Volume 8 (2007), Issue 2, Article 44, 9 pp.
A REFINEMENT OF HÖLDER’S INEQUALITY AND APPLICATIONS
XUEMEI GAO, MINGZHE GAO, AND XIAOZHOU SHANG
D EPARTMENT OF M ATHEMATICS AND C OMPUTER S CIENCE
N ORMAL C OLLEGE , J ISHOU U NIVERSITY
J ISHOU H UNAN 416000, P EOPLE ’ S R EPUBLIC OF C HINA
mingzhegao@163.com
Received 10 February, 2007; accepted 10 June, 2007
Communicated by S.S. Dragomir
A BSTRACT. In this paper, it is shown that a refinement of Hölder’s inequality can be established
using the positive definiteness of the Gram matrix. As applications, some improvements on
Minkowski’s inequality, Fan Ky’s inequality and Hardy’s inequality are given.
Key words and phrases: Inner product space, Gram matrix, Variable unit-vector, Minkowski’s inequality, Fan Ky’s inequality,
Hardy’s inequality.
2000 Mathematics Subject Classification. 26D15, 46C99.
1. I NTRODUCTION
For convenience, we need to introduce the following notations which will be frequently used
throughout the paper:
! r1
∞
∞
X
X
, kak2 = kak ,
(ar , bs ) =
arn bsn , kakr =
arn
n=1
r
s
Z
(f , g ) =
n=1
∞
r
s
Z
f (x) g (x) dx,
0
∞
kf kr =
r
f (x) dx
r1
,
kf k2 = kf k ,
0
and
Sr (α, y) = αr/2 , y kαk−r/2
,
r
where a = (a1 , a2 , . . . ) are sequences of real numbers, f : [0, ∞) → [0, ∞) are measurable
functions and α and y are elements of an inner product space E of real sequences.
Let a = (a1 , a2 , . . . ) and b = (b1 , b2 , . . . ) be sequences of real numbers in Rn . Then Hölder’s
inequality can be written in the form:
(1.1)
(a, b) ≤ kakp kbkq .
The equality in (1.1) holds if and only if api = kbqi , i = 1, 2, . . . , where k is a constant.
The authors would like to thank the anonymous referee for valuable comments that have been implemented in the final version of this paper.
A Project Supported by scientific Research Fund of Hunan Provincial Education Department (06C657).
100-07
2
G AO X UEMEI , M INGZHE G AO , AND X IAOZHOU S HANG
This inequality is important in function theory, functional analysis, Fourier analysis and analytic number theory, etc. However, there are drawbacks in this inequality. For example, let
a = (a1 , a2 , . . . , an , 0, . . . , 0) ,
b = (0, 0, . . . , bn+1 , bn+2 , . . . , b2n ) ,
a, b ∈ R2n .
If we let ai = bj = 1, i = 1, 2, . . . , n; j = n + 1, n + 2, . . . , 2n, and substitute them into (1.1),
then we have 0 ≤ n. In this case, Hölder’s inequality is meaningless.
In the present paper we establish a new inequality that improves Hölder’s inequality and
remedies the defect pointed out above. At the same time, some significant refinements for
a number of the classical inequalities can be established. As space is limited, only several
applications of the new inequality are given.
2. M AIN R ESULTS
Let α and β be elements of an inner product space E. p
Then the inner product of α and β is
denoted by (α, β) and the norm of α is given by kαk = (α, α). In our previous papers ([1],
[2]), the following result has been obtained by means of the positive definiteness of the Gram
matrix.
Lemma 2.1. Let α, β and γ be three arbitrary vectors of E. If kγk = 1, then
(2.1)
|(α, β)|2 ≤ kαk2 kβk2 − (kαk |x| − kβk |y|)2 ,
where x = (β, γ) , y = (α, γ). The equality in (2.1) holds if and only if α and β are linearly dependent, or γ is a linear combination of α and β, and xy = 0 but x and y are not simultaneously
equal to zero.
For the sake of completeness, we give here a short proof of (2.1), which can also be found in
[2].
Proof of Lemma 2.1. Consider the Gram determinant constructed by the vectors α, β and γ:
(α, α) (α, β) (α, γ) G(α, β, γ) = (β, α) (β, β) (β, γ) .
(γ, α) (γ, β) (γ, γ) According to the positive definiteness of Gram matrix we have G(α, β, γ) ≥ 0, and G(α, β, γ) =
0 if and only if the vectors α, β and γ are linearly dependent.
Expanding this determinant and using the condition kγk = 1 we obtain
G(α, β, γ) = kαk2 kβk2 − (α, β)2 − kαk2 x2 − 2(α, β)xy + kβk2 y 2
≤ kαk2 kβk2 − (α, β)2 − kαk2 x2 − 2 |(α, β)xy| + kβk2 y 2
≤ kαk2 kβk2 − (α, β)2 − {kαk |x| − kβk |y|}2
where x = (β, γ) and y = (α, γ). It follows that the equality holds if and only if the vectors α
and β are linearly dependent; or the vector γ is a linear combination of the vector α and β, and
xy = 0 but x and y are not simultaneously equal to zero.
Applying Lemma 2.1, we can now establish the following refinement of Hölder’s inequality.
Theorem 2.2. Let an, bn ≥ 0, (n = 1, 2, . . . ),
0 < kbkq < +∞, then
(2.2)
1
p
+
1
q
= 1 and p > 1. If 0 < kakp < +∞ and
(a, b) ≤ kakp kbkq (1 − r)m ,
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 44, 9 pp.
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A R EFINEMENT OF H ÖLDER ’ S I NEQUALITY AND A PPLICATIONS
3
where
2
r = (Sp (a, c) − Sq (b, c)) ,
m = min
1 1
,
p q
kck = 1
,
and
ap/2 , c
bq/2 , c ≥ 0.
The equality in (2.2) holds if and only if ap/2 and bq/2 are
linearly
dependent; or if the vector
c is a linear combination of ap/2 and bq/2 , and ap/2 , c bq/2 , c = 0, but the vector c is not
simultaneously orthogonal to ap/2 and bq/2 .
Proof. Firstly, we consider the case p 6= q. Without loss of generality, we suppose that p > q >
p
1. Since p1 + 1q = 1, we have p > 2. Let R = p2 , Q = p−2
, then R1 + Q1 = 1. By Hölder’s
inequality we obtain
(2.3)
(a, b) =
=
∞
X
ak b k
k=1
∞ X
q/p
ak b k
1−q/p
bk
k=1
≤
∞ X
q/p
ak b k
k=1
R
! R1
∞ X
1−q/p
bk
Q
! Q1
k=1
2/p
= ap/2 , bq/2
kbkq(1−2/p)
.
q
The equality in (2.3) holds if and only if ap/2 and bq/2 are linearly dependent. In fact, the equality
in (2.3) holds if and only if for any k, there exists c0 (c0 6= 0) such that
R
Q
q/p
1−q/p
ak b k
= c 0 bk
.
p/2
q/2
It is easy to deduce that ak = c0 bk .
If α, β and γ in (2.1) are replaced by ap/2 , bq/2 and c respectively, then we have
2
(2.4)
ap/2 , bq/2 ≤ kakpp kbkqq (1 − r) ,
where r = (Sp (a, c) − Sq (b, c))2 . Substituting (2.4) into (2.3), we obtain after simplifications
(2.5)
1
(a, b) ≤ kakp kbkq (1 − r) p .
It is known from Lemma 2.1 that the equality in (2.5) holds if and only if ap/2 and bq/2 are lin
early dependent; or if the vector c is a linear combination of ap/2 and bq/2 , and ap/2 , c bq/2 , c =
0, but the vector c is not simultaneously orthogonal to ap/2 and bq/2 .
Note the symmetry of p and q. The inequality (2.2) follows from (2.5).
Secondly, consider the case p = 2. By Lemma 2.1, we obtain:
(2.6)
1
(a, b) ≤ kak kbk (1 − r) 2 ,
2
(b,c)
−
, kck = 1 and (a, c) (b, c) ≥ 0. The equality in (2.6) holds if and
where r = (a,c)
kak
kbk
only if a and b are linearly dependent, or the vector c is a linear combination of a and b, and
(a, c)(b, c) = 0, but (a, c) and (b, c) are not simultaneously equal to zero.
The proof of the theorem is thus completed.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 44, 9 pp.
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4
G AO X UEMEI , M INGZHE G AO , AND X IAOZHOU S HANG
Consider the example given in the Introduction. Let c = (c1 , c2 , . . . , c2n ), c ∈ R2n , where
ci = √1n , i = 1, 2, . . . , n and cj = 0, j = n + 1, n + 2, . . . , 2n. It is easy to deduce that kck = 1
and r = 1. Substituting them into (2.2), it follows that the equality is valid.
The following theorem provides a similar result to Theorem 2.2.
1
p
Theorem 2.3. Let f (x) , g (x) ≥ 0 (x ∈ (0, +∞)),
and 0 < kgkq < +∞, then
1
q
+
= 1 and p > 1. If 0 < kf kp < +∞
(f, g) ≤ kf kp kgkq (1 − r)m ,
(2.7)
where
2
r = (Sp (f, h) − Sq (g, h)) ,
m = min
Z
khk = 1,
khk =
i.e.
1 1
,
p q
,
12
∞
2
h (x)dx
=1
0
and
f p/2 , h g q/2 , h ≥ 0.
The equality in (2.3) holds if and only if f p/2 and g q/2 are linearly
dependent; or the vector h
p/2
q/2
p/2
q/2
is a linear combination of f and g , and f , h g , h = 0, but the vector h is not
simultaneously orthogonal to f p/2 and g q/2 .
Its proof is similar to that of Theorem 2.2. Hence it is omitted.
3. A PPLICATIONS
3.1. A Refinement of Minkowski’s Inequality. We firstly give a refinement of Minkowski’s
inequality for the discrete form.
Theorem 3.1. Let ak , bk ≥ 0, p > 1. If 0 < kakp < +∞ and 0 < kbkp < +∞, then
(3.1)
ka + bkp < kakp + kbkp (1 − r)m ,
where
ka + bkp =
∞
X
! p1
(ak + bk )p
,
k=1
r = min {r (a) , r (b)} ,
(
r (x) =
xp/2 , c
−
m = min
(a + b)p/2 , c
1
1
,1 −
p
p
,
)2
, x = a, b;
ka + bkp/2
p
∞
X
(a + b)p/2 , c =
(ak + bk )p/2 ck ,
kxkp/2
p
k=1
and c is a variable unit-vector.
n
o
Proof. Let m = min p1 , 1 − p1 ,
ka + bkp =
∞
X
! p1
p
(ak + bk )
.
k=1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 44, 9 pp.
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A R EFINEMENT OF H ÖLDER ’ S I NEQUALITY AND A PPLICATIONS
5
By Theorem 2.2, we have
∞
X
(3.2)
p−1
ak (ak + bk )
!1− p1
∞
X
≤ kakp
k=1
k=1
∞
X
∞
X
p
(1 − r(a))m
(ak + bk )
and
(3.3)
bk (ak + bk )p−1 ≤ kbkp
k=1
!1− p1
(1 − r(b))m ,
(ak + bk )p
k=1
where
(
r (x) =
xp/2 , c
kxkp/2
p
(a + b)p/2 , c
−
)2
,
ka + bkp/2
p
∞
X
ka + bkp/2
=
p
(a + b)p/2 , c =
x = a, b,
! 12
(ak + bk )p
k=1
∞
X
,
(ak + bk )p/2 ck ,
k=1
and c is a variable unit-vector.
Adding (3.5) and (3.3) we obtain, after simplifying:
ka + bkp ≤ kakp (1 − r(a))m + kbkp (1 − r(b))m .
(3.4)
Let r = min {r (a) , r (b)}, then the inequality (3.1) follows. This completes the proof of
Theorem 3.1.
If we choose a unit-vector c such that its ith component is 1 and the rest is zero, i.e. c =
(0, 0, . . . , 0, 1 , 0, . . . ), then
(i)
(
r (x) =
p/2
xi
kxkp/2
p
−
(ai + bi )p/2
)2
x = a, b.
ka + bkp/2
p
Similarly, we can establish a refinement of Minkowski’s integral inequality.
Theorem 3.2. Let f (x) , g (x) ≥ 0, p > 1. If 0 < kf kp < +∞ and 0 < kgkp < +∞, then
(3.5)
kf + gkp < kf kp + kgkp (1 − r)m ,
where
∞
Z
p
kf + gkp =
p1
(f (x) + g (x)) dx ,
1
1
r = min {r (f ) , r (g)} , m = min
,1 −
,
p
p
(
)2
tp/2 , h
(f + g)p/2 , h
r (t) =
−
, t = f, g,
ktkp/2
kf + gkp/2
p
p
Z ∞
p/2
(f + g) , h =
(f (x) + g (x))p/2 h (x) dx,
0
0
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6
G AO X UEMEI , M INGZHE G AO , AND X IAOZHOU S HANG
and h is a variable unit-vector, i.e.
Z
12
∞
khk =
2
h (x) dx
= 1.
0
Its proof is similar to that of Theorem 3.1. Hence it is omitted.
Remark 3.3. The variable unit-vector h can be chosen in accordance with our requirements. For
example, we may choose h such that
s
2
h (x) =
.
π (1 + x2 )
3.2. A Strengthening of Fan Ky’s Inequality.
Theorem 3.4. Let A, B and C be three positive definite matrices of order n, 0 ≤ λ ≤ 1. Then


2 m
1
1
|AC| 4
|BC| 4
(3.6)
|A|λ |B|1−λ ≤ |λA + (1 − λ) B| 1 −  1 − 1   ,
1 (A + C) 2
1 (B + C) 2
2
n
where |C| = π ,
2
m = min {λ, 1 − λ}.
Proof. When λ = 0, 1, the inequality (3.3) is obviously valid. Hence we need only consider the
case 0 < λ < 1.
If D is a positive definite matrix of order n, then it is known from [4] that
Z +∞
Z +∞
π n/2
(3.7)
Jn =
···
e−(x,Dx) dx =
1 ,
−∞
−∞
|D| 2
where x = (x1 , x2 , . . . , xn ), and dx = dx1 dx2 · · · dxn .
Let F (x) = e−λ(x,Ax) and G (x) = e−(1−λ)(x,Bx) . If p =
and (2.7) we have
1
λ
and q =
1
,
1−λ
according to (3.4)
π n/2
(3.8)
1
|λA + (1 − λ) B| 2
Z +∞
Z +∞
=
···
F (x) G (x) dx
−∞
Z
−∞
+∞
≤
Z
+∞
F p (x) dx
···
−∞
n/2
p1 Z
+∞
Z
+∞
···
−∞
m
−∞
Gq (x) dx
1q
(1 − r)m
−∞
π (1 − r)
=
1 ,
λ
1−λ 2
|A| |B|
where
2
r = S (F, H) − S (G, H)
1
1
1
− 2(1−λ)
1
− 2λ
2(1−λ)
2λ
=
F , H kF k 1 − G
, H kGk 1
,
1
λ
1
1−λ
λ
− 12 (x,Cx)
where H = e
1−λ
, C is a positive definite matrix of order n, and
Z +∞
12
Z +∞
kHk =
···
H 2 (x) dx
= 1.
−∞
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−∞
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A R EFINEMENT OF H ÖLDER ’ S I NEQUALITY AND A PPLICATIONS
7
By the definition of the variable unit-vector H, it is easy to deduce that |C| = π n . Hence we
have
Z +∞
1
Z +∞
1
2λ
F ,H =
···
F 2λ (x) H (x) dx
−∞
−∞
(
π n/2
=
1 =
1 (A + C) 2
2
|C|
1
(A + C)
2
and
1/2λ
kF k1/λ
Z
+∞
Z
···
=
−∞
12
+∞
F
1/λ
(x) dx
(
=
) 12
|A|1/2
−∞
whence
π n/2
) 12
=
|C|
|A|
14
,
1
|AC| 4
S1/λ (F, H) = 1 .
1 (A + C) 2
2
Similarly,
1
|BC| 4
S1/(1−λ) (G, H) = 1 ,
1 (B + C) 2
2
therefore we obtain

(3.9)
1
4
2
1
4
|AC|
|BC|
r = 
1 − 1  .
1 (A + C) 2
1 (B + C) 2
2
2
It follows from (3.8) and (3.9) that the inequality (3.3) is valid.
3.3. An Improvement of Hardy’s Inequality. We give firstly a refinement of Hardy’s inequality for the discrete form.
P
Theorem 3.5. Let an ≥ 0, βn = n1 nk=1 ak , p1 + 1q = 1 and p > 1. If 0 < kakp < +∞, then
p
(3.10)
kβkp ≤
kakp (1 − r)m ,
p−1
where
(ap/2 , c)
(β p/2 , c)
−
kakp/2
kβkp/2
p
p
n
o
1 1
c is a variable unit-vector and m = min p , q .
r=
!2
,
Proof. Firstly, we estimate the difference of the following two terms:
p
p
(3.11)
βnp −
βnp−1 an = βnp −
(nβn − (n − 1)βn−1 ) βnp−1
p−1
p−1
p1
np
(n − 1)p
p
p
= βn 1 −
+
(βnp )p−1 βn−1
.
p−1
p−1
Applying the arithmetic-geometric mean inequality to the second term on the right-hand side of
(3.11) we get
p1
1
p
p
≤
(p − 1)βnp + βn−1
.
(3.12)
(βnp )p−1 βn−1
p
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8
G AO X UEMEI , M INGZHE G AO , AND X IAOZHOU S HANG
It follows from (3.11) and (3.12) that
p
np
(n − 1)
p
p−1
p
−
βn an ≤ βn 1 −
+
(p − 1)βnp + βn−1
p−1
p−1
p−1
1
p
=
(n − 1)βn−1
− nβnp .
p−1
Summing the above inequality with respect to n, we have
βnp
N
X
N
βnp
n=1
p X p−1
1
p
−
βn an ≤ −
(N βN
) ≤ 0.
p − 1 n=1
p−1
Hence
N
X
N
βnp
p X p−1
≤
β an .
p − 1 n=1 n
βnp
p X p−1
β an .
≤
p − 1 n=1 n
n=1
Letting N → ∞, we get
∞
X
(3.13)
∞
n=1
Applying the inequality (2.2) to the right-hand side of (3.13) we obtain
! p1 ∞
! 1q
∞
∞
X
X
p X
p
an βnp−1 ≤
ap
(3.14)
βn(p−1)q
(1 − r)m
p − 1 n=1
p − 1 n=1 n
n=1
1
p
q
=
kakp kβkpp (1 − r)m ,
p−1
n
o
2
where r = (Sp (a, c) − Sq (β p−1 , c)) , c is a variable unit-vector and m = min p1 , 1q .
We obtain from (3.13) and (3.14) after simplification
p
(3.15)
kβkp ≤
kakp (1 − r)m .
p−1
It is easy to deduce that
Sp (a, c) =
ap/2 , c
and
kakp/2
p
Sq (β p−1 , c) =
(β (p−1)q/2 , c)
kβ p−1 kq/2
q
Hence
r=
a
p/2
,c
kak−p/2
p
− β
p/2
,c
=
kβk−p/2
p
2
(β p/2 , c)
kβkp/2
p
.
,
where c is a variable unit-vector. The proof of the theorem is completed.
A variable unit-vector c can be chosen in accordance with our requirements. For example, we
may choose c ∈ R∞ such that c = (1, 0, 0, . . . ). Obviously, kck = 1 and
2
−p/2
−p/2
p
r = a1 kakp
− kβkp
.
Similarly, we can establish a refinement of Hardy’s integral inequality.
Rx
R∞
Theorem 3.6. Let f (x) ≥ 0, g(x) = x1 0 f (t)dt, p1 + 1q = 1 and p > 1. If 0 < 0 f (t)dt <
+∞, then
p
(3.16)
kgkp <
kf kp (1 − r)m ,
p−1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 44, 9 pp.
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A R EFINEMENT OF H ÖLDER ’ S I NEQUALITY AND A PPLICATIONS
where
r=
(f p/2 , h)
kf kp/2
p
−
(g p/2 , h)
kgkp/2
p
h is a variable unit-vector, i.e.
Z ∞
21
2
khk =
h (t)dt
= 1 and
0
9
!2
,
m = min
1 1
,
p q
.
Proof. Using integration by parts and then applying (2.2) we obtain that
Z ∞
p
p
(3.17)
kgkp =
f, g p−1
g p (t)dt =
p−1
0
p−1 p
≤
kf kp g q (1 − r)m
p−1
p
=
kf kp kgkp−1
(1 − r)m ,
p
p−1
n
o
2
where r = (Sp (f, h) − Sq (g p−1 , h)) , m = min p1 , 1q and h is a variable unit-vector. It is
easy to deduce that
p/2
f p/2 , h
g
,
h
Sp (f, h) =
and
Sq g p−1 , h =
.
p/2
kf kp
kgkp/2
p
It follows that the inequality (3.16) is valid. The theorem is thus proved.
A variable unit-vector h can be chosen in accordance with our requirements. For example,
we may choose h such that h (x) = e−x/2 . Obviously, we then have
Z ∞
12
2
khk =
h (t)dt
= 1.
0
R EFERENCES
[1] MINGZHE GAO, On Heisenberg’s inequality, J. Math. Anal. Appl., 234(2) (1999), 727–734.
[2] TIAN-XIAO HE, J.S. SHIUE AND ZHONGKAI LI, Analysis, Combinatorics and Computing, Nova
Science Publishers, Inc. New York, 2002, 197-204.
[3] JICHANG KUANG, Applied Inequalities, Hunan Education Press, 2nd ed. 1993. MR 95j: 26001.
[4] E.F. BECKENBACK AND R. BELLMAN, Inequalities, 2nd ed., Springer, 1965.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 44, 9 pp.
http://jipam.vu.edu.au/