A Refinement of Jensen's Inequality for a Class of Increasing and Concave Functions

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Volume 2008, Article ID 717614,14pages doi:10.1155/2008/717614

Research Article

A Refinement of Jensen’s Inequality for a Class of

Increasing and Concave Functions

Ye Xia

Department of Computer and Information Science and Engineering, University of Florida, Gainesville, FL 32611-6120, USA

Correspondence should be addressed to Ye Xia,[email protected]

Received 23 January 2008; Accepted 9 May 2008

Recommended by Ondrej Dosly

Suppose thatfxis strictly increasing, strictly concave, and twice continuously diﬀerentiable on a nonempty intervalI , andfxis strictly convex onI. Suppose thatxk a, bI, where 0< a < b, andpk 0 fork 1, , n, and suppose that n

k1pk 1. Letx n

k1pkxk, and 2 n

k1pkxk x 2

. We show n_k₁pkfxkfx 1 2, _kn₁pkfxkfx 2 2, for suitably chosen 1and 2. These results can

be viewed as a refinement of the Jensen’s inequality for the class of functions specified above. Or they can be viewed as a generalization of a refined arithmetic mean-geometric mean inequality introduced by Cartwright and Field in 1978. The strength of the above result is in bringing the variations of thexk’s into consideration, through 2_.

Copyrightq2008 Ye Xia. 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. Main theorem

The goal is to generalize the following refinement of the arithmetic mean-geometric mean inequality introduced in 1. The result in this paper can also be viewed as a refinement of Jensen’s inequality for a class of increasing and concave functions. Many other refinements can be found in2.

Theorem 1.1 see 1. Suppose that xk ∈ a, band pk ≥ 0 for k 1, . . . , n, where a > 0, and

suppose thatn_k₁pk 1. Then, writingxn_k₁pkxk,

1 2b

n

k1 pk

xk−x

2_≤ x−

n

k1 xpk

k ≤

1 2a

n

k1 pk

xk−x

2

. 1.1

For notational simplicity, define

σ2 n

k1 pk

xk−x

2n k1

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The vector p p1, . . . , pnsatisfying pk ≥ 0 for k 1, . . . , nandn_k₁pk 1 will be called a

weight vector. Sometimes, we writexpandσ2_p_{to emphasize the dependency on the weight}

vectorp. We have the following main theorem.

Theorem 1.2. Suppose that fx is strictly increasing, strictly concave, and twice continuously diﬀerentiable on a nonempty intervalI ⊆ R, and suppose that fxis strictly convex on I. Suppose thatxk ∈a, b⊆I, where0< a < b, andpk ≥0fork 1, . . . , n, and suppose thatnk1pk 1. Then,

writingxn_k₁pkxk,

a

n

k1 pkf

xk

≤fx−θ1σ2

, 1.3

whereθ1min{μ1, μ2}. Here,

μ1min

q,t,x

f1qtx−f1tx

21−qtxf1qtx , 1.4

where the minimization is taken overq ∈0,1,t∈0,b−x/x, andx∈a, b, and

μ2 min

x∈a,b−

fx−fb

2x−bfx, 1.5

b

n

k1 pkf

xk

≥fx−θ2σ2

, 1.6

whereθ2 max{π1, π2}, provided 0 < π1 <∞ andxp−θ2σ2p ∈I for the givenx1, . . . , xn

and for all possible weight vectorsp. Here,

π1

min

t,q,x,θ

2f1qtx−θq1−qt2_x2

−f1qtx−θq1−qt2_x2−qtx

−1

, 1.7

where the minimization is taken overθ ∈0,1/21−qtx,q ∈ 0,1,t∈0,b−x/x, and x∈a, b we will see later that1qtx1−θq1−qt2x2₁is an alternative expression forx−θσ2

whenn2. And

π2 max

x∈a,b

fa−fx

2x−afx. 1.8

Proof. The proof is similar to the proof forTheorem 1.11, based on induction onn. We first demonstrate partaof the theorem. The fact that the functionf is always defined atx−θ1σ2

will be proved inLemma 1.3.

The case of n 1 is trivial because σ2 _{0. When}_n _{2, write} _x

2 1tx1, where

0 ≤t≤b−x1/x1,p1 1−q, andp2 q. With these definitions, we havex 1qtx1, and σ2q1−qt2x2₁. Define

gt fx−θ1σ2

−2

k1 pkf

xk

f1qtx1−θ1q1−qt2x12

−1−qfx1

−qf1tx1

.

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Clearly,g0 0. We will show that, for anyθ1 satisfying 0 ≤ θ1 ≤ μ1,gt≥ 0 fort ≥ 0, and

hence,gt≥0 fort≥0:

gt fx−θ1σ2

qx1−2θ1q1−qtx21

−qx1f

1tx1

qx1

fx−θ1σ2

1−2θ11−qtx1

−f1tx1

.

1.10

Since x1 > 0, let us ignore the factor qx1. We wish to show, for all admissible q, t, and x1

Admissible parameters are those that make x1, . . . , xn fall on a, b. In this case, q ∈ 0,1, x1∈a, bandt∈0,b−x1/x1.,

fx−θ1σ2

1−2θ11−qtx1

−f1tx1

≥0. 1.11

Equation1.11is true if and only if

θ1≤

1−f1tx1

/fx−θ1σ2

21−qtx1

. 1.12

The right-hand side The casesq 1 or t 0 do not pose problems because the right-hand side is still finite.is an increasing function ofθ1. Substitutingθ1 0 into the right-hand side,

it suﬃces to show

θ1 ≤

fx−f1tx1

21−qtx1f

x

f1qtx1

−f1tx1

21−qtx1f

1qtx1

. 1.13

Sinceμ1as in1.4achieves the minimum of the right-hand side above, we can choose anyθ1

satisfying 0≤θ1≤μ1.

Forn >2, let us suppose that1.3has been proved for up ton−1, and consider the case of n. Fix x1, . . . , xn. We may assume that allxk are distinct. Otherwise, we can combine those

identicalxktogether by combining the correspondingpktogether, and we are back to the case

ofn−1 or less.

Letp p1, . . . , pn, and let

hpfx−θ1σ2

−n

k1 pkf

xk

. 1.14

Define the setSby

Sp1, . . . , pn

| pk ≥0, ∀k

. 1.15

We wish to showhp≥ 0 on the setSsubject to the constraintp1· · ·pn 1. Suppose the

minimum ofhpis in the interior ofS. By the Lagrange multiplier method, any such minimum

pmust satisfy the following set of equations for some real numberλ:

∂h

∂pkp λ

∂ ∂pk

n

k1

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This gives, for allk,

fx−θ1σ2

xk−θ1

x2_k−2xxk

−fxk

λ. 1.17

From this, we deduce that each of thexkmust satisfy the following equation with variabley:

fx−θ1σ2

y−θ1y22θ1xy

−fy−λ0. 1.18

We will consider the critical points of the left-hand side above, that is, the zeros of its derivative with respect toy. By taking the derivative, the critical points are the solutions to the equation

fx−θ1σ2

2θ1

x−y1fy. 1.19

The left-hand side of 1.19 is a decreasing linear function ofy. Under the assumption that

fyis strictly convex, there can be at most two solutions to the equation. We will show that, under suitable conditions, there is at most one solution. The situation is illustrated inFigure 1. We will find conditions for the following to hold, for any admissiblexandσ2_,

fx−θ1σ2

2θ1

x−b1> fb. 1.20

When thexkare not all identical,σ2/0. Hence, forθ1>0,fx−θ1σ2> fx>0, it is enough

to consider

fx2θ1

x−b1≥fb 1.21

which is the same as

θ1≤ −

fx−fb

2x−bfx. 1.22

We can chooseθ1as in the theorem, which satisfiesθ1≤μ1and

θ1≤μ2 min x∈a,b−

fx−fb

2x−bfx. 1.23

By Rolle’s theorem, we conclude that there can be at most two distinct roots to 1.18 on the intervala, b. This contradicts our assumption that allx1, . . . , xn are distinct. Hence, it

must be true that the minimum ofhpinSsubject to the constraintp1· · ·pn1 occurs on

the boundary ofS, where some of thepk must be zero. At the minimum, saypo, we are back

to the case of n−1 or less, and by the induction hypothesis, hpo _≥ _{0. Hence,} _h_p _≥ _{0 for}

arbitraryp∈Ssubject to the constraintp1· · ·pn1.

We now proceed to show1.6in partb. For the case ofn 2, let us replaceθ1byθ2

in1.9, and rename the functiongt. Againg0 0. We will find appropriateθ2 for which

gt≤0 fort≥0. Following similar steps as before, we get

gt qx1

fx−θ2σ2

1−2θ21−qtx1

−f1tx1

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[image:5.600.203.400.75.238.2]

a _b y

Figure 1:Illustration of functionsfx−θ1σ22θ1x−y 1andfy.

To showgt≤0, it suﬃces to show fx−θ2σ2

1−2θ21−qtx1

−f1tx1

≤0. 1.25

Observe that if 1−2θ21−qtx1≤0, or equivalently,θ2≥1/21−qtx1, the above automatically

holds.We assume the convention 1/0∞.We sketch our subsequent strategy. For any fixed

q,t, andx1, findθ2q, t, x1that satisfies both1.25and

θ2

q, t, x1

≤ ₂₁₋1

qtx1

. 1.26

Suchθ2q, t, x1must exist because 1/21−qtx1qualifies. Then, we can take supθ2q, t, x1

over all admissibleq,t, andx1.

By the mean value theorem, there existsη∈x−θ2σ2,1tx1such that fx−θ2σ2

1−2θ21−qtx1

−f1tx1

−fη1−qtx1

1θ2qtx1

−fx−θ2σ2

2θ21−qtx1.

1.27

Note that, for1.25to hold, it suﬃces if −fη1θ2qtx1

−fx−θ2σ2

2θ2≤0. 1.28

Since−fxis decreasing and positive,1.28is implied by −fx−θ2σ2

1θ2qtx1

−fx−θ2σ2

2θ2≤0, 1.29

which is equivalent to

1

θ2 ≤

2fx−θ2σ2

−fx−θ2σ2

−qtx1. 1.30

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For any fixedq,t, andx1, suppose we obtain

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−fx−θσ2−qtx1

−1

1.31

and suppose 0 < θ2q, t, x1< ∞. Let us considerθ2 ≥ θ2q, t, x1. Ifθ2 >1/21−qtx1, then

1.25holds trivially. Ifθ2≤1/21−qtx1, then

1

θ2

≤ 1

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−fx−θσ2−qtx1≤

2fx−θ2σ2

−fx−θ2σ2

−qtx1, 1.32

that is,1.30holds, which implies that1.28, and hence,1.25hold. Hence, we can choose

θ2≥sup_q,t,x₁θ2q, t, x1, where the supremum is over all admissibleq,t, andx1. To summarize,

we can choose the followingθ2for the theorem, provided 0< π1<∞,

θ2≥π1

min

t,q,x1,θ

2fx−θσ2

−1

, 1.33

where the minimization is taken overθ∈ 0,1/21−qtx1, and over all admissibleq,x1,t,

that is,q∈0,1,x1∈a, b, andt∈0,b−x1/x1. This minimization can be solved easily in

a number of cases, which we will show later.

Forn >2, the proof is nearly identical to that for parta. Let us suppose1.6has been proved for up ton−1, and consider the case ofn. Fixx1, . . . , xn. We may assume that allxkare

distinct as before. Let

hpfx−θ2σ2

−n

k1 pkf

xk

. 1.34

We wish to showhp≤0 on the setSdefined before, subject to the constraintp1· · ·pn1.

Suppose the maximum ofhpis in the interior ofS. Applying the Lagrange multiplier method for finding the constrained maximum of hpon S, we deduce that any maximum p must satisfy the following set of equations for some real numberλ:

∂h ∂pk

p λ ∂

∂pk

n

k1 pk−1

∀k. 1.35

This gives, for allk,

fx−θ2σ2

xk−θ2

x2_k−2xxk

−fxk

λ. 1.36

Each of thexk must satisfy the following equation with variabley:

fx−θ2σ2

y−θ2y22θ2xy

−fy−λ0. 1.37

We will consider the critical points of the left-hand side above, which are the solutions to the equation

fx−θ2σ2

2θ2

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[image:7.600.203.400.75.247.2]

a _b y

Figure 2:Illustration of functionsfx−θ2σ22θ2x−y 1andfy.

Under the assumption thatfyis strictly convex, there can be at most two solutions to the equation. We will show that, under suitable conditions, there is at most one solution. The situation is illustrated in Figure 2. We will find conditions for the following to hold, for any admissiblexandσ2_:

fx−θ2σ2

2θ2

x−a1> fa. 1.39

When thexkare not all identical,σ2/0. Hence, forθ2>0,fx−θ2σ2> fx>0, it is enough

to consider

fx2θ2

x−a1≥fa 1.40

which is the same as

θ2 ≥ −

fx−fa

2x−afx. 1.41

We can chooseθ2as in the theorem, which satisfiesθ2≥π1and

θ2≥π2 max x∈a,b−

fx−fa

2x−afx. 1.42

By Rolle’s theorem, we conclude that there can be at most two distinct roots to 1.37 on the intervala, b. This contradicts our assumption that allx1, . . . , xn are distinct. Hence, it

must be true that the maximum ofhpinSsubject to the constraintp1· · ·pn1 occurs on

the boundary ofS, where some of thepk must be zero. At the maximum, saypo, we are back

to the case of n−1 or less, and by the induction hypothesis, hpo _≤ _{0. Hence,} _h_p _≤ _{0 for}

arbitraryp∈Ssubject to the constraintp1· · ·pn1.

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Lemma 1.3. For any integer n ≥ 1, and for all x1, . . . , xn with each xk ∈ a, band allp1, . . . , pn,

where eachpk ≥0andn_k₁pk 1,

x−θ1σ2≥a, 1.43

whereθ1is as given inTheorem 1.2.

Proof. It suﬃces to show

x−μ1σ2≥a. 1.44

The case of n 1 is trivial, since σ2 _{0. For the case}_n _{2, as before, write} _x

2 1tx1,

where 0 ≤ t ≤ b−x1/x1,p1 1−q, andp2 q. With these, we havex 1qtx1, and σ2 _q₁₋_q_t2_x2

1. In the cases whereq 0,q 1 ort 0,1.44holds trivially. For 0 < q < 1

andt >0,1.44holds if and only if

μ1≤ψq

1qtx1−a q1−qt2_x2

1

. 1.45

We will show that this is indeed true for all admissibleq,t, andx1. For the casex1 a,ψq

1/1−qta. Becauseta≤b−a,ψq≥1/b−a. Whena < x1≤b, the value ofψqapproaches

∞asqapproaches 0 or 1. The minimum value must be on0,1. The derivative ofψqis

ψq q1−qtx1

1qtx1−a

2q−1

1−q2q2_t2_x2 1

. 1.46

Forqothat satisfiesψqo 0, we have the following identity:

1qot

x1−a qo

1−qo

tx1

1−2qo . 1.47

Hence,

ψqo

1

1−2qo

tx1

. 1.48

Becausetx1≤b−a, we getψqo≥1/b−a. By the definition ofμ1, for all admissibleq,t, and x1,

μ1≤

fx−f1tx1

21−qtx1f

x ≤

1 21−qtx1

. 1.49

Hence,μ1≤1/2b−a. Therefore,1.45holds for all admissibleq,t, andx1.

Consider the case ofn≥3. Fixx1, . . . , xnand suppose they are all distinct. Let

φp x−μ1σ2−a n

k1

pkxk−μ1 n

k1

pkx2_kμ1

_n

k1 pkxk

2

−a. 1.50

Consider minimizing φpover all possible weight vectors and supposepo _{is the minimum.}

Then, there exists a constantλsuch that, for anykwithpo_k >0, we must have∂φ/∂pkpo λ.

That is,

xk−μ1

x2_k−2xpoxk

λ, ∀k with po_k >0. 1.51

For the given po_{, there can be at most two distinct} _x

k satisfying the equation y −μ1y2 −

2xpo_y _λ_{in variable} _y_{. Hence, there can be at most two nonzero components in}_po_{. This}

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For partbofTheorem 1.2, the proof requiresxp−θ2σ2pto be within the domain of

the functionffor various unknown weight vectorsp. This is why the statement ofTheorem 1.2 makes the assumption that this is true for all possible weight vectors. The following lemma gives a simple suﬃcient condition for this assumption to hold.

Lemma 1.4. Fix x1, . . . , xn on a, b, n ≥ 2. Let θ2 be as given in Theorem 1.2. Without loss of

generality, assumex1< x2<· · ·< xn. Then,

awhenθ2≤1/xn−x1,

xp−θ2σ2p≥x1 for all weight vectors p, 1.52

and hence,xp−θ2σ2p∈a, bfor all weight vectorsp;

bwhenθ2>1/xn−x1,

xp−θ2σ2p≥x1−

θ2

xn−x1

−12

4θ2 for all weight vectorsp. 1.53

Hence, ifmin{a, x1−θ2xn−x1−12/4θ2}, b ⊆ I, thenxp−θ2σ2p ∈ Ifor all weight

vectorsp.

Proof. Writep p1, . . . , pn. Consider the minimization problem,

min

p≥0, nk1pk1

xp−θ2σ2p. 1.54

Using the same argument as in the proof ofLemma 1.3, we can conclude that the minimum is achieved at some p with at most two nonzero components. Hence, it suﬃces to consider

minimization problems of the following form:

min

pi≥0, pj≥0, pipj1

pixipjxj−θ2

pix_i2pjx_j2−

pixipjxj

2

. 1.55

It remains to be decided which i and j should be used in the above minimization. Suppose xi < xj here,i and j are unknown indices. We claim that i 1. To see this, the

partial derivative of the objective function with respect to xi is pi −θ22pixi−2xpi, where x pixipjxj. Sincexi ≤ x, the partial derivative is nonnegative, and hence, the function is

nondecreasing inxi.

Oncexiis chosen to bex1, the second partial derivative of the function with respect to xjis−2θ2pj−p_j2, which is nonpositive. The minimum is achieved at eitherxj x1orxj xn.

To summarize, the original minimization problem 1.54is achieved either at p1 1,

in which case the minimum value isx1, or it has the same minimum value as the following

problem:

min

p1≥0, pn≥0, p1pn1

p1x1pnxn−θ2

p1x2₁pnx2n−

p1x1pnxn

2

. 1.56

It is easy to show that, ifθ2 ≤1/xn−x1, the minimum of1.60is achieved atp11 and the

minimum value isx1. Otherwise, the minimum is achieved at

p1 1

2

1 1

θ2

xn−x1

, p2 1

2

1− 1

θ2

xn−x1

, 1.57

and the minimum value is x1 −θ2xn−x1−12/4θ2, which is no greater than x1 for all θ2>0.

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Remark 1.5. For partaof the theorem, we can chose a smaller value,−u/2fa, forμ1than

that in1.4. Letu≤0 be an upper bound offxona, b. Note that

1qtx1−1tx1−1−qtx1. 1.58

By the mean value theorem, there exists someη∈1qtx1,1tx1such that f1qtx1

−f1tx1

21−qtx1f

1qtx1

−fη 2f1qtx1

≥ −u

2fa. 1.59

Therefore, we can choose−u/2faforμ1.

Remark 1.6. For part bof the theorem, two simpler but less widely applicable choices forθ2

can be deduced from1.33. Sinceqtx1≤b−a,π1can be relaxed to

π₁1 1

minx∈a,b2fx/−fx−b−a

. 1.60

We must require minx∈a,b2fx/−fx−b−a > 0 forπ₁1 to be useful. An even simpler

choice is, when 2fb fab−a>0,

π₁2 1

2fb/−fa−b−a

−fa

2fb fab−a. 1.61

Note that 0 < π₁2 < ∞ implies 0 < π₁1 < ∞, which, in turn, implies 0 < π1 < ∞. In this case, π1≤π₁1≤π₁2. Similarly, if 0< π₁1<∞, thenπ1≤π₁1.

π2as in1.8can also be relaxed. Since, for someξ∈a, b,

−₂fx−fa

x−afx − fξ

2fx, 1.62

a relaxation is

π₂1 −f

_a

2fb. 1.63

Hence, for partbof the theorem, we can chooseθ2 max{π₁1, π₂1}, provided 0 < π₁1 < ∞.

Alternatively, we can chooseθ2max{π₁2, π₂1}π₁2, provided 0< π₁2<∞.

Remark 1.7. The strength ofTheorem 1.2is in bringing the variations ofxk’s into consideration,

through σ2. There are alternative methods for finding tighter upper bound of n_k₁pkfxk

than that given by Jensen’s inequality, n_k₁pkfxk ≤ fx. For instance, we can apply the

arithmetic mean-geometric mean inequality. For anyα,

n

k1

eαfxkpk ≤

n

k1

pkeαfxk. 1.64

Hence, for anyα >0,

n

k1 pkf

xk

≤ 1

αlog

n

k1

pkeαfxk

. 1.65

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2. Special cases and examples

The following theorem is needed.

Theorem 2.1see3, page 4. LetIbe a nonempty interval ofR. A functionh:I → Ris convex if and only if, for allxo ∈I,hx−hxo/x−xois a nondecreasing function ofxonI\ {xo}. Corollary 2.2. Let I be a nonempty interval of R. Suppose the function h : I → Ris convex, and

hx/0for allx∈I. If1/hxis a convex function, then, for allxo ∈I,hx−hxo/x−xohx

is a nonincreasing function of xon I\ {xo}. If 1/hx is a concave function, then, for all xo ∈ I,

hx−hxo/x−xohxis a nondecreasing function ofxonI\ {xo}.

Proof. Consider

hx−hxo

x−xo

hx

1−hxo

/hx

x−xo

−gx−g

xo

x−xo , 2.1

where we definegx hxo/hx. If 1/hxis convex, thengxis convex. ByTheorem 2.1,

hx−hxo/x−xohxis nonincreasing. If 1/hxis concave, thengxis concave, and

hence,hx−hxo/x−xohxis nondecreasing.

Theorem 2.3. Consider the functionfxas specified in Theorem 1.2. In addition, supposefxis convex. If1/fxis concave, then

θ1 min

x∈a,b

−fx

2fx. 2.2

If1/fxis convex, then

θ1

fa−fb

2b−afa. 2.3

Proof. If 1/fxis a concave function, then 1/f1qtx1is a concave function inq forq ∈

0,1. ByCorollary 2.2,f1qtx1−f1tx1/21−qtx1f1qtx1is nonincreasing

inq. Its minimum with respect toqoccurs atq1. Substitutingq1, we get

μ1 min

t,x1

−f1tx1

2f1tx1

min

q,t,x1

f1qtx1

−f1tx1

21−qtx1f

1qtx1

min

x∈a,b

−fx

2fx. 2.4

On the other hand, byCorollary 2.2,−fx−fb/2x−bfxis a nonincreasing function ofx. Hence,

μ2− fb

2fb ≥μ1. 2.5

If 1/fxis a convex function,f1qtx1−f1tx1/21−qtx1f1qtx1

achieves its minimum with respect toqatq0. Substitutingq0, we get

μ1min

t,x1

fx1

−f1tx1

2tx1f

x1

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ByTheorem 2.1,fx1−f1tx1/2tx1is a nonincreasing function oft, and hence, its

minimum occurs att b−x1/x1. Hence,

μ1 min

x1∈a,b

fx1

−fb

2b−x1

fx1

μ2. 2.7

By Corollary 2.2, fx1 − fb/2b − x1fx1 is a nondecreasing function of x1. Its

minimum, which occurs atx1a, isfa−fb/2b−afa.

Example 2.4fx logx. Here,fx 1/xandfx −1/x2. Hence,fxand 1/fxare both convex functions. We can choose

θ1 min q,t,x1

f1qtx1

−f1tx1

21−qtx1f

1qtx1

min

t,x1 1 21tx1

1 2b,

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−fx−θσ2 −qtx1

−1

min

θ∈0,1/21−qtx1

2x−θσ2−qtx1

−1

2

x−₂₁₋1 qtx1σ

2 ₋_qtx

1 −1 2

x1qtx1−

q1−qt2_x2 1

21−qtx1 − qtx1

−1

2x1

−1 .

2.8

Maximizing overx1, we getπ1 1/2a. Also,

π2 max

x∈a,b−

fx−fa

2x−afx

1

2a. 2.9

Example 2.5 fx xα, where 0 < α < 1. Here,fx αxα−1 is convex, andfx αα−

1xα−2. Because 1/xα−1is a concave function, we can applyTheorem 2.3, and get

θ1 min

x∈a,b

−fx

2fx xmin∈a,b

1−α

2x

1−α

2b ,

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−1

min

θ∈0,1/21−qtx1

2x−θσ2

1−α −qtx1

−1

₂_x

1qtx1

1−α −qtx1

−1

2x1

1−α α

1−αqtx1

(13)

Maximizing the last expression overq, we get1−α/2x1. Maximizing the result overx1, we

get

π1 1−α

2a . 2.10

Also, by Corollary 2.2,−fx−fa/2x−afxis a nonincreasing function ofx, and achieves its maximum atxa. Hence,

π2 max

x∈a,b−

fx−fa

2x−afx − fa

2fa

1−α

2a . 2.11

Example 2.6fx −e−x_{. Here,}_f_x _e−x_and_f_x ₋_e−x_{. Because 1}_/f_x_{is convex, we}

can applyTheorem 2.3, and get

θ1

fa−fb

2b−afa

1−e−b−a 2b−a ,

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−fx−θσ2 −qtx1

−1

2−qtx1

−1 .

Maximizing the last expression overqtx1, we get

π1

1

2−b−a. 2.12

We must require b − a < 2. Also, by Corollary 2.2, −fx − fa/2x − afx is a nondecreasing function ofx, and achieves its maximum atxb. Hence,

π2 max

x∈a,b−

fx−fa

2x−afx −

fb−fa

2b−afb

eb−a₋₁

2b−a. 2.13 Lemma 2.7. For0< b−a <2,

1 2−b−a≥

eb−a₋₁

2b−a. 2.14

Proof. Note that, for 0< b−a <2,2.14is equivalent to

2b−a≥2eb−a−1−b−aeb−a b−a. 2.15

Letvb−a, and assumev ∈0,2. It suﬃces to show

v−2ev−1vev ≥0. 2.16

Since at v 0, the left hand above is 0. We need only to show that v −2ev −1 vev is nondecreasing function ofv forv ≥ 0. Its derivative is 1−evvev, which is equal to zero at

v0. But, forv≥0,

1−evvevvev≥0. 2.17

(14)

Example 2.8fx −1/xk _where_{k >} _{0. Here,}_f_x _k/xk1_,_f_x ₋_k_k₁_/xk1_{. Since}

1/fx xk1_/k_{is convex, by}_{Theorem 2.3,}

θ1

fa−fb

2b−afa

1−a/bk1

2b−a ,

θ2

q, t, x1

min

θ∈0,1/21−qtx1

2fx−θσ2

−1

min

θ∈0,1/21−qtx1

2x−θσ2

k1 −qtx1

−1

2x1qtx1

k1 −qtx1

−1

2x1

k1 −

k k1qtx1

−1 .

Maximizing the last expression overqandt, we get 1/k2x1/k1−bk/k1. Maximizing

the result overx1, we get

π1 k1

k2a−bk. 2.18

We must requirek2a−bk >0, ork <2a/b−a, for the aboveπ1to be useful forTheorem 1.2.

Also, byCorollary 2.2,−fx−fa/2x−afxis a nondecreasing function ofx, and achieves its maximum atxb:

π2 max

x∈a,b−

fx−fa

2x−afx−

fb−fa

2b−afb

b/ak1−1

2b−a . 2.19

References

1 D. I. Cartwright and M. J. Field, “A refinement of the arithmetic mean-geometric mean inequality,”

Proceedings of the American Mathematical Society, vol. 71, no. 1, pp. 36–38, 1978.

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