1. Introduction
Recently Chunaev [1] obtained Hölder and Minkowski type inequalities with alternating signs. His results are a supplement to Jensen type inequalities with alternating signs obtained earlier by Szegö [2], Bellman [3, 4], Brunk [5], and others (see [6–11], [12, Section 5.38], and also Remark 7).
In this paper, we intend to give inequalities of Hölder and Minkowski type with more general weights, including both the case of weights with alternating signs and the classical case of nonnegative weights (see, e.g., [12, Section 4.2] and [1, 13]). Namely, weights
p
k
,
k
=
1
,
…
,
n
, satisfying the property
(1)
P
k
⩾
0
,
where
P
k
:
=
∑
m
=
1
k
p
m
,
k
=
1
,
…
,
n
,
are considered. We follow proofs in [1] with several changes in order to obtain our results.
In what follows, we denote nonnegative sequences of real numbers in bold print; for example,
a
=
{
a
k
}
k
=
1
n
or
b
=
{
b
k
}
k
=
1
n
, where
n
is a positive integer or infinity. Expressions like
a
≡
1
mean that all elements of
a
equal
1
. In proofs we use several wellknown inequalities for
α
,
β
⩾
0
and
p
⩾
1
:
(2)
(
α
+
β
)
p
⩽
2
p

1
(
α
p
+
β
p
)
(
Jensen
′
s
inequality
)
,
(3)
α
β
⩽
α
p
p
+
β
q
q
,
1
p
+
1
q
=
1
(
Young
′
s
inequality
)
,
(4)
p
β
p

1
⩽
α
p

β
p
α

β
⩽
p
α
p

1
,
α
>
β
(
see
[
14
]
)
.
2. Hölder Type Inequalities
In this section, we show that there is not a direct analog of Hölder’s inequality in the case of our weights, but one of reverse Hölder’s inequalities exists. Note that reverse Hölder’s inequalities for nonnegative weights are well studied (see [13]).
Theorem 1.
Let
a
and
b
be nonincreasing such that
(5)
0
<
a
⩽
a
k
⩽
A
<
∞
,
0
<
b
⩽
b
k
⩽
B
<
∞
,
where
k
=
1
,
…
,
n
. If, moreover,
P
k
⩾
0
, and
p
,
q
>
1
,
1
/
p
+
1
/
q
=
1
, then
(6)
0
⩽
(
∑
k
=
1
n
p
k
a
k
q
)
1
/
q
(
∑
k
=
1
n
p
k
b
k
p
)
1
/
p
∑
k
=
1
n
p
k
a
k
b
k
⩽
(
p
A
/
a
)
1
/
p
(
q
B
/
b
)
1
/
q
.
The left hand side of (6) should be read as there exists no positive constant, depending on
a
,
A
,
b
,
B
,
p
, or
q
, which bounds the fraction in (6) from below.
Before the proof of Theorem 1, we establish the following fact.
Lemma 2.
Let
a
be nonincreasing and
b
nondecreasing such that
b
k
⩽
B
for
k
=
1
,
…
,
n
. If, moreover,
P
k
⩾
0
for
k
=
1
,
…
,
n
, then
(7)
∑
k
=
1
n
p
k
a
k
b
k
⩽
B
∑
k
=
1
n
p
k
a
k
.
Proof.
Applying the Abel transformation, we have
(8)
B
∑
k
=
1
n
p
k
a
k

∑
k
=
1
n
p
k
a
k
b
k
=
∑
k
=
1
n

1
P
k
(
a
k
(
B

b
k
)

a
k
+
1
(
B

b
k
+
1
)
)
+
P
n
a
n
(
B

b
n
)
,
where the latter expression is nonnegative since the sequences
a
and
{
B

b
k
}
are nonincreasing, and
P
k
⩾
0
. The equality holds, for example, if
b
≡
B
.
Proof.
We denote the fraction in (6) by
F
H
. Applying the Abel transformation to the numerator and the denominator of
F
H
easily yields
F
H
⩾
0
. But we prove even more, namely, that there exist no positive constants bounding
F
H
from below. Following [1], let
p
k
=
(

1
)
k
+
1
,
k
=
1
,
…
,
n
, where
n
is even, and let
a
=
{
a
1
,
a
1
,
a
3
,
a
3
,
…
,
a
n
,
a
n
,
…
}
be positive and nondecreasing. The sequence
b
is arbitrary except such that
b
2
k

1

b
2
k
=
0
for all
k
=
1
,
…
,
n
/
2
. It follows that
(9)
F
H
=
0
·
(
∑
k
=
1
n
(

1
)
k
+
1
b
k
p
)
1
/
p
∑
k
=
1
n
/
2
a
2
k

1
(
b
2
k

1

b
2
k
)
=
0
.
Thus
F
H
cannot be bounded from below by a positive absolute constant or a constant depending on
p
,
q
, maximum or minimum elements of
a
and
b
.
Now we prove the right hand side of (6). Here
N
H
denotes the numerator of
F
H
. First we apply the Abel transformation:
(10)
N
H
=
(
∑
k
=
1
n

1
P
k
(
a
k
q

a
k
+
1
q
)
+
P
n
a
n
q
)
1
/
q
×
(
∑
k
=
1
n

1
P
k
(
b
k
p

b
k
+
1
p
)
+
P
n
b
n
p
)
1
/
p
.
By the right hand side of (4) and the Abel transformation
(11)
N
H
⩽
(
q
A
q

1
)
1
/
q
(
p
B
p

1
)
1
/
p
C
1
/
q
D
1
/
p
×
(
∑
k
=
1
n
C
p
k
a
k
)
1
/
q
(
∑
k
=
1
n
D
p
k
b
k
)
1
/
p
,
where
C
and
D
are arbitrary positive constants. Therefore, (3) after several simplifications gives
(12)
N
H
⩽
(
p
A
)
1
/
p
(
q
B
)
1
/
q
C
1
/
q
D
1
/
p
(
∑
k
=
1
n
p
k
(
C
q
b
k
+
D
p
a
k
)
a
k
b
k
)
.
In the latter expression,
{
C
/
(
q
b
k
)
+
D
/
(
p
a
k
)
}
is nondecreasing and
{
a
k
b
k
}
is nonincreasing, because
a
and
b
are nonincreasing. Hence by Lemma 2
(13)
N
H
⩽
(
p
A
)
1
/
p
(
q
B
)
1
/
q
C
1
/
q
D
1
/
p
max
k
{
C
q
b
k
+
D
p
a
k
}
∑
k
=
1
n
p
k
a
k
b
k
⩽
(
p
A
)
1
/
p
(
q
B
)
1
/
q
(
1
q
b
(
C
D
)
1
/
p
+
1
p
a
(
D
C
)
1
/
q
)
×
∑
k
=
1
n
p
k
a
k
b
k
.
It is easily seen that, in order to get the smallest constant in the latter inequality, we must choose
C
/
D
=
b
/
a
. It gives the right hand side of (6). Note that the constant there belongs to
(
1
;
∞
)
.
Remark 3.
From Theorem 1, it is seen that the constant in the right hand side of (6) tends to infinity as
a
→
0
or
b
→
0
(note that this constant is better than in [1]). Now we give an example of sequences confirming this [1]. In Theorem 1 we suppose that
p
k
=
(

1
)
k
+
1
,
n
=
2
m
+
1
,
a
≡
1
, and
b
=
b
2
m
+
1
→
0
in
b
. It gives
(14)
F
H
=
(
∑
k
=
1
2
m
+
1
(

1
)
k
+
1
a
k
q
)
1
/
q
(
∑
k
=
1
2
m
+
1
(

1
)
k
+
1
b
k
p
)
1
/
p
∑
k
=
1
2
m
+
1
(

1
)
k
+
1
a
k
b
k
=
(
∑
k
=
1
2
m
(

1
)
k
+
1
b
k
p
)
1
/
p
∑
k
=
1
2
m
(

1
)
k
+
1
b
k
.
From the left hand side of (4) we deduce
(15)
F
H
=
(
∑
k
=
1
m
(
b
2
k

1
p

b
2
k
p
)
)
1
/
p
∑
k
=
1
m
(
b
2
k

1

b
2
k
)
⩾
p
1
/
p
(
b
2
m
∑
k
=
1
m
(
b
2
k

1

b
2
k
)
)
1

1
/
p
,
where
1

1
/
p
>
0
. Therefore, for a fixed positive
b
2
m
the sum in the denominator can be made sufficiently small by an appropriate choice of
b
. Consequently,
F
H
can be arbitrarily large. The same is for
a
=
a
2
m
+
1
→
0
.
It is clear that if
p
=
q
=
2
then the constant in the right hand side of (6) equals
2
A
B
(
a
b
)

1
⩾
2
. Now we give a more precise constant belonging to
[
1
;
∞
)
for the case when
a
and
b
satisfy several additional conditions.
Proposition 4.
Let
a
and
b
be nonincreasing such that the sequence
{
a
k
/
b
k
}
is monotone and
0
<
m
⩽
a
k
/
b
k
⩽
M
<
∞
. If, moreover,
P
k
⩾
0
for
k
=
1
,
…
,
n
, then
(16)
0
⩽
∑
k
=
1
n
p
k
a
k
2
∑
k
=
1
n
p
k
b
k
2
(
∑
k
=
1
n
p
k
a
k
b
k
)
2
⩽
1
4
(
m
M
+
M
m
)
2
.
The left hand side of (16) should be read as there exists no positive constant, depending on
m
and
M
, which bounds the fraction in (16) from below.
Proof.
The left hand side inequality follows by the same method as in the proof of Theorem 1. To prove the right hand side we denote the numerator of the fraction in (16) by
N
C
. First we suppose
{
a
k
/
b
k
}
to be nondecreasing, so
1
⩽
a
k
/
(
m
b
k
)
⩽
M
/
m
. Applying (3) with
p
=
q
=
2
yields
(17)
N
C
⩽
1
4
m
2
(
∑
k
=
1
n
p
k
(
a
k
2
+
(
m
b
k
)
2
)
)
2
=
1
4
(
∑
k
=
1
n
p
k
(
a
k
m
b
k
+
m
b
k
a
k
)
a
k
b
k
)
2
.
In the latter expression, the sequence
{
c
k
+
1
/
c
k
}
, where
c
k
=
a
k
/
(
m
b
k
)
, is nondecreasing. Indeed,
{
c
k
}
is nondecreasing and moreover
c
1
⩾
1
. Since
f
(
x
)
=
x
+
1
/
x
is convex for
x
∈
(
0
;
∞
)
and has a minimum at
x
=
1
, the sequence
{
f
(
c
k
)
}
is nondecreasing. From this by Lemma 2
(18)
N
C
⩽
1
4
(
max
k
{
f
(
c
k
)
}
)
2
(
∑
k
=
1
n
(

1
)
k
+
1
a
k
b
k
)
2
,
where
max
k
{
f
(
c
k
)
}
=
m
/
M
+
M
/
m
. Supposing
{
a
k
/
b
k
}
to be nonincreasing and taking into account that
m
/
M
⩽
a
k
/
(
M
b
k
)
⩽
1
, we obtain the right hand side of (16) by the same technique.
It is easily seen that equality in (16) holds, for example, if
a
≡
b
. The fact that the constant in the right hand side of (16) belongs to
[
1
;
∞
)
is obvious.
From the wellknown weighted inequality of arithmetic and geometric means (see, e.g., [14, Chapter 2]) supposing
a
m
⩾
0
and
v
m
>
0
, we have
(19)
∏
m
=
1
M
a
m
⩽
∑
m
=
1
M
v
m
a
m
1
/
v
m
,
∑
m
=
1
M
v
m
=
1
.
This is a multivariable version of Young’s inequality (3). From this we obtain a multivariable version of Theorem 1 (but with less precise constant).
Proposition 5.
Let
x
m
:
=
{
x
m
,
k
}
k
=
1
n
be nonincreasing sequences such that
0
<
a
m
⩽
x
m
,
k
⩽
A
m
<
∞
, where
m
=
1
,
…
,
M
. If, moreover,
P
k
⩾
0
,
k
=
1
,
…
,
n
, and
w
k
>
0
,
k
=
1
,
…
,
n
, are such that
∑
m
=
1
M
w
m
=
1
, then
(20)
0
⩽
∏
m
=
1
M
(
∑
k
=
1
n
p
k
x
m
,
k
1
/
w
m
)
w
m
∑
k
=
1
n
p
k
∏
m
=
1
M
x
m
,
k
⩽
∑
m
=
1
M
A
m
1
/
w
m

1
∏
j
=
1
,
j
≠
m
M
A
j
1
/
w
j

1
w
j
a
j
.
The left hand side of (20) should be read as there exists no positive constant, depending on
a
m
,
A
m
, and
w
m
, which bounds the fraction in (20) from below.
Proof.
Set
F
H
is the fraction in (20). Nonexistence of a positive constant bounding
F
H
from below follows from Theorem 1. To prove the right hand side we denote the numerator of
F
H
by
N
H
. By the Abel transformation
(21)
N
H
=
∏
m
=
1
M
(
∑
k
=
1
n

1
P
k
(
x
m
,
k
1
/
w
m

x
m
,
k
+
1
1
/
w
m
)
+
P
n
x
m
,
n
1
/
w
m
)
w
m
.
The right hand side of (4) and the Abel transformation yields
(22)
N
H
⩽
∏
m
=
1
M
A
m
1
/
w
m

1
w
m
∏
m
=
1
M
(
∑
k
=
1
n
p
k
x
m
,
k
)
w
m
.
Supposing
v
m
=
w
m
in (19), we obtain
(23)
N
H
⩽
∏
m
=
1
M
A
m
1
/
w
m

1
w
m
f
f
f
×
(
∑
k
=
1
n
p
k
(
∑
m
=
1
M
w
m
∏
m
=
1
,
m
≠
k
M
x
m
,
k

1
)
∏
m
=
1
M
x
m
,
k
)
,
where it is obvious that
{
∑
m
=
1
M
w
m
∏
m
=
1
,
m
≠
k
M
x
m
,
k

1
}
k
=
1
n
is nondecreasing and that
{
∏
m
=
1
M
x
m
,
k
}
k
=
1
n
is nonincreasing. Thus by Lemma 2
(24)
N
H
⩽
∏
m
=
1
M
A
m
1
/
w
m

1
w
m
max
k
{
∑
m
=
1
M
w
m
∏
m
=
1
,
j
≠
m
M
x
j
,
k

1
}
f
f
f
×
∑
k
=
1
n
p
k
∏
m
=
1
M
x
m
,
k
.
Several simplifications give the right hand side of (20).
3. Minkowski Type Inequalities
In this section we prove precise Minkowski type inequalities with our weights. As we have already mentioned, these generalize both the case of weight with alternating signs and the case of nonnegative weights (see [1]).
Theorem 6.
Let
a
and
b
be nonnegative nonincreasing sequences, and
P
k
⩾
0
for
k
=
1
,
…
,
n
. Then for
p
⩾
1
(25)
0
⩽
(
∑
k
=
1
n
p
k
a
k
p
)
1
/
p
+
(
∑
k
=
1
n
p
k
b
k
p
)
1
/
p
(
∑
k
=
1
n
p
k
(
a
k
+
b
k
)
p
)
1
/
p
⩽
2
1

1
/
p
,
The constant
2
1

1
/
p
is best possible. The left hand side of (25) should be read as there exists no positive constant, depending on only
p
, which bounds the fraction in (25) from below.
Proof.
Throughout the proof,
F
M
denotes the fraction in (25). Applying the Abel transformation for the numerator and the denominator of
F
M
easily yields
F
M
⩾
0
. Moreover, there exists no positive constant depending on
p
only that bounds
F
M
from below. Indeed [1], for each
p
>
1
there exists a sequence such that
F
M
tends to zero. Supposing that
p
k
=
(

1
)
k
+
1
,
n
⩾
2
,
a
=
{
1,1
,
0
,
…
,
0
,
…
}
and
b
=
{
b
,
0
,
…
,
0
,
…
}
with some
b
>
0
, from the left hand side of (4) we deduce
(26)
F
M
=
b
(
(
1
+
b
)
p

1
)
1
/
p
⩽
b
(
p
b
)
1
/
p
<
b
1

1
/
p
.
In this way
F
M
→
0
as
b
→
0
since
1

1
/
p
>
0
for all
p
>
1
.
Now we prove the right hand side of (25). From (2) we have
(27)
(
(
∑
k
=
1
n
p
k
a
k
p
)
1
/
p
+
(
∑
k
=
1
n
p
k
b
k
p
)
1
/
p
)
p
⩽
2
p

1
(
∑
k
=
1
n
p
k
(
a
k
p
+
b
k
p
)
)
.
Now, before extraction of the
p
th root, it is enough to show that
(28)
∑
k
=
1
n
p
k
(
a
k
p
+
b
k
p
)
⩽
∑
k
=
1
n
p
k
(
a
k
+
b
k
)
p
,
p
⩾
1
.
Inequality (28) by the Abel transformation is equivalent to
(29)
∑
k
=
1
n
p
k
c
k
=
∑
k
=
1
n

1
P
k
(
c
k

c
k
+
1
)
+
P
n
c
n
⩾
0
,
where
c
k
:
=
(
a
k
+
b
k
)
p

(
a
k
p
+
b
k
p
)
. The latter inequality holds since
P
k
⩾
0
for all
k
and
c
k
⩾
c
k
+
1
for
k
=
1
,
…
,
n

1
. Indeed, for the function
f
(
x
,
y
)
=
(
x
+
y
)
p

(
x
p
+
y
p
)
, where
x
⩾
0
,
y
⩾
0
, and
p
⩾
1
, we have
f
x
′
⩾
0
and
f
y
′
⩾
0
. Therefore,
(30)
f
(
a
k
,
y
)
⩾
f
(
a
k
+
1
,
y
)
,
f
(
x
,
b
k
)
⩾
f
(
x
,
b
k
+
1
)
⟹
f
(
a
k
,
b
k
)
⩾
f
(
a
k
+
1
,
b
k
+
1
)
.
This completes the proof of (28).
The precision of the constant
2
1

1
/
p
comes out from the following observation from [1]. If
p
k
=
1
for all
k
,
a
=
{
1
,
…
,
1,0
,
…
,
0
}
(first
n
elements are units) and
b
=
{
n
1
/
p
,
0
,
…
,
0
}
, then after several simplifications we get
(31)
(
∑
k
=
1
n
a
k
p
)
1
/
p
+
(
∑
k
=
1
n
b
k
p
)
1
/
p
(
∑
k
=
1
n
(
a
k
+
b
k
)
p
)
1
/
p
=
2
(
1

1
n
+
(
1
+
1
n
1
/
p
)
p
)

1
/
p
=
2
1

1
/
p

ε
n
,
where positive
ε
n
→
0
as
n
→
∞
.
Remark 7.
The following JensenSteffensen type statement was proved in [15] (see also [12, Section 2.2]).
Let
a
be a nonincreasing positive sequence and
φ
a function convex on
[
a
n
;
a
1
]
such that
φ
(
0
)
=
0
. Then the necessary and sufficient condition on weights
p
k
in order that
(32)
φ
(
∑
k
=
1
n
p
k
a
k
)
⩽
∑
k
=
1
n
p
k
φ
(
a
k
)
,
P
k
=
∑
m
=
1
k
p
m
,
is
0
⩽
P
k
⩽
1
,
k
=
1
,
…
,
n
.
From this point of view, the sufficient condition
P
k
⩾
0
,
k
=
1
,
…
,
n
, in Theorems 1 and 6 seems to be quite close to the necessary one.
4. Further Generalizations
Now we give integral versions of Lemma 2 and Theorems 1 and 6. In what follows, we use the notation
(33)
P
(
x
)
:
=
∫
α
x
p
(
t
)
d
t
,
x
∈
[
α
;
β
]
,
and suppose that all functions of
x
are integrable and differentiable on
[
α
;
β
]
.
Lemma 8.
For
x
∈
[
α
;
β
]
, let
f
(
x
)
be nonnegative and nonincreasing, and let
g
(
x
)
be nondecreasing such that
0
⩽
g
(
x
)
⩽
B
, and
P
(
x
)
⩾
0
. Then
(34)
∫
α
β
f
(
x
)
g
(
x
)
d
P
(
x
)
⩽
B
∫
α
β
f
(
x
)
d
P
(
x
)
.
Proof.
Applying integration by parts gives
(35)
B
∫
α
β
f
(
x
)
d
P
(
x
)

∫
α
β
f
(
x
)
g
(
x
)
d
P
(
x
)
=
P
(
x
)
f
(
x
)
(
B

g
(
x
)
)

α
β

∫
α
β
P
(
x
)
d
(
f
(
x
)
(
B

g
(
x
)
)
)
=
P
(
β
)
f
(
β
)
(
B

g
(
β
)
)
+
∫
α
β
P
(
x
)
(
f
(
x
)
g
′
(
x
)

f
′
(
x
)
(
B

g
(
x
)
)
)
d
x
⩾
0
.
Here we took into account that
P
(
α
)
=
0
;
P
(
x
)
,
f
(
x
)
,
g
′
(
x
)
,
B

g
(
x
)
are nonnegative and
f
′
(
x
)
is nonpositive for
x
∈
[
α
;
β
]
. It is easily seen that equality holds, for example, if
g
(
x
)
≡
B
.
Using Lemma 8 and integration by parts instead of the Abel transformation, we obtain the following results by essential repeating proofs of Theorems 1 and 6. We emphasize that
d
P
(
x
)
may be negative here in contrast to the classical case.
Theorem 9.
For
x
∈
[
α
;
β
]
, let
f
(
x
)
and
g
(
x
)
be nonincreasing and let
(36)
0
<
a
⩽
f
(
x
)
⩽
A
<
∞
,
0
<
b
⩽
g
(
x
)
⩽
B
<
∞
.
If, moreover,
P
(
x
)
⩾
0
,
x
∈
[
α
;
β
]
, and
p
,
q
>
1
,
1
/
p
+
1
/
q
=
1
, then
(37)
0
⩽
(
∫
α
β
f
q
(
x
)
d
P
(
x
)
)
1
/
q
(
∫
α
β
g
p
(
x
)
d
P
(
x
)
)
1
/
p
∫
α
β
f
(
x
)
g
(
x
)
d
P
(
x
)
⩽
(
p
A
/
a
)
1
/
p
(
q
B
/
b
)
1
/
q
.
The left hand side of (37) should be read as there exists no positive constant, depending on
a
,
A
,
b
,
B
,
p
, and
q
, which bounds the fraction in (37) from below.
Theorem 10.
For
x
∈
[
α
;
β
]
, let
f
(
x
)
and
g
(
x
)
be nonnegative and nonincreasing, and
P
(
x
)
⩾
0
. Then
(38)
0
⩽
(
∫
α
β
f
p
(
x
)
d
P
(
x
)
)
1
/
p
+
(
∫
α
β
g
p
(
x
)
d
P
(
x
)
)
1
/
p
(
∫
α
β
(
f
(
x
)
+
g
(
x
)
)
p
d
P
(
x
)
)
1
/
p
⩽
2
1

1
/
p
,
p
⩾
1
.
The constant
2
1

1
/
p
is best possible. The left hand side of (38) should be read as there exists no positive constant, depending only on
p
, which bounds the fraction in (38) from below.
In conclusion we give several examples concerning Theorems 9 and 10. Let
p
(
t
)
=
sin
t
and
x
∈
[
0
;
∞
)
in (33); then
P
(
x
)
=
1

cos
x
⩾
0
, and thus
(39)
0
⩽
(
∫
0
∞
f
q
(
x
)
sin
x
d
x
)
1
/
q
(
∫
0
∞
g
p
(
x
)
sin
x
d
x
)
1
/
p
∫
0
∞
f
(
x
)
g
(
x
)
sin
x
d
x
⩽
(
p
A
/
a
)
1
/
p
(
q
B
/
b
)
1
/
q
,
0
⩽
(
∫
0
∞
f
p
(
x
)
sin
x
d
x
)
1
/
p
+
(
∫
0
∞
g
p
(
x
)
sin
x
d
x
)
1
/
p
(
∫
0
∞
(
f
(
x
)
+
g
(
x
)
)
p
sin
x
d
x
)
1
/
p
⩽
2
1

1
/
p
,
p
⩾
1
.
Appropriate discretization yields inequalities with alternating signs obtained earlier in [1] (the case
p
k
=
(

1
)
k
+
1
in Theorems 1 and 6).
If
P
(
x
)
is nondecreasing for
x
∈
[
α
;
β
]
(i.e.,
d
P
(
x
)
is nonnegative), Theorems 9 and 10 give the classical case of nonnegative weights, for which we can put
1
instead of
0
in the left hand sides of (37) and (38) due to Hölder’s and Minkowski’s inequalities.