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IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. VII (Mar-Apr. 2014), PP 01-04 www.iosrjournals.org

Dual Results of Opial’s Inequality Y.O. Anthonio1, S.O. Salawu2 and S.O. Sogunro3 1

Department of Mathematics, Lagos State University, Ojo, Nigeria. Department of Mathematics, University of Ilorin, Ilorin, Nigeria. 3 Department of Mathematics, Lagos State Polytechnic, Ikorodu, Nigeria. 2

Abstract: In this paper, we used Jensen’s inequality for the case of convex functions, firstly to obtain a Calvert’s generalization and second, to obtain a Maron’s generalization of Opial’s inequality. The main tool was adaptation of Jensen’s inequality for convex functions. Keywords: Integral inequalities, Maron’s inequality and Calvert’s generalization of Opial’s inequality Jensen’s inequality, convex functions.

I.

Introduction:

Opial [8] established the following interesting integral inequality: Let ( x, y)  C [0, b] be such x(0)  x(b)  0 and x(t )  0 in (0, b), then b

x(t ) x (t ) dt 

a

Where

b

b ( x (t )) 2 dt  4a

(1)

b in the best possible constant. 4

In 1067 Maroni [5] obtained a generalized Opial’s inequality by using Holder inequality with indices The result obtained is the following: Theorem 1:

 and  .

Let p(t ) be positive and continuous on

 ,   with  p1 (t )dt  , where   1, x(t ) be absolutely 

function on

 ,   and x(0)  0 . The following inequality holds. 2

x(t ) x (t ) dt 

Where

1

1

2

   1  1    p (t )dt    p(t ) x (t )  dt     2      

 1. equality holds in (2) in and only if

p 

1 

(2)

( s)ds

Calvert [2] also established the following result: Theorem 2: [2] assume that (i) x(t ) is absolutely continuous in  , and x( )  0

f (t ) is continuous, complex-valued, defined in the range of x(t ) and for all real for t of the form

(ii) s

t ( s)   x (u ) du : f ( t ) for all t and f (t ) is real t  0 and is increasing there, 

(iii)

p(t ) is positive, continuous and

p 

1 

(t )dt  , where

1

1

 1. then the following inequality

holds. 2

2

     f (t ) x (t ) dt  F   p1  (t )dt    p(t ) x (t ) dt      www.iosrjournals.org

(3) 1 | Page


Dual Results Of Opial’s Inequality Where F (t )

t

t

0

  f (t )ds, t  0. Equality holds in (3) if and only if x(t )   p1  ( s)ds.

The aim of this paper is to generalize Maroni and Calvert results using Jensen’s inequality.

II.

Some Adaptation of Jensen’s inequalities:

 be continuous and convex function and let h( s, t ) be a non negative function and  be non 1 decreasing function. Let     (t )   (t )   and suppose  has a continuous inverse  (which is Let

necessarily concave). Then,

  (t )    (t )   h( s, t )d ( s)   ( ) 1 ( h( s, t ) )d ( s )       (t )  1  ( t ) (4)    (t )    (t )     d ( s)  d ( s )       (t )    (t )   With the inequality reserved if  is concave. The inequality (4) above is known as Jensen’s inequality for convex function. Setting  (u)  u ,  (t )  t and  (t )  0 in (4), then we obtain.  t  h( s, t )d ( s)  0  ( f (t ))  f  t  0 d (s)   III.

1

1   t     ( h( s, t ) ) l d ( s )   0     t    d  ( s ) 0      

(5)

Main Result:

Before stating our main result in this section, we shall need the following useful Lemma: Lemma 1: Let x(t ),  (t ) and f (u ) be absolutely continuous and non decreasing functions on a, b for with f (t )  0. Let l , k , o,  and and measurable function on

 

0ab

 be real numbers such that   0, o  0 and also let R(t) be non negative

a, b such that

t  x (t )  f   x (t ) R(t )d (t )    (t ) l  y (t )   R(t ) 1  (t ) 1 y (t ).   0  (6) Then the following inequality holds: b

b

x (t ) f (t ) dt   f ( y (t )) y (t )dt

a

a

(7 )

Proof: Setting h( s, t )  x(t ) R(t ) in (5) , we have

 t  x (t ) R(t )d (t )  0  ( f (t ))  f  t  0 d (t )   l By setting f ( (t ))   (t ) in (8) yields

1

1   t     ( x (t ) R(t ) ) l d (t )   0     t    0 d (t )      

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(8)

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Dual Results Of Opial’s Inequality

t  f   x (t ) R(t )d (t )    0   (t ) l

1

t l   f x (t ) R(t )  d (t )    0   (t ) 

(9)

Hence, 

1 t  t  l    f  x(t ) R(t )d (t )   (t )   f  x(t ) R(t ) l d (t )    (t ) l  y(t )    0  0 

(10)

Now let t

y (t )   f  x (t ) R(t ) l  (t ) 1

(11)

0

y (t )  f  x(t ) R(t ) l  (t ) 1

then

y (t )  f  x(t ) R(t )  (t ) using the fact that f (u )  u  to have y (t ) l  x(t ) R(t ) l  (t ) l l

that is

(12) l

(13) (14)

x(t )  R(t )  (t ) y (t ) Combining both (10) and (15) yields, inequality (6) and the proof is complete. 1

1

(15)

Remarks 1: By setting

f (u)  u , R(t )  P(t )

1  k 1

,  (t )

1 k 1

,   l in lemma 1 yields

1 1 1 1 t    x(t )  f   x(t ) P(t ) k 1 P(t ) k 1 dt    (t ) l 1 y(t ) l  P(t ) k 1 P(t ) k 1 y (t ).   0  Integrating both sides of inequality (16) over [a, b] with the respect to t, to get b

 a

(16)

t  b x (t )  f   x (t ) dt    y (t ) y (t )dt 0  a

(17)

That is l

b t  l     a x (t )   0 x (t ) dt   a y(t ) y (t )dt  F ( y(b))  F ( y(a)). If y(a)  0, then inequality (18) becomes b

(18)

l

b t  l      x ( t )  x ( t ) dt  a    y (t ) y (t )dt  F ( y (b)). a 0  By using Holders inequality with o and  we obtain b

(19)

1 1 o

1 b b  b  b   y (b)   x (t ) dt   R o (t ) R  x (t ) (t )dt    R1o (t )dt    R(t ) x (t ) dt  a a a  a  (20) Combing inequality (19) and (20) to obtain inequality (3) if in inequality (20)   o and    which is 1

our desired result. Furthermore, we need the following Lemma to obtain a generalization of Maroni.

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Dual Results Of Opial’s Inequality Lemma 2: Let x(t ),  (t ), f (u), R(t ), l , k , o and

 be as in Lemma 1 such that

1  t  l     x (t )  f  x (t ) R(t ) d (t )   (t ) y (t ) l  R(t ) 1  (t ) 1 y (t ).   0 

(21) Then, the following inequality holds: b

 a

b  1 x (t ) f (t ) dt   y (t ) y (t )dt    y (t ) dt  2a a  b

2

(22)

Proof: The proof is similar to the proof of lemma 1. Since

f (u)  u l , inequality (10) becomes l

t   x (t ) R(t )d (t )    (t ) l  y (t )     0  l  t    x (t ) R(t )d (t )    (t ) l y (t )   0  Combining (15) and (24) to obtain the inequality (21)

(23)

(24)

This completes the proof of the Lemma. Consider all conditions of remark 1 1 1 l l l 1 1 t    k  1 k  1 l l k  1 k   x(t )   x(t ) P(t ) P(t ) dt   (t ) y(t )  P(t ) P(t ) 1 y (t ).   0  t  x (t )   x (t ) dt   y (t ) y (t ) 0 

(25)

(26)

t

Putting

 x(t ) dt  x(t ) and integrate both side of inequality (26) over [a, b] with the respect to t obtain 0 b

 a

b  1 x(t ) x(t ) dt   y(t ) y (t )dt    x(t ) dt  2a a  b

1 1 t   1  o     x (t ) P(t ) P(t ) o dt  20 

2

(27)

2

(28) 2

2

1 b o   b 1    P(t )1  dt    x(t ) P(t ) o dt  2a  a  This is the generalization of inequality (2).

(29)

Acknowledgements: The authors gratefully acknowledged with thanks, Professor J.A Oguntuase for his contributions which have improved the final version of this paper.

References [1]. [2]. [3]. [4]. [5].

Adeagbo-Shelkh A.G and Imoru C.O: An integral inequality of the Hardy’s-Type. Kragujevac J Math. 29(2006), 57-61. Calvert J.: Some generalization of Opial’s Inequality, Pron. Amer. Math. Soc. 18(1967) 72-75 Maroni P.M,: Surl’ingelite d’Opial-Bessack, C.R Aca. Sci. Paris A264 (1967), 62-64. C.O Fabulurin, A.G Adeagbo-Shelkh and Y.O Anthonio,: On an inequality relates to Opial, Octogon Mathematics Magazine, 18(2010), No 1, 32-41. Obial Z.: Surune integalite. Ann. Polon. Math. 8(1960), 29-32.

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