Báo cáo hóa học: " Research Article Cauchy Means of the Popoviciu Type" pot

Pe ˇcari ´c1, 2 1 Abdus Salam School of Mathematical Sciences, 68-B New Muslim Town, GC University, Lahore 54000, Pakistan 2 Faculty of Textile Technology, University of Zagreb, Pierotti

Volume 2009, Article ID 628051, 16 pages

doi:10.1155/2009/628051

Research Article

Cauchy Means of the Popoviciu Type

Matloob Anwar,1 Naveed Latif,1 and J Pe ˇcari ´c1, 2

1 Abdus Salam School of Mathematical Sciences, 68-B New Muslim Town, GC University,

Lahore 54000, Pakistan

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

Correspondence should be addressed to Naveed Latif,sincerehumtum@yahoo.com

Received 9 October 2008; Accepted 2 February 2009

Recommended by Wing-Sum Cheung

We discuss log-convexity for the differences of the Popoviciu inequalities and introduce some mean value theorems and related results Also we give the Cauchy means of the Popoviciu type and we show that these means are monotonic

Copyrightq 2009 Matloob Anwar et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction and Preliminaries

Let fx and px be two positive real valued functions withb

a pxdx  1, then from theory

of convex meanscf 1 3, the well-known Jensen inequality gives that for t < 0 or t > 1,

b

a pxfx t dx ≥

b a pxfxdx

t

and vise versa for 0 < t < 1 In 4, Simic has considered the difference

D s  D s a, b, f, p 

b

a pxfx s dx −

b

a pxfxdx

s 1.2

The following result was given in4 see also 5

Theorem 1.1 Let fx, px be nonnegative and integrable functions for x ∈ a, b, with

b

a pxdx  1, then for 0 < r < s < t; r, s, t /  1, one has



D s ss − 1

t−r

≤



D r rr − 1

t−s

D t tt − 1

s−r

Remark 1.2 For extension ofTheorem 1.1seecf 4

Popoviciu6 8, 9, pages 214-215 has proved the following results

Theorem 1.3 Let φ : a, b → R be convex and f : 0, 1 → R be continuous, increasing, and

convex such that a ≤ fx ≤ b for x0, 1 Then

1

0

φfxdx ≤ b  a − 2  f

b − a φa 

2 f − a

b − a2

b

a

where



f 

1

0

If φ is strictly convex, then the equality in 1.4 holds if and only if

fx  a  b − a x − λ  |x − λ|21 − λ , where λ 

b  a − 2  f

Theorem 1.4 Let φ : a, b → R be continuous and convex, and let f : 0, 1 → R be convex of

order 1, , n  1 such that a ≤ fx ≤ b for x0, 1.

Then

1

0

φfxdx ≤

1

0

φ

U j f, x

dx for ja  b

j  1 ≤ f ≤ j − 1a  b

j , 2 ≤ j ≤ n, 1.7

1

0

φfxdx ≤ V   f for a ≤  f ≤ na  b

where

U j t, x  a  jj  1



t − ja  b

j  1



j − 1a  b



x



x j−1 , 1.9

V x  b  na − n  1x

b − a φa 

n  1x − a

nb − a n1/n

b

a φxdx

x − a n−1/n . 1.10

If φ is strictly convex, then equality in 1.7 holds if and only if

and equality in1.8 holds if

fx  a  b − a



x − λ  |x − λ|

21 − λ

n

, where λ  b  na − n  1  f

With the help of the following useful lemmas we prove our results

Lemma 1.5 Define the function

ϕ s x 

⎧

⎪

⎪

⎪

⎪

x s ss − 1 , s /  0, 1;

− log x, s  0;

x log x, s  1.

1.13

Then ϕs x  x s−2 , that is, ϕ s x is convex for x > 0.

The following lemma is equivalent to definition of convex functionsee 9, page 2

Lemma 1.6 If φ is a convex function on I for all s1 , s2, s3 ∈ I for which s1 < s2< s3, the following

is valid

φ

s1



s3− s2



 φs2



s1− s3



 φs3



s2− s1



≥ 0. 1.14

We quote here another useful lemma from log-convexity theorycf 4

Lemma 1.7 A positive function f is log-convex in the Jensen-sense on an open interval I, that is, for

each s, t ∈ I,

fsft ≥ f2



s  t

2



if and only if the relation

u2fs  2uwf



s  t

2



holds for each real u, w and s, t ∈ I.

The following lemma given in 10 gives the relation between Beta function β and

Hypergeometric function F.

Lemma 1.8 Suppose a, b, c, α, γ ∈ R are such that a  c > b > 0 and 0 < α < 2γ, β and F are Beta

and Hypergeometric functions, respectively Then

∞

0

x b−1

1  αx a 1  γx c dx  γ −b βb, a  c − bF



a, b

a  c

γ − α γ



. 1.17

The paper is organized in the following way After this introduction, in the second section we discuss the log-convexity of differences of the Popoviciu inequalities 1.4, 1.7, and1.8 In the third section we introduce some mean value theorems and the Cauchy means

of the Popoviciu-type and discuss its monotonicity

2 Main Results

Theorem 2.1 Let f : 0, 1 → R be continuous, increasing, and convex such that 0 < a ≤ fx ≤ b

for x ∈ 0, 1, and let  f be defined in 1.5 and

Ωs f 

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

1

ss − 1



b  a − 2  f

b − a a

s 2 f − a

b − a2s  1



b s1 − a s1

−

1

0

fx s dx



, s /  0, 1;

2 f − a

b − a 

1

0

logfxdx − b  a − 2  f

b − a log a −

2 f − a

b − a2b log b − a log a, s  0;

b  a − 2  f

b − a a log a 



f − a

b − a2



b2log b − a2log a

− f − ab  a

2b − a −

1

0

2.1

and letΩs f be positive.

One has thatΩs f is log-convex and the following inequality holds for −∞ < r < s < t < ∞,

Ωt−r

s f ≤ Ω t−s

r fΩ s−r

Proof Consider the function defined by

ωx  u2ϕ s x  2uwϕ r x  w2ϕ t x, 2.3

where r  s  t/2, ϕ sis defined by1.13 and u, w ∈ R We have

ωx  u2x s−2  2uwx r−2  w2x t−2

ux s/2−1  wx t/2−12

> 0, x > 0.

2.4

Therefore, ωx is convex for x > 0 UsingTheorem 1.3,

b  a − 2  f

b − a



u2ϕ s a  2uwϕ r a  w2ϕ t a

 2 f − a

b − a2

b

a



u2ϕ s x  2uwϕ r x  w2ϕ t xdx

≥

1

0



u2ϕ s fx  2uwϕ r fx  w2ϕ t fxdx,

u2



b  a − 2  f

b − a ϕ s a 2 f − a

b − a2

b

a

ϕ s xdx −

1

0

ϕ s fxdx



 2uw



b  a − 2  f

b − a ϕ r a 2 f − a

b − a2

b

a

ϕ r xdx −

1

0

ϕ r fxdx



 w2



b  a − 2  f

b − a ϕ t a 2 f − a

b − a2

b

a

ϕ t xdx −

1

0

ϕ t fxdx



≥ 0,

2.5

since

Ωs f  b  a − 2  f

b − a ϕ s a 2 f − a

b − a2

b

a

ϕ s xdx −

1

0

ϕ s fxdx, f 

1

0

fxdx, 2.6

we have

u2Ωs f  2uwΩ r f  w2Ωt f ≥ 0. 2.7

ByLemma 1.7, we have

Ωs fΩ t f ≥ Ω2

r f  Ω2

that isΩs f is log-convex in the Jensen-sense for s ∈ R Since

lim

s → 0Ωs f  Ω0f, lim

s → 1Ωs f  Ω1f. 2.9

This impliesΩs f is continuous, therefore it is log-convex.

Since Ωs f is log-convex, that is, log Ω s f is convex, therefore by Lemma 1.6 for

−∞ < r < s < t < ∞ and taking φ s log Ωs, we get

logΩt−r

s f ≤ log Ω t−s

r f  log Ω s−r

t f, 2.10 which is equivalent to2.2

Theorem 2.2 Let f, Ω s f be defined in Theorem 2.1 and let t, s, u, v be real numbers such that

s ≤ u, t ≤ v, s /  t, u / v, one has

 Ωt f

Ωs f

1/tưs

≤Ωv f

Ωu f

1/vưu

Proof Incf 9, page 2, we have the following result for convex function f with x1 ≤ y1,

x2≤ y2, x1/  x2, y1/  y2:

f

x2



ư fx1



x2ư x1 ≤ f



y2



ư fy1



y2ư y1 . 2.12

Since byTheorem 2.1,Ωs f is log-convex, we can set in 2.12:

fx  log Ω x and x1 s, x2 t, y1 u, y2 v We get

logΩt f ư log Ω s f

t ư s ≤ logΩv f ư log Ω u f

log Ωt f

Ωs f

1/tưs

≤ logΩv f

Ωu f

1/vưu

,

2.13

and after applying exponential function, we get2.11

Theorem 2.3 Let f : 0, 1 → R be convex of order 1, , n  1 such that 0 < a ≤ fx ≤ b for

x ∈ 0, 1, and let  f be defined in 1.5 and

Λs f 

1

0

ϕ s



U j f, x

dx ư

1

0

ϕ s fxdx, 2.14

where

U j f, x

 a  jj  1





f ư ja  b

j  1



j ư 1a  b

j ư f



x



x jư1 , 2.15

for

ja  b

j  1 ≤ f ≤ j ư 1a  b

j , 2≤ j ≤ n, 2.16

and letΛs f be positive.

One has thatΛs f is log-convex and the following inequality holds for ư∞ < r < s < t < ∞,

Λtưr

s f ≤ Λ tưs

r fΛ sưr

t f. 2.17

Proof As in the proof ofTheorem 2.1, we useTheorem 1.4instead ofTheorem 1.3

Theorem 2.4 Let f, Λ s f be defined in Theorem 2.3 and t, s, u, v be real numbers such that s ≤ u,

t ≤ v, s /  t, u / v, one has

 Λt f

Λs f

1/tưs

≤Λv f

Λu f

1/vưu

Proof Similar to the proof ofTheorem 2.2

Lemma 2.5 Let f : 0, 1 → R be convex of order 1, , n  1 such that 0 < a ≤ fx ≤ b for x ∈

0, 1,  f be defined in 1.5 and V be defined in 1.10, and let β and F are Beta and Hypergeometric

functions respectively, and

Γs f  V f

ư

1

0

ϕ s fxdx. 2.19

Then

Γs f 

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

1

ss ư 1



b  na ư n  1  f

s n  1  f ư a

nb ư a n1/n a

s1/n



b ư a b

1/n

×β

 1

n , 1



F

⎛

⎜

⎝

s1

n 1,1

n

1

n 1

bưa b

⎞

⎟

⎠ư

1

0

ft s dt

⎤

⎥

⎦, s /  0, 1;

b  na ư n  1  f

b ư a ưlog a ư n  1  f ư a

nb ư a n1/n

×

⎡

⎢

⎣n log bb ư a 1/n ư na 1/n



b ư a b

1/n1

β

 1

n  1, 1



F

⎛

⎜ n1  1,1n 1 1

n 2

b ư a b

⎞

⎟

⎤

⎥

⎦



1

0

b  na ư n  1  f

b ư a a log a 

n  1  f ư a

nb ư a n1/n

×



nabưa 1/n log b n

n  1 log bbưa

1/n1 ưa 1/n



bưa b

1/n1

β

 1

n 1, 1



×F

⎛

⎜ n1  1, n1  1 1

n 2

b ư a b

⎞

⎟ n

n  1  na

⎤

⎥

⎦ư

1

0

ftlogftdt, s  1,

2.20

a ≤  f ≤ na  b

Proof First, we solve these three integrals

I1

b

a

t s

t − a n−1/n dt, I2

b

a

log t

t − a n−1/n dt, I3

b

a

t log t

t − a n−1/n dt. 2.22

Take

I1

b

a

t s

t − a n−1/n dt. 2.23

Substitute

t  a  bx

1 x , dt 

b − a

1  x2dx, A

and limits, when t → a then x → 0, when t → b then x → ∞ So,

I1

b

a

t s

t − a n−1/n dt 

∞

0

a  bx/1  x s

a  bx/1  x − a1−1/n ·

b − a

1  x2dx

 b − a 1/n

∞

0

a  bx s 1  x1−1/n

1  x s2 x1−1/n dx,

I1 b − a 1/n a s

∞

0

x 1/n−1

1  x s1/n1 1  b/ax −s dx.

2.24

By usingLemma 1.8with a  s  1/n  1, b  1/n, c  −s, α  1, γ  b/a such that 1/n  1 > 1/n > 0 and 0 < 1 < 2b/a, we get

I1  a s1/n b − a 1/n β

 1

n , 1



F

⎛

⎜

⎝

s 1

n  1,1

n

1

n 1

b − a b

⎞

⎟

⎠. 2.25

Take second integral

I2

b

a

log t

t − a n−1/n dt, 2.26

using integration by parts, we have

I2 n log bb − a 1/n − n

b

a

t−1t − a 1/n

dt. 2.27 Let

I4

b

a

t−1t − a 1/n dt. 2.28

By using same substitutionA as above, we get

I4

∞

0

1 x

a  bx·



a  bx

1 x − a

1/n

· b − a

1  x2dx  b − a 1/n1

a

∞

0

x 1/n1−1

1  x 1/n1 1  b/ax dx.

2.29

By usingLemma 1.8with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0 and 0 < 1 < 2b/a, we get

I2 n log bb − a 1/n − na 1/n



b − a b

1/n1

β

 1

n  1, 1



F

⎛

⎜

⎝

1

n  1,1

n 1 1

n 2

b − a b

⎞

⎟

⎠. 2.30

Now, take third integral

I3

b

a

t log t

t − a n−1/n dt. 2.31 Using integration by parts, we get

I3 nab − a 1/n

log b  n

n  1 log bb − a

1/n1− n

n  1

b

a

t − a 1/n1

t dt − na

b

a

t − a 1/n

t dt.

2.32 Let

I5

b

a

t − a 1/n1

By using same substitutionA as above, we get

I5

∞

0

a  bx/1  x − a 1/n1

a  bx/1  x ·

b − a

1  x2dx  b − a 1/n1

a

∞

0

x 1/n1−1

1  x 1/n1 1  b/ax dx.

2.34

By usingLemma 1.8with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0 and 0 < 1 < 2b/a, we get

I5 a 1/n



b − a b

1/n1

β

 1

n  1, 1



F

⎛

⎜

⎝

1

n  1,1

n 1 1

n 2

b − a b

⎞

⎟

⎠. 2.35

Let

I6

b

a

t − a 1/n

By using same substitutionA as above, we have

I6 

∞

0

a  bx/1  x − a 1/n

a  bx/1  x ·

b − a

1  x2dx  b − a 1/n1

a

∞

0

x 1/n1−1

1  x 1/n1 1  b/ax dx.

2.37

By usingLemma 1.8with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0 and 0 < 1 < 2b/a, we get

I6 a 1/n



b − a a

1/n1

β

 1

n  1, 1



F

⎛

⎜

⎝

1

n  1,1

n 1 1

n 2

b − a b

⎞

⎟

⎠. 2.38

Then

I3 nab − a 1/n

log b  n

n  1 log bb − a

1/n1 − a 1/n



b − a a

1/n1

× β

 1

n  1, 1



F

 1

n  1,1

n  1,1

n 2 b − a

b



n

n  1  na



.

2.39

For s /  0, 1,

Γs f  1

ss − 1



b  na − n  1  f

s n  1  f − a

nb − a n1/n

b

a

t s

t − a n−1/n dt −

1

0

ft s dt



.

2.40

Using I1, we have

Γs f  1

ss − 1



b  na − n  1  f

s n  1  f − a

nb − a n1/n a

s1/n



b − a b

1/n

×β

 1

n , 1



F

⎛

⎜

⎝

s 1

n  1,1

n

1

n 1

b − a b

⎞

⎟

⎠ −

1

0

ft s dt

⎤

⎥

⎦.

2.41

For s  0,

Γ0f bna−n1  f

b − a −log a− n1  f − a

nb−a n1/n

b

a

log t

t−a n−1/n dt

1

0

logftdt 2.42

Using I2, we have

Γ0f  b  na − n  1  f

b − a −log a − n  1  f − a

nb − a n1/n

×

⎡

⎢

⎣n log bb − a 1/n − na 1/n



b − a b

1/n1

β

 1

n  1, 1



F

⎛

⎜

⎝

1

n  1,1

n 1 1

n 2

b − a b

⎞

⎟

⎠

⎤

⎥

⎦



1

0

logftdt

2.43

For s  1,

Γ1f  b  na − n  1  f

b − a a log a 

n  1  f − a

nb − a n1/n

b

a

t log t

t − a n−1/n dt −

1

0

ftlogftdt.

2.44

Using I3, we have

Γ1f  b  na − n  1  f

b − a a log a 

n  1  f − a

nb − a n1/n

×



nab−a 1/n log b n

n1 log bb−a

1/n1 −a 1/n



b−a b

1/n1

β

 1

n 1, 1



×F

⎛

⎜

⎝

1

n  1,1

n 1 1

n 2

b − a b

⎞

⎟

⎠



n

n  1  na

⎤⎥

⎦ −

1

0

ftlogftdt.

2.45

Theorem 2.6 Let f : 0, 1 → R be convex of order 1, , n  1 such that 0 < a ≤ fx ≤ b for

x ∈ 0, 1,  f be defined in 1.5 and let Γs f defined in 2.20 be positive.

One has thatΓs f is log-convex and the following inequality holds for ư∞ < r < s < t < ∞,

Γtưr

s f ≤ Γ tưs

r fΓ sưr

Proof As in the proof ofTheorem 2.1, we useTheorem 1.4instead ofTheorem 1.3

Theorem 2.7 Let f, Γ s f be defined in Theorem 2.6 and t, s, u, v be real numbers such that s ≤ u,

t ≤ v, s /  t, u / v, one has

 Γt f

Γs f

1/tưs

≤Γv f

Γu f

1/vưu

Proof Similar to the proof ofTheorem 2.2

3 Cauchy Means

Let us note that 2.11 has the form of some known inequalities between means e.g., Stolarsky’s means, etc. Here we prove that expressions on both sides of 2.11 are also means

Lemma 3.1 Let h ∈ C2I be such that h is bounded, that is, m ≤ h ≤ M Then the functions

φ1, φ2defined by

φ1t  M

2 t2ư ht, φ2t  ht ư m

2t2, 3.1

are convex functions.

Theorem 3.2 Let h ∈ C2I1, I1 is a compact interval in R and f be a continuous, increasing and

convex such that a ≤ fx ≤ b for x ∈ 0, 1,  f be defined in 1.5 then ∃ ξ ∈ a, b, ξ / 0 such that

b  a ư 2  f

b ư a ha 

2 f ư a

b ư a2

b

a hxdx ư

1

0

hfxdx

 hξ

2



b  a ư 2  f

b ư a a

22 f ư a

b2 ba  a2 3b ư a ư

1

0

fx2dx



.

3.2

Proof Suppose m  min hx ≤ hx ≤ M  max hx for x ∈ I1 Then by applying φ1and

φ2defined inLemma 3.1for φ in 1.4, we have

b  a ư 2  f

b ư a φ1a 2 f ư a

b ư a2

b

a

φ1xdx ≥

1

0

φ1fxdx,

b  a ư 2  f

b ư a φ2a 2 f ư a

b ư a2

b

a

φ2xdx ≥

1

0

φ2fxdx,

3.3

that is,

M

2



b  a − 2  f

b − a a

22 f − a

b2 ba  a2 3b − a −

1

0

fx2

dx



≥ b  a − 2  f

b − a ha 

2 f − a

b − a2

b

a hxdx −

1

0

hfxdx,

3.4

b  a − 2  f

b − a ha 

2 f − a

b − a2

b

a hxdx −

1

0

hfxdx

≥ m 2



b  a − 2  f

b − a a

22 f − a

b2 ba  a2 3b − a −

1

0

fx2dx



.

3.5

By combining3.4 and 3.5 and using the fact that for m ≤ ρ ≤ M there exists ξ ∈ I1such

that hξ  ρ we get 3.2.

Theorem 3.3 Let k, l ∈ C2I1 and satisfy 3.2, f be a continuous, increasing and convex such that

a ≤ fx ≤ b for x ∈ 0, 1,  f be defined in 1.5, and

fx /  a  b − a x − λ  |x − λ|21 − λ , where λ  b  a − 2  f

then there exists ξ ∈ I1such that

kξ

lξ 



b  a − 2  f/b − a

ka 

2 f − a/b − a2b

a kxdx −1

0kfxdx



b  a − 2  f/b − a

la  2 f − a/b − a2b

a lxdx −1

0lfxdx . 3.7

Provided that denominators are non-zero.

Proof Consider the linear functionals Ψ and η such that Ψm  ηmξ for some function

m ∈ C2I1 and ξ ∈ I1 Consider the following linear combination

where c1and c2are defined as follows:

c1 Ψl  b  a − 2  f

b − a la 

2 f − a

b − a2

b

a lxdx −

1

0

lfxdx,

c2 Ψk  b  a − 2  f

b − a ka 

2 f − a

b − a2

b

a kxdx −

1

0

kfxdx.

3.9

Since k, l ∈ C2I1 and satisfy 3.2, therefore m as linear combination of k and l should also satisfy3.2