Ann. Funct. Anal. 2 (2011), no. 1, 72–83
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
REFINEMENTS OF HÖLDER’S INEQUALITY DERIVED FROM
FUNCTIONS ψp,q,λ AND φp,q,λ
LUDMILA NIKOLOVA1 AND SANJA VAROŠANEC2∗
Communicated by C. P. Niculescu
Abstract. We investigate a convex function ψp,q,λ = max{ψp , λψq }, (1 ≤
q < p ≤ ∞), and its corresponding absolute normalized norm k.kψp,q,λ . We
determine a dual norm and use it for getting refinements of the classical Hölder
inequality. Also, we consider a related concave function φp,q,λ = min{ψp , λψq },
(0 < p < q ≤ 1).
1. Preliminaries
Since the end of 20th century several mathematicians have intensively researched properties of the absolute normalized norms on C2 (see [2], [6], [9],
[10], [11]). In this section we give some properties of it which have impact to the
classical Hölder inequality for two pairs of numbers.
Let us recall that a norm k.k on C2 is said to be absolute if k(x, y)k = k(|x|, |y|)k
for all x, y ∈ C and it is normalized if k(1, 0)k = k(0, 1)k = 1. The set of all
absolute normalized norms on C2 is denoted by Na . Let Ψ denotes the family of
all convex functions ψ on [0, 1] with ψ(0) = ψ(1) = 1 satisfying
max{1 − t, t} ≤ ψ(t) ≤ 1,
(0 ≤ t ≤ 1).
Classes Na and Ψ are in one-to-one correspondence, (see [1]). Namely, if k.k ∈ Na ,
then the function ψ(t) = k(1 − t, t)k belongs to Ψ. Conversely, if ψ ∈ Ψ, then the
Date: Received: 26 March 2011; Accepted: 8 April 2011.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 46B20; Secondary 46B99, 26D15.
Key words and phrases. Hölder’s inequality, absolute normalized norm, concave function,
ψp,q,λ function.
72
REFINEMENTS OF HÖLDER’S INEQUALITY
73
mapping
(
k(x, y)kψ =
(|x| + |y|)ψ
0
|y|
|x|+|y|
, (x, y) 6= (0, 0);
(x, y) = (0, 0)
(1.1)
is a norm and k.kψ ∈ Na .
The mostly investigated examples of absolute normalized norms are lp -norm
k.kp and the norm k.kω,q of the two-dimensional Lorentz sequence space d(2) (ω, q)
(see [3], [4], [7]). The corresponding convex function ψp for the norm k.kp is
1
p
p p
[(1
−
t)
+
t
]
, 1 ≤ p < ∞;
ψp (t) =
max{t, 1 − t},
p = ∞.
A norm k.kω,q of the two-dimensional Lorentz sequence space d(2) (ω, q) is defined
as:
k(x, y)kω,q = (x∗q + ωy ∗q )1/q
where 0 < ω < 1, q ≥ 1, (x∗ , y ∗ ) is a non-increasing rearrangement of (|x|, |y|).
Its corresponding convex function ψω,q is equal to
(
1
((1 − t)q + ωtq ) q , 0 ≤ t ≤ 12 ;
ψω,q (t) =
1
(tq + ω(1 − t)q ) q , 12 ≤ t ≤ 1.
Another example of particular interest is the following. Let 1 ≤ q < p ≤ ∞
1
1
and 2 p − q < λ < 1. Then the mapping
k.kp,q,λ = max{k.kp , λk.kq }
is a norm from Na and the corresponding convex function is defined by
ψp,q,λ (t) = max{ψp (t), λψq (t)}, t ∈ [0, 1].
For ψ ∈ Ψ, the dual of the norm k.kψ is denoted by k.k∗ψ . In [5] the following
results about dual of the norm were proved: The mapping k.k∗ψ is an absolute
normalized norm and the corresponding function ψ ∗ ∈ Ψ is given by
(1 − s)(1 − t) + st
ψ(s)
s∈[0,1]
ψ ∗ (t) = sup
(1.2)
for t with 0 ≤ t ≤ 1. Also, the following generalized Hölder inequality holds for
convex function ψ ∈ Ψ:
|x1 x2 | + |y1 y2 | ≤ k(x1 , y1 )kψ k(x2 , y2 )k∗ψ , (x1 , y1 ), (x2 , y2 ) ∈ C2 .
(1.3)
For example, the dual norm of the space d(2) (ω, q) was completely determined
by Mitani and Saito in [7]. In the same paper they stated that in formula (1.2)
the conditions ”s ∈ [0, 1] and t ∈ [0, 1]” can be replaced with ”s ∈ [0, 1/2] and
t ∈ [0, 1/2]” respectively.
In [8] authors considered a family Ψ̃ of concave functions ψ̃ on [0, 1] with ψ̃(0) =
ψ̃(1) = 1. The mapping k.kψ̃ defined by (1.1) satisfies the inverse Minkowski
74
L. NIKOLOVA, S. VAROŠANEC
inequality, i.e. for u, v, z, w ∈ C the following is valid
k(|u| + |z|, |v| + |w|)kψ̃ ≥ k(|u|, |v|)kψ̃ + k(|z|, |w|)kψ̃ .
Furthermore, if ψ̃ ∈ Ψ̃, then ψ̃(t)
is non-increasing on h0, 1] and
t
decreasing on [0, 1i. Moreover, if 0 ≤ p ≤ r, 0 ≤ q ≤ s, we have
ψ̃(t)
1−t
is non-
k(p, q)kψ ≤ k(r, s)kψ .
For a concave function ψ ∈ Ψ̃ let us define a function ψ∗ by
(1 − s)(1 − t) + st
0≤s≤1
ψ(s)
ψ∗ (t) = inf
for 0 ≤ t ≤ 1. The corresponding map k.k∗ψ is defined by (1.1). Similar concluding
as in [7] gives us that if ψ is symmetric with respect to t = 1/2, then ψ∗ is also
symmetric with respect to t = 1/2 and
ψ∗ (t) =
(1 − s)(1 − t) + st
0≤s≤1/2
ψ(s)
inf
for 1/2 ≤ t ≤ 1, ([8]). Also, the inverse generalized Hölder inequality holds, i.e.
k(x1 , y1 )kψ k(x2 , y2 )k∗ψ ≤ |x1 x2 | + |y1 y2 |,
(1.4)
where x1 , x2 , y1 , y2 ∈ C, ψ ∈ Ψ̃.
Examples of concave functions from the family Ψ̃ are the following:
(i) ψp for p ∈ h0, 1i.
(ii) ψω,q for q ∈ h0, 1i and ω ≥ 1, (see [8]).
1
1
(iii) Let 0 < p < q ≤ 1 and λ ∈ h1, 2 p − q i. The function φp,q,λ = min{ψp , λψq }
belongs to the family Ψ̃.
In this paper we consider functions ψp,q,λ and φp,q,λ . Firstly, in the following
∗
section we consider the function ψp,q,λ . We will determine the dual function ψp,q,λ
and the corresponding map k.k∗ψp,q,λ . Also, we will state and analyse the generalized Hölder inequality and the Cauchy inequality which appear in that case
and find out some refinements of the classical Hölder and the Cauchy inequalities. The last section is devoted to the function φp,q,λ which belongs to Ψ̃. We
will calculate k.kφp,q,λ , φp,q,λ ∗ , the corresponding map k.k∗φp,q,λ and investigate
inequalities which arise from the inverse generalized Hölder inequality (1.4).
2. Function ψp,q,λ
2.1. Case p < ∞. Let 1 ≤ q < p < ∞ and λ ∈ h21/p−1/q , 1i. Let us consider a
function ψp,q,λ (s) = maxs∈[0,1] {ψp (s), λψq (s)}. It is a function from Ψ.
Let s0 ∈ [0, 1/2] be a point such that ψp (s0 ) = λψq (s0 ), i.e. [(1 − s0 )p + sp0 ]1/p =
λ[(1 − s0 )q + sq0 ]1/q . Then we have
ψp (t)
for t ∈ [0, s0 ] ∪ [1 − s0 , 1];
ψp,q,λ (t) =
λψq (t) for t ∈ [s0 , 1 − s0 ]
REFINEMENTS OF HÖLDER’S INEQUALITY
75
and the corresponding absolute normalized norm is
(|x|p + |y|p )1/p
k(x, y)kψp,q,λ = max{k(x, y)kp , λk(x, y)kq } =
λ(|x|q + |y|q )1/q
for
for
y∗
x∗∗
y
x∗
≤k
≥k
s0
where k = 1−s
.
0
The function ψp,q,λ is symmetric, so by Propositions 2 and 3 from [7], the
∗
function ψp,q,λ
is also symmetric and
1
(1 − s)(1 − t) + st
, t ∈ [0, ].
ψp,q,λ (s)
2
s∈[0,1/2]
∗
ψp,q,λ
(t) = sup
∗
Let us calculate the function ψp,q,λ
.
max{A, B} where
(2.1)
∗
From (2.1) we have that ψp,q,λ
(t) =
(1 − s)(1 − t) + st
(1 − s)(1 − t) + st
, B = max
.
s∈[0,s0 ]
s∈[s0 ,1/2]
ψp (s)
λψq (s)
A = max
Firstly, we consider
((1 − s)(1 − t) + st)p
ht (s) =
(1 − s)p + sp
p1
1
= (ft (s)) p .
The first derivative of ft (s) is equal
ft0 (s) =
p(1 − s − t + 2st)p−1 (t(1 − s)p−1 − (1 − t)sp−1 )
((1 − s)p + sp )2
and the unique stationary point on h0, 1/2i is
1
s1 =
t p−1
t
1
p−1
+ (1 − t)
1
p−1
1
=
1+
1−t
t
1 .
p−1
The function ht increases for s ≤ s1 and decreases for s ≥ s1 . Consider now
another function
1
1 ((1 − s)(1 − t) + st)q q
gt (s) =
.
λ
(1 − s)q + sq
The point of the local maximum of gt on interval h0, 1/2i is
1
s2 =
t q−1
t
1
q−1
+ (1 − t)
1
q−1
1
=
1+
1−t
t
1 .
q−1
Since 0 ≤ t ≤ 12 , q < p we have s2 ≤ s1 and there are exist three different
orders: s0 ≤ s2 ≤ s1 , s2 ≤ s1 ≤ s0 and s2 ≤ s0 ≤ s1 .
Case a) If s0 ≤ s2 ≤ s1 , then:
A = max ht (s) = ht (s0 ) =
s∈[0,s0 ]
(1 − s0 )(1 − t) + s0 t
1
((1 − s0 )p + sp0 ) p
= Cs0 (t),
76
L. NIKOLOVA, S. VAROŠANEC
1
1 ((1 − s2 )(1 − t) + s2 t)q q
B = max gt (s) = gt (s2 ) =
s∈[s0 ,1/2]
λ
(1 − s2 )q + sq2
q−1
q
q
1
= ((1 − t) q−1 + t q−1 ) q .
λ
The Hölder inequality for pairs (1 − s0 , s0 ) and (1 − t, t) with exponents q and
q
states:
1−q
q
1
q
(1 − s0 )(1 − t) + s0 t ≤ ((1 − s0 )q + sq0 ) q ((1 − t) 1−q + t 1−q )
1−q
q
.
Using the previous result we get that
Cs0 (t) =
(1 − s0 )(1 − t) + s0 t
1
λ((1 − s0 )q + sq0 ) q
and in this case
∗
ψp,q,λ
(t) =
≤
q−1
q
q
1
((1 − t) q−1 + t q−1 ) q
λ
q−1
q
q
1
((1 − t) q−1 + t q−1 ) q .
λ
Case b) If s2 ≤ s1 ≤ s0 , then
A = max ht (s) = ht (s1 ) =
(1 − s1 )(1 − t) + s1 t
1
((1 − s1 )p + sp1 ) p
s∈[0,s0 ]
B = max gt (s) = gt (s0 ) =
p
(1 − s0 )(1 − t) + s0 t
1
λ((1 − s0 )q + sq0 ) q
As above, using once more the Hölder inequality we see that
s∈[s0 ,1/2]
p
p
∗
ψp,q,λ
(t) = max{[(1 − t) p−1 + t p−1 ]
p−1
p
p
= [(1 − t) p−1 + t p−1 ]
= Cs0 (t).
p
p
, Cs0 (t)} = [(1 − t) p−1 + t p−1 ]
Case c) If s2 ≤ s0 ≤ s1 , then
A = max ht (s) = ht (s0 ) = Cs0 (t),
s∈[0,s0 ]
B = max gt (s) = gt (s0 ) = Cs0 (t)
s∈[s0 ,1/2]
and
∗
ψp,q,λ
(t) = Cs0 (t).
Let us express conditions of above cases in terms of the variable t.
a) If s2 ≥ s0 , then
1
≥ s0
1
q−1
1 + 1−t
t
which is equivalent to the inequality
1
t ≥ t2 =
q−1 .
1−s0
1 + s0
b) If s1 ≤ s0 , then
1
1+
1−t
t
≤ s0
1
p−1
p−1
p
p−1
p
.
REFINEMENTS OF HÖLDER’S INEQUALITY
77
or equivalently
1
t ≤ t1 =
1+
1−s0
s0
p−1 .
Note that since q < p, 0 < s0 < 1/2 we have 0 < t1 < t2 ≤ 12 .
c) And finally, if s2 ≤ s0 ≤ s1 , then t1 ≤ t ≤ t2 .
The following theorem is a result of the above discussion.
Theorem 2.1. Let 1 ≤ q < p < ∞ and λ ∈
∗
ψp,q,λ
is equal

0
0
0
((1 − t)p + tp )1/p ,





 Cs0 (t)
1
0
0
0
∗
ψp,q,λ
(t) =
((1 − t)q + tq )1/q ,

λ



Cs0 (1 − t)


0
0
0
((1 − t)p + tp )1/p ,
h21/p−1/q , 1i. Then the function
0 ≤ t ≤ t1 ;
t1 ≤ t ≤ t2 ;
t2 ≤ t ≤ 1 − t2 ;
1 − t2 ≤ t ≤ 1 − t1 ;
1 − t1 ≤ t ≤ 1,
where p0 and q 0 are conjugate exponents of p and q, i.e.
1
t1 =
1+
and
Cs0 (t) =
1−s0
s0
p−1 , t2 =
(1 − s0 )(1 − t) + s0 t
((1 − s0 )p +
1
sp0 ) p
=
1
p
+ p10 = 1 and 1q + q10 = 1,
1
1+
1−s0
s0
q−1
(1 − s0 )(1 − t) + s0 t
1
λ((1 − s0 )q + sq0 ) q
.
∗
Let us determine the norm which corresponds to the function ψp,q,λ
.
|y|
1
∗
∗
Let t = |x|+|y| ∈ [0, t1 ]. Since t1 ≤ 2 , then |y| ≤ |x|, i.e. x = |x|, y = |y| and the
p−1
y∗
y∗
s0
1
inequality 0 ≤ t ≤ t1 gives x∗ +y∗ ≤ 1−s0 p−1 or x∗ ≤ 1−s0
. Using formula
1+
s0
0
0
0
(1.1) we get that in this case k(x, y)k∗ψp,q,λ = (|x|p + |y|p )1/p . Calculations for
other cases are similar. So, we have the following result.
The norm k(x, y)k∗ψp,q,λ is equal to

0
0
0
y∗

(|x|p + |y|p )1/p ,
≤ k p−1 ;

x∗


1
∗

∗
∗
k p−1 ≤ xy ∗ ≤ k q−1 ;
1 (ky + x ),
p
k(x, y)k∗ψp,q,λ =
(kλ + 1) p



1
0
0
0
y∗

 (|x|q + |y|q |)1/q ,
≥ k q−1
x∗
λ
s0
where k = 1−s
and (x∗ , y ∗ ) is a non-increasing rearrangement of (|x|, |y|).
0
Example 2.2. Let us investigate
the simplest case √when q = 1 and p = 2. Then
√
2
1
1− 2λ2 −1
2λ2 −1
√
λ ∈ h 2 , 1i, t1 = s0 =
, t2 = 12 and k = λ −1−λ
. In that case
2
2
((1 − t)2 + t2 )1/2 , 0 ≤ t ≤ s0 or 1 − s0 ≤ t ≤ 1;
ψ2,1,λ (t) =
1,
s0 ≤ t ≤ 1 − s 0 ,
78
L. NIKOLOVA, S. VAROŠANEC


((1 − t)2 + t2 )1/2 ,




 2s0 − 1 t + 1 − s0 ,
∗
λ
λ
ψ2,1,λ
(t) =
s0
1 − 2s0


t+ ,


λ
λ


((1 − t)2 + t2 )1/2 ,
0 ≤ t ≤ s0 ;
s0 ≤ t ≤ 12 ;
1
2
≤ t ≤ 1 − s0 ;
1 − s0 ≤ t ≤ 1.
The corresponding norms are
k(x, y)kψ2,1,λ =
(
k(x, y)k∗ψ2,1,λ =
(x2 + y 2 )1/2 ,
λ(|x| + |y|),
y∗
x∗∗
y
x∗
(x2 + y 2 )1/2 ,
1 − s0
(ky ∗ + x∗ ),
λ
≤ k;
≥ k,
y∗
x∗
y∗
x∗
≤ k;
≥ k.
1
2.2. Case p = ∞. If p = ∞ and λ ∈ h2− q , 1i, then
k(x, y)kψ∞,q,λ = max{kx, yk∞ , λk(x, y)kq }
and

 1 − t, t ∈ [0, s0 ];
λψq (t), t ∈ [s0 , 1 − s0 ];
ψ∞,q,λ (t) =
 t,
t ∈ [1 − s0 , 1]
where s0 is a point from [0, 1] such that 1 − s0 = λ((1 − t)q + tq )1/q . An easy
calculation, similar to the calculation in the first part of this section gives us that

0 ≤ t ≤ t2 ;

 Cs0 (t)
1
0
0
0
∗
ψ∞,q,λ (t) =
((1 − t)q + tq )1/q , t2 ≤ t ≤ 1 − t2 ;

 λ
Cs0 (1 − t)
1 − t2 ≤ t ≤ 1
and
∗
y
ky ∗ + x∗ ,
≤ k q−1 ;
x∗
=
1
∗
0
0
0
(|x|q + |y|q |)1/q , xy ∗ ≥ k q−1
λ
where q 0 , t2 and k are defined as in the case when p 6= ∞.
∗
∗
Let us observe that ψ∞,q,λ
= limp→∞ ψp,q,λ
and k.k∗ψ∞,q,λ = limp→∞ k.k∗ψp,q,λ . For
∗
q = 1 the function ψ∞,1,λ
was calculated in [5].
(
k(x, y)k∗ψ∞,q,λ
2.3. Refinements of the Hölder inequality and the Cauchy inequality.
As a consequence of the generalized Hölder inequality (1.3) and results about
norms k.kψp,q,λ and k.k∗ψp,q,λ we have six inequalities which are valid in the different
regions. As we will see two of them have a form of the classical Hölder inequality
and some of them are better than the classical Hölder inequality. Because of
simplicity, let x1 , x2 , y1 , y2 be non-negative real numbers. These six inequalities
are the following:
y∗
y∗
(1) If x1∗ ≤ k and x2∗ ≤ k p−1 , then
1
2
0
0
0
x1 x2 + y1 y2 ≤ (xp1 + y1p )1/p (xp2 + y2p )1/p .
REFINEMENTS OF HÖLDER’S INEQUALITY
(2) If
y1∗
x∗1
≤ k and k p−1 ≤
y2∗
x∗2
x1 x2 + y1 y2 ≤
(3) If
y1∗
x∗1
y2∗
x∗2
≤ k and
≤ k q−1 , then
1
(xp + y1p )1/p (ky2∗ + x∗2 ).
(k p + 1)1/p 1
≥ k q−1 , then
x1 x2 + y1 y2 ≤
(4) If
y1∗
x∗1
y2∗
x∗2
≥ k and
79
0
0
1 p
0
(x1 + y1p )1/p (xq2 + y2q )1/q .
λ
≤ k p−1 , then
0
0
0
x1 x2 + y1 y2 ≤ λ(xq1 + y1q )1/q (xp2 + y2p )1/p .
(5) If
y1∗
x∗1
≥ k and k p−1 ≤
y2∗
x∗2
x1 x2 + y1 y2 ≤
(6) If
y1∗
x∗1
y2∗
x∗2
≥ k and
≤ k q−1 , then
(k p
λ
(xq + y1q )1/q (ky2∗ + x∗2 ).
+ 1)1/p 1
≥ k q−1 , then
0
0
0
x1 x2 + y1 y2 ≤ (xq1 + y1q )1/q (xq2 + y2q )1/q .
In the following theorem we give a refinement of the classical Hölder inequality.
s0
Theorem 2.3. Let x1 , x2 , y1 , y2 > 0, 1 ≤ q < p and k = 1−s
. Let p0 and q 0 be a
0
conjugate exponents of p and q respectively, i.e. p1 + p10 = 1 and 1q + q10 = 1.
If
y1∗
x∗1
≤ k and k p−1 ≤
x1 x2 + y1 y2 ≤
If
y1∗
x∗1
(k p
≤ k q−1 , then
0
0
1
0
(xp1 + y1p )1/p (ky2∗ + x∗2 ) ≤ (xp1 + y1p )1/p (xp2 + y2p )1/p . (2.2)
1/p
+ 1)
≥ k and k p−1 ≤
x1 x2 + y1 y2 ≤
y2∗
x∗2
y2∗
x∗2
≤ k q−1 , then
λ
q
q 1/q
q
q 1/q
q0
q 0 1/q 0
∗
∗
(x
+
y
)
(ky
+
x
)
≤
(x
+
y
)
(x
+
y
) . (2.3)
1
1
1
1
2
2
2
2
(k p + 1)1/p
y∗
y∗
Proof. Assume that x1∗ ≤ k and k p−1 ≤ x2∗ ≤ k q−1 . The first inequality in (2.2) is
1
2
a consequence of the generalized Hölder inequality (1.3). The second inequality
p0
p0 1/p0
1
∗
∗
is equivalent to (kp +1)
. Furthemore, the previous
1/p (ky2 + x2 ) ≤ (x2 + y2 )
inequality is equivalent to
0
1
(1 + kt)p
≤ (1 + k p ) p−1
0
p
1+t
for t ∈ [k p−1 , k q−1 ]. Function y(t) =
1
(1+kt)p
1+tp0
0
(2.4)
is non-increasing on [k p−1 , k q−1 ] and
y(k p−1 ) = (1 + k p ) p−1 . So, inequality (2.4) holds and refinement (2.2) is proved.
Inequality (2.3) is proved similarly.
80
L. NIKOLOVA, S. VAROŠANEC
When p = 2 and q = 1, then using the generalized Hölder inequality (1.3) we
have the following inequalites:
y∗
y∗
(1) If x1∗ ≤ k and x2∗ ≤ k, then
1
2
x1 x2 + y1 y2 ≤ (x21 + y12 )1/2 (x22 + y22 )1/2 .
(2) If
y1∗
x∗1
y∗
≥ k, x2∗ ≤ k, then
2
x1 x2 + y1 y2 ≤ λ(|x1 | + |y1 |)(x22 + y22 )1/2 .
(3) If
y1∗
x∗1
y∗
≤ k, x2∗ ≥ k, then
2
x1 x2 + y1 y2 ≤
(4) If
y1∗
x∗1
1 − s0 2
(x1 + y12 )1/2 (x∗2 + ky2∗ ).
λ
y∗
≥ k, x2∗ ≥ k, then
2
x1 x2 + y1 y2 ≤ (1 − s0 )(|x1 | + |y1 |)(x∗2 + ky2∗ ).
The first inequality is the classical Cauchy inequality. As a consequence of
(2.2), the third one is a refinement of the Cauchy inequality, i.e. we have
x1 x2 + y1 y2 ≤
for
y1∗
x∗1
1 − s0 2
(x1 + y12 )1/2 (x∗2 + ky2∗ ) ≤ (x21 + y12 )1/2 (x22 + y22 )1/2
λ
(2.5)
y∗
≤ k, x2∗ ≥ k.
2
y1∗
x∗1
y∗
But for
≥ k, x2∗ ≤ k (conditions of case (2)) we can put better inequality.
2
Let us prove that
1 − s0 ∗
(x1 + ky1∗ ) ≤ λ(|x1 | + |y1 |).
(2.6)
λ
y∗
Putting t = x1∗ the previous inequality becomes (1 − s0 )(1 + kt) ≤ λ2 (1 + t) or
1
t(λ2 − (1 − s0 )k) ≥ 1 − s0 − λ2 . Since s0 ≤ 1/2, then λ2 − (1 − s0 )k = λ2 − s0 =
2
s0
0 −λ
= 1−s
= k. The last inequality
2s20 − 3s0 + 1 ≥ 0 and we can write t ≥ 1−s
λ2 −s0
0
t ≥ k is true in the case (2), so we prove (2.6).
y∗
y∗
If we change the places of (x1 , y1 ) and (x2 , y2 ) in (2.5), then for x1∗ ≥ k, x2∗ ≤ k
1
2
we get
x1 x2 + y1 y2 ≤
1 − s0 2
(x2 + y22 )1/2 (x∗1 + ky1∗ ) ≤ (x21 + y12 )1/2 (x22 + y22 )1/2
λ
This and (2.6) shows that we have improved the inequality from the case (2)
and this improved inequality
1 − s0 2
(x2 + y22 )1/2 (x∗1 + ky1∗ )
λ
is also an refinement of the classical Cauchy inequality.
So in the case (1) we have the classical Cauchy inequality and in cases (2) and
(3) we have refinements of the classical Cauchy inequality.
x1 x2 + y1 y2 ≤
REFINEMENTS OF HÖLDER’S INEQUALITY
81
3. Function φp,q,λ
Let 0 < p < q < 1 and λ ∈ h1, 21/p−1/q i. The function φp,q,λ is defined as
φp,q,λ (s) = mins∈[0,1] {ψp (s), λψq (s)}. It is a function from Ψ̃ and it is easy to see
that
ψp (t)
for t ∈ [0, s0 ];
φp,q,λ =
λψq (t) for t ∈ [s0 , 12 ]
which is expanded to the whole interval [0, 1] by symmetrization with respect to
the point 1/2, where s0 ∈ [0, 1/2] is a point such that ψp (s0 ) = λψq (s0 ). Also,
∗
(|x|p + |y|p )1/p
for xy ∗ ≤ k;
∗
k(x, y)kφp,q,λ = min{k(x, y)kp , λk(x, y)kq } =
λ(|x|q + |y|q )1/q for xy ∗ ≥ k
s0
where k = 1−s
. In the following text we will use notation φ = φp,q,λ .
0
Using the same method as in the Section 2, we state the following theorem:
Theorem 3.1. Let 0 < p < q < 1 and λ ∈ h1, 21/p−1/q i. Then the function φ∗ is
equal

0
0
0
((1 − t)p + tp )1/p ,
0 ≤ t ≤ 1 − t1 ;




C
(1
−
t)
1
− t1 ≤ t ≤ 1 − t2 ;

 s0
1
0
0
0
φ∗ (t) =
((1 − t)q + tq )1/q , 1 − t2 ≤ t ≤ t2 ;

λ



C (t)
t2 ≤ t ≤ t1 ;

 s0
p0
p0 1/p0
((1 − t) + t ) ,
t1 ≤ t ≤ 1,
0
0
where p and q are conjugate exponents of p and q, i.e. p1 + p10 = 1 and 1q + q10 = 1,
1
t1 =
1+
and
Cs0 (t) =
1−s0
s0
p−1 , t2 =
(1 − s0 )(1 − t) + s0 t
=
1
1+
1−s0
s0
q−1
(1 − s0 )(1 − t) + s0 t
1 .
((1 − s0 )p + sp0 )
λ((1 − s0 )q + sq0 ) q
The corresponding map k.k∗φ is equal

y∗
p0
p0 1/p0
1−p
(|x|
+
|y|
)
,
;
∗ ≤ k

x


1
∗

∗
∗
k 1−p ≤ xy ∗ ≤ k 1−q ;
1 (ky + x ),
k(x, y)k∗φ =
p
(k + 1) p



y∗
 1 (|x|q0 + |y|q0 |)1/q0 ,
≥ k 1−q
x∗
λ
s0
where k = 1−s
and (x∗ , y ∗ ) is a non-increasing rearrangement of (|x|, |y|).
0
1
p
As a consequence of the inverse generalized Hölder inequality (1.4) and results
about k.kφ and k.k∗φ we have six inequalities which are valid in the different
regions. As we will see two of them have a form of the classical inverse Hölder
inequality. Because of simplicity, let x1 , x2 , y1 , y2 be non-negative real numbers.
These inequalities are the following:
y∗
y∗
(1) If x1∗ ≤ k and x2∗ ≤ k 1−p , then
1
2
0
0
0
x1 x2 + y1 y2 ≥ (xp1 + y1p )1/p (xp2 + y2p )1/p .
82
L. NIKOLOVA, S. VAROŠANEC
(2) If
y1∗
x∗1
≤ k and k 1−p ≤
y2∗
x∗2
x1 x2 + y1 y2 ≥
(3) If
y1∗
x∗1
y2∗
x∗2
≤ k and
≤ k 1−q , then
(k p
≥ k 1−q , then
0
0
1 p
0
(x1 + y1p )1/p (xq2 + y2q )1/q .
λ
x1 x2 + y1 y2 ≥
(4) If
y1∗
x∗1
y2∗
x∗2
≥ k and
1
(xp1 + y1p )1/p (ky2∗ + x∗2 ).
1/p
+ 1)
≤ k 1−p , then
0
0
0
x1 x2 + y1 y2 ≥ λ(xq1 + y1q )1/q (xp2 + y2p )1/p .
(5) If
y1∗
x∗1
≥ k and k 1−p ≤
y2∗
x∗2
x1 x2 + y1 y2 ≥
(6) If
y1∗
x∗1
≥ k and
y2∗
x∗2
≤ k 1−q , then
(k p
λ
(xq + y1q )1/q (ky2∗ + x∗2 ).
+ 1)1/p 1
≥ k 1−q , then
0
0
0
x1 x2 + y1 y2 ≥ (xq1 + y1q )1/q (xq2 + y2q )1/q .
In the following theorem we give a refinement of the classical inverse Hölder
inequality.
Theorem 3.2. Let x1 , x2 , y1 , y2 > 0, 0 < p < q < 1 and k =
be a conjugate exponents of p and q respectively.
y∗
y∗
If x1∗ ≤ k and k 1−p ≤ x2∗ ≤ k 1−q , then
1
s0
.
1−s0
Let p0 and q 0
2
0
0
1
0
x1 x2 + y1 y2 ≥ p
(xp1 + y1p )1/p (ky2∗ + x∗2 ) ≥ (xp1 + y1p )1/p (xp2 + y2p )1/p .
1/p
(k + 1)
If
y1∗
x∗1
≥ k and k 1−p ≤
x1 x2 + y1 y2 ≥
(k p
y2∗
x∗2
≤ k 1−q , then
0
0
λ
0
(xq1 + y1q )1/q (ky2∗ + x∗2 ) ≥ (xq1 + y1q )1/q (xq2 + y2q )1/q .
1/p
+ 1)
Proof. The proof is similar to the proof of Theorem 2.3.
Acknowledgement The research of the first author was partially supported
by the Sofia University SRF under contract No 150/2011. The research of the
second author was supported by the Ministry of Science, Education and Sports
of the Republic of Croatia under grant 058-1170889-1050.
References
[1] F.F. Bonsall and J. Duncan, Numerical Ranges II, London Math. Soc. Lecture Notes Ser.,
Vol. 10, Cambridge Univ. Press, Cambridge, 1973.
[2] H. Cui and F. Wang, Gao’s constants of Lorentz sequence spaces, Soochow J. Math. 33
(2007), 707–717.
[3] M. Kato, On Lorentz spaces lp,q (E), Hiroshima Math. J. 6 (1976), 73–93.
[4] M. Kato and L. Maligranda, On James and Jordan-von Neumann constants of Lorentz
sequences spaces, J. Math. Anal. Appl. 258 (2001), 457–465.
REFINEMENTS OF HÖLDER’S INEQUALITY
83
[5] K.-I. Mitani, S. Oshiro and K.-S. Saito, Smoothness of Ψ-direct sums of Banach spaces,
Math. Inequal. Appl. 8 (2005), 147–157.
[6] K.-I. Mitani and K.-S. Saito, The James constant of absolute norms on R2 , J. Nonlinear
Convex Anal. 4 (2003), 399–410.
[7] K.-I. Mitani and K.-S. Saito, Dual of two dimensional Lorentz sequence spaces, Nonlinear
Anal. 71 (2009), 5238–5247.
[8] L. Nikolova, L.E. Persson and S. Varošanec, On the Beckenbach-Dresher inequality in the
ψ-direct sums of spaces and related results, submitted
[9] K-S. Saito, M. Kato and Y. Takahashi, Von Neumann-Jordan constant of absolute normalized norms on C2 , J. Math. Anal. Appl. 244 (2000), 515–532.
[10] K.-S. Saito, M. Kato and Y. Takahashi, Absolute norms on Cn , J. Math. Anal. Appl. 252
(2000), 879–905.
[11] Y. Takahashi, M. Kato and K.-S. Saito, Strict convexity of absolute norms on C2 and
direct sums of Banach spaces, J. Inequal. Appl. 7 (2002), 179–186.
1
Department of Mathematics and Informatics, Sofia University, Sofia, Bulgaria.
E-mail address: ludmilan@fmi.uni-sofia.bg
2
Department of Mathematics, University of Zagreb, Zagreb, Croatia.
E-mail address: varosans@math.hr