
⎤ ⎦
1/pi
Proof. We find that
A:
n
i1
⎡
⎣xiλi−11−pi1−λi
n
j1 xλj−1
j
⎤ ⎦
1/pi
n
i1
⎡ ⎣x1−λipi
i n
j1 xλj−1
j
⎤ ⎦
1/pi
n
i1 x1−λi
i
⎛
⎝n
j1 xλj−1
j
⎞ ⎠
n i11/pi
,
2.2
and then2.1is valid.
Definition 2.2. Ifn∈N,Rn
:{x1, . . . , xnxi >0 i 1, . . . , n}, λ∈R,andkλx1, . . . , xnis a
measurable function inRn such that for anyu >0 andx1, . . . , xn∈ Rn, kλux1, . . . , uxn
u−λk
λx1, . . . , xn,then callkλx1, . . . , xnthe homogeneous function of−λ-degree inRn.
Lemma 2.3. As for the assumption of Lemma2.1, ifni1λi λ, kλx1, . . . , xn≥0is a homogeneous
function of−λ-degree inRn
,
Hi:
Rn−1
kλu1, . . . , ui−1,1, ui1, . . . , un n
j1j /i
uλj−1
j du1· · ·dui−1dui1· · ·dun 2.3
i 1, . . . , n, and Hn kλ ∈ R,then each Hi Hn kλi 1, . . . , n,and for any
i1, . . . , n, it follows that
ωixi:xλii
Rn−1
kλx1, . . . , xn n
j1j /i
xλj−1
j dx1· · ·dxi−1dxi1· · ·dxnkλ. 2.4
Proof. Settinguj unvj j /i, nin the integralHi,we find that
Hi
Rn−1
kλ
v1, . . . , vi−1, u−n1, vi1, . . . , vn−1,1
n−1
j1j /i
vλj−1
j u−
1−λi
n dv1· · ·dvi−1dvi1· · ·dvn−1dun.
2.5
Settingvi u−n1in the above integral, we obtain Hi Hn.Settinguj xj/xi j /iin
2.4, we find thatωixi Hi Hn kλ.
Lemma 2.4. As for the assumption of Lemma2.3, setting
kλ1, . . . ,λn−1
:
Rn−1
kλu1, . . . , un−1,1
n−1
j1 uλj−1
then there existδ0 > 0 andI {λ1, . . . ,λn−1 | λi λiδi,|δi| ≤ δ0 i 1, . . . , n−1},such that for anyλ1, . . . ,λn−1 ∈ I, kλ1, . . . ,λn−1 ∈R,if and only ifkλ1, . . . ,λn−1is continuous at
λ1, . . . , λn−1.
Proof. The sufficiency property is obvious. We prove the necessary property of the condition by mathematical induction. Forn2, since
kλ1δ1
1
0
kλu1,1u1λ1δ1−1du1
∞
1
kλu1,1u1λ1δ1−1du1,
kλu1,1u1λ1δ1−1≤kλu1,1uλ11−δ0−1du1, u1∈0,1,
kλu1,1u1λ1δ1−1≤kλu1,1uλ11δ0−1du1, u1∈1,∞,
2.7
andkλ1−δ0 kλ1δ0<∞,then by Lebesgue control convergence theoremcf.13, it
follows thatkλ1 δ1 kλ1 o1δ1 → 0.Assuming that forn≥2, kλ1, . . . ,λn−1is
continuous atλ1, . . . , λn−1,then forn1,in view of the result forn2,we have that
lim
δn→0
kλ1δ1, . . . , λnδn
lim
δn→0
∞
0
⎛
⎝
Rn−1
kλu1, . . . , un,1 n−1
j1 uλjδj−1
j du1· · ·dun−1
⎞ ⎠uλnδn−1
n dun
∞
0
⎛
⎝
Rn−1
kλu1, . . . , un,1 n−1
j1 uλjδj−1
j du1· · ·dun−1
⎞ ⎠uλn−1
n dun
Rn−1
∞
0
kλu1, . . . , un,1uλnn−1dun
n−1
j1
uλjjδj−1du1· · ·dun−1,
2.8
then by the assumption forn,it follows that
lim
δn→0
kλ1δ1, . . . , λnδn kλ1, . . . , λn o1 δi−→0, i1, . . . , n−1. 2.9
By mathematical induction, we prove that forn ∈ N\ {1}, kλ1, . . . ,λn−1is continuous at
λ1, . . . , λn−1.
Lemma 2.5. As for the assumption of Lemma2.4, if0< ε <min1≤i≤n{|pi|}δ0, then forε → 0,
Iε:ε
∞
1 · · ·
∞
1
kλx1, . . . , xn n
j1
xλj−ε/pj−1
Proof. Settinguj xj/xn j 1, . . . , n−1,we find that
Iεε
∞
1 xn−1−ε
⎡ ⎣∞
x−1 n
· · ·
∞
x−1 n
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j du1· · ·dun−1
⎤
⎦dxn. 2.11
SettingDj:{u1, . . . , un−1|uj ∈0, x−n1, uk∈0,∞ k /j}and
Ajxn:
· · ·
Dj
kλu1, . . . , un−1,1
n−1
j1
uλjj−ε/pj−1du1· · ·dun−1, 2.12
then by2.11, it follows that
Iε≥
Rn−1
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j du1· · ·dun−1−ε n−1
j1
∞
1
xn−1Ajxndxn. 2.13
Without loses of generality, we estimate that∞1 x−1
n An−1xndxn O1.In fact, settingα >0
such that|ε/pn−1 α|< δ0,since−uαn−1lnun−1 → 0un−1 → 0,there existsM >0,such
that−uαn−1lnun−1≤Mun−1∈0,1,and then by Fubini theorem, it follows that
0≤
∞
1
x−n1An−1xndxn
∞
1 x−n1
⎡
⎣
Rn−2
x−1 n
0
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j dun−1du1· · ·dun−2
⎤ ⎦dxn
1
0
Rn−2
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j
u−1 n−1
1
x−n1dxn
du1· · ·dun−1
1
0
Rn−2
kλu1, . . . , un−1,1
n−1
j1
uλjj−ε/pj−1−lnun−1du1· · ·dun−1
≤M
1
0
Rn−2
kλu1, . . . , un−1,1
n−2
j1
uλj−ε/pj−1
j u
λn−1−ε/pn−1α−1
n−1 du1· · ·dun−1
≤M
Rn−1
kλu1, . . . , un−1,1
n−2
j1
uλj−ε/pj−1
j u
λn−1−ε/pn−1α−1
n−1 du1· · ·dun−1
M·k
λ1− ε p1
, . . . , λn−2− ε pn−2
, λn−1−
ε pn−1
α
<∞.
Hence by2.13, we have that
Iε≥
Rn−1
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j du1· · ·dun−1−o11. 2.15
By Lemma2.4, we find that
Iε≤ε
∞
1 x−1−ε
n
⎡ ⎣∞
0 · · ·
∞
0
kλu1, . . . , un−1,1
×n−1
j1
uλj−ε/pj−1
j du1· · ·dun−1
⎤ ⎦dxn
∞
0 · · ·
∞
0
kλu1, . . . , un−1,1
n−1
j1
uλj−ε/pj−1
j du1· · ·dun−1
k
λ1− ε p1
, . . . , λn−1− ε pn−1
kλo21,
2.16
Then by combination with2.15, we have2.10.
Lemma 2.6. Suppose that n ∈ N\ {1}, p1 ∈ R\ {1},
n
i11/pi 1,1/qn 1−1/pn,λ1, . . . , λn∈Rn,ni1λi λ,thenkλx1, . . . , xn≥0is a measurable function of−λ-degree inRnsuch that
kλ
Rn−1
kλu1, . . . , un−1,1
n−1
j1 uλj−1
j du1· · ·dun−1∈R. 2.17
Iffi≥0are measurable functions inRi1, . . . , n−1,then (1) forpi >1i1, . . . , n,
J:
∞
0 xqnλn−1
n
Rn−1
kλx1, . . . , xn n−1
i1
fixidx1· · ·dxn−1
qn dxn
1/qn
≤kλ n−1
i1
∞
0
xpi1−λi−1fpixdx
1/pi ,
2.18
Proof. 1Forpi>1 i1, . . . , n,by H ¨older’s inequalitycf.14 and2.4, it follows that
Rn−1
kλx1, . . . , xn n−1
i1
fixidx1· · ·dxn−1
qn ⎧ ⎪ ⎨ ⎪ ⎩
Rn−1
kλx1, . . . , xn n−1
i1
⎡
⎣xiλi−11−pi
n
j1j /i
xλj−1
j
⎤ ⎦
1/pi fixi
×
⎡
⎣xnλn−11−pn n−1
j1 xλj−1
j
⎤ ⎦
1/pn
dx1· · ·dxn−1
⎫ ⎪ ⎬ ⎪ ⎭ qn ≤
Rn−1
kλx1, . . . , xn n−1
i1
⎡
⎣xiλi−11−pi
n
j1j /i
xλjj−1
⎤ ⎦
qn/pi
×fqn
i xidx1· · ·dxn−1
×
⎧ ⎨ ⎩
Rn−1
kλx1, . . . , xnx
λn−11−pn
n
n−1
j1 xλj−1
j dx1· · ·dxn−1
⎫ ⎬ ⎭
qn−1
kλqn−1x1n−qnλn
Rn−1
kλx1, . . . , xn
×n−1
i1
⎡
⎣xiλi−11−pi
n
j1j /i
xλjj−1
⎤ ⎦
qn/pi fqn
i xidx1· · ·dxn−1,
2.19
J≤kλ1/pn
⎧ ⎪ ⎨ ⎪ ⎩ ∞ 0
Rn−1
kλx1, . . . , xn
×n−1
i1
⎡
⎣xiλi−11−pi
n
j1j /i
xλj−1
j
⎤ ⎦
qn/pi fqn
i xidx1· · ·dxn−1dxn
⎫ ⎪ ⎬ ⎪ ⎭
1/qn
kλ1/pn
⎧ ⎪ ⎨ ⎪ ⎩
Rn−1
∞
0
kλx1, . . . , xnxλnn−1dxn
×n−1
i1
⎡
⎣xiλi−11−pi
n−1
j1j /i
xλj−1
j
⎤ ⎦
qn/pi fqn
i xidx1· · ·dxn−1
⎫ ⎪ ⎬ ⎪ ⎭
1/qn .
Forn≥3,by H ¨older’s inequality again, it follows that
J≤kλ1/pn
⎧ ⎪ ⎨ ⎪ ⎩
n−1
i1
Rn−1
∞
0
kλx1, . . . , xnxnλn−1dxn
×xiλi−11−pi
n−1
j1j /i
xλj−1
j f pi
i xidx1· · ·dxn−1
⎤ ⎦
qn/pi⎫⎪⎬
⎪ ⎭
1/qn
kλ1/pn n−1
i1
⎧ ⎨ ⎩
∞
0
⎡
⎣
Rn−1
kλx1, . . . , xn
×xλi
i n
j1j /i
xλj−1
j dx1· · ·dxi−1dxi1· · ·dxn
⎤
⎦xpi1−λi−1
i f pi
i xidxi
⎫ ⎬ ⎭
1/pi
kλ1/pn n−1
i1
∞
0
ωixixipi1−λi−1fipixidxi
1/pi .
2.21
Then by2.4, we have2.18 note that forn2,we do not use H ¨older’s inequality again.
2For 0< p1 <1, pi <0i2, . . . , n,by the reverse H ¨older’s inequality and the same way,
we obtain the reverses of2.18.
3. Main Results and Applications
As for the assumption of Lemma2.6, settingφix: xpi1−λi−1x ∈0,∞;i 1, . . . , n,then
we find thatφ1/1−pn
n x xqnλn−1.Ifpi >1i1, . . . , n,then define the following real function
spaces:
Lpi
φi0,∞:
f;fp i,φi
∞
0
φix((fx((pidx
1/pi
<∞ i1, . . . , n,
n−1
i1 Lpi
φi0,∞:
)
f1, . . . , fn−1
;fi∈Lpφii0,∞, i1, . . . , n−1
*
,
3.1
and a multiple Hilbert-type integral operatorT : +ni−11Lpi
φi0,∞ → L
qn
φ1/1n −pn
as follows: for
f f1, . . . , fn−1∈
+n−1
i1L
pi
φi0,∞,
Tfxn:
Rn−1
kλx1, . . . , xn n−1
i1
Then by2.18, it follows thatTf ∈ Lqn
φ1/1n −pn
,T is bounded,Tfq
n,φ1/1n −pn ≤ kλ
+n−1
i1fipi,φi,
andT ≤kλ,where
T: sup
f∈+n−1
i1Lpiφi0,∞fi/θ,i1,...,n−1
Tf
qn,φ1/n1−pn
+n−1
i1fipi,φi
. 3.3
Define the formal inner product ofTf1, . . . , fn−1andfnas
Tf1, . . . , fn−1
, fn
:
Rn
kλx1, . . . , xn n
i1
fixidx1· · ·dxn. 3.4
Theorem 3.1. Suppose that n ∈ N\ {1}, p1 ∈ R \ {1},ni11/pi 1,1/qn 1−1/pn,
thenkλx1, . . . , xn≥0is a measurable function of−λ-degree inRn,and for anyλ1, . . . , λn∈Rn,
it satisfiesni1λi λand
kλ
Rn−1
kλu1, . . . , un−1,1
n−1
j1 uλj−1
j du1· · ·dun−1>0. 3.5
Iffi≥0∈Lφpii0,∞,fpi,φi >0i1, . . . , n,then (1) forpi >1 i1, . . . , n,Tkλand the
following equivalent inequalities are obtained:
T
f1, . . . , fn−1q n,φ
1/1−pn n < kλ
n−1
i1
fi
pi,φi, 3.6
Tf1, . . . , fn−1
, fn
< kλ n
i1
fi
pi,φi, 3.7
where the constant factorkλis the best possible; (2) for0 < p1 <1, pi < 0 i2, . . . , n,using the
formal symbols of the case inpi>1 i1, . . . , n,the equivalent reverses of 3.6and3.7with the
best constant factor are given.
Proof. 1Forpi>1i1, . . . , n,if2.18takes the form of equality, then forn≥3 in2.21,
there exist constantsCiandCk i /ksuch that they are not all zero and
Cix
λi−11−pi
i
n−1
j1j /i
xλj−1
j f pi
i xi
Ckx
λk−11−pk
k
n−1
j1j /k
xλj−1
j f pk
k xka.e.in R n
,
3.8
viz. Cixpii1−λifipixi Ckxkpk1−λkfkpkxk C a.e. in Rn. Assuming that Ci > 0, then
xpi1−λi−1
i f pi
consider2.19forfpk
k xk 1 in the above. Hence we have3.6. By H ¨older’s inequality, it
follows that
Tf, fn
∞
0
xλn−1/qn
n
Rn−1
kλx1, . . . , xn n−1
i1
fixidx1· · ·dxn−1
×x1/qn−λn
n fnxn
dxn≤T
f1, . . . , fn−1qn,φ1/1−pn n
fnpn,φn,
3.9
and then by3.6, we have3.7. Assuming that3.7is valid, setting
fnxn:xqnnλn−1
Rn−1
kλx1, . . . , xn n−1
i1
fixidx1· · ·dxn−1
qn−1
, 3.10
thenJ )∞0 xpn1−λn−1
n fnpnxndxn
*1/qn
.By2.18, it follows thatJ <∞.IfJ 0,then3.6is naturally valid. Assuming that 0< J <∞,by3.7, it follows that
∞
0
xpn1−λn−1
n f
pn
n xndxnJqn
Tf, fn
< kλ n
i1
fi
pi,φi,
∞
0
xpn1−λn−1
n fnpnxndxn
1/qn
J < kλ n−1
i1
fipi,φi,
3.11
and then3.6is valid, which is equivalent to3.7.
For ε > 0 small enough, setting fixas follows: fix 0, x ∈ 0,1; fix
xλi−ε/pi−1, x ∈ 1,∞ i 1, . . . , n, if there existsk ≤ k
λ, such that3.7is still valid as we
replacekλbyk,then in particular, by Lemma2.5, we have that
kλo1 Iεε
Tf1, . . . ,fn−1
,fn
< εk
n
i1
f
i pi,φi
k, 3.12
andkλ ≤ kε → 0.Hencek kλis the best value of3.7. We conform that the constant
factorkλin3.6is the best possible; otherwise, we can get a contradiction by3.9that the
constant factor in3.7is not the best possible. ThereforeTkλ.
2For 0 < p1 < 1, pi <0 i 2, . . . , n,by using the reverse H ¨older’s inequality and
Example 3.2. Forλ >0, λi λ/ri i1, . . . , n,ni11/ri 1, kλx1, . . . , xn 1/ni1xi
λ
,
we obtainkλ 1/Γλ
+n
i1Γλ/ri cf.7,9.1.19. By Theorem3.1, it follows thatT
kλ 1/Γλ
+n
i1Γλ/ri, and then by3.7, we find that
Rn
1
n
i1xi
λ n
i1
Fixidx1· · ·dxn
< 1
Γλ
n
i1
Γ
λ ri
∞
0
xpi1−λ/ri−1
i Fipixidxi
1/pi .
3.13
SettingFixi xβ/ni fixiandλiλ/ri−β/ni1, . . . , nin3.13, we obtainni1λi
λ−β, min1≤i≤n{λi}>−β/nand
Rn
n
,+n
i1xi
β
n
i1xi
λ n
i1
fixidx1· · ·dxn
< 1
Γλ
n
i1
Γ
λi
β n
∞
0
xpi1−λi−1
i fi pix
idxi
1/pi .
3.14
It is obvious that3.13and3.14are equivalent in which the constant factors are all the best possible. Hence forkλ−βx1, . . . , xn
n
,+n
i1xi
β
/ni1xi
λ
λ > 0,min1≤i≤n{λi} > −β/n,
we can show thatTkλ−β 1/Γλ+ni1Γλiβ/n.
Example 3.3. For λ > 0, λi λ/rii 1, . . . , n,
n
i11/ri 1, kλx1, . . . , xn
1/max1≤i≤n{xi}λ,we obtainkλ 1/λn−1+ni1ricf.7,9.1.24. By Theorem3.1, it follows
thatTkλ 1/λn−1+ni1ri, and then by3.7, we find that
Rn
1
max1≤i≤n{xi}λ n
i1
Fixidx1· · ·dxn
< 1 λn−1
n
i1 ri
∞
0
xpi1−λ/ri−1
i Fi pix
idxi
1/pi .
3.15
SettingFixi xβ/ni fixiandλiλ/ri−β/ni1, . . . , nin3.15, we obtain
n
i1λi
λ−β, min1≤i≤n{λi}>−β/nand
Rn
n
,+n
i1xi
β
max1≤i≤n{xi}λ n
i1
fixidx1· · ·dxn
< λ
n
i1
1
λiβ/n
∞
0
xpi1−λi−1
i fi pix
idxi
1/pi .
Hence forkλ−βx1, . . . , xn
n
,+n
i1xi
β
/max1≤i≤n{xi}λ λ > 0,min1≤i≤n{λi} > −β/n, we
can show thatTkλ−βλ+ni11/λiβ/n.
Example 3.4. For λ > 0, λi −λ/ri,
n
i11/ri 1 i 1, . . . , n, k−λx1, . . . , xn
min1≤i≤n{xi}λ,by mathematical induction, we can show that
k−λ
Rn−1
min{u1, . . . , un−1,1}λ
n−1
j1
u−jλ/rj−1du1· · ·dun−1
+n i1ri
λn−1 . 3.17
In fact, forn2,we obtain
k−λ
1
0 uλ/r2−1
1 du1
∞
1
u−λ/r1−1
1 du1
1
λr1r2. 3.18
Assuming that forn≥2 3.17is valid, then forn1, it follows that
k−λ
Rn−1
n
j2 u−λ/rj−1
j
-∞
0
min{u1, . . . , un,1}λu−1λ/r1−1du1
.
du2· · ·dun
Rn−1
n
j2 u−λ/rj−1
j
min{u2,...,un,1}
0
uλ1u−λ/r1−1 1 du1
∞
min{u2,...,un,1}
min{u2, . . . , un,1}λu−1λ/r1−1du1
du2· · ·dun
r12
λr1−1
Rn−1
min{u2, . . . , un,1}λ1−1/r1 n
j2
u−λ1−1/r1/1−1/r1rj−1
j du2· · ·dun
r12
λr1−1
1
λ1−1/r1n−1
n1
i2
1− 1
r1
ri 1
λn n1
i1 ri.
3.19
Then by mathematical induction,3.17is valid forn∈N\ {1}. By Theorem3.1, it follows thatTk−λ 1/λn−1
+n
i1ri, and by3.7, we find that
Rn
min
1≤i≤n{xi}
λ n
i1
Fixidx1· · ·dxn
< 1 λn−1
n
i1 ri
∞
0
xpi1λ/ri−1
i Fipixidxi
1/pi .
SettingFixi xi−β/nfixiand λi −λ/riβ/ni 1, . . . , nin 3.20, we obtain
n
i1λiβ−λ,max1≤i≤n{λi}< β/nand
Rn
min1≤i≤n{xi}λ
n
,+n
i1xi
β n
i1
fixidx1· · ·dxn
< λ
n
i1
1
β/n−λi
∞
0
xpi1−λi−1
i fi pix
idxi
1/pi .
3.21
Hence forkβ−λx1, . . . , xn min1≤i≤n{xi}λ/
n
,+n
i1xi
β
λ >0,max1≤i≤n{λi}< β/n, we can
show thatTkβ−λλ
+n
i11/β/n−λi.
Acknowledgments
This work is supported by the Emphases Natural Science Foundation of Guangdong Institution, Higher Learning, College and Universityno. 05Z026, and Guangdong Natural Science Foundationno. 7004344.
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