Journal of Inequalities in Pure and
Applied Mathematics
ON AN INTEGRAL INEQUALITY
J. PEČARIĆ AND T. PEJKOVIĆ
Faculty of Textile Technology
University of Zagreb
Pierottijeva 6, 10000 Zagreb
Croatia.
volume 5, issue 2, article 47,
2004.
Received 20 October, 2003;
accepted 13 March, 2004.
Communicated by: A. Lupaş
EMail: pecaric@mahazu.hazu.hr
EMail: pejkovic@student.math.hr
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
148-03
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Abstract
In this article we give different sufficient conditions for inequality
Rb
γ
a f(x) dx to hold.
R
β
b
α
f(x)
dx
a
≥
2000 Mathematics Subject Classification: 26D15.
Key words: Integral inequality, Inequalities between means.
Contents
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Conditions Based on Inequalities Between Means . . . . . . . . . 5
Conditions Associated with the Functions with Bounded
Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
References
On an Integral Inequality
J. Pečarić and T. Pejković
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1.
Introduction
In this paper we wish to investigate some sufficient conditions for the following
inequality:
Z b
β Z b
α
(1.1)
f (x) dx ≥
f (x)γ dx.
a
a
This is a generalization of the inequalities that appear in the papers [4, 5, 6, 7,
8, 9].
F. Qi in [7] considered inequality (1.1) for α = n + 2, β = 1/(n + 1), γ = 1,
n ∈ N. He proved that under conditions
f ∈ Cn ([a, b]);
f (i) (a) ≥ 0, 0 ≤ i ≤ n − 1;
f (n) (x) ≥ n!, x ∈ [a, b]
the inequality is valid.
Later, S. Mazouzi and F. Qi gave what appeared to be a simpler proof of
the inequality under the same conditions (Corollary 3.6 in [1]). Unfortunately
their proof was incorrect. Namely, they made a false substitution and arrived
at the condition f (x) ≥ (n + 1)(x − a)n which is not true, e.g. for function
f (x) = x − a, whereas this function obviously satisfies the conditions of the
theorem if n = 1.
K.-W. Yu and F. Qi ([9]) and N. Towghi ([8]) gave other conditions for the
inequality (1.1) to hold under this special choice of constants α, β, γ.
T.K. Pogány in [6], by avoiding the assumption of differentiability used in
[7, 8, 9], and instead using the inequalities due to Hölder, Nehari (Lemma 2.4)
and Barnes, Godunova and Levin (Lemma 2.5) established some inequalities
which are a special case of (1.1) when α = 1 or γ = 1.
On an Integral Inequality
J. Pečarić and T. Pejković
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To obtain some conditions for the inequality (1.1) we will first proceed similarly to T.K. Pogány ([6]) and in the second part of this article we will be using
a method from the paper [4].
On an Integral Inequality
J. Pečarić and T. Pejković
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2.
Conditions Based on Inequalities Between Means
We want to transform inequality (1.1) to a form more suitable for us. It can
easily be seen that if f (x) ≥ 0, for all x ∈ [a, b] and γ > 0, inequality (1.1) is
equivalent to
(2.1)
 R
! α1  αβ
γ
b
α
β−1
f (x) dx
a
 (b − a) γ ≥

b−a
Rb
a
f (x)γ dx
b−a
! γ1
.
On an Integral Inequality
Definition 2.1. Let f be a nonnegative and integrable function on the segment
[a, b]. The r-mean (or the r-th power mean) of f is defined as
 R b
r1
r

a f (x) dx




b−a
Rb

a ln f (x)dx
[r]
exp
M (f ) :=
b−a



m



M
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(r 6= 0, +∞, −∞),
(r = 0),
(r = −∞),
(r = +∞).
where m = inf f (x) and M = sup f (x) for x ∈ [a, b].
According to the previous definition inequality (2.1) can be written as
(2.2)
J. Pečarić and T. Pejković
αβ
β−1
M[α] (f ) γ (b − a) γ ≥ M[γ] (f )
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We will be using the following inequalities:
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Lemma 2.1 (power mean inequality, [2]). If f is a nonnegative function on
[a, b] and −∞ ≤ r < s ≤ +∞, then
M[r] (f ) ≤ M[s] (f ).
Lemma 2.2 (Berwald inequality, [3, 5]). If f is a nonnegative concave function on [a, b], then for 0 < r < s we have
M[s] (f ) ≤
(r + 1)1/r [r]
M (f ).
(s + 1)1/s
On an Integral Inequality
Lemma 2.3 (Thunsdorff inequality, [3]). If f is a nonnegative convex function
on [a, b] with f (a) = 0, then for 0 < r < s we have
M[s] (f ) ≥
(r + 1)1/r [r]
M (f ).
(s + 1)1/s
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Lemma 2.4 (Nehari inequality, [2]). Let f , g be nonnegative concave functions on [a, b]. Then, for p, q > 0 such that p−1 + q −1 = 1, we have
Z b
p1 Z b
1q
Z b
p
q
f (x) dx
g(x) dx
≤ N(p, q)
f (x)g(x)dx,
a
a
where N(p, q) =
p
p1 Z
f (x) dx
a
b
q
g(x) dx
a
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.
(1+p)1/p (1+q)1/q
b
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a
Lemma 2.5 (Barnes-Godunova-Levin inequality, [3, 2]). Let f , g be nonnegative concave functions on [a, b]. Then, for p, q > 1 we have
Z
J. Pečarić and T. Pejković
1q
Z
≤ B(p, q)
b
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J. Ineq. Pure and Appl. Math. 5(2) Art. 47, 2004
f (x)g(x)dx,
a
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where B(p, q) =
6(b−a)1/p+1/q−1
.
(1+p)1/p (1+q)1/q
Let us first state our results in a clear table. Each result is an independent set
of conditions that guarantee the inequality (1.1) is valid.
Result
Conditions on
constants α, β, γ, a, b
Conditions on function f
(holds for all x ∈ [a, b])
1.
α ≥ γ > 0, αβ > γ
f (x) ≥ (b − a) αβ−γ
2.
3 (i).
3 (ii).
4 (i).
4 (ii).
5.
6.
α ≤ γ > 0, αβ < 0
α ≥ γ > 0, αβ ≥ γ,
β−1
(b − a) γ ≥ 1
α ≥ γ > 0, αβ ≤ γ,
β−1
(b − a) γ ≥ 1
0 < α ≤ γ, αβ ≥ γ
β−1
1/α
(b − a) γ ≥ (α+1)
(γ+1)1/γ
0 < α ≤ γ, αβ ≤ γ
β−1
1/α
(b − a) γ ≥ (α+1)
1/γ
(γ+1)
0 < α ≤ γ, αβ > γ
0 < γ ≤ α, β < 0
−β+1
0 ≤ f (x) ≤ (b − a)
f (x) ≥ 1
−β+1
αβ−γ
Lemma for
the proof
Lemma 2.1
Lemma 2.1
Lemma 2.1
0 ≤ f (x) ≤ 1
Lemma 2.1
f concave
f (x) ≥ 1
f concave
0 ≤ f (x) ≤ 1
f concave, f (x) ≥
γ
1−β
(α+1)1/α αβ−γ
αβ−γ
(b − a)
(γ+1)1/γ
Lemma 2.2
f concave, 0 ≤ f (x) ≤
αβ
αβ−γ
1−β
1/α
(b − a) αβ−γ (α+1)
(γ+1)1/γ
Lemma 2.2
Lemma 2.2
On an Integral Inequality
J. Pečarić and T. Pejković
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Result
7.
Conditions on
constants α, β, γ, a, b
0 < γ ≤ α, αβ > γ
Conditions on function f
Lemma for
(holds for all x ∈ [a, b])
the proof
f convex, f (a) = 0, f (x) ≥ Lemma 2.3
α−γ
β
(b−a) α(αβ−γ)
(bα+1 −aα+1 )1/α
8.
9.
0 < α ≤ γ, β < 0
0 < γ < α, β < 0
·
(α+1) αβ−γ
1
x
(γ+1) αβ−γ
f convex, f (a) = 0,
0 ≤ f (x) ≤
αβ
αβ−γ
1−β
1/α
(b − a) αβ−γ (α+1)
(γ+1)1/γ
Lemma 2.3
f concave,
1−β
0 ≤ f (x) ≤ (b − a) αβ−γ
Lemma 2.4
On an Integral Inequality
J. Pečarić and T. Pejković
αβ
6 γ(γ−αβ)
×
( 2α−γ
α−γ )
β(α−γ)
γ(γ−αβ)
β
γ−αβ
( α+γ
γ )
Remark 2.1. Observe that in the results 4(i). and 5. it is enough for the condition on f to hold in endpoints of segment [a, b] (ie., for f (a) and f (b)).
Remark 2.2. There is only one result in the table obtained with the help of
Lemma 2.4 and none with the Lemma 2.5 because the constants in the conditions are quite complicated.
Remark 2.3. If we make the substitution γ 7→ 1, β 7→
2.1 in [6] is acquired.
1
β
in Result 1, Theorem
We will prove only a few results after which the method of proving the others
will become clear.
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Proof of Result 1. Lemma 2.1 implies that
M[α] (f ) ≥ M[γ] (f ).
(2.3)
Also
−β+1
M[α] (f ) ≥ M[−∞] (f ) ≥ (b − a) αβ−γ ,
so by raising this inequality to a power, we get
αβ−γ
−β+1
M[α] (f ) γ ≥ (b − a) γ .
(2.4)
On an Integral Inequality
Multiplying (2.3) and (2.4) we get (2.2).
J. Pečarić and T. Pejković
Proof of Result 3 (i). M[α] (f ) ≥ 1 because f (x) ≥ 1, so from
Lemma 2.1 it follows
αβ
(2.5)
M[α] (f ) γ ≥ M[α] (f ) ≥ M[γ] (f ).
By multiplication of (2.5) and the condition (b − a)
β−1
γ
αβ
γ
≥ 1 we get (2.2).
Proof of Result 5. From
[α]
M (f ) ≥ M
and
αβ−γ
γ
(2.6)
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[−∞]
(f ) ≥ (b − a)
1−β
αβ−γ
(α + 1)1/α
(γ + 1)1/γ
> 0 we obtain
αβ−γ
β−1
(α + 1)1/α
M (f ) γ (b − a) γ ≥
.
(γ + 1)1/γ
[α]
≥ 1 and
γ
αβ−γ
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According to Lemma 2.2:
(2.7)
(α + 1)1/α
M (f )
≥ M[γ] (f ).
1/γ
(γ + 1)
[α]
From (2.6) and (2.7), by multiplication, we arrive at (2.2).
Proof of Result 7. Since
α−γ
f (x) ≥
(b − a) α(αβ−γ)
β
·
(α + 1) αβ−γ
(bα+1 − aα+1 )1/α (γ + 1)
1
αβ−γ
On an Integral Inequality
x
J. Pečarić and T. Pejković
by integration it follows that
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(2.8)
1−β
M[α] (f ) ≥ (b − a) αβ−γ
γ
1/α αβ−γ
(α + 1)
(γ + 1)1/γ
.
However, Lemma 2.3 implies inequality (2.7). Thus, from (2.8) and (2.7) we
finally find that inequality (2.2) is valid.
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3.
Conditions Associated with the Functions with
Bounded Derivative
In this section we will prove inequality (1.1) under different assumptions including the differentiability of f and boundedness of its derivative.
J. Pečarić and W. Janous proved in [4] the following theorem.
Theorem 3.1. Let 1 < p ≤ 2 and r ≥ 3. The differentiable function f : [0, c] →
R satisfies f (0) = 0 and 0 ≤ f 0 (x) ≤ M for all 0 ≤ x ≤ c, c subject to
1
p(p − 1)22−p M p−r r−2p+1
(3.1)
0<c≤
.
r−1
Then
Z
p
c
f (x)dx
0
Z
≥
On an Integral Inequality
J. Pečarić and T. Pejković
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c
f (x)r dx.
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0
(If f 0 (x) ≥ M the reverse inequality holds true under the condition that the
second inequality in (3.1) is reversed.)
Remark 3.1. The emphasized words were left out of [4].
The following generalization will be proved:
Theorem 3.2. Let α > 0, 1 < β ≤ 2 and γ ≥ 2α + 1. The differentiable
function f : [0, c] → R satisfies f (0) = 0 and 0 ≤ f 0 (x) ≤ M for all 0 ≤ x ≤
c, c subject to
1
β(β − 1)(α + 1)2−β M αβ−γ γ−αβ−β+1
(3.2)
0<c≤
.
γ−α
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Then
Z
c
α
f (x) dx
0
β
c
Z
f (x)γ dx.
≥
0
Remark 3.2. For α = 1, β = p, γ = r, we get Theorem 3.1.
Proof. From f (0) = 0 and 0 ≤ f 0 (x) ≤ M we obtain
Z x
M α xα+1
α
α α
0 ≤ f (x) ≤ M x and 0 ≤
f (t)α dt ≤
α+1
0
for 0 ≤ x ≤ c.
On an Integral Inequality
Now we define
Z
x
F (x) :=
β Z
α
f (t) dt −
0
J. Pečarić and T. Pejković
x
f (t)γ dt.
0
0
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α
Then F (0) = 0 and F (x) = f (x) g(x), where
Z x
β−1
α
g(x) := β
f (t) dt
− f (x)γ−α .
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Clearly, g(0) = 0 and g 0 (x) = f (x)α h(x), where
β−2
Z x
α
h(x) := β(β − 1)
f (t) dt
− (γ − α)f (x)γ−2α−1 f 0 (x).
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From the conditions of the theorem we have
α α+1 β−2
M x
− (γ − α)(M x)γ−2α−1 M
h(x) ≥ β(β − 1)
α+1
=M
γ−2α (α+1)(β−2)
x
β(β − 1)(α + 1)
2−β
M
αβ−γ
− (γ − α)x
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γ−αβ−β+1
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Thus, since (3.2) is equivalent to
β(β − 1)(α + 1)2−β M αβ−γ ≥ (γ − α)xγ−αβ−β+1 ,
x ∈ [0, c],
we have h(x) ≥ 0, g 0 (x) ≥ 0, g(x) ≥ 0, F 0 (x) ≥ 0 and finally F (x) ≥ 0. So
F (c) ≥ 0.
Substituting c = a − b and translating function f a units to the right (f (x) 7→
f (x − a)) we obtain the following theorem.
On an Integral Inequality
Theorem 3.3. Let α > 0, 1 < β ≤ 2 and γ ≥ 2α + 1. The differentiable
function f : [a, b] → R satisfies f (a) = 0 and 0 ≤ f 0 (x) ≤ M for all a ≤ x ≤
b, where
(3.3)
0<b−a≤
β(β − 1)(α + 1)2−β M αβ−γ
γ−α
Then the inequality (1.1) holds.
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1
γ−αβ−β+1
.
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References
[1] S. MAZOUZI AND F. QI, On an open problem regarding an integral inequality, J. Inequal. Pure Appl. Math., 4(2)(2003), Art 31. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=269].
[2] D.S. MITRINOVIĆ, Analitičke Nejednakosti, Gradjevinska knjiga,
Beograd, 1970. (English edition: Analyitic Inequalities, Springer-Verlag,
Berlin-Heidelberg-New York, 1970.)
[3] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc., BostonSan Diego-New York, 1992.
[4] J.E. PEČARIĆ AND W. JANOUS, On an integral inequality, El. Math., 46,
1991.
[5] J.E. PEČARIĆ, O jednoj nejednakosti L. Berwalda i nekim primjenama,
ANU BIH Radovi, Odj. Pr. Mat. Nauka, 74(1983), 123–128.
[6] T.K. POGÁNY, On an open problem of F. Qi, J. Inequal. Pure Appl.
Math., 3(4) (2002), Art 54. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=206].
[7] F. QI, Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2)
(2000), Art 19. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=113].
[8] N. TOWGHI, Notes on integral inequalities, RGMIA Res. Rep. Coll., 4(2)
(2001), 277–278.
On an Integral Inequality
J. Pečarić and T. Pejković
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[9] K.-W. YU AND F. QI, A short note on an integral inequality, RGMIA Res.
Rep. Coll., 4(1) (2001), 23–25.
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J. Pečarić and T. Pejković
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