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volume 5, issue 2, article 47, 2004.

Received 20 October, 2003;

accepted 13 March, 2004.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

ON AN INTEGRAL INEQUALITY

J. PE ˇCARI ´C AND T. PEJKOVI ´C

Faculty of Textile Technology University of Zagreb Pierottijeva 6, 10000 Zagreb Croatia.

EMail:pecaric@mahazu.hazu.hr EMail:pejkovic@student.math.hr

c

2000Victoria University ISSN (electronic): 1443-5756 148-03

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On an Integral Inequality J. Peˇcari´c and T. Pejkovi´c

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J. Ineq. Pure and Appl. Math. 5(2) Art. 47, 2004

http://jipam.vu.edu.au

Abstract

In this article we give different sufficient conditions for inequality Rb

af(x)^αdxβ

≥ Rb

af(x)^γdxto hold.

2000 Mathematics Subject Classification:26D15.

Key words: Integral inequality, Inequalities between means.

1. Introduction

In this paper we wish to investigate some sufficient conditions for the following inequality:

(1.1)

Z b a

f(x)^αdx ^β

≥ Z b

a

f(x)^γdx.

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) forα=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 ifn = 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 constantsα,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 (Lemma2.4) and Barnes, Godunova and Levin (Lemma 2.5) established some inequalities which are a special case of (1.1) whenα= 1orγ = 1.

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To obtain some conditions for the inequality (1.1) we will first proceed sim- ilarly to T.K. Pogány ([6]) and in the second part of this article we will be using a method from the paper [4].

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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 allx ∈ [a, b]andγ > 0,inequality (1.1) is equivalent to

(2.1)



 Rb

a f(x)^αdx b−a

!_α¹



αβ γ

(b−a)^β−1^γ ≥ Rb

af(x)^γdx b−a

!¹_γ .

Definition 2.1. Letf be a nonnegative and integrable function on the segment [a, b]. Ther-mean (or ther-th power mean) off is defined as

M^[r](f) :=









 ^Rb

af(x)^rdx b−a

¹_r

(r6= 0,+∞,−∞), exp^R^b

alnf(x)dx b−a

(r= 0),

m (r=−∞),

M (r= +∞).

wherem= inff(x)andM = supf(x)forx∈[a, b].

According to the previous definition inequality (2.1) can be written as

(2.2) M^[α](f)

αβ

γ (b−a)^β−1^γ ≥M^[γ](f)

We will be using the following inequalities:

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Lemma 2.1 (power mean inequality, [2]). Iff is a nonnegative function on [a, b]and−∞ ≤r < s ≤+∞, then

M^[r](f)≤M^[s](f).

Lemma 2.2 (Berwald inequality, [3, 5]). Iff is a nonnegative concave func- tion on[a, b], then for0< r < swe have

M^[s](f)≤ (r+ 1)^1/r

(s+ 1)^1/sM^[r](f).

Lemma 2.3 (Thunsdorff inequality, [3]). Iffis a nonnegative convex function on[a, b]withf(a) = 0, then for0< r < swe have

M^[s](f)≥ (r+ 1)^1/r

(s+ 1)^1/sM^[r](f).

Lemma 2.4 (Nehari inequality, [2]). Letf, g be nonnegative concave func- tions on[a, b]. Then, forp, q >0such thatp⁻¹+q⁻¹ = 1, we have

Z b a

f(x)^pdx

¹pZ b a

g(x)^qdx ¹q

≤N(p, q) Z b

a

f(x)g(x)dx,

whereN(p, q) = _(1+p)1/p⁶(1+q)^1/q.

Lemma 2.5 (Barnes-Godunova-Levin inequality, [3,2]). Letf,gbe nonneg- ative concave functions on[a, b]. Then, forp, q >1we have

Z b a

f(x)^pdx

¹_pZ b a

g(x)^qdx

1 q

≤B(p, q) Z b

a

f(x)g(x)dx,

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whereB(p, q) = _(1+p)^6(b−a)_1/p^1/p+1/q−1_(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 Conditions on functionf Lemma for constantsα,β,γ,a,b (holds for allx∈[a, b]) the proof 1. α≥γ >0,αβ > γ f(x)≥(b−a)^−β+1^αβ−γ Lemma2.1 2. α≤γ >0,αβ <0 0≤f(x)≤(b−a)^−β+1^αβ−γ Lemma2.1 3 (i). α≥γ >0,αβ ≥γ, f(x)≥1 Lemma2.1

(b−a)^β−1^γ ≥1

3 (ii). α≥γ >0,αβ ≤γ, 0≤f(x)≤1 Lemma2.1 (b−a)^β−1^γ ≥1

4 (i). 0< α≤γ,αβ ≥γ f concave Lemma2.2 (b−a)^β−1^γ ≥ ^(α+1)_(γ+1)1/γ^1/α f(x)≥1

4 (ii). 0< α≤γ,αβ ≤γ f concave Lemma2.2 (b−a)^β−1^γ ≥ ^(α+1)_(γ+1)_1/γ^1/α 0≤f(x)≤1

5. 0< α≤γ,αβ > γ f concave,f(x)≥ Lemma2.2 (b−a)^αβ−γ^1−β

(α+1)^1/α (γ+1)^1/γ

_αβ−γ^γ

6. 0< γ ≤α,β <0 f concave,0≤f(x)≤ Lemma2.2 (b−a)^αβ−γ^1−β

(α+1)^1/α (γ+1)^1/γ

_αβ−γ^αβ

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Result Conditions on Conditions on functionf Lemma for constantsα,β,γ,a,b (holds for allx∈[a, b]) the proof 7. 0< γ ≤α,αβ > γ f convex,f(a) = 0,f(x)≥ Lemma2.3

(b−a)

α−γ α(αβ−γ)

(b^α+1−a^α+1)^1/α · ^(α+1)

β αβ−γ

(γ+1)

1 αβ−γx

8. 0< α≤γ,β <0 f convex,f(a) = 0, Lemma2.3 0≤f(x)≤

(b−a)^αβ−γ^1−β _(α+1)1/α

(γ+1)^1/γ

_αβ−γ^αβ

9. 0< γ < α,β <0 f concave, Lemma2.4 0≤f(x)≤(b−a)^αβ−γ^1−β

× ⁶

αβ γ(γ−αβ)

(^2α−γα−γ)

β(α−γ) γ(γ−αβ)(^α+γγ )

β γ−αβ

Remark 2.1. Observe that in the results4(i). and5. it is enough for the condi- tion onf to hold in endpoints of segment[a, b](ie., forf(a)andf(b)).

Remark 2.2. There is only one result in the table obtained with the help of Lemma 2.4 and none with the Lemma2.5 because the constants in the condi- tions are quite complicated.

Remark 2.3. If we make the substitution γ 7→ 1, β 7→ _β¹ in Result 1, Theorem 2.1 in [6] is acquired.

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. Lemma2.1implies that

(2.3) M^[α](f)≥M^[γ](f).

Also

M^[α](f)≥M^[−∞](f)≥(b−a)

−β+1 αβ−γ, so by raising this inequality to a power, we get

(2.4) M^[α](f)^αβ−γ_γ

≥(b−a)^−β+1^γ .

Multiplying (2.3) and (2.4) we get (2.2).

Proof of Result 3 (i). M^[α](f) ≥ 1 because f(x) ≥ 1, so from ^αβ_γ ≥ 1 and Lemma2.1it follows

(2.5) M^[α](f)^αβ_γ

≥M^[α](f)≥M^[γ](f).

By multiplication of (2.5) and the condition(b−a)^β−1^γ ≥1we get (2.2).

Proof of Result 5. From

M^[α](f)≥M^[−∞](f)≥(b−a)^αβ−γ^1−β

(α+ 1)^1/α (γ+ 1)^1/γ

γ αβ−γ

and ^αβ−γ_γ >0we obtain

(2.6) M^[α](f)^αβ−γ_γ

(b−a)^β−1^γ ≥ (α+ 1)^1/α (γ+ 1)^1/γ.

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According to Lemma2.2:

(2.7) M^[α](f)(α+ 1)^1/α

(γ+ 1)^1/γ ≥M^[γ](f).

From (2.6) and (2.7), by multiplication, we arrive at (2.2).

Proof of Result 7. Since

f(x)≥ (b−a)

α−γ α(αβ−γ)

(b^α+1−a^α+1)^1/α · (α+ 1)^αβ−γ^β (γ+ 1)^αβ−γ¹

x

by integration it follows that

(2.8) M^[α](f)≥(b−a)^αβ−γ^1−β

(α+ 1)^1/α (γ+ 1)^1/γ

γ αβ−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 in- cluding the differentiability off and boundedness of its derivative.

J. Peˇcari´c and W. Janous proved in [4] the following theorem.

Theorem 3.1. Let1< p ≤2andr≥3. The differentiable functionf : [0, c]→ Rsatisfiesf(0) = 0and0≤f⁰(x)≤M for all0≤x≤c,csubject to

(3.1) 0< c ≤

p(p−1)2^2−pM^p−r r−1

_r−2p+1¹ .

Then Z c

0

f(x)dx p

≥ Z c

0

f(x)^rdx.

(If f⁰(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 functionf : [0, c] →Rsatisfiesf(0) = 0and0≤f⁰(x)≤ M for all0 ≤x≤ c,csubject to

(3.2) 0< c≤

β(β−1)(α+ 1)^2−βM^αβ−γ γ −α

_{γ−αβ−β+1}¹ .

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Then

Z c 0

f(x)^αdx β

≥ Z c

0

f(x)^γdx.

Remark 3.2. Forα = 1,β =p,γ =r, we get Theorem3.1.

Proof. Fromf(0) = 0and0≤f⁰(x)≤M we obtain 0≤f(x)^α ≤M^αx^α and 0≤

Z x 0

f(t)^αdt≤ M^αx^α+1

α+ 1 for 0≤x≤c.

Now we define

F(x) :=

Z x 0

f(t)^αdt β

− Z x

0

f(t)^γdt.

ThenF(0) = 0andF⁰(x) =f(x)^αg(x), where g(x) := β

Z x 0

f(t)^αdt β−1

−f(x)^γ−α.

Clearly,g(0) = 0andg⁰(x) =f(x)^αh(x), where h(x) := β(β−1)

Z x 0

f(t)^αdt β−2

−(γ−α)f(x)^γ−2α−1f⁰(x).

From the conditions of the theorem we have h(x)≥β(β−1)

M^αx^α+1 α+ 1

β−2

−(γ−α)(M x)^γ−2α−1M

=M^γ−2αx^{(α+1)(β−2)} β(β−1)(α+ 1)^2−βM^αβ−γ−(γ−α)x^{γ−αβ−β+1}

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Thus, since (3.2) is equivalent to

β(β−1)(α+ 1)^2−βM^αβ−γ ≥(γ−α)x^{γ−αβ−β+1}, x∈[0, c],

we have h(x)≥ 0, g⁰(x) ≥ 0, g(x) ≥ 0,F⁰(x) ≥ 0and finallyF(x) ≥ 0. So F(c)≥0.

Substitutingc=a−band translating functionf aunits to the right (f(x)7→

f(x−a)) we obtain the following theorem.

Theorem 3.3. Let α > 0, 1 < β ≤ 2 and γ ≥ 2α + 1. The differentiable functionf : [a, b] →Rsatisfiesf(a) = 0and0≤f⁰(x)≤M for alla ≤x≤ b, where

(3.3) 0< b−a≤

β(β−1)(α+ 1)^2−βM^αβ−γ γ−α

γ−αβ−β+1¹

.

Then the inequality (1.1) holds.

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References

[1] S. MAZOUZI AND F. QI, On an open problem regarding an integral in- equality, J. Inequal. Pure Appl. Math., 4(2)(2003), Art 31. [ONLINE:

http://jipam.vu.edu.au/article.php?sid=269].

[2] D.S. MITRINOVI ´C, Analitiˇcke Nejednakosti, Gradjevinska knjiga, Beograd, 1970. (English edition: Analyitic Inequalities, Springer-Verlag, Berlin-Heidelberg-New York, 1970.)

[3] J.E. PE ˇCARI ´C, F. PROSCHANANDY.L. TONG, Convex Functions, Par- tial Orderings, and Statistical Applications, Academic Press, Inc., Boston- San Diego-New York, 1992.

[4] J.E. PE ˇCARI ´CANDW. JANOUS, On an integral inequality, El. Math., 46, 1991.

[5] J.E. PE ˇCARI ´C, 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.

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[9] K.-W. YU ANDF. QI, A short note on an integral inequality, RGMIA Res.

Rep. Coll., 4(1) (2001), 23–25.