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A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor
Accepted Manuscript A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor
Bing He, Qiru Wang
PII: DOI: Reference:
S0022-247X(15)00562-4 http://dx.doi.org/10.1016/j.jmaa.2015.06.019 YJMAA 19572
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
11 February 2015
Please cite this article in press as: B. He, Q. Wang, A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor, J. Math. Anal. Appl. (2015), http://dx.doi.org/10.1016/j.jmaa.2015.06.019
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A Multiple Hilbert-Type Discrete Inequality with a New Kernel and Best Possible Constant Factor$ Bing Hea , Qiru Wangb,∗ a Department
of Mathematics, Guangdong University of Education, Guangzhou 510303, PR China b School of Mathematics and Computational Science, Sun Yat-sen University, Guangzhou 510275, PR China
Abstract By introducing a distinctive kernel and using the way of weight functions, a brand new multiple Hilbert-type discrete inequality with a best constant factor is presented. The operator expressions, equivalent forms, reverse inequalities and particular cases are discussed. Keywords: multiple Hilbert-type discrete inequality, weight function, kernel, operator expression, best constant factor 2010 MSC: 26D15, 47A07 1. Introduction p Suppose that p > 1, (1/p) + (1/q) = 1, am , bn ≥ 0, a = {am }∞ m=1 ∈ l , b = ∞ q p 1/p {bn }∞ > 0, ||b||q > 0, the well-known Hardyn=1 ∈ l , ||a||p = { m=1 am }
Hilbert’s discrete inequality is given by ([1, p.226]): ∞ ∞ π a m bn < ||a||p ||b||q , m + n sin(π/p) n=1 m=1 5
and an equivalent form is ∞ ∞ n=1
am m+n m=1
p 1/p <
π ||a||p , sin(π/p)
(1)
(2)
$ Supported
by the NNSF of China (Nos. 11301090 and 11271379). author Email addresses: [email protected] (Bing He), [email protected] (Qiru Wang)
∗ Corresponding
Preprint submitted to Journal of Mathematical Analysis and Applications
June 15, 2015
where the constant factor π/ sin(π/p) is the best possible. Define Hardy-Hilbert’s ∞ am discrete operator T : lp → lp by: (T a) (n) := m=1 m+n , where a = {am }∞ m=1 ∈ lp , n ∈ N. Then (2) can be written as ||T a||p < [π/ sin(π/p)] ||a||p and ||T || ≤ π/ sin(π/p). Since the constant factor in (2) is the best possible, we have 10
||T || = π/ sin(π/p). Inequalities (1) and (2) have been studied extensively, which are important in analysis and applications ([2]-[7]). Hilbert-type inequalities may be classified into several types (integral, discrete and half-discrete, etc.). On discrete inequalities, by introducing an independent parameter λ (2 − min{p, q} < λ ≤ 2),
15
Yang [9] gave an extension of (1) in 2002 as follows: ∞ ∞ p+λ−2 q+λ−2 a m bn , ||a||p,ξ ||b||q,ξ , 0, ||b||q,ξ > 0. In the same year, another similar form was established by Yang [10]:
∞ ∞
π a m bn ||a||p,φ ||b||q,ψ , < μ + nμ m μ sin(π/p) n=1 m=1 20
(4)
where 0 < μ ≤ min{p, q}, φ(n) = n(p−1)(1−μ) , ψ(n) = n(q−1)(1−μ) , ||a||p,φ := ∞ { n=1 φ(n)apn }1/p > 0, ||b||q,ψ > 0. It is significant to generalize Hilbert-type inequalities into multiple versions. In recent years, some results have been obtained ([11]-[18]). In 2009, the following multiple Hilbert-type discrete inequality had been studied by Yang [3, Example 9.1.1]
25
n n If n ∈ N\{1}, pi , ri > 1, i=1 p1i = i=1 r1i = 1, 0 < λ ≤ min1≤i≤n {ri },
p i ∞ p (1−λ/ri )−1 (i) 0 < mi =1 mi i < ∞ (i = 1, · · · , n), then am i ∞
···
mn =1
<
∞ m1
n
1 Γ Γ(λ) i=1
where the constant factor
( =1
1 Γ(λ)
λ ri
n
n
1
i=1
mi )λ
∞ mi =1
n
i=1 Γ
i=1
a(i) mi
pi (1− rλ )−1 i mi
a(i) mi
pi
1/pi ,
(5)
λ ri
is the best possible. It is not an easy
job to find new forms of such inequalities. Exploiting these multiple inequalities 2
will greatly increase the scope of our work. This paper will include a completely 30
new multiple discrete Hilbert-type inequality with an interesting kernel. The setup of this paper is as follows. In Section 2, a number of basic lemmas are obtained. To calculate the constant factor, we herein utilize mathematical induction to do an excellent computing. To prove the best of the constant factor, we apply some analytical skills to do a good estimate for some formula. Section
35
3 presents our main results, the techniques that will be used in the proof are mainly based on classical real analysis, especially on the well-known H¨ older’s inequality. In addition, the equivalent forms, operator expressions as well as some reverse inequalities are also considered.
2. Basic lemmas 40
Lemma 1. ([3], (9.1.1)). If n ∈ N\{1} = {2, 3, 4, 5, · · ·}, pi = 0, 1(i = n 1, 2, · · · , n), i=1 p1i = 1, then n i=1
⎡ ⎣mpi i −1
n
⎤ p1
i
⎦ m−1 j
= 1.
(6)
j=1(j=i)
Lemma 2. ([16]). If c > 0, n ∈ N, then
c 0
n n! i (ln c) . (ln x) dx = c (−1)n−i · i! i=0 n
(7)
Lemma 3. ([16]). Suppose that s, t ∈ (0, +∞), a = min{s, 1}, b = max{s, 1}, k ∈ N ∪ {0}, then k min{s, t, 1} min{s, t, 1} ln t−1 dt max{s, t, 1} max{s, t, 1} 0 k
a k+1
a i a k−i k! +2 (−1) · . − ln ln b b i! b i=0
= 45
∞
Lemma 4. For all n ∈ N\{1}, we have the following identity λn (u1 )
∞
:= 0
···
∞ 0
n min1≤j≤n {uj , 1} −1 u du2 · · · dun max1≤j≤n {uj , 1} j=2 j
3
(8)
=
n−1 (−1)n−1 min{u1 , 1} min{u1 , 1} 1+ ln n! max{u1 , 1} n! max{u1 , 1} i+1 n−3 (−1)i+1 1 1 min{u1 , 1} − ln + . i! i+1 n max{u1 , 1} i=0
(9)
Proof. We will prove (9) by induction. If n = 2, then min{u1 , 1} min{u1 , 1} − ln +2 λ2 (u1 ) = max{u1 , 1} max{u1 , 1} holds by applying Lemma 3 with k = 0. Now we assume that (9) holds for n = k ∈ N\{1}. Then, by (8), we obtain ∞ ∞ k+1 min1≤j≤k+1 {uj , 1} −1 ··· u du2 · · · duk+1 λk+1 (u1 ) = max1≤j≤k+1 {uj , 1} j=2 j 0 0 ⎛ ⎞ ∞ ∞ ∞ k+1 min {u , 1} 1≤j≤k+1 j ⎝ ⎠ du2 = ··· u−1 j du3 · · · duk+1 max {u , 1} u2 1≤j≤k+1 j 0 0 0 j=3 k−1 ∞ min{u1 , u2 , 1} (−1)k−1 min{u1 , u2 , 1} = k! 1+ ln max{u1 , u2 , 1} k! max{u1 , u2 , 1} 0 i+1 k−3 (−1)i+1 1 1 min{u1 , u2 , 1} du2 − ln + i! i + 1 k max{u , u , 1} u2 1 2 i=0 min{u1 , 1} (−1)k−1 min{u1 , 1} 2 − ln + = k! max{u1 , 1} max{u1 , 1} k! k i k−1 min{u1 , 1} min{u1 , 1} k−1−i (k − 1)! ln × − ln +2 (−1) max{u1 , 1} i! max{u1 , 1} i=0 i+2 k−3 (−1)i+1 1 1 min{u1 , 1} − − ln + i! i+1 k max{u1 , 1} i=0 ⎤⎫ j ⎬ i+1 (i + 1)! min{u1 , 1} ⎦ . +2 (−1)i+1−j ln ⎭ j! max{u1 , 1} j=0
Let x =
min{u1 ,1} max{u1 ,1}
in the above, then
λk+1 (u1 )
=
k−1 (−1)k−1 k i k−1−i (k − 1)! − (ln x) + 2 (ln x) (−1) k!x 2 − ln x + k! i! i=0 ⎡ ⎤⎫ k−3 i+1 ⎬ (−1)i+1 1 1 ⎣ (i + 1)! i+2 j ( − ) − (ln x) (ln x) ⎦ +2 (−1)i+1−j + ⎭ i! i+1 k j! i=0 j=0 4
=
k!x 2 +
k−3 (−1)k 2 i+1 k +2 + (ln x) − ln x 1− k k k! i=0
k−1 k−3 k−3 (−1)i (ln x)i+2 1 (−1)i+1 2 (−1)i i i+2 (ln x) + + (ln x) k i=1 i! (i + 1)! k i! i=0 ⎫ i=0 k−3 i+1 ⎬ 1 (−1)j 1 j − (ln x) +2 (i + 1) ⎭ i + 1 k j=1 j! i=0 k−3 (−1)k i + 1 2 k k!x k + 1 + (ln x) − 1 + + 2 ln x 1− k! k k i=0 k−1 2 (−1)s (−1)s 1 (−1)s−1 + + + k s! (s − 1)! k (s − 2)! s=2 k−3 i + 1 (−1)s s +2 ) (ln x) (1 − k s! i=s−1 (−1)k k (ln x) − k ln x k!x k + 1 + k! k−1 (−1)s k + 1 s − 1 (ln x) + (s − 1)! s s=2 k min{u1 , 1} min{u1 , 1} (−1)k (k + 1)! ln 1+ max{u1 , 1} (k + 1)! max{u1 , 1} i+1 k−2 (−1)i+1 1 1 min{u1 , 1} − ln + . i! i+1 k+1 max{u1 , 1} i=0
+
=
=
=
50
Thus (9) holds for n = k + 1. By mathematical induction, (9) holds for all n ∈ N\{1}.
Lemma 5. For n ∈ N\{1} and ωi (i = 1, · · · , n) defined by ωi (xi ) :=
∞ 0
···
∞ 0
min1≤j≤n {xj } max1≤j≤n {xj }
n
x−1 j dx1 · · · dxi−1 dxi+1 · · · dxn ,
j=1(j=i)
(10) we have ωi (xi ) = n!.
(11)
Proof. If we make the substitution vj = xj /xi (j = i), i = 1, · · · , n, then ∞ ∞ min{v1 , · · · vi−1 , 1, vi+1 , · · · vn } ··· ωi (xi ) = max{v 1 , · · · vi−1 , 1, vi+1 , · · · vn } 0 0 5
n
×
vj−1 dv1 · · · dvi−1 dvi+1 · · · dvn .
j=1(j=i) 55
Furthermore, by using the substitution uj =
vj vn (j
= i, n), ui =
1 vn
in the above,
we find that ωi (xi ) = ωn (xn ) =
∞ 0
···
∞ 0
In view of (9) we have ⎛ ∞ ∞ ⎝ ··· ωn (xn ) =
=
Setting x =
n−1 min1≤j≤n−1 {uj , 1} −1 u du1 · · · dun−1 . max1≤j≤n−1 {uj , 1} j=1 j
⎞ n−1 min1≤j≤n−1 {uj , 1} −1 u du2 · · · dun−1 ⎠ u−1 1 du1 max1≤j≤n−1 {uj , 1} j=2 j 0 0 0 n−2 ∞ min{u1 , 1} min{u1 , 1} (−1)n−2 ln (n − 1)! 1+ max{u1 , 1} (n − 1)! max{u1 , 1} 0 i+1 n−4 (−1)i+1 1 1 min{u1 , 1} du1 − ln + . i! i + 1 n − 1 max{u , 1} u1 1 i=0
min{u1 ,1} max{u1 ,1} ,
∞
we obtain
ωn (xn )
=
=
1
(−1)n−2 n−2 (ln x) (n − 1)! 0 n−4 (−1)i+1 1 1 i+1 − (ln x) + dx i! i+1 n−1 i=0 n−4 i+1 1 + = n!. 1− 2(n − 1)! 1 + n − 1 i=0 n−1
2(n − 1)!
1+
Hence ωi (xi ) = ωn (n) = n!. 60
Lemma 6. Let n ∈ N\{1}, pi = 0, 1(i = 1, · · · , n),
n
1 i=1 pi
= 1, and let
weight functions i (mi )(i = 1, · · · , n) be defined by i (mi ) :=
∞ mn =1
···
∞
∞
mi+1 =1 mi−1 =1
···
∞ min1≤j≤n {mj } max 1≤j≤n {mj } m =1 1
n
m−1 j . (12)
j=1(j=i)
Then i (mi ) < n!(i = 1, · · · , n),
(13)
n (mn ) ≥ n!(1 − θ(mn )) > 0,
(14)
6
where θ(mn )
65
:=
1−
=
O(
1 n!
∞ 1 mn
···
∞ 1 mn
n−1 min1≤j≤n−1 {tj , 1} −1 t dt1 · · · dtn−1 max1≤j≤n−1 {tj , 1} j=1 j
1 )(0 < α < 1). mα n
(15)
Proof. By the decreasing property and Lemma 5, we find for any i = 1, · · · , n i (mi ) <
∞
···
0
∞
min1≤j≤n {xj } max1≤j≤n {xj }
0
n
x−1 j dx1 · · · dxi−1 dxi+1 · · · dxn = n!.
j=1(j=i)
On the other hand, ≥
n (mn )
∞ 1
=
··· ···
1 mn
n−1 min1≤j≤n {xj } −1 x dx1 · · · dxn−1 (Let uj = xj /mn ) max1≤j≤n {xj } j=1 j
∞
n−1 min1≤j≤n−1 {uj , 1} −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
1
∞
∞
1 mn
= n! (1 − θ(mn )) > 0, where θ(mn ) is indicated by (15). Moreover, we have ⎡ ⎤ ∞ ∞ n−1 1 ⎣ min1≤j≤n−1 {uj , 1} ⎦ n! − ··· u−1 0 < θ(mn ) = j du1 · · · dun−1 1 1 n! max {u , 1} 1≤j≤n−1 j m m j=1 n
≤
1 n!
n−1 k=1
1 mn
0
n
u−1 k An (uk )duk ,
where An (uk ) = 70
∞
0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
n−1
u−1 j du1 · · · duk−1 duk+1 · · · dun−1 .
j=1(j=k)
Define F (u1 , · · · , un−1 ) = u−α n−1
min{u1 , · · · , un−1 , 1} min{u1 , · · · , un−2 , 1} / , max{u1 , · · · , un−1 , 1} max{u1 , · · · , un−2 , 1}
where 0 < α < 1. Since lim
un−1 →0+
F (u1 , · · · , un−1 ) =
lim
un−1 →∞
7
F (u1 , · · · , un−1 ) = 0,
there exists a constant L > 0, such that for any (u1 , · · · , un−1 ) ∈ Rn−1 + , F (u1 , · · · , un−1 ) ≤ L, that is, min{u1 , · · · , un−1 , 1} min{u1 , · · · , un−2 , 1} ≤ Luα . n−1 max{u1 , · · · , un−1 , 1} max{u1 , · · · , un−2 , 1} 1 Without loss of generality, estimating 0mn u−1 n−1 An (un−1 )dun−1 gives m1 n u−1 n−1 An (un−1 )dun−1 0 ⎡ ⎤ m1 ∞ ∞ n−2 n min {u , 1} 1≤j≤n−1 j ⎣ = u−1 ··· u−1 du1 · · · , dun−2 ⎦ dun−1 n−1 max1≤j≤n−1 {uj , 1} j=1 j 0 0 0
≤
L ×
1 mn
0 n−2 j=1
= 75
uα−1 n−1
∞ 0
···
∞ 0
⎤
min1≤j≤n−2 {uj , 1} max1≤j≤n−2 {uj , 1}
⎦ u−1 j du1 · · · dun−2 dun−1
L · (n − 1)!
1 mn
uα−1 n−1 dun−1
0
=O
1 mα n
.
Thus θ(mn ) = O( m1α ). n
Lemma 7. If n ∈ N\{1}, pi = 0, 1(i = 1, · · · , n), for any ε > 0 small enough, ∞ I(ε) := ε ··· 1
∞ 1
n
1 i=1 pi
= 1, then we have
n min1≤j≤n {xj } − pεj −1 x dx1 · · · dxn max1≤j≤n {xj } j=1 j
n! + o(1)(ε → 0+ ).
=
(16)
Proof. Let uj = xj /xn (j = 1, · · · , n − 1), then
=
I(ε) ⎡ ∞ −ε−1 ⎣ xn ε 1
≥
∞ 0
−ε
···
0
n−1 ∞ i=1
=
∞
1
∞ 1 xn
···
∞ 1 xn
⎤ n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 ⎦ dxn max1≤j≤n−1 {uj , 1} j=1 j
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
x−1 n Bi (xn )dxn
[n! + o(1)] − ε
n−1 ∞ i=1
1
+ x−1 n Bi (xn )dxn (ε → 0 ),
8
and
Bi (xn )
:
∞
= ×
0 n−1 j=1
80
···
∞
0
0
− pε j
−1
uj
1 xn
∞ 0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
du1 · · · dui−1 dui dui+1 · · · dxn−1 .
∞ Without loss of generality, evaluating 1 x−1 n Bn−1 (xn )dxn gives ∞ x−1 n Bn−1 (xn )dxn 1 1 ∞ ∞ ∞ xn min1≤j≤n−1 {uj , 1} −1 = xn ··· max1≤j≤n−1 {uj , 1} 0 1 0 0 ⎤ n−1 − pε −1 uj j du1 · · · dun−2 dun−1 ⎦ dxn × j=1
1
1 un−1
= 0
=
x−1 n dxn
1
×du1 · · · dun−1 1 (− ln un−1 ) 0
∞ 0
∞ 0
···
∞ 0
···
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u max1≤j≤n−1 {uj , 1} j=1 j
∞ 0
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 , max1≤j≤n−1 {uj , 1} j=1 j
where we have used Fubini’s theorem. Let 0 < β < 1, then limun−1 →0+ uβn−1 (− ln un−1 ) = 0. Hence there exists a constant M > 0, such that 0 < uβn−1 (− ln un−1 ) < M, un−1 ∈ (0, 1], and therefore 0
<
∞
x−1 n Bn−1 (xn )dxn ≤ M
1 n−2
×
j=1
= 85
namely,
− pε j
uj
−1
−β− p ε n−1
un−1
−1
∞ 0
∞ 0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
du1 · · · dun−1
M [n! + o(1)] < ∞(ε, β → 0+ ),
n−1 ∞ i=1
1
x−1 n Bi (xn )dxn = O(1). Thus I(ε) ≥ n! + o(1) − εO(1)(ε → 0+ ).
On the other hand, still by substituting uj = xj /xn (j = 1, · · · , n − 1), it follows that I(ε) 9
≤
ε
∞ 1
∞
= 0
=
⎡ ⎣ x−ε−1 n
···
∞ 0
∞ 0
···
∞ 0
⎤ n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 ⎦ dxn max1≤j≤n−1 {uj , 1} j=1 j
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
n! + o(1)(ε → 0+ ),
and the proof is complete. In what follows, for the sake of simplicity, we set
⎧ ⎛ ⎞qn ⎫1/qn ∞ ∞ ∞ n−1 1−qn ⎨ ⎬ [ n (mn )] min1≤j≤n {mj } ⎝ ⎠ ··· a(i) . J1 := mi ⎩ ⎭ mn max1≤j≤n {mj } i=1 m =1 m =1 m =1 n
90
n−1
1
Lemma 8. Suppose that n ∈ N\{1}, pi = 0, 1 (i = 1, · · · , n), 1, q1n = 1 −
(i) 1 p n , am i
n
1 i=1 pi
=
≥ 0.
(a) If pi > 1(i = 1, · · · , n), then J1 ≤
n−1 i=1
∞ mi =1
i (mi )mipi −1
a(i) mi
pi
1/pi .
(17)
(b) If pi < 0(i = 1, · · · , n − 1), 0 < pn < 1, then the reverse of (17) is obtained. 95
Proof. (a) By using H¨ older’s inequality ([20]) for pi > 1(i = 1, · · · , n), in view of (6) and (12), we have ⎞qn ⎛ ∞ ∞ n−1 min {m } 1≤j≤n j ⎝ ··· a(i) ⎠ max1≤j≤n {mj } i=1 mi mn−1 =1 m1 =1 ⎧ ⎡ ⎤1/pi ⎪ ∞ ∞ n−1 n ⎨ min1≤j≤n {mj } ⎣mpi i −1 ⎦ = ··· m−1 a(i) mi j ⎪ max {m } 1≤j≤n j ⎩mn−1 =1 m1 =1 i=1 j=1(j=i) ⎤1/pn ⎫qn ⎡ ⎪ n−1 ⎬ −1 ⎦ pn −1 ⎣ × mn mj ⎪ ⎭ j=1 ≤
⎡ ∞ n−1 min1≤j≤n {mj } ⎣mpi i −1 ··· max {m } 1≤j≤n j =1 m =1 i=1
∞ mn−1
1
10
n j=1(j=i)
⎤qn /pi 1 ⎦ mj
a(i) mi
q n
×
=
⎧ ∞ ⎨ ⎩
mn−1
⎫qn −1 n−1 ∞ ⎬ min1≤j≤n {mj } pn −1 mn ··· m−1 j ⎭ max1≤j≤n {mj } =1 m =1 j=1 1
[ n (mn )]
×
n−1 i=1
qn −1
n
⎣mpi i −1
j=1(j=i)
∞
≤
×
n−1 i=1
⎩
×
i=1
n−1
J1
1 i=1 pi /qn
≤
a(i) mi
⎣mpi i −1
⎦ m−1 j
mi =1
n
i=1
a(i) mi
j=1(j=i)
∞
···
m1 =1
⎡ ⎣mpi i −1
⎫1/qn ⎬ qn ⎪ ⎪ ⎭
∞ min1≤j≤n {mj } −1 mn max 1≤j≤n {mj } m =1 n
⎤qn /pi
n−1
⎦ m−1 j
a(i) mi
j=1(j=i)
⎫1/qn ⎬ qn ⎪ ⎪ ⎭
.
mn =1
⎤
mi+1 =1 mi−1
pi −1 ⎦ a(i) m m−1 mi i j
∞ mi =1
i (mi )mpi i −1
∞ min1≤j≤n {mj } ··· max 1≤j≤n {mj } =1 m =1
∞
1
⎫1/pi pi ⎬
j=1(j=i)
=
= 1, applying H¨ older’s inequality once more and (12), gives
i=1
n−1
.
⎤qn /pi
n
⎧ ⎡ n−1 ∞ ∞ ∞ ⎨ ⎣ ··· ⎩ ×
qn
1
mn−1 =1 n−1
1
⎤qn /pi
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1
⎡
⎧ ∞ ⎨
=
∞ min1≤j≤n {mj } max 1≤j≤n {mj } m =1
1 ⎦ mj
···
mn =1
Since
···
mn−1 =1
⎡
J1
∞
mn
a(i) mi
⎭ p i
1/pi ,
and (13) follows. 100
(b) It is similar to the proof of (a) with the reverse H¨ older’s inequality for pi < 0(i = 1, · · · , n), 0 < pn < 1.
11
3. Main results In this section, lp,ϕ denotes the space of real-valued sequences a = {ami }∞ mi =1 satisfying ||a||p,ϕ < ∞, where ⎧ ⎫ 1/p ∞ ⎨ ⎬ p−1 p , lp,ϕ := a = {ami }∞ : a ∈ R, ||a|| = m |a | < ∞ mi p,ϕ mi mi =1 i ⎩ ⎭ mi =1
105
and ϕ(mi ) = mp−1 (i = 1, · · · , n). For convenience, we set i ∞
I :=
···
mn =1
J :=
∞ n min1≤j≤n {mj } (i) a , max1≤j≤n {mj } i=1 mi m =1 1
⎧ ∞ ⎨ ⎩
mn
⎛ ⎞qn ⎫1/qn ∞ ∞ n−1 1 ⎝ min1≤j≤n {mj } (i) ⎠ ⎬ ··· a , ⎭ mn m =1 max1≤j≤n {mj } i=1 mi =1 m =1 n−1
1
and ⎧ ⎛ ⎞qn ⎫1/qn ∞ ∞ ∞ n−1 1−qn ⎬ ⎨ [1 − θ(m )] min {m } n 1≤j≤n j ⎝ ⎠ ··· a(i) . Jˆ := ⎭ ⎩ mn max1≤j≤n {mj } i=1 mi m =1 m =1 m =1 n
n−1
1
Theorem 1. If n ∈ N\{1}, pi > 1(i = 1, · · · , n),
n
1 i=1 pi
= 1, q1n = 1 −
1 pn ,
a(i) ∈ lpi ,ϕi , ||a(i) ||pi ,ϕi > 0(i = 1, · · · , n), then the following inequalities hold 110
and are equivalent: I < n!
n
||a(i) ||pi ,ϕi ,
(18)
i=1
J < n!
n−1
||a(i) ||pi ,ϕi ,
(19)
i=1
where the constant factor n! in the above inequalities is the best possible. Proof. By (13) and (17), (19) follows. Since inequality gives I=
∞
⎡
mn =1 115
×
∞
n ⎣m−1/q n
···
mn−1 =1 n−1 i=1
a(i) mi
#
1 pn
+ q1n = 1, applying H¨ older’s
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1 1
$ n (n) m1/q amn ≤ J||a(n) ||pn ,ϕn . n
12
(20)
From (19), that (18) follows. Assuming (18) holds and setting a(n) mn
⎛ ⎞qn −1 ∞ ∞ n−1 1 ⎝ min1≤j≤n {mj } (i) ⎠ := ··· a , mn m =1 max1≤j≤n {mj } i=1 mi m =1 n−1
1
then we find J qn −1 = ||a(n) ||pn ,ϕn . By (17), we have J < ∞. If J = 0, (19) is trivial so we assume that J > 0, applying (18) gives ||a(n) ||ppnn ,ϕn = J qn = I < n!
||a(n) ||ppnn −1 ,ϕn = J < n! 120
n
||a(i) ||pi ,ϕi ,
i=1
n−1
||a(i) ||pi ,ϕi ,
i=1
so that (19) holds, which is equivalent to (18). (i)
− pε −1
To see that the constant n! is the best possible, we take a ˜mi = mi
i
(i =
1, · · · , n), where ε is small and positive. If there exists a constant K ≤ n!, such that (18) is still valid when n! is replaced by K, applying (16) gives εI˜ = ≥ =
∞
∞ n min1≤j≤n {mj } (i) ··· a ˜ ε max1≤j≤n {mj } i=1 mi mn =1 m1 =1 ∞ ∞ n min1≤j≤n {xj } − pεj −1 ε ··· x dx1 · · · dxn max1≤j≤n {xj } j=1 j 1 1
n! + o(1)(ε → 0+ ).
On the other hand, εI˜ <
εK
n
||˜ a(i) ||pi ,ϕi = εK
i=1
≤ 125
εK 1 +
∞ 1
dx x1+ε
∞ m=1
1 m1+ε
= K(1 + ε).
Letting ε → 0+ , it follows that n! ≤ K. Hence n! = K is the best possible constant factor of (18). By the equivalence, the constant factor n! in (19) is still the best possible. Otherwise, we would achieve a contradiction by (20) that the constant in (18)
is not the best possible. 13
130
Under the hypothesis of Theorem 1, we define a Multiple
n−1 Hilbert-type discrete operator T : i=1 lpi ,ϕi → lqn ,ϕ1/(1−pn ) by Remark 1.
n
(T a)(mn ) :=
∞
···
mn−1 =1
∞ n−1 min1≤j≤n {mj } (i) a , max1≤j≤n {mj } i=1 mi m =1 1
% & n−1 where a = a(1) , · · · a(n−1) ∈ i=1 lpi ,ϕi . (19) may be re-written as ||T a||qn ,ϕ1/(1−pn ) < n! n
n−1
||a(i) ||pi ,ϕi ,
(21)
i=1
and therefore T a ∈ lqn ,ϕ1/(1−pn ) . Hence T is a bounded linear operator with n
||T || ≤ n!. Since the constant factor in (19) is the best possible, we have ||T || := a(=θ)∈
sup
n−1 i=1
lpi ,ϕi
||T a||qn ,ϕ1/(1−pn ) n = n!.
n−1 (i) ||a ||pi ,ϕi i=1
Remark 2. The following inequality is obtained by taking n = 2 in (18)
135
1/p ∞ 1/q ∞ ∞ ∞ min{m, n} a m bn < 2 np−1 apn nq−1 bqn , max{m, n} n=1 m=1 n=1 n=1
(22)
which is given by [8]. Therefore we give a multiple generalization of [8]. In order to obtain reverse inequalities corresponding to those in Theorem 1, we define the function Φn (mn ) := (1 − θ(mn ))ϕn (mn ). For pi < 1(pi = 0), the space lpi ,ϕi is not normal space. But we still use them as the formal symbols in 140
the following. Theorem 2. If n ∈ N\{1}, pi < 0(i = 1, · · · , n − 1), 0 < pn < 1, 1,
1 qn
= 1−
1 pn ,
a
(i)
n
1 i=1 pi
=
(i)
∈ lpi ,ϕi , ||a ||pi ,ϕi > 0(i = 1, · · · , n), then the following
inequalities hold and are equivalent: I > n!
n−1
||a(i) ||pi ,ϕi ||a(n) ||pn ,Φn ,
(23)
i=1
Jˆ > n!
n−1
||a(i) ||pi ,ϕi ,
i=1 145
where the constant factor n! in the above inequalities is the best possible.
14
(24)
Proof. For 0 < pn < 1, we have qn < 0. By (13), (14), the reverse of (17) and the assumptions, we get (24). By using reverse H¨ older’s inequality ([20]), we obtain I
=
⎡
∞ mn =1
×
n−1 i=1
≥
∞
n ⎣( n (mn ))−1/pn m−1/q n
a(i) mi
···
mn−1 =1
# ( n (mn ))
1/pn
n (n) m1/q am n n
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1 1
$
Jˆ · ||a(n) ||pn ,Φn .
(25)
Then we get (23) from (24). 150
On the other hand, assuming (23) holds and setting a(n) mn
[1 − θ(mn )] := mn
⎛
1−qn
⎝
⎞qn −1 ∞ n−1 min1≤j≤n {mj } (i) ⎠ ··· a , max1≤j≤n {mj } i=1 mi =1 m =1
∞
mn−1
1
then we find Jˆqn −1 = ||a(n) ||pn ,Φn . By the reverse of (17), we have Jˆ > 0. If Jˆ = ∞, (24) is trivial so we assume that J < ∞, applying (23) gives ||a(n) ||ppnn ,Φn = Jˆqn = I > n!
n−1
||a(i) ||pi ,ϕi ||a(n) ||pn ,Φn ,
i=1
ˆ ||a(n) ||ppnn −1 ,Φn = J > n!
n−1
||a(i) ||pi ,ϕi ,
i=1
so that (24) holds, which is equivalent to (23). 155
(i)
− pε −1
To see that the constant n! is the best possible, we take a ˜mi = mi
i
(i =
1, · · · , n), where ε is small and positive. If there exists a constant K ≥ n!, such that (23) is still valid when n! is replaced by K, applying (13) gives I˜ :=
∞ mn =1
= ≤
···
∞ n min1≤j≤n {mj } (i) a ˜ max1≤j≤n {mj } i=1 mi m =1 1
∞
1
∞
mn =1
m1+ε n
mn−1 =1
(n! + o(1))
···
∞ n−1 min1≤j≤n {mj } − pεi −1 m max1≤j≤n {mj } j=1 j m =1 1
∞
1 1+ε . m n mn =1 15
On the other hand, in view of (15), we obtain I˜ >
K
n−1
||˜ a(i) ||pi ,ϕi ||˜ a(n) ||pn ,Φn
i=1
=
K
∞ mn =1
1
⎧ ⎨
⎩ m1+ε n
∞ 1−
1 mn =1 O m1+ε+α n ∞ 1 mn =1 m1+ε n
⎫1/pn ⎬ ⎭
.
Consequently, (n! + o(1)) > K 160
⎧ ⎨ ⎩
∞ 1−
mn =1 O ∞
⎫1/pn ⎬ 1+ε+α 1
mn 1 mn =1 m1+ε n
⎭
.
Letting ε → 0+ , it follows that n! ≤ K. Hence n! = K is the best possible constant factor of (23). By the equivalence, the constant factor n! in (24) is still the best possible. Otherwise, we would reach a contradiction by (25) that the constant in (23) is
not the best possible. 165
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18
Bing He, Qiru Wang
PII: DOI: Reference:
S0022-247X(15)00562-4 http://dx.doi.org/10.1016/j.jmaa.2015.06.019 YJMAA 19572
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
11 February 2015
Please cite this article in press as: B. He, Q. Wang, A multiple Hilbert-type discrete inequality with a new kernel and best possible constant factor, J. Math. Anal. Appl. (2015), http://dx.doi.org/10.1016/j.jmaa.2015.06.019
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A Multiple Hilbert-Type Discrete Inequality with a New Kernel and Best Possible Constant Factor$ Bing Hea , Qiru Wangb,∗ a Department
of Mathematics, Guangdong University of Education, Guangzhou 510303, PR China b School of Mathematics and Computational Science, Sun Yat-sen University, Guangzhou 510275, PR China
Abstract By introducing a distinctive kernel and using the way of weight functions, a brand new multiple Hilbert-type discrete inequality with a best constant factor is presented. The operator expressions, equivalent forms, reverse inequalities and particular cases are discussed. Keywords: multiple Hilbert-type discrete inequality, weight function, kernel, operator expression, best constant factor 2010 MSC: 26D15, 47A07 1. Introduction p Suppose that p > 1, (1/p) + (1/q) = 1, am , bn ≥ 0, a = {am }∞ m=1 ∈ l , b = ∞ q p 1/p {bn }∞ > 0, ||b||q > 0, the well-known Hardyn=1 ∈ l , ||a||p = { m=1 am }
Hilbert’s discrete inequality is given by ([1, p.226]): ∞ ∞ π a m bn < ||a||p ||b||q , m + n sin(π/p) n=1 m=1 5
and an equivalent form is ∞ ∞ n=1
am m+n m=1
p 1/p <
π ||a||p , sin(π/p)
(1)
(2)
$ Supported
by the NNSF of China (Nos. 11301090 and 11271379). author Email addresses: [email protected] (Bing He), [email protected] (Qiru Wang)
∗ Corresponding
Preprint submitted to Journal of Mathematical Analysis and Applications
June 15, 2015
where the constant factor π/ sin(π/p) is the best possible. Define Hardy-Hilbert’s ∞ am discrete operator T : lp → lp by: (T a) (n) := m=1 m+n , where a = {am }∞ m=1 ∈ lp , n ∈ N. Then (2) can be written as ||T a||p < [π/ sin(π/p)] ||a||p and ||T || ≤ π/ sin(π/p). Since the constant factor in (2) is the best possible, we have 10
||T || = π/ sin(π/p). Inequalities (1) and (2) have been studied extensively, which are important in analysis and applications ([2]-[7]). Hilbert-type inequalities may be classified into several types (integral, discrete and half-discrete, etc.). On discrete inequalities, by introducing an independent parameter λ (2 − min{p, q} < λ ≤ 2),
15
Yang [9] gave an extension of (1) in 2002 as follows: ∞ ∞ p+λ−2 q+λ−2 a m bn , ||a||p,ξ ||b||q,ξ , 0, ||b||q,ξ > 0. In the same year, another similar form was established by Yang [10]:
∞ ∞
π a m bn ||a||p,φ ||b||q,ψ , < μ + nμ m μ sin(π/p) n=1 m=1 20
(4)
where 0 < μ ≤ min{p, q}, φ(n) = n(p−1)(1−μ) , ψ(n) = n(q−1)(1−μ) , ||a||p,φ := ∞ { n=1 φ(n)apn }1/p > 0, ||b||q,ψ > 0. It is significant to generalize Hilbert-type inequalities into multiple versions. In recent years, some results have been obtained ([11]-[18]). In 2009, the following multiple Hilbert-type discrete inequality had been studied by Yang [3, Example 9.1.1]
25
n n If n ∈ N\{1}, pi , ri > 1, i=1 p1i = i=1 r1i = 1, 0 < λ ≤ min1≤i≤n {ri },
p i ∞ p (1−λ/ri )−1 (i) 0 < mi =1 mi i < ∞ (i = 1, · · · , n), then am i ∞
···
mn =1
<
∞ m1
n
1 Γ Γ(λ) i=1
where the constant factor
( =1
1 Γ(λ)
λ ri
n
n
1
i=1
mi )λ
∞ mi =1
n
i=1 Γ
i=1
a(i) mi
pi (1− rλ )−1 i mi
a(i) mi
pi
1/pi ,
(5)
λ ri
is the best possible. It is not an easy
job to find new forms of such inequalities. Exploiting these multiple inequalities 2
will greatly increase the scope of our work. This paper will include a completely 30
new multiple discrete Hilbert-type inequality with an interesting kernel. The setup of this paper is as follows. In Section 2, a number of basic lemmas are obtained. To calculate the constant factor, we herein utilize mathematical induction to do an excellent computing. To prove the best of the constant factor, we apply some analytical skills to do a good estimate for some formula. Section
35
3 presents our main results, the techniques that will be used in the proof are mainly based on classical real analysis, especially on the well-known H¨ older’s inequality. In addition, the equivalent forms, operator expressions as well as some reverse inequalities are also considered.
2. Basic lemmas 40
Lemma 1. ([3], (9.1.1)). If n ∈ N\{1} = {2, 3, 4, 5, · · ·}, pi = 0, 1(i = n 1, 2, · · · , n), i=1 p1i = 1, then n i=1
⎡ ⎣mpi i −1
n
⎤ p1
i
⎦ m−1 j
= 1.
(6)
j=1(j=i)
Lemma 2. ([16]). If c > 0, n ∈ N, then
c 0
n n! i (ln c) . (ln x) dx = c (−1)n−i · i! i=0 n
(7)
Lemma 3. ([16]). Suppose that s, t ∈ (0, +∞), a = min{s, 1}, b = max{s, 1}, k ∈ N ∪ {0}, then k min{s, t, 1} min{s, t, 1} ln t−1 dt max{s, t, 1} max{s, t, 1} 0 k
a k+1
a i a k−i k! +2 (−1) · . − ln ln b b i! b i=0
= 45
∞
Lemma 4. For all n ∈ N\{1}, we have the following identity λn (u1 )
∞
:= 0
···
∞ 0
n min1≤j≤n {uj , 1} −1 u du2 · · · dun max1≤j≤n {uj , 1} j=2 j
3
(8)
=
n−1 (−1)n−1 min{u1 , 1} min{u1 , 1} 1+ ln n! max{u1 , 1} n! max{u1 , 1} i+1 n−3 (−1)i+1 1 1 min{u1 , 1} − ln + . i! i+1 n max{u1 , 1} i=0
(9)
Proof. We will prove (9) by induction. If n = 2, then min{u1 , 1} min{u1 , 1} − ln +2 λ2 (u1 ) = max{u1 , 1} max{u1 , 1} holds by applying Lemma 3 with k = 0. Now we assume that (9) holds for n = k ∈ N\{1}. Then, by (8), we obtain ∞ ∞ k+1 min1≤j≤k+1 {uj , 1} −1 ··· u du2 · · · duk+1 λk+1 (u1 ) = max1≤j≤k+1 {uj , 1} j=2 j 0 0 ⎛ ⎞ ∞ ∞ ∞ k+1 min {u , 1} 1≤j≤k+1 j ⎝ ⎠ du2 = ··· u−1 j du3 · · · duk+1 max {u , 1} u2 1≤j≤k+1 j 0 0 0 j=3 k−1 ∞ min{u1 , u2 , 1} (−1)k−1 min{u1 , u2 , 1} = k! 1+ ln max{u1 , u2 , 1} k! max{u1 , u2 , 1} 0 i+1 k−3 (−1)i+1 1 1 min{u1 , u2 , 1} du2 − ln + i! i + 1 k max{u , u , 1} u2 1 2 i=0 min{u1 , 1} (−1)k−1 min{u1 , 1} 2 − ln + = k! max{u1 , 1} max{u1 , 1} k! k i k−1 min{u1 , 1} min{u1 , 1} k−1−i (k − 1)! ln × − ln +2 (−1) max{u1 , 1} i! max{u1 , 1} i=0 i+2 k−3 (−1)i+1 1 1 min{u1 , 1} − − ln + i! i+1 k max{u1 , 1} i=0 ⎤⎫ j ⎬ i+1 (i + 1)! min{u1 , 1} ⎦ . +2 (−1)i+1−j ln ⎭ j! max{u1 , 1} j=0
Let x =
min{u1 ,1} max{u1 ,1}
in the above, then
λk+1 (u1 )
=
k−1 (−1)k−1 k i k−1−i (k − 1)! − (ln x) + 2 (ln x) (−1) k!x 2 − ln x + k! i! i=0 ⎡ ⎤⎫ k−3 i+1 ⎬ (−1)i+1 1 1 ⎣ (i + 1)! i+2 j ( − ) − (ln x) (ln x) ⎦ +2 (−1)i+1−j + ⎭ i! i+1 k j! i=0 j=0 4
=
k!x 2 +
k−3 (−1)k 2 i+1 k +2 + (ln x) − ln x 1− k k k! i=0
k−1 k−3 k−3 (−1)i (ln x)i+2 1 (−1)i+1 2 (−1)i i i+2 (ln x) + + (ln x) k i=1 i! (i + 1)! k i! i=0 ⎫ i=0 k−3 i+1 ⎬ 1 (−1)j 1 j − (ln x) +2 (i + 1) ⎭ i + 1 k j=1 j! i=0 k−3 (−1)k i + 1 2 k k!x k + 1 + (ln x) − 1 + + 2 ln x 1− k! k k i=0 k−1 2 (−1)s (−1)s 1 (−1)s−1 + + + k s! (s − 1)! k (s − 2)! s=2 k−3 i + 1 (−1)s s +2 ) (ln x) (1 − k s! i=s−1 (−1)k k (ln x) − k ln x k!x k + 1 + k! k−1 (−1)s k + 1 s − 1 (ln x) + (s − 1)! s s=2 k min{u1 , 1} min{u1 , 1} (−1)k (k + 1)! ln 1+ max{u1 , 1} (k + 1)! max{u1 , 1} i+1 k−2 (−1)i+1 1 1 min{u1 , 1} − ln + . i! i+1 k+1 max{u1 , 1} i=0
+
=
=
=
50
Thus (9) holds for n = k + 1. By mathematical induction, (9) holds for all n ∈ N\{1}.
Lemma 5. For n ∈ N\{1} and ωi (i = 1, · · · , n) defined by ωi (xi ) :=
∞ 0
···
∞ 0
min1≤j≤n {xj } max1≤j≤n {xj }
n
x−1 j dx1 · · · dxi−1 dxi+1 · · · dxn ,
j=1(j=i)
(10) we have ωi (xi ) = n!.
(11)
Proof. If we make the substitution vj = xj /xi (j = i), i = 1, · · · , n, then ∞ ∞ min{v1 , · · · vi−1 , 1, vi+1 , · · · vn } ··· ωi (xi ) = max{v 1 , · · · vi−1 , 1, vi+1 , · · · vn } 0 0 5
n
×
vj−1 dv1 · · · dvi−1 dvi+1 · · · dvn .
j=1(j=i) 55
Furthermore, by using the substitution uj =
vj vn (j
= i, n), ui =
1 vn
in the above,
we find that ωi (xi ) = ωn (xn ) =
∞ 0
···
∞ 0
In view of (9) we have ⎛ ∞ ∞ ⎝ ··· ωn (xn ) =
=
Setting x =
n−1 min1≤j≤n−1 {uj , 1} −1 u du1 · · · dun−1 . max1≤j≤n−1 {uj , 1} j=1 j
⎞ n−1 min1≤j≤n−1 {uj , 1} −1 u du2 · · · dun−1 ⎠ u−1 1 du1 max1≤j≤n−1 {uj , 1} j=2 j 0 0 0 n−2 ∞ min{u1 , 1} min{u1 , 1} (−1)n−2 ln (n − 1)! 1+ max{u1 , 1} (n − 1)! max{u1 , 1} 0 i+1 n−4 (−1)i+1 1 1 min{u1 , 1} du1 − ln + . i! i + 1 n − 1 max{u , 1} u1 1 i=0
min{u1 ,1} max{u1 ,1} ,
∞
we obtain
ωn (xn )
=
=
1
(−1)n−2 n−2 (ln x) (n − 1)! 0 n−4 (−1)i+1 1 1 i+1 − (ln x) + dx i! i+1 n−1 i=0 n−4 i+1 1 + = n!. 1− 2(n − 1)! 1 + n − 1 i=0 n−1
2(n − 1)!
1+
Hence ωi (xi ) = ωn (n) = n!. 60
Lemma 6. Let n ∈ N\{1}, pi = 0, 1(i = 1, · · · , n),
n
1 i=1 pi
= 1, and let
weight functions i (mi )(i = 1, · · · , n) be defined by i (mi ) :=
∞ mn =1
···
∞
∞
mi+1 =1 mi−1 =1
···
∞ min1≤j≤n {mj } max 1≤j≤n {mj } m =1 1
n
m−1 j . (12)
j=1(j=i)
Then i (mi ) < n!(i = 1, · · · , n),
(13)
n (mn ) ≥ n!(1 − θ(mn )) > 0,
(14)
6
where θ(mn )
65
:=
1−
=
O(
1 n!
∞ 1 mn
···
∞ 1 mn
n−1 min1≤j≤n−1 {tj , 1} −1 t dt1 · · · dtn−1 max1≤j≤n−1 {tj , 1} j=1 j
1 )(0 < α < 1). mα n
(15)
Proof. By the decreasing property and Lemma 5, we find for any i = 1, · · · , n i (mi ) <
∞
···
0
∞
min1≤j≤n {xj } max1≤j≤n {xj }
0
n
x−1 j dx1 · · · dxi−1 dxi+1 · · · dxn = n!.
j=1(j=i)
On the other hand, ≥
n (mn )
∞ 1
=
··· ···
1 mn
n−1 min1≤j≤n {xj } −1 x dx1 · · · dxn−1 (Let uj = xj /mn ) max1≤j≤n {xj } j=1 j
∞
n−1 min1≤j≤n−1 {uj , 1} −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
1
∞
∞
1 mn
= n! (1 − θ(mn )) > 0, where θ(mn ) is indicated by (15). Moreover, we have ⎡ ⎤ ∞ ∞ n−1 1 ⎣ min1≤j≤n−1 {uj , 1} ⎦ n! − ··· u−1 0 < θ(mn ) = j du1 · · · dun−1 1 1 n! max {u , 1} 1≤j≤n−1 j m m j=1 n
≤
1 n!
n−1 k=1
1 mn
0
n
u−1 k An (uk )duk ,
where An (uk ) = 70
∞
0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
n−1
u−1 j du1 · · · duk−1 duk+1 · · · dun−1 .
j=1(j=k)
Define F (u1 , · · · , un−1 ) = u−α n−1
min{u1 , · · · , un−1 , 1} min{u1 , · · · , un−2 , 1} / , max{u1 , · · · , un−1 , 1} max{u1 , · · · , un−2 , 1}
where 0 < α < 1. Since lim
un−1 →0+
F (u1 , · · · , un−1 ) =
lim
un−1 →∞
7
F (u1 , · · · , un−1 ) = 0,
there exists a constant L > 0, such that for any (u1 , · · · , un−1 ) ∈ Rn−1 + , F (u1 , · · · , un−1 ) ≤ L, that is, min{u1 , · · · , un−1 , 1} min{u1 , · · · , un−2 , 1} ≤ Luα . n−1 max{u1 , · · · , un−1 , 1} max{u1 , · · · , un−2 , 1} 1 Without loss of generality, estimating 0mn u−1 n−1 An (un−1 )dun−1 gives m1 n u−1 n−1 An (un−1 )dun−1 0 ⎡ ⎤ m1 ∞ ∞ n−2 n min {u , 1} 1≤j≤n−1 j ⎣ = u−1 ··· u−1 du1 · · · , dun−2 ⎦ dun−1 n−1 max1≤j≤n−1 {uj , 1} j=1 j 0 0 0
≤
L ×
1 mn
0 n−2 j=1
= 75
uα−1 n−1
∞ 0
···
∞ 0
⎤
min1≤j≤n−2 {uj , 1} max1≤j≤n−2 {uj , 1}
⎦ u−1 j du1 · · · dun−2 dun−1
L · (n − 1)!
1 mn
uα−1 n−1 dun−1
0
=O
1 mα n
.
Thus θ(mn ) = O( m1α ). n
Lemma 7. If n ∈ N\{1}, pi = 0, 1(i = 1, · · · , n), for any ε > 0 small enough, ∞ I(ε) := ε ··· 1
∞ 1
n
1 i=1 pi
= 1, then we have
n min1≤j≤n {xj } − pεj −1 x dx1 · · · dxn max1≤j≤n {xj } j=1 j
n! + o(1)(ε → 0+ ).
=
(16)
Proof. Let uj = xj /xn (j = 1, · · · , n − 1), then
=
I(ε) ⎡ ∞ −ε−1 ⎣ xn ε 1
≥
∞ 0
−ε
···
0
n−1 ∞ i=1
=
∞
1
∞ 1 xn
···
∞ 1 xn
⎤ n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 ⎦ dxn max1≤j≤n−1 {uj , 1} j=1 j
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
x−1 n Bi (xn )dxn
[n! + o(1)] − ε
n−1 ∞ i=1
1
+ x−1 n Bi (xn )dxn (ε → 0 ),
8
and
Bi (xn )
:
∞
= ×
0 n−1 j=1
80
···
∞
0
0
− pε j
−1
uj
1 xn
∞ 0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
du1 · · · dui−1 dui dui+1 · · · dxn−1 .
∞ Without loss of generality, evaluating 1 x−1 n Bn−1 (xn )dxn gives ∞ x−1 n Bn−1 (xn )dxn 1 1 ∞ ∞ ∞ xn min1≤j≤n−1 {uj , 1} −1 = xn ··· max1≤j≤n−1 {uj , 1} 0 1 0 0 ⎤ n−1 − pε −1 uj j du1 · · · dun−2 dun−1 ⎦ dxn × j=1
1
1 un−1
= 0
=
x−1 n dxn
1
×du1 · · · dun−1 1 (− ln un−1 ) 0
∞ 0
∞ 0
···
∞ 0
···
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u max1≤j≤n−1 {uj , 1} j=1 j
∞ 0
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 , max1≤j≤n−1 {uj , 1} j=1 j
where we have used Fubini’s theorem. Let 0 < β < 1, then limun−1 →0+ uβn−1 (− ln un−1 ) = 0. Hence there exists a constant M > 0, such that 0 < uβn−1 (− ln un−1 ) < M, un−1 ∈ (0, 1], and therefore 0
<
∞
x−1 n Bn−1 (xn )dxn ≤ M
1 n−2
×
j=1
= 85
namely,
− pε j
uj
−1
−β− p ε n−1
un−1
−1
∞ 0
∞ 0
···
∞ 0
min1≤j≤n−1 {uj , 1} max1≤j≤n−1 {uj , 1}
du1 · · · dun−1
M [n! + o(1)] < ∞(ε, β → 0+ ),
n−1 ∞ i=1
1
x−1 n Bi (xn )dxn = O(1). Thus I(ε) ≥ n! + o(1) − εO(1)(ε → 0+ ).
On the other hand, still by substituting uj = xj /xn (j = 1, · · · , n − 1), it follows that I(ε) 9
≤
ε
∞ 1
∞
= 0
=
⎡ ⎣ x−ε−1 n
···
∞ 0
∞ 0
···
∞ 0
⎤ n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 ⎦ dxn max1≤j≤n−1 {uj , 1} j=1 j
n−1 min1≤j≤n−1 {uj , 1} − pεj −1 u du1 · · · dun−1 max1≤j≤n−1 {uj , 1} j=1 j
n! + o(1)(ε → 0+ ),
and the proof is complete. In what follows, for the sake of simplicity, we set
⎧ ⎛ ⎞qn ⎫1/qn ∞ ∞ ∞ n−1 1−qn ⎨ ⎬ [ n (mn )] min1≤j≤n {mj } ⎝ ⎠ ··· a(i) . J1 := mi ⎩ ⎭ mn max1≤j≤n {mj } i=1 m =1 m =1 m =1 n
90
n−1
1
Lemma 8. Suppose that n ∈ N\{1}, pi = 0, 1 (i = 1, · · · , n), 1, q1n = 1 −
(i) 1 p n , am i
n
1 i=1 pi
=
≥ 0.
(a) If pi > 1(i = 1, · · · , n), then J1 ≤
n−1 i=1
∞ mi =1
i (mi )mipi −1
a(i) mi
pi
1/pi .
(17)
(b) If pi < 0(i = 1, · · · , n − 1), 0 < pn < 1, then the reverse of (17) is obtained. 95
Proof. (a) By using H¨ older’s inequality ([20]) for pi > 1(i = 1, · · · , n), in view of (6) and (12), we have ⎞qn ⎛ ∞ ∞ n−1 min {m } 1≤j≤n j ⎝ ··· a(i) ⎠ max1≤j≤n {mj } i=1 mi mn−1 =1 m1 =1 ⎧ ⎡ ⎤1/pi ⎪ ∞ ∞ n−1 n ⎨ min1≤j≤n {mj } ⎣mpi i −1 ⎦ = ··· m−1 a(i) mi j ⎪ max {m } 1≤j≤n j ⎩mn−1 =1 m1 =1 i=1 j=1(j=i) ⎤1/pn ⎫qn ⎡ ⎪ n−1 ⎬ −1 ⎦ pn −1 ⎣ × mn mj ⎪ ⎭ j=1 ≤
⎡ ∞ n−1 min1≤j≤n {mj } ⎣mpi i −1 ··· max {m } 1≤j≤n j =1 m =1 i=1
∞ mn−1
1
10
n j=1(j=i)
⎤qn /pi 1 ⎦ mj
a(i) mi
q n
×
=
⎧ ∞ ⎨ ⎩
mn−1
⎫qn −1 n−1 ∞ ⎬ min1≤j≤n {mj } pn −1 mn ··· m−1 j ⎭ max1≤j≤n {mj } =1 m =1 j=1 1
[ n (mn )]
×
n−1 i=1
qn −1
n
⎣mpi i −1
j=1(j=i)
∞
≤
×
n−1 i=1
⎩
×
i=1
n−1
J1
1 i=1 pi /qn
≤
a(i) mi
⎣mpi i −1
⎦ m−1 j
mi =1
n
i=1
a(i) mi
j=1(j=i)
∞
···
m1 =1
⎡ ⎣mpi i −1
⎫1/qn ⎬ qn ⎪ ⎪ ⎭
∞ min1≤j≤n {mj } −1 mn max 1≤j≤n {mj } m =1 n
⎤qn /pi
n−1
⎦ m−1 j
a(i) mi
j=1(j=i)
⎫1/qn ⎬ qn ⎪ ⎪ ⎭
.
mn =1
⎤
mi+1 =1 mi−1
pi −1 ⎦ a(i) m m−1 mi i j
∞ mi =1
i (mi )mpi i −1
∞ min1≤j≤n {mj } ··· max 1≤j≤n {mj } =1 m =1
∞
1
⎫1/pi pi ⎬
j=1(j=i)
=
= 1, applying H¨ older’s inequality once more and (12), gives
i=1
n−1
.
⎤qn /pi
n
⎧ ⎡ n−1 ∞ ∞ ∞ ⎨ ⎣ ··· ⎩ ×
qn
1
mn−1 =1 n−1
1
⎤qn /pi
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1
⎡
⎧ ∞ ⎨
=
∞ min1≤j≤n {mj } max 1≤j≤n {mj } m =1
1 ⎦ mj
···
mn =1
Since
···
mn−1 =1
⎡
J1
∞
mn
a(i) mi
⎭ p i
1/pi ,
and (13) follows. 100
(b) It is similar to the proof of (a) with the reverse H¨ older’s inequality for pi < 0(i = 1, · · · , n), 0 < pn < 1.
11
3. Main results In this section, lp,ϕ denotes the space of real-valued sequences a = {ami }∞ mi =1 satisfying ||a||p,ϕ < ∞, where ⎧ ⎫ 1/p ∞ ⎨ ⎬ p−1 p , lp,ϕ := a = {ami }∞ : a ∈ R, ||a|| = m |a | < ∞ mi p,ϕ mi mi =1 i ⎩ ⎭ mi =1
105
and ϕ(mi ) = mp−1 (i = 1, · · · , n). For convenience, we set i ∞
I :=
···
mn =1
J :=
∞ n min1≤j≤n {mj } (i) a , max1≤j≤n {mj } i=1 mi m =1 1
⎧ ∞ ⎨ ⎩
mn
⎛ ⎞qn ⎫1/qn ∞ ∞ n−1 1 ⎝ min1≤j≤n {mj } (i) ⎠ ⎬ ··· a , ⎭ mn m =1 max1≤j≤n {mj } i=1 mi =1 m =1 n−1
1
and ⎧ ⎛ ⎞qn ⎫1/qn ∞ ∞ ∞ n−1 1−qn ⎬ ⎨ [1 − θ(m )] min {m } n 1≤j≤n j ⎝ ⎠ ··· a(i) . Jˆ := ⎭ ⎩ mn max1≤j≤n {mj } i=1 mi m =1 m =1 m =1 n
n−1
1
Theorem 1. If n ∈ N\{1}, pi > 1(i = 1, · · · , n),
n
1 i=1 pi
= 1, q1n = 1 −
1 pn ,
a(i) ∈ lpi ,ϕi , ||a(i) ||pi ,ϕi > 0(i = 1, · · · , n), then the following inequalities hold 110
and are equivalent: I < n!
n
||a(i) ||pi ,ϕi ,
(18)
i=1
J < n!
n−1
||a(i) ||pi ,ϕi ,
(19)
i=1
where the constant factor n! in the above inequalities is the best possible. Proof. By (13) and (17), (19) follows. Since inequality gives I=
∞
⎡
mn =1 115
×
∞
n ⎣m−1/q n
···
mn−1 =1 n−1 i=1
a(i) mi
#
1 pn
+ q1n = 1, applying H¨ older’s
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1 1
$ n (n) m1/q amn ≤ J||a(n) ||pn ,ϕn . n
12
(20)
From (19), that (18) follows. Assuming (18) holds and setting a(n) mn
⎛ ⎞qn −1 ∞ ∞ n−1 1 ⎝ min1≤j≤n {mj } (i) ⎠ := ··· a , mn m =1 max1≤j≤n {mj } i=1 mi m =1 n−1
1
then we find J qn −1 = ||a(n) ||pn ,ϕn . By (17), we have J < ∞. If J = 0, (19) is trivial so we assume that J > 0, applying (18) gives ||a(n) ||ppnn ,ϕn = J qn = I < n!
||a(n) ||ppnn −1 ,ϕn = J < n! 120
n
||a(i) ||pi ,ϕi ,
i=1
n−1
||a(i) ||pi ,ϕi ,
i=1
so that (19) holds, which is equivalent to (18). (i)
− pε −1
To see that the constant n! is the best possible, we take a ˜mi = mi
i
(i =
1, · · · , n), where ε is small and positive. If there exists a constant K ≤ n!, such that (18) is still valid when n! is replaced by K, applying (16) gives εI˜ = ≥ =
∞
∞ n min1≤j≤n {mj } (i) ··· a ˜ ε max1≤j≤n {mj } i=1 mi mn =1 m1 =1 ∞ ∞ n min1≤j≤n {xj } − pεj −1 ε ··· x dx1 · · · dxn max1≤j≤n {xj } j=1 j 1 1
n! + o(1)(ε → 0+ ).
On the other hand, εI˜ <
εK
n
||˜ a(i) ||pi ,ϕi = εK
i=1
≤ 125
εK 1 +
∞ 1
dx x1+ε
∞ m=1
1 m1+ε
= K(1 + ε).
Letting ε → 0+ , it follows that n! ≤ K. Hence n! = K is the best possible constant factor of (18). By the equivalence, the constant factor n! in (19) is still the best possible. Otherwise, we would achieve a contradiction by (20) that the constant in (18)
is not the best possible. 13
130
Under the hypothesis of Theorem 1, we define a Multiple
n−1 Hilbert-type discrete operator T : i=1 lpi ,ϕi → lqn ,ϕ1/(1−pn ) by Remark 1.
n
(T a)(mn ) :=
∞
···
mn−1 =1
∞ n−1 min1≤j≤n {mj } (i) a , max1≤j≤n {mj } i=1 mi m =1 1
% & n−1 where a = a(1) , · · · a(n−1) ∈ i=1 lpi ,ϕi . (19) may be re-written as ||T a||qn ,ϕ1/(1−pn ) < n! n
n−1
||a(i) ||pi ,ϕi ,
(21)
i=1
and therefore T a ∈ lqn ,ϕ1/(1−pn ) . Hence T is a bounded linear operator with n
||T || ≤ n!. Since the constant factor in (19) is the best possible, we have ||T || := a(=θ)∈
sup
n−1 i=1
lpi ,ϕi
||T a||qn ,ϕ1/(1−pn ) n = n!.
n−1 (i) ||a ||pi ,ϕi i=1
Remark 2. The following inequality is obtained by taking n = 2 in (18)
135
1/p ∞ 1/q ∞ ∞ ∞ min{m, n} a m bn < 2 np−1 apn nq−1 bqn , max{m, n} n=1 m=1 n=1 n=1
(22)
which is given by [8]. Therefore we give a multiple generalization of [8]. In order to obtain reverse inequalities corresponding to those in Theorem 1, we define the function Φn (mn ) := (1 − θ(mn ))ϕn (mn ). For pi < 1(pi = 0), the space lpi ,ϕi is not normal space. But we still use them as the formal symbols in 140
the following. Theorem 2. If n ∈ N\{1}, pi < 0(i = 1, · · · , n − 1), 0 < pn < 1, 1,
1 qn
= 1−
1 pn ,
a
(i)
n
1 i=1 pi
=
(i)
∈ lpi ,ϕi , ||a ||pi ,ϕi > 0(i = 1, · · · , n), then the following
inequalities hold and are equivalent: I > n!
n−1
||a(i) ||pi ,ϕi ||a(n) ||pn ,Φn ,
(23)
i=1
Jˆ > n!
n−1
||a(i) ||pi ,ϕi ,
i=1 145
where the constant factor n! in the above inequalities is the best possible.
14
(24)
Proof. For 0 < pn < 1, we have qn < 0. By (13), (14), the reverse of (17) and the assumptions, we get (24). By using reverse H¨ older’s inequality ([20]), we obtain I
=
⎡
∞ mn =1
×
n−1 i=1
≥
∞
n ⎣( n (mn ))−1/pn m−1/q n
a(i) mi
···
mn−1 =1
# ( n (mn ))
1/pn
n (n) m1/q am n n
∞ min1≤j≤n {mj } max1≤j≤n {mj } m =1 1
$
Jˆ · ||a(n) ||pn ,Φn .
(25)
Then we get (23) from (24). 150
On the other hand, assuming (23) holds and setting a(n) mn
[1 − θ(mn )] := mn
⎛
1−qn
⎝
⎞qn −1 ∞ n−1 min1≤j≤n {mj } (i) ⎠ ··· a , max1≤j≤n {mj } i=1 mi =1 m =1
∞
mn−1
1
then we find Jˆqn −1 = ||a(n) ||pn ,Φn . By the reverse of (17), we have Jˆ > 0. If Jˆ = ∞, (24) is trivial so we assume that J < ∞, applying (23) gives ||a(n) ||ppnn ,Φn = Jˆqn = I > n!
n−1
||a(i) ||pi ,ϕi ||a(n) ||pn ,Φn ,
i=1
ˆ ||a(n) ||ppnn −1 ,Φn = J > n!
n−1
||a(i) ||pi ,ϕi ,
i=1
so that (24) holds, which is equivalent to (23). 155
(i)
− pε −1
To see that the constant n! is the best possible, we take a ˜mi = mi
i
(i =
1, · · · , n), where ε is small and positive. If there exists a constant K ≥ n!, such that (23) is still valid when n! is replaced by K, applying (13) gives I˜ :=
∞ mn =1
= ≤
···
∞ n min1≤j≤n {mj } (i) a ˜ max1≤j≤n {mj } i=1 mi m =1 1
∞
1
∞
mn =1
m1+ε n
mn−1 =1
(n! + o(1))
···
∞ n−1 min1≤j≤n {mj } − pεi −1 m max1≤j≤n {mj } j=1 j m =1 1
∞
1 1+ε . m n mn =1 15
On the other hand, in view of (15), we obtain I˜ >
K
n−1
||˜ a(i) ||pi ,ϕi ||˜ a(n) ||pn ,Φn
i=1
=
K
∞ mn =1
1
⎧ ⎨
⎩ m1+ε n
∞ 1−
1 mn =1 O m1+ε+α n ∞ 1 mn =1 m1+ε n
⎫1/pn ⎬ ⎭
.
Consequently, (n! + o(1)) > K 160
⎧ ⎨ ⎩
∞ 1−
mn =1 O ∞
⎫1/pn ⎬ 1+ε+α 1
mn 1 mn =1 m1+ε n
⎭
.
Letting ε → 0+ , it follows that n! ≤ K. Hence n! = K is the best possible constant factor of (23). By the equivalence, the constant factor n! in (24) is still the best possible. Otherwise, we would reach a contradiction by (25) that the constant in (23) is
not the best possible. 165
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