Volume 2008, Article ID 546829,14pages doi:10.1155/2008/546829
Research Article
On Some Extensions of Hardy-Hilbert’s Inequality and Applications
Laith Emil Azar
Department of Mathematics, Al Al-Bayt University, P.O. Box. 130095, Mafraq 25113, Jordan
Correspondence should be addressed to Laith Emil Azar,azar [email protected] Received 23 October 2007; Accepted 2 January 2008
Recommended by Shusen Ding
By introducing some parameters we establish an extension of Hardy-Hilbert’s integral inequality and the corresponding inequality for series. As an application, the reverses, some particular results and their equivalent forms are considered.
Copyrightq2008 Laith Emil Azar. 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.
1. Introduction Iffx,gx≥0, 0<∞
0 f2xdx <∞, and 0<∞
0 g2xdx <∞, thensee1 ∞
0
fxgy
xy dx dy < π
∞ 0
f2xdx
1/2 ∞ 0
g2xdx 1/2
, 1.1
∞
0
fxgy
max{x, y}dx dy <4
∞ 0
f2xdx1/2 ∞ 0
g2xdx1/2
, 1.2
where the constant factorsπ and 4 are the best possible in1.1and1.2, respectively. In- equality 1.1 is called Hilbert’s integral inequality and1.2 is called Hilbert’s type which have been extended by Hardysee2as follows: ifp > 1, 1/p1/q 1,fx,gx > 0, 0<∞
0 fpxdx <∞, and 0<∞
0 gqxdx <∞, then ∞
0
fxgy
xy dx dy < π sinπ/p
∞ 0
fpxdx
1/p ∞ 0
gqxdx 1/q
, 1.3
∞
0
fxgy
max{x, y}dx dy < pq
∞ 0
fpxdx
1/p ∞ 0
gqxdx 1/q
, 1.4
where the constant factorsπ/sinπ/pandpqare the best possible in1.3and1.4, respec- tively. Hardy-Hilbert’s inequality and its applications are important in analysissee3. Re- cently, Yang4gave some generalizations and the reverse form of1.3as follows: ifp > 1, 1/p1/q 1,r >1, 1/r1/s1,λ > 0,fx, gx ≥0, 0<∞
0 xp1−λ/r−1fpxdx <∞, and 0<∞
0 xq1−λ/s−1gqxdx <∞, then ∞
0
fxgy
xλyλ dx dy < π λsinπ/r
∞ 0
xp1−λ/r−1fpxdx
1/p ∞ 0
xq1−λ/s−1gqxdx 1/q
, 1.5
where the constant factorπ/λsinπ/ris the best possible.
The corresponding inequalities for series1.3and1.4are
∞ n1
∞ m1
ambn
mn < π sinπ/p
∞
n0
apn
1/p∞
n0
bqn 1/q
,
∞ n1
∞ m1
ambn
max{m, n} < pq ∞
n0
apn
1/p∞ n0
bqn 1/q
,
1.6
where the sequences{an}and{bn}are such that 0 <∞
n0apn <∞, 0 <∞
n0aqn <∞,and the constant factorπ/sinπ/pandpqare the best possible. By introducing a parameter 0< λ≤2, some extensions of1.6 pq2were given by Yang5,6as
∞ n1
∞ m1
ambn
mλnλ <π λ
∞ n0
n1−λa2n
1/2∞ n0
n1−λbn2 1/2
,
∞ n1
∞ m1
ambn max
mλ, nλ < 4 λ
∞
n0
n1−λa2n
1/2∞
n0
n1−λb2n 1/2
.
1.7
Very recently, in7the following extensions were given:
∞
0
fxgx
Amin{x, y}Bmax{x, y}dx dy < DA, B ∞
0
f2xdx
1/2∞ 0
g2xdx 1/2
,
∞ n1
∞ m1
ambn
Amin{m, n}Bmax{m, n} < DA, B∞ n1
a2n
1/2∞ n1
bn2 1/2
,
1.8
where the constant factorDA, B see7, Lemma 2.1is the best possible in both inequalities.
For more information related to this subject see, for example,8,9.
In this paper by introducing some parameters, we generalize1.8and we obtain the reverse form for each of them. Some particular results and the equivalent form are also consid- ered.
2. Main results
Lemma 2.1. Suppose that λ > 0, A ≥ 0, B > 0. Define the weight coefficients ωλA, B, x and ωλA, B, yby
ωλA, B, x: ∞
0
xλ/2y−1λ/2 Amin
xλ, yλ Bmax
xλ, yλ dy, 2.1
ωλA, B, y: ∞
0
x−1λ/2yλ/2 Amin
xλ, yλ Bmax
xλ, yλ dx, 2.2
thenωλA, B, x ωλA, B, y CλA, Bis a constant defined by
CλA, B
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ 4 λ√
ABarctan A
B forA >0, B >0, 4
λB forA0, B >0.
2.3
Proof. Settingt y/xλ, we get
ωλA, B, x ∞
0
xλ/2y−1λ/2 Amin
xλ, yλ Bmax
xλ, yλ dy, 1
λ ∞
0
t−1/2
Amin{1, t}Bmax{1, t}dt:I,
2.4
iforA,B >0, we obtain
I 1 λ
1 0
t−1/2 AtBdt
∞
1
t−1/2 ABtdt
1 λ
2
√AB √
A/B 0
dt t21 2
√AB √
A/B 0
dt t21
4 λ√
ABarctan
A B;
2.5
ii forA0,B >0, we find
I 1 λ
1 0
t−1/2 B dt
∞
1
t−1/2 Bt dt
4
λB. 2.6
Hence,ωλA, B, x CλA, B. By the symmetry we still haveωλA, B, y CλA, B.
The lemma is proved.
Lemma 2.2. Forp >1or 0< p <1, 1/p1/q1,λ >0,A≥0,B >0 and 0< ε < pλ/2, setting
Jε ∞
1
xλ/2−1−ε/pyλ/2−1−ε/q Amin
xλ, yλ Bmax
xλ, yλ dx dy, 2.7 then forε→0,
1 ε
CλA, B o1
−O1< Jε<1 ε
CλA, B o1
. 2.8
Proof. Settingt x/yλ, we find
Jε ∞
1
xλ/2−1−ε/pyλ/2−1−ε/q Amin
xλ, yλ Bmax
xλ, yλ dx dy 1
λ ∞
1
y−1−ε ∞
y−λ
t−1/2−ε/λp
Amin{t,1}Bmax{t,1}dt dy 1
λε ∞
0
t−1/2−ε/λp
Amin{t,1}Bmax{t,1}dt dy
−1 λ
∞
1
y−1−ε y−λ
0
t−1/2−ε/λp
Amin{t,1}Bmax{t,1}dt dy
1 ε
CλA, B o1
−1 λ
∞
1
y−1−ε y−λ
0
t−1/2−ε/λp AtB dt dy
≥ 1 ε
CλA, B o1
−1 λ
∞
1
y−1 y−λ
0
t−1/2−ε/λp B dt dy 1
ε
CλA, B o1
−O1.
2.9
On the other hand,
Jε ∞
1
xλ/2−1−ε/pyλ/2−1−ε/q Amin
xλ, yλ Bmax
xλ, yλ dx dy
<
∞
1
∞ 0
xλ/2−1−ε/p Amin
xλ, yλ Bmax
xλ, yλ dx
yλ/2−1−ε/qdy
1 ε
CλA, B o1 .
2.10
Hence,2.8is valid. The lemma is proved.
Theorem 2.3. If p > 1, 1/p1/q 1,λ > 0, A ≥ 0,B > 0, fx,gx ≥ 0 such that 0 <
∞
0 xp1−λ/2−1fpxdx <∞, 0<∞
0 xq1−λ/2−1gqxdx <∞, then
S: ∞
0
fxgx Amin
xλ, yλ Bmax
xλ, yλ dx dy
< CλA, B ∞
0
xp1−λ/2−1fpxdx1/p ∞ 0
xq1−λ/2−1gqxdx1/q
,
2.11
where the constant factorCλA, Bdefined in2.3is the best possible. In particular,
iforλAB1,C11,1 π, and inequality2.11reduces to Hardy-Hilbert’s inequality ∞
0
fxgx
xy dx dy < π
∞ 0
xp/2−1fpxdx
1/p ∞ 0
xq/2−1gqxdx 1/q
; 2.12
ii forA0,λB1,C10,1 4 and2.11reduces to Hardy-Hilbert’s-type inequality ∞
0
fxgx
max{x, y}dx dy <4
∞ 0
xp/2−1fpxdx
1/p ∞ 0
xq/2−1gqxdx 1/q
. 2.13
Proof. By the Holder inequality, taking into account2.1, we get
S ∞
0
1 Amin
xλ, yλ Bmax xλ, yλ
1/p
x1−λ/2/q y1−λ/2/pfx
×
1 Amin
xλ, yλ}Bmax xλ, yλ}
1/qy1−λ/2/p x1−λ/2/qgy
dx dy
≤ ∞
0
x1−λ/2p−1yλ/2−1 Amin
xλ, yλ}Bmax
xλ, yλ}fpxdx1/p
× ∞
0
y1−λ/2q−1xλ/2−1 Amin
xλ, yλ Bmax
xλ, yλ gqydy1/q
∞
0
ωλA, B, xxp1−λ/2−1fpxdx
1/p ∞ 0
ωλA, B, yyq1−λ/2−1gqydy 1/q
≤CλA, B ∞
0
xp1−λ/2−1fpxdx1/p ∞ 0
yq1−λ/2−1gqydy1/q
.
2.14
If2.14takes the form of equality, then there exist constantsMandNwhich are not all zero such that
M x1−λ/2p−1yλ/2−1
Amin
xλ, yλ Bmax
xλ, yλ fpx N y1−λ/2q−1xλ/2−1 Amin
xλ, yλ Bmax
xλ, yλ gqy, Mxp1−λ/2fpx Nyq1−λ/2gqy, a.e. in0,∞×0,∞
2.15
Hence, there exists a constantcsuch that
Mxp1−λ/2fpx Nyq1−λ/2gqy c a.e. in0,∞. 2.16
We claim thatM0. In fact, ifM /0, then
xp1−λ/2−1fpx c
Mx a.e. in0,∞ 2.17
which contradicts the fact that 0<∞
0 xp1−λ/2−1fpxdx <∞. Hence, by2.14we get2.11.
If the constant factorCλA, Bis not the best possible, then there exists a positive constant KwithK < CλA, B, thus2.11is still valid if we replaceCλA, BbyK. For 0< ε < pλ/2, setting fand g asfx gx 0 forx ∈ 0,1,fx xλ/2−1−ε/p; gx xλ/2−1−ε/q for x∈1,∞, then we have
K
∞ 0
xp1−λ/2−1fpxdx1/p ∞ 0
xq1−λ/2−1gqxdx1/q
K
∞ 0
x−1−εdx
1/p ∞ 0
x−1−εdx 1/q
K ε.
2.18
By using2.8, we find ∞
0
fx gxdx dy Amin
xλ, yλ Bmax
xλ, yλ ∞
1
∞ 1
xλ/2−1−ε/pdx Amin
xλ, yλ Bmax xλ, yλ
yλ/2−1−ε/qdy
> 1 ε
CλA, B o1
−O1.
2.19
Therefore, we get
1 ε
CλA, B o1
−O1< K
ε 2.20
or
1 λ
CλA, B o1
−εO1< K. 2.21
Forε→ 0, it follows thatCλA, B ≤ Kwhich contradicts the fact thatK < CλA, B.
Hence, the constant factorCλA, Bin2.11is the best possible. The theorem is proved.
Theorem 2.4. If 0 < p < 1, 1/p1/q 1,λ > 0, A ≥ 0, B > 0, fx, gx ≥ 0 such that 0<∞
0 xp1−λ/2−1fpxdx <∞, 0<0<∞
0 xq1−λ/2−1gqxdx <∞, then ∞
0
fxgx Amin
xλ, yλ Bmax
xλ, yλ dx dy
> CλA, B ∞
0
xp1−λ/2−1fpxdx1/p ∞ 0
xq1−λ/2−1gqxdx1/q
,
2.22
where the constant factorCλA, Bdefined in2.3is the best possible. In particular,
iforλAB1,C11,1 π, and inequality2.22reduces to Hardy-Hilbert’s inequality ∞
0
fxgx
xy dx dy > π
∞ 0
xp/2−1fpxdx
1/p ∞
0
xq/2−1gqxdx 1/q
, 2.23
ii forA0,λB1,C10,1 4 and2.22reduces to Hardy-Hilbert’s-type inequality ∞
0
fxgx
max{x, y}dx dy >4
∞ 0
xp/2−1fpxdx
1/p ∞ 0
xq/2−1gqxdx 1/q
. 2.24
Proof. By reverse Holder’s inequality, and the same way, we have2.22. If the constant factor CλA, Bin2.22is not the best possible, then there exists a positive constantH withH >
CλA, Bsuch that2.22is still valid if we replaceCλA, BbyH. For 0< ε < pλ/2, settingf andgas inTheorem 2.3, then we have
H
∞ 0
xp1−λ/2−1fpxdx1/p ∞ 0
xq1−λ/2−1gqxdx1/q
H
∞ 0
x−1−εdx
1/p ∞ 0
x−1−εdx 1/q
H ε .
2.25
By using2.8, we find ∞
0
fx gxdx dy Amin
xλ, yλ Bmax
xλ, yλ ∞
1
∞ 1
xλ/2−1−ε/pdx Amin
xλ, yλ Bmax xλ, yλ
yλ/2−1−ε/qdy
< 1 ε
CλA, B o1 .
2.26
Therefore, we get
1 ε
CλA, B o1
> H
ε, 2.27
or
CλA, B o1≥H. 2.28 Forε→0, it follows thatCλA, B≥Hwhich contradicts the fact thatH > CλA, B. Hence, the constant factorCλA, Bin2.22is the best possible. The theorem is proved.
Theorem 2.5. Under the assumption ofTheorem 2.3, ∞
0
yλp/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dxp
dy <
CλA, Bp∞
0
xp1−λ/2−1fpxdx, 2.29 where the constant factorCλA, Bpis the best possible. Inequalities2.11and2.29are equivalent.
Proof. Setting
gy yλp/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx p−1
, 2.30
then by2.11we have ∞
0
yq1−λ/2−1gqydy ∞
0
yλp/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx p
dy
∞
0
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx
×
yλp/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx p−1
dy
∞
0
fxgy
Amin
xλ, yλ Bmax
xλ, yλ dx dy
≤CλA, B ∞
0
xp1−λ/2−1fpxdx1/p ∞ 0
yq1−λ/2−1gqydy1/q
. 2.31 Hence, we obtain
∞
0
yq1−λ/2−1gqydy≤
CλA, Bp∞
0
xp1−λ/2−1fpxdx. 2.32
Thus, by2.11, both2.31and2.32keep the form of strict inequalities, then we have2.29.
By Holder’s inequality, we find ∞
0
fxgy Amin
xλ, yλ Bmax
xλ, yλ dx dy
∞
0
yλ/2−1/p
∞
0
fx Amin
xλ, yλ Bmax
xλ, yλ dx
y1/p−λ/2gy dy
≤ ∞
0
ypλ/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx
p1/p ∞ 0
yq1−λ/2−1gqydy 1/q
. 2.33 Therefore, by2.29we have2.11, and inequalities2.29and2.11are equivalent. If the constant factor in2.29is not the best possible, then by2.33we can get a contradiction that the constant factor in2.11is not the best possible. The theorem is proved.
Theorem 2.6. Under the assumption ofTheorem 2.4, ∞
0
yλp/2−1
∞ 0
fx Amin
xλ, yλ Bmax
xλ, yλ dx p
dy >
CλA, Bp∞
0
xp1−λ/2−1fpxdx, 2.34 where the constant factorCλA, Bpis the best possible. Inequalities2.22and2.34are equivalent.
The proof ofTheorem 2.6is similar to that ofTheorem 2.5, so we omit it.
3. Discrete analogous
Lemma 3.1. Suppose that 0 < λ ≤ 2,A≥ 0,B > 0. Then the weight coefficientsλA, B, mand λA, B, n,defined, respectively, by
λA, B, m:∞
n1
mλ/2n−1λ/2 Amin
mλ, nλ Bmax
mλ, nλ m∈N, 3.1
λA, B, n:∞
m1
nλ/2m−1λ/2 Amin
mλ, nλ Bmax
mλ, nλ n∈N, 3.2
satisfy the following inequalities:
CλA, B
1−θλA, B, m
< λA, B, m< CλA, B, 3.3 CλA, B
1−θλA, B, n
< λA, B, n< CλA, B, 3.4
whereθλA, B, r: 1/C1A, Br−λ
0 t−1/2/AtBdtO1/rλ/2∈0,1 r ∈N r → ∞,
andCλA, Bis defined by2.3.
Proof. Since 0< λ≤2,A≥0,B >0, byLemma 2.1we get
λA, B, m<
∞
0
mλ/2y−1λ/2 Amin
mλ, yλ Bmax
mλ, yλ dy, ωλA, B, m CλA, B.
3.5
On the other hand, we have
λA, B, m>
∞
1
mλ/2y−1λ/2 Amin
mλ, yλ Bmax
mλ, yλ dy 1
λ ∞
m−λ
t−1/2
Amin{1, t}Bmax{1, t}dt I−1
λ m−λ
0
t−1/2 AtBdt I
1−θλA, B, m ,
3.6
whereI 1/λC1A, Band
0< θλA, B, m 1 C1A, B
m−λ
0
t−1/2
AtBdt <1. 3.7
Since
m−λ
0
t−1/2 AtBdt≤
m−λ
0
t−1/2
B dt 2
Bmλ/2, 3.8
thenθλA, B, m O1/mλ/2. Therefore,3.3is valid. By the symmetry,3.4is still valid. The lemma is proved.
Lemma 3.2. Ifp >0p /1, 1/p1/q1, 0< λ≤2,A≥0,B >0, and 0< ε < pλ/2,setting
Lε ∞
n1
∞ m1
mλ/2−1−ε/pnλ/2−1−ε/q Amin
mλ, nλ Bmax
mλ, nλ , 3.9
then forε→0,
CλA, B−o1∞
n1
1
n1ε < Lε<
CλA, B o1 ∞
n1
1
n1ε. 3.10
Proof. Settingt x/nλin the following, by3.4, we have
Lε<
∞ n1
∞
0
xλ/2−1−ε/pnλ/2−1−ε/q Amin
xλ, nλ Bmax
xλ, nλ dx ∞
n1
1 n1ε
1 λ
∞
0
t−1/2−ε/λp
Amin{t,1}Bmax{t,1}dt
CλA, B o1 ∞
n1
1 n1ε
ε−→0 ,
Lε>
∞ n1
∞
1
xλ/2−1−ε/pnλ/2−1−ε/q Amin
xλ, nλ Bmax
xλ, nλ dx ∞
n1
1 n1ε
1 λ
∞
n−λ
t−1/2−ε/λp
Amin{t,1}Bmax{t,1}dt
∞
n1
1 n1ε
CλA, B o1 −1 λ
n−λ
0
t−1/2−ε/λp AtB dt
>
∞ n1
1 n1ε
CλA, B o1
−1 λ
∞ n1
1 n
n−λ
0
t−1/2−ε/λp
B dt
∞
n1
1 n1ε
CλA, B o1
− 1
λB/2−εB/p ∞ n1
1 n1λ/2−ε/p
∞
n1
1 n1ε
CλA, B o1 − 1 λB/2−εB/p
∞ n1
1 n1λ/2−ε/p
∞ n1
1 n1ε
−1
∞
n1
1 n1ε
CλA, B−o1 ε−→0 .
3.11
Thus, inequality3.10holds. The lemma is proved.
Theorem 3.3. If p > 1, 1/p1/q 1, 0 < λ ≤ 2,A ≥ 0, B > 0,an, bn ≥ 0 such that 0 <
∞
n1np1−λ/2−1apn<∞, 0<∞
n1nq1−λ/2−1bqn<∞, then
D:∞
n1
∞ m1
ambn Amin
mλ, nλ Bmax mλ, nλ
< CλA, B ∞
n1
np1−λ/2−1apn
1/p∞
n1
nq1−λ/2−1bqn 1/q
,
3.12
where the constant factorCλA, Bdefined in2.3is the best possible. In particular,
iforλAB1,C11,1 π, and inequality3.12reduces to Hardy-Hilbert’s inequality ∞
n1
∞ m1
ambn
mn< π ∞
n1
np/2−1apn
1/p∞ n1
nq/2−1bqn 1/q
; 3.13
ii forA0,λB1,C10,1 4 and3.12reduces to Hardy-Hilbert’s-type inequality ∞
n1
∞ m1
ambn
max{m, n} <4 ∞
n1
np/2−1apn
1/p∞ n1
nq/2−1bqn 1/q
. 3.14
Proof. By the Holder inequality, taking into account3.1, we get
D∞
n1
∞ m1
1 Amin
mλ, nλ Bmax mλ, nλ
1/p
m1−λ/2/q n1−λ/2/pam
×
1 Amin
mλ, nλ Bmax mλ, nλ
1/q
n1−λ/2/p m1−λ/2/qbn
≤ ∞
n1
∞ m1
1 Amin
mλ, nλ Bmax mλ, nλ
m1−λ/2p−1 n1−λ/2 apm
1/p
× ∞
n1
∞ m1
1 Amin
mλ, nλ Bmax mλ, nλ
n1−λ/2q−1 m1−λ/2 bnq
1/q
∞
m1
λA, B, mmp1−λ/2−1apm
1/p∞ n1
λA, B, nnq1−λ/2−1bqn 1/q
.
3.15
Then, by3.3and3.4we obtain3.12.
It remains to show that the constant factorCλA, Bis the best possible, to do that we set for 0< ε < pλ/2,ammλ/2−1−ε/p;bnnλ/2−1−ε/q, by3.9we have
∞ n1
∞ m1
ambn Amin
mλ, nλ Bmax
mλ, nλ Lε. 3.16
If there exists a constant 0< K≤CλA, Bsuch that3.12is still valid if we replaceCλA, B byK, then in particular by3.10we find
CλA, B−o1∞
n1
1
n1ε < Lε< K ∞
n1
np1−λ/2−1apn
1/p∞ n1
nq1−λ/2−1bnq 1/q
K ∞ n1
1 n1ε,
3.17
it follows thatCλA, B−o1< Kand thenCλA, B≤Kε→0. Therefor,KCλA, Bis the best constant factor in3.12. The theorem is proved.
Theorem 3.4. If 0 < p < 1, 1/p1/q 1, 0 < λ ≤ 2,A ≥ 0, B > 0,an,bn ≥ 0 such that 0<∞
n1np1−λ/2−1apn<∞, 0<∞
n1nq1−λ/2−1bnq<∞, then ∞
n1
∞ m1
ambn
Amin
mλ, nλ Bmax mλ, nλ
> CλA, B∞ n1
1−θλA, B, n
np1−λ/2−1apn
1/p∞ n1
nq1−λ/2−1bnq 1/q
,
3.18
where the constant factorCλA, Bdefined in2.4is the best possible. In particular, iforλAB1,C11,1 π, and
∞ n1
∞ m1
ambn mn> π
∞ n1
1− 2
πarctan 1 n1/2
np/2−1apn
1/p∞ n1
nq/2−1bqn 1/q
; 3.19
ii forA0,λB1,C10,1 4, and ∞
n1
∞ m1
ambn max{m, n}>4
∞
n1
1− 1
2n1/2
np/2−1apn
1/p∞
n1
nq/2−1bnq 1/q
. 3.20
Proof. By reverse Holder’s inequality, we get D∞
n1
∞ m1
ambn
Amin
mλ, nλ Bmax mλ, nλ ∞
n1
∞ m1
1 Amin
mλ, nλ Bmax mλ, nλ
1/p
m1−λ/2/q n1−λ/2/pam
×
1 Amin
mλ, nλ Bmax mλ, nλ
1/q
n1−λ/2/p m1−λ/2/qbn
≥∞ m1
λA, B, mmp1−λ/2−1apm
1/p∞ n1
λA, B, nnq1−λ/2−1bqn
1/q
.
3.21
Then by3.3and3.4, in view ofq <0, we have3.18. For 0< ε < pλ/2, settingam mλ/2−1−ε/p,bnnλ/2−1−ε/qm, n∈N. If there exists a constantK≥CλA, Bsuch that3.18is still valid if we replaceCλA, BbyK, then in particular by3.9and3.10we find
CλA, B o1 ∞
n1
1
n1ε > Lε
> K ∞
n1
1−θλA, B, n
np1−λ/2−1apn
1/p∞ n1
nq1−λ/2−1bqn 1/q
K ∞
n1
1 n1ε−∞
n1
O
1 nλ/2
1 n1ε
1/p∞
n1
1 n1ε
1/q
K ∞ n1
1 n1ε
1−∞
n1
1 n1ε
−1∞ n1
O
1 nλ/2
1 n1ε
1/p ,
3.22
it follows that
CλA, B o1 > K
1− ∞
n1
1 n1ε
−1∞ n1
O
1 nλ/2
1 n1ε
1/p
. 3.23
Hence, ifε → 0, we get CλA, B ≥ K. Thus,K CλA, Bis the best constant factor in 3.18.
Theorem 3.5. Under the assumption ofTheorem 3.3, ∞
n1
nλp/2−1 ∞
m1
am Amin
mλ, nλ Bmax mλ, nλ
p
<
CλA, Bp∞ m1
mp1−λ/2−1apm, 3.24 where the constant factorCλA, Bpis the best possible. Inequalities3.12and3.24are equivalent.
Proof. Setting
bnnλp/2−1 ∞
m1
am
Amin
mλ, nλ Bmax mλ, nλ
p−1
, 3.25
we get
∞ n1
nq1−λ/2−1bnq∞
n1
∞ m1
ambn
Amin
mλ, nλ Bmax
mλ, nλ . 3.26 By3.12and using the same method ofTheorem 2.5, we obtain3.24. We may show that the constant factor in3.24is the best possible and inequality3.12is equivalent to3.24.
Theorem 3.6. Under the assumption ofTheorem 3.4, ∞
n1
nλp/2−1 ∞
m1
am Amin
mλ, nλ Bmax mλ, nλ
p
>
CλA, Bp∞ m1
mp1−λ/2−1apm, 3.27 where the constant factorCλA, Bpis the best possible. Inequalities3.18and3.27are equivalent.
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