1. Trang chủ
  2. » Khoa Học Tự Nhiên

Báo cáo hóa học: " Research Article On Logarithmic Convexity for Power Sums and Related Results" potx

9 272 0
Tài liệu đã được kiểm tra trùng lặp

Đang tải... (xem toàn văn)

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 9
Dung lượng 487,47 KB

Các công cụ chuyển đổi và chỉnh sửa cho tài liệu này

Nội dung

Pe ˇcari ´c 1, 2 and Atiq Ur Rehman 2 1 Faculty of Textile Technology, University of Zagreb, 10000 Zagreb, Croatia 2 Abdus Salam School of Mathematical Sciences, GC University, Lahore 54

Trang 1

Volume 2008, Article ID 389410, 9 pages

doi:10.1155/2008/389410

Research Article

On Logarithmic Convexity for Power Sums and

Related Results

J Pe ˇcari ´c 1, 2 and Atiq Ur Rehman 2

1 Faculty of Textile Technology, University of Zagreb, 10000 Zagreb, Croatia

2 Abdus Salam School of Mathematical Sciences, GC University, Lahore 54660, Pakistan

Correspondence should be addressed to Atiq Ur Rehman, mathcity@gmail.com

Received 28 March 2008; Revised 23 May 2008; Accepted 29 June 2008

Recommended by Martin j Bohner

We give some further consideration about logarithmic convexity for differences of power sums inequality as well as related mean value theorems Also we define quasiarithmetic sum and give some related results.

Copyright q 2008 J Peˇcari´c and A U Rehman 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 x  x1, , x n , p  p1, , p n denote two sequences of positive real numbers with

n

i1 p i  1 The well-known Jensen Inequality 1, page 43 gives the following, for t < 0 or

t > 1:

n



i1

p i x t

i

n

i1

p i x i

t

1.1

and vice versa for 0 < t < 1.

Simi´c2 has considered the difference of both sides of 1.1 He considers the function defined as

λ t 

n

i1 p i x t

i− n i1 p i x it

tt − 1 , t / 0, 1;

log

n

i1

p i x i



−n

i1

p i log x i , t  0;

n



i1

p i x i log x i

n

i1

p i x i

 log

n

i1

p i x i



, t  1;

1.2

and has proved the following theorem

Trang 2

Theorem 1.1 For −∞ < r < s < t < ∞, then

λ t−r

Anwar and Peˇcari´c 3 have considerd further generalization ofTheorem 1.1 Namely, they introduced new means of Cauchy type in4 and further proved comparison theorem for these means

In this paper, we will give some results in the case where instead of means we have power sums

Letx be positive n-tuples The well-known inequality for power sums of order s and r,

for s > r > 0see 1, page 164, states that

 n



i1

x s i

1/s

<

 n



i1

x i r

1/r

Moreover, if p  p1, , p n  is a positive n-tuples such that p i ≥ 1 i  1, , n, then for

s > r > 0 see 1, page 165, we have

n

i1

p i x s i

1/s

<

n

i1

p i x r i

1/r

Let us note that1.5 can also be obtained from the following theorem 1, page 152

Theorem 1.2 Let x and p be two nonnegative n-tuples such that xi ∈ 0, a i  1, , n and

n



i1

p i x i ≥ x j , for j  1, , n, n

i1

p i x i ∈ 0, a. 1.6

If fx/x is an increasing function, then

f

n

i1

p i x i



≥n

i1

Remark 1.3 Let us note that if f x/x is a strictly increasing function, then equality in 1.7 is valid if we have equalities in1.6 instead of inequalities, that is, x1 · · ·  x nandn

1p i  1.

The following similar result is also valid1, page 153

Theorem 1.4 Let fx/x be an increasing function If 0 < x1≤ · · · ≤ x n and if the following hold.

i there exists an m≤ n such that

P1≥ P2≥ · · · ≥ P m ≥ 1, P m1  · · ·  P n  0, 1.8

where P k k i1 p i , P k  P n − P k−1 k  2, , n and P1 P n , then1.7 holds.

ii If there exists an m≤ n such that

0≤ P1≤ P2≤ · · · ≤ P m ≤ 1, P m1  · · ·  P n  0, 1.9

then the reverse of inequality in1.7 holds.

In this paper, we will give some applications of power sums That is, we will prove results similar to those shown in2,3, but for power sums

Trang 3

2 Main results

Lemma 2.1 Let

ϕ t x 

x t

t − 1 , t / 1;

x log x, t  1.

2.1

Then ϕ t x/x is a strictly increasing function for x > 0.

Proof Since ϕ t x/x  x t−2 > 0, for x > 0, therefore ϕ t x/x is a strictly increasing function for x > 0.

Lemma 2.2 2 A positive function f is log convex in Jensen’s 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

 w2ft ≥ 0 2.3

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

The following lemma is equivalent to the definition of convex functionsee 1, page 2

Lemma 2.3 If f is continuous and convex for all x1, x2, x3of an open interval I for which x1< x2< x3, then

x3− x2fx1  x1− x3fx2  x2− x1fx3 ≥ 0. 2.4

Theorem 2.4 Let x and p be two positive n-tuples n ≥ 2 and let

φ t  φ t ( x; p  ϕt

 n



i1

p i x i



−n

i1

such that condition1.6 is satisfied and all x i ’s are not equal Then φ t is log-convex Also for r < s < t where r, s, t ∈ R, we have

φ st−r ≤ φ rt−s φ ts−r 2.6

Proof Since ϕ t x/x is a strictly increasing function for x > 0 and all x i’s are not equal, therefore

byTheorem 1.2with f  ϕ t, we have

ϕ t

 n



i1

p i x i



>n

i1

p i ϕ t x i  ⇒ φ t  ϕ t

 n



i1

p i x i



−n

i1

p i ϕ t x i  > 0, 2.7

that is, φ tis a positive-valued function

Trang 4

Let fx  u2ϕ s x  2uwϕ r x  w2ϕ t x, where r  s  t/2 and u, w ∈ R:

x



 u2x s−2  2uwx r−2  w2x t−2 ,

 ux s−2/2  wx t−2/22≥ 0.

2.8

This implies that f x/x is monotonically increasing.

ByTheorem 1.2, we have

f

 n



i1

p i x i



−n

i1

p i fx i ≥ 0

⇒ u2



ϕ s

 n



i1

p i x i



−n

i1

p i ϕ s x i



 2uw



ϕ r

 n



i1

p i x i



−n

i1

p i ϕ r x i



 w2



ϕ t

 n



i1

p i x i



−n

i1

p i ϕ t x i



≥ 0

⇒ u2φ s  2uwφ r  w2φ t ≥ 0.

2.9

Now byLemma 2.2, we have that φ tis log-convex in Jensen sense

Since limt→1 φ t  φ1, it follows that φ tis continuous, therefore it is a log-convex function

1, page 6

Since φ t is log-convex, that is, log φ tis convex, we have byLemma 2.3that, for r < s < t with f  log φ,

t − s log φ r  r − t log φ s  s − r log φ t ≥ 0, 2.10 which is equivalent to2.6

Similar application ofTheorem 1.4gives the following

Theorem 2.5 Let x and p be two positive n-tuples n ≥ 2 such that 0 < x1≤ · · · ≤ x n , all x i ’s are not equal and

i if φ t  φ t ( x; p  ϕtn i1 p i x i −n i1 p i ϕ t x i  such that condition 1.8 is satisfied, then φ t

is log-convex, also for r < s < t, we have

φ st−r ≤ φ rt−s φ ts−r; 2.11

ii moreover if φ t  −φ t and1.9 is satisfied, then we have that φ t is log-convex and

φ st−r ≤ φ rt−s φ ts−r 2.12

We will also use the following lemma

Trang 5

Lemma 2.6 Let f be a log-convex function and assume that if x1 ≤ y1, x2 ≤ y2, x1 / x2, y1 / y2 Then the following inequality is valid:

2

fx1

1/x2−x1 

fy1

1/y2−y1 

Proof In1, page 2, we have the following result for convex function f, with x1 ≤ y1, x2 ≤

y2, x1 / x2, y1 / y2:

fx2 − fx1

x2− x1 ≤ fy y2 − fy1

Putting f  log f, we get

log 2

fx1

1/x2−x1 

≤ log fy2

1

1/y2−y1 

from which2.13 immediately follows

Let us introduce the following

Definition 2.7 Let x and p be two nonnegative n-tuples n ≥ 2 such that pi ≥ 1 i  1, , n, then for t, r, s∈ R, we define

A s

t,rx; p 

r − s

t − s

n i1 p i x s

it/s−n i1 p i x t

i

n i1 p i x s

ir/s−n i1 p i x r

i

1/t−r

, t / r, r / s, t / s,

A s

s,r x; p  A s

r,sx; p 

r − s s

n i1 p i x s

i logn i1 p i x s

i − sn i1 p i x s

i log x i

n i1 p i x s

ir/s−n i1 p i x r

i

1/s−r

, s / r,

A s

r,rx; p  exp

 1

s − r 

n i1 p i x s

ir/slogn

i1 p i x s

i − sn i1 p i x r

i log x i

s{n i1 p i x s

ir/s−n i1 p i x r

i}



, s / r,

A s

s,sx; p  exp



n i1 p i x s

ilogn i1 p i x s

i2− s2n

i1 p i x s

i log x i2

2s{n i1 p i x s

i logn i1 p i x s

i  − sn i1 p i x s

i log x i}



.

2.16

Remark 2.8 Let us note that A s s,r x; p  A s

r,sx; p  limt→s A s

t,rx; p  limt→s A s

r,tx; p,

A s

r,rx; p  limt→r A s

t,r x; p and A s

s,sx; p  limr→s A s

r,rx; p.

Theorem 2.9 Let r, t, u, v ∈ Rsuch that r < u, t < v, r / t, u / v Then we have

A s t,r ( x; p ≤ A s

Proof Let

φ t  φ tx; p 

1

t − 1

 n



i1

p i x i

t

−n

i1

p i x t i



, t / 1;

n



i1

p i x ilog

n



i1

p i x i−n

i1

p i x i log x i , t  1.

2.18

Trang 6

Now taking x1 r, x2  t, y1  u, y2 v, where r, t, u, v / 1, and ft  φ tinLemma 2.6, we have



r ư 1

t ư 1

n i1 p i x itưn i1 p i x t

i

n i1 p i x irưn i1 p i x r

i



u ư 1

v ư 1

n i1 p i x vưn i1 p i x v

i

n i1 p i x s

iuưn i1 p i x u

i

. 2.19

Since s > 0 by substituting x i  x s

i , t  t/s, r  r/s, u  u/s and v  v/s, where r, t, u, v / s,

in above inequality, we get



r ư s

t ư s

n i1 p i x s

it/sưn i1 p i x t

i

n i1 p i x s

ir/sưn i1 p i x r

i

s/tưr



u ư s

v ư s

n i1 p i x s

iv/sưn i1 p i x v

i

n i1 p i x s

iu/sưn i1 p i x u

i

s/vưu

. 2.20

By raising power 1/s, we get2.17 for r, t, u, v / s.

FromRemark 2.8, we get2.17 is also valid for r  s or t  s or r  t or t  r  s.

Corollary 2.10 Let

Φs

t 

1

t ư s

n

i1

p i x s i

t/s

ưn

i1

p i x t

1

s

n

i1

p i x s i

 log

n

i1

p i x s i



ư sn

i1

p i x s

i log x i , t  s.

2.21

Then for t, r, u ∈ Rand t < r < u, we have

s

ruưt≤ Φs

tuưrs

Proof Taking v  r in 2.17, we get 2.22

3 Mean value theorems

Lemma 3.1 Let f ∈ C1I, where I  0, a such that

m ≤ xfx ư fx

Consider the functions φ1and φ2defined as

φ1x  Mx2ư fx,

Then φ i x/x for i  1, 2 are monotonically increasing functions.

Proof We have that

φ1x

x  Mx ư fx x ⇒ 1x

x



 M ư xfx ư fx

x2 ≥ 0,

φ2x

x  fx x ư mx ⇒ 2x x



 xfx ư fx x2 ư m ≥ 0,

3.3

that is, φ i x/x for i  1, 2 are monotonically increasing functions.

Trang 7

Theorem 3.2 Let x and p be two positive n-tuples n ≥ 2 satisfy condition 1.6, all x i ’s are not equal and let f ∈ C1I, where I  0, a Then there exists ξ ∈ 0, a such that

f

 n



i1

p i x i



−n

i1

p i fx i  ξfξ − fξ

ξ2

 n



i1

p i x i

2

−n

i1

p i x2

Proof InTheorem 1.2, setting f  φ1and f  φ2, respectively, as defined inLemma 3.1, we get the following inequalities:

f

 n



i1

p i x i



−n

i1

p i fx i  ≤ M

 n



i1

p i x i

2

−n

i1

p i x2

i , f

n

i1

p i x i



−n

i1

p i fx i  ≥ m

n

i1

p i x i

2

−n

i1

p i x2

i

3.5

Now by combining both inequalities, we get,

m ≤ f

n

i1 p i x i −n

i1 p i fx i

n i1 p i x i2−n i1 p i x2

i

n i1 p i x i2−n i1 p i x2

i is nonzero, it is zero if equalities are given in conditions1.6, that is,

x1 · · ·  x nandn

i1 p i 1

Now by condition3.1, there exist ξ ∈ I, such that

fn i1 p i x i −n i1 p i fx i

n i1 p i x i2−n

i1 p i x2

i

 ξfξ − fξ

and3.7 implies 3.4

Theorem 3.3 Let x and p be two positive n-tuples n ≥ 2 satisfy condition 1.6, all x i ’s are not equal and let f, g ∈ C1I, where I  0, a Then there exists ξ ∈ I such that the following equality is true:

fn i1 p i x i −n

i1 p i fx i

gn i1 p i x i −n

i1 p i gx i 

ξfξ − fξ

provided that the denominators are nonzero.

Proof Let a function k ∈ C1I be defined as

k  c1f − c2g, 3.9

where c1and c2are defined as

c1 g

 n



i1

p i x i



−n

i1

p i gx i ,

c2 f

 n



i1

p i x i



−n

i1

p i fx i .

3.10

Trang 8

Then, usingTheorem 3.2with f  k, we have

0 c1

ξfξ − fξ

ξ2 − c2

ξgξ − gξ

ξ2

n i1

p i x i

2

−n

i1

p i x2

Since

 n



i1

p i x i

2

−n

i1

p i x2

therefore,3.11 gives

c2

c1  ξf ξgξ − fξ ξ − gξ 3.13 After putting values, we get3.8

Let α be a strictly monotone continuous function then quasiarithmetic sum is defined as

follows:

S α x; p  α−1

n

i1

p i αx i



Theorem 3.4 Let x and p be two positive n-tuples n ≥ 2, all xi ’s are not equal and let α, β, ∈ C1I be

strictly monotonic continuous functions, γ ∈ C1I be positive strictly increasing continuous function,

where I  0, a and

n



i1

p i γx i  ≥ γx j , for j  1, , n, n

i1

p i γx i  ∈ 0, γa. 3.15

Then there exists η from 0, γa such that

αS γ ( x; p − αSα (x; p

βS γ ( x; p − βSβ (x; p 

γηαη − γηαη

γηβη − γηβη 3.16

is valid, provided that all denominators are not zero.

Proof If we choose the functions f and g so that f  α ◦ γ−1, g  β ◦ γ−1, and x i → γx i Substituting these in3.8,

αS γ x; p − αS αx; p

βS γ x; p − βS βx; p 

ξα ◦ γ−1ξ − γ◦ γ−1ξα ◦ γ−1ξ

ξβ ◦ γ−1ξ − γ◦ γ−1ξβ ◦ γ−1ξ . 3.17 Then by setting γ−1η  ξ, we get 3.16

Corollary 3.5 Let x and p be two nonnegative n-tuples and let t, r, s ∈ R Then

A s

Proof If t, r, and s are pairwise distinct, then we put α x  x t , βx  x r , and γx  x s in

3.16 to get 3.18

For other cases, we can consider limit as in Remark2.8

Acknowledgment

The authors are really very thankful to Mr Martin J Bohner for his useful suggestions

Trang 9

1 J E Peˇcari´c, F Proschan, and Y L Tong, Convex Functions, Partial Orderings, and Statistical Applications, vol 187 of Mathematics in Science and Engineering, Academic Press, Boston, Mass, USA, 1992.

2 S Simi´c, “On logarithmic convexity for differences of power means,” Journal of Inequalities and

Applications, vol 2007, Article ID 37359, 8 pages, 2007.

3 M Anwar and J E Peˇcari´c, “On logarithmic convexity for differences of power means,” to appear in

Mathematical Inequalities & Applications.

4 M Anwar and J E Peˇcari´c, “New means of Cauchy’s type,” Journal of Inequalities and Applications, vol.

2008, Article ID 163202, 10 pages, 2008.

... x/x for i  1, are monotonically increasing functions.

Trang 7

Theorem 3.2 Let x and p... Bohner for his useful suggestions

Trang 9

1 J E Peˇcari´c, F Proschan, and Y L Tong, Convex...

2 S Simi´c, ? ?On logarithmic convexity for differences of power means,” Journal of Inequalities and< /small>

Applications, vol 2007, Article ID 37359, pages, 2007.

Ngày đăng: 22/06/2014, 02:20

TỪ KHÓA LIÊN QUAN

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN

🧩 Sản phẩm bạn có thể quan tâm