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# A new interpretation of Jensen's inequality and geometric properties of φ -means

Yasuo Nakasuji1, Keisaku Kumahara1 and Sin-Ei Takahasi2*

Author Affiliations

1 The Open University of Japan, Chiba 261-8586, Japan

2 Toho University, Yamagata University (Professor Emeritus), Chiba 273-0866, Japan

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Journal of Inequalities and Applications 2011, 2011:48 doi:10.1186/1029-242X-2011-48

 Received: 23 November 2010 Accepted: 5 September 2011 Published: 5 September 2011

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

We introduce a mean of a real-valued measurable function f on a probability space induced by a strictly monotone function φ. Such a mean is called a φ-mean of f and written by Mφ(f). We first give a new interpretation of Jensen's inequality by φ-mean. Next, as an application, we consider some geometric properties of Mφ(f), for example, refinement, strictly monotone increasing (continuous) φ-mean path, convexity, etc.

Mathematics Subject Classification (2000): Primary 26E60; Secondary 26B25, 26B05.

##### Keywords:
Jensen's inequality; Mean; Refinement; Convexity; Concavity

### 1. Introduction

We are interested in means of real-valued measurable functions induced by strictly monotone functions. These means are somewhat different from continuously differentiable means, i.e., C1-means introducing by Fujii et al. [1], but they include many known numerical means. Here we first give a new interpretation of Jensen's inequality by such a mean and we next consider some geometric properties of such means, as an application of it.

Throughout the paper, we denote by (Ω, μ), I and f a probability space, an interval of ℝ and a real-valued measurable function on Ω with f(ω) ∈ I for almost all ω ∈ Ω, respectively. Let C(I) be the real linear space of all continuous real-valued functions defined on I. Let (resp. ) be the set of all φ C(I) which is strictly monotone increasing (resp. decreasing) on I. Then (resp. ) is a positive (resp. negative) cone of C(I). Put . Then Csm(I) denotes the set of all strictly monotone continuous functions on I.

Let Csm,f(I) be the set of all φ Csm (I) with φ f L1 (Ω, μ). Let φ be an arbitrary function of Csm,f(I). Since φ(I) is an interval of ℝ and μ is a probability measure on Ω, it follows that

Then there exists a unique real number Mφ(f) ∈ I such that . Since φ is one-to-one, it follows that

We call Mφ(f) a φ-quasi-arithmetic mean of f with respect to μ (or simply, φ-mean of f). A φ-mean of f has the following invariant property:

for each a, b ∈ ℝ with a ≠ 0.

Assume that μ(Ω\{ω1, ..., ωn}) = 0 for some ω1, ..., ωn ∈ Ω, f is a positive measurable function on Ω and I = ℝ. Then Mφ(f) will denote a weighted arithmetic mean, a weighted geometric mean, a weighted harmonic mean, etc. of {f(ω1), ..., f(ωn)} if φ(x) = x, φ(x) = log x, , etc., respectively.

In Section 2, we prepare some lemmas which we will need in the proof of our main results.

In Section 3, we first see that a φ-mean function: ∇φ Mφ(f) is order-preserving as a new interpretation of Jensen's inequality (see Theorem 1). We next see that there is a strictly monotone increasing φ-mean (continuous) path between two φ-means (see Theorem 2). We next see that the φ-mean function is strictly concave (or convex) on a suitable convex subset of Csm,f(I) (see Theorem 3). We also observe a certain boundedness of φ-means, more precisely,

under some conditions (see Theorem 4).

In Section 4, we treat a special φ-mean in which φ is a C2-functions with no stationary points.

In Section 5, we will give a new refinement of the geometric-arithmetic mean inequality as an application of our results.

### 2. Lemmas

This section is devoted to collecting some lemmas which we will need in the proof of our main results. The first lemma is to describe geometric properties of convex function, but this will be standard, so we will omit the proof (cf. [[2], (13.34) Exercise: Convex functions].

Lemma 1. Let φ be a real-valued function on I. Then the following three assertions are pairwise equivalent:

(i) φ is convex (resp. strictly convex).

(ii) For any c I°, a function λc,φ defined by

is monotone increasing (resp. strictly monotone increasing) on I\{c}.

(iii) For any c I°, there is a real constant mc ∈ ℝ such that

for all x I\{c}, i.e., the line through (c, φ(c)) having slope mc is always below or on (resp. below) the graph of φ.

Here I° denotes the interior of I.

For φ, ψ Csm(I) and c I°, put

This function has the following invariant property:

for each a, b ∈ ℝ with a ≠ 0. In this case, we have the following

Lemma 2. Let . Then, the following three assertions are pairwise equivalent:

(i) For any c I°, λc,φ,ψ is monotone increasing (resp. strictly monotone increasing) on I\{c}.

(ii) For any c I°, there is a real constant mc ∈ ℝ such that

for all x I\{c}.

(iii) ψ φ-1 is convex (resp. strictly convex) on φ(I).

Proof. (i) ⇒ (ii). Fix c I° arbitrarily. For any x I\{c}, put u = φ(x) and then

(1)

If λc,φ,ψ is monotone increasing (resp. strictly monotone increasing) on I\{c}, then λc,φ,ψ φ-1 is also monotone increasing (resp. strictly monotone increasing) on φ(I)\{φ(c)} and hence by (1) and Lemma 1, we can find a real constant mc ∈ ℝ which is independent of x such that

Since u = φ(x), we have

(ii) ⇒ (iii). Take u φ(I) and d ∈ (φ(I))arbitrarily. Put x = φ-1 (u) and c = φ-1 (d). Then x I and c I°. If we can find a real constant mc ∈ ℝ which is independent of u such that

then

and hence ψ φ-1 is convex (resp. strictly convex) on φ(I) by Lemma 1.

(iii) ⇒ (i). Take c I° and x I\{c} arbitrarily. Put u = φ(x) and d = φ(c).

Then u φ(I)\{d} and d ∈ (φ(I)), hence

(2)

If ψ φ-1 is convex (resp. strictly convex) on φ(I), then by (2) and Lemma 11, λc,φ,ψ φ-1 and hence λc,φ,ψ is monotone increasing (resp. strictly monotone increasing) on I\{c}. □

For each φ Csm(I), t ∈ [0, 1] and x, y I, put

This can be regarded as a φ-mean of {x, y} with respect to a probability measure which represents a weighted arithmetic mean (1-t) x + ty.

For each φ Csm(I), denote by ∇φ a three variable real-valued function:

on (0, 1) × {(x, y) ∈ I2 : x y}. For each φ, ψ Csm (I), we write ∇φ ≤ ∇ψ (resp. ∇φ < ∇ψ) if

for all t ∈ (0, 1) and x, y I with x y.

Remark. The continuity of φ implies that ∇φ ≤ ∇ψ (resp. ∇φ < ∇ψ) if and only if

for all x, y I with x y.

These order relations "≤" and "<" play an important role in our discussion.

Lemma 3. Let φ, ψ ∈ Csm (I). Then

(i) ∇φ = ∇ψ holds if and only if ψ = + b for some a, b ∈ ℝ with a ≠ 0.

(ii) If , then φ ≤ ∇ψ (resp. φ < ∇ψ) holds if and only if ψ φ-1 is convex (resp. strictly convex) on φ(I).

(iii) If , then φ ≤ ∇ψ (resp. φ < ∇ψ) holds if and only if ψ φ-1 is concave (resp. strictly concave) on φ(I).

Proof. (i) Suppose that ∇φ = ∇ψ holds. Take u, v φ(I) with u v arbitrarily and put x = φ-1(u) and y = φ-1(v), hence x y. By hypothesis,

for all t ∈ (0, 1). This means that ψ φ-1 is convex and concave on φ(I) and hence we can write ψ(φ-1(u)) = au + b for all u φ(I) and some a, b ∈ ℝ. Therefore, ψ(x) = (x) + b for all x I. Since ψ is non-constant, it follows that a ≠ 0.

The reverse assertion is straightforward.

(ii) Assume that ψ is monotone increasing. Take u, v φ(I) with u v arbitrarily and put x = φ-1(u) and y = φ-1(v), hence x y. If ∇φ ≤ ∇ψ holds, then

and hence

for all t ∈ (0, 1). This means that ψ φ-1 is convex.

Conversely, if ψ φ-1 is convex, we see that ∇φ ≤ ∇ψ holds by observing the reverse of the above proof.

Also a similar observation implies that ∇φ < ∇ψ holds if and only if ψ φ-1 is strictly convex on I.

(iii) Assume that ψ is monotone decreasing. Then -ψ is monotone increasing. Hence, by (ii), we have that ∇φ ≤ ∇-ψ (resp. ∇φ < ∇) holds if and only if (-ψ) ∘ φ-1 is convex (resp. strictly convex) on φ(I). However, since ∇ψ = ∇-ψ holds by (i) and (-ψ) ∘ φ-1 is convex (resp. strictly convex) on φ(I) iff ψ φ-1 is concave (resp. strictly concave) on φ(I), we obtain the desired result. □

Lemma 4. Let (or ) with φ < ∇ψ. For each s ∈ [0, 1], define ξs = (1 - s) φ + . Then

(i) Each ξs belongs to (resp. ) when (resp. ).

(ii) For each t ∈ (0, 1) and x, y I with x y, a function is strictly monotone increasing on [0, 1].

Proof. (i) Straightforward.

(ii) Assume with ∇φ < ∇ψ. Take t ∈ (0, 1) and x, y I with x y arbitrarily. To show that a function is strictly monotone increasing on [0, 1], let 0 ≤ s1 < s2 ≤ 1. Take c Iarbitrarily. Since ∇φ < ∇ψ holds, it follows from Lemmas 2 and 3 that λc,φ,ψ is strictly monotone increasing on I\{c}. Moreover, we have

for each x ∈ I\{c}. Therefore, we have

(3)

and

(4)

for each x ∈ I\{c}. If s1 = 0, then it is trivial by (3) that is strictly monotone increasing on I\{c}. If s1 ≠ 0, then

for all x I\{c}. So, by (4), is also strictly monotone increasing on I\{c}. Hence we see that holds by (i), Lemmas 2 and 3. This implies that . Then a function is strictly monotone increasing on [0, 1], as required.

For the case of , since , it follows from the above discussion that a function is strictly monotone increasing on [0, 1]. However, by Lemma 3-(i), , where t ∈ (0, 1) and x, y I with x y, and then we obtain the desired result. □

Lemma 5. Let φ and ψ be two functions on I such that ψ - φ is strictly monotone increasing (resp. decreasing) on I and ψ is convex (resp. concave) on I. Then

holds for all t ∈ (0, 1) and x, y I with x < y.

Proof. Let x, y I with x < y and t ∈ (0, 1). Put z = (1 - t)x + ty. Then, we must show that (1 - t) φ(x) + (y) - ((1 - t)φ + )(z) > 0 (resp. < 0). Since x < z < y and ψ- φ is strictly monotone increasing (resp. decreasing) on I, it follows that

Also since ψ is convex (resp. concave) on I, it follows from Lemma 1 that λz,ψ is monotone increasing (resp. decreasing). Therefore, we have

so that (1 - t) φ(x) + (y) - ((1 - t)φ + )(z) > 0 (resp. < 0), as required. □

The following lemma gives an equality condition of Jensen's inequality. For the sake of completeness, we will give a proof.

Lemma 6. Let δ be a strictly convex or strictly concave function on I. Suppose that g is a real-valued integrable function on Ω such that g(ω) ∈ I for almost all ω ∈ Ω and δ g L1 (Ω, μ). Then if and only if g is a constant function.

Proof. We first consider the strictly convex case. Put . If c = inf I, then c g(ω) for almost all ω ∈ Ω and so g(ω) = c must hold for almost all ω ∈ Ω. Similarly, if c = max I, then g(ω) = c for almost all ω ∈ Ω. Therefore, we can without loss of generality assume that c belongs to I. Since δ is strictly convex, we can from Lemma 1 find a real constant mc ∈ ℝ such that

(5)

for all x I\{c}. Replacing x by g(ω) in (5), we obtain

for almost all ω ∈ Ω. Integrating both sides of this equation, we have

(6)

Now assume that . Then (6) implies that

for almost all ω ∈ Ω. If μ({g c}) > 0, then we can find ωc ∈ Ω such that δ(g(ωc)) = mc (g(ωc) - c) + δ(c) and g(ωc) ≠ c. This contradicts (5) and hence g(ω) = c for almost all ω ∈ Ω.

Conversely, assume that g is a constant function on Ω. Then it is trivial that .

For the strictly concave case, since -δ is strictly convex on I, it follows from the above discussion that iff g is a constant function on Ω. However, since iff , we obtain the desired result. □

Lemma 7. Suppose that f is non-constant and φ, ψ Csm,f (I). Then

(i) If either ψ φ-1 is convex (resp. strictly convex) on φ(I) and or ψ φ-1 is concave (resp. strictly concave) on φ(I) and , then

holds.

(ii) If either ψ φ-1 is convex (resp. strictly convex) on φ(I) and or ψ φ-1 is concave (resp. strictly concave) on φ(I) and , then

holds.

Proof. (i) Put δ = ψ φ-1 and g = φ f. Assume that g is convex on φ(I) and . Since g and δ g are integrable functions on Ω, we have

(7)

by Jensen's inequality. This means Mφ (f) ≤ Mψ (f) because ψ is monotone increasing on I.

Next assume that g is concave on φ(I) and . Then

(8)

by Jensen's inequality. This also means Mφ (f) ≤ Mψ (f) because ψ is monotone decreasing on I.

For the strict case, since g is a non-constant function on Ω, we obtain the desired results from (7), (8), Lemma 6 and the above argument. □

(ii) Similarly.

### 3. Main results

In this section, we first give a new interpretation of Jensen's inequality by φ-mean. Next, as an application, we consider some geometric properties of φ-means of a real-valued measurable function f on Ω.

The first result asserts that a φ-mean function: ∇φ Mφ (f) is well defined and order preserving, and this assertion simultaneously gives a new interpretation of Jensen's inequality. However, this assertion also teaches us that a simple inequality yields a complicated inequality.

Theorem 1. Suppose that f is non-constant and φ, ψ Csm,f (I). Then

(i) If φ ≤ ∇ψ holds, then Mφ (f) ≤ Mψ (f).

(ii) If φ < ∇ψ holds, then Mφ (f) < Mψ (f).

Proof. (i) Suppose that ∇φ ≤ ∇ψ holds. If ψ is monotone increasing on I, then ψ φ-1 is convex on φ(I) by Lemma 3-(ii). Therefore, we have Mφ (f) ≤ Mψ (f) by Lemma 7-(i). If ψ is monotone decreasing on I, then ψ φ-1 is concave on φ(I) by Lemma 3-(iii). Therefore, we have Mφ (f) ≤ Mψ (f) by Lemma 7-(i).

(ii) Similarly. □

Let φ, ψ Csm,f (I) and t ∈ (0, 1). Then, we can easily see that if either both φ and ψ are monotone increasing or both φ and ψ are monotone decreasing, then (1 - t)φ + is also an element of Csm,f (I) [cf. Lemma 4-(i)]. The next result asserts that there is a strictly monotone increasing φ-mean (continuous) path between two φ-means.

Theorem 2. Suppose that f is non-constant and φ, ψ Csm,f (I) with φ < ∇ψ.

(i) If [or ], then a function: s → M(1-s)φ+(f) is strictly monotone increasing on [0, 1].

(ii) If [resp. ] and ψ(x) - φ(x) ≥ 0 (resp. ≤ 0) for all x I, then a function: s → M(1-s)φ+(f) is strictly monotone increasing and continuous on [0.1].

Proof. (i) Suppose that [or ]. For each s ∈ [0, 1], define ξs = (1 - s)φ + sψ. Let 0 ≤ s1 < s2 ≤ 1. Then, we must show that . By Lemma 4-(ii), a function is strictly monotone increasing on [0, 1] for each t ∈ (0, 1) and x, y I with x y, and hence we see that holds. Therefore, we have from Theorem 1-(ii) that , as required.

(ii) Suppose that and φ(x) ≤ ψ(x) for all x I. Since ψ = φ + (ψ- φ), it follows that . For each s ∈ [0, 1], put αs = M(1-s)φ+ (f). Then, we must show that a function s αs is continuous on [0, 1]. To do this, take 0 ≤ s < t ≤ 1 arbitrarily. By (i), we have αs < αt. Note that

and

Therefore, we have

(9)

Since and φ(x) ≤ ψ(x) for all x I by hypothesis, it follows that

Hence, after taking the limit with respect to s in the Eq. (9), we obtain

However, since φ-1 is continuous on φ(I), we conclude that

Similarly, after taking the limit with respect to t in the Eq. (9), we obtain

These observations imply that a function s αs is continuous on [0, 1], as required.

For the case that and φ(x) ≥ ψ(x) for all x I, a similar argument above implies that a function s αs is also continuous on [0, 1]. □

The next result asserts that the φ-mean function is strictly concave (or convex) on a suitable convex subset of Csm,f(I).

Theorem 3. Suppose that f is non-constant and φ, ψ Csm,f(I) with φ < ∇ψ. Then

(i) If (resp. ) and ψ is convex (resp. concave) on I, then

holds for all t ∈ (0, 1).

(ii) If (resp. ) and ψ is convex (resp. concave) on I, then

holds for all t ∈ (0, 1).

Proof. (i) Suppose that [resp. ] and ψ is convex [resp. concave] on I. Since ψ = φ + (ψ- φ), it follows from hypothesis that [resp. ]. Put x = Mφ(f) and y = Mψ(f), and so x < y by Theorem 1-(ii). Also, we have from definition that

Let 0 < t < 1 and put u = M(1-t)φ+(f). Then, we have

by definition. Therefore,

Put z = (1 - t)x + ty. Then, by the above equality and Lemma 5, we have

Since (1 - t)φ + is strictly increasing (resp. decreasing), it follows that z < u, that is,

This means that (1 - t)Mφ(f) + tMψ(f) < M(1-t)φ+(f).

(ii) Similarly.

Remark. It seems that Theorem 3 is slightly related to [3,4] which discuss a comparison between a convex linear combination of the arithmetic and geometric means and the generalized logarithmic mean.

The following result describes a certain boundedness of φ-means.

Theorem 4. Suppose that f is non-constant and φ, ψ Csm,f(I) with φ < ∇ψ.

(i) If [or ], then a function: s M(1-s)φ+(f) is strictly monotone increasing on [0, ∞) and

(ii) If [resp. ] and ψ(x) - φ(x) ≥ 0 (resp. ≤ 0) for all x I, then a function: s M(1-s)φ+(f) is strictly monotone increasing and continuous on [0, ∞).

Proof. (i) Suppose that . For each s ≥ 1, put ξs = (1 - s)φ + . Since ξs = φ + s(ψ - φ), it follows from hypothesis that each ξs is in , and then ξs Csm,f(I). Since ψ = φ + (ψ- φ), it follows from hypothesis that ψ is also in . Then by Lemmas 2 and 3, we have that λc,φ,ψ is strictly monotone increasing on I\{c} for any c I°. Let 1 ≤ s1 < s2 < ∞ and take c I° arbitrarily. In this case, we obtain the equality (4), as observe in the proof of Lemma 4-(ii). Note that

for all x I\{c}. So, by (4), is also strictly monotone increasing on I\{c}. Then by Lemmas 2 and 3, we conclude that . Therefore, we have from Theorem 1-(ii) that and then a function: s M(1-s)φ+(f) is strictly monotone increasing on [1, ∞) and hence [0, ∞) by Theorem 2-(i).

Moreover, we can easily see that

and

for all s ≥ 1, x I\{c} and c I°. This implies that is strictly monotone increasing on I\{c} for each s ≥ 1 and c I°. Then by Lemmas 2 and 3, we conclude that for each s ≥ 1. Therefore, we have from Theorem 1-(ii) that for each s ≥ 1.

Now take s ≥ 1 arbitrarily and put and α = Mψ-φ(f), so αs < α. Since , it follows that φ(αs) < φ(α) and (ψ- φ)(αs) < (ψ- φ)(α). By definition, we have

Also since , it follows from an invariant property of φ-mean that and then

Therefore, we have

Hence, after taking the limit with respect to s, we obtain

However, since (ψ- φ)-1 is continuous on (ψ- φ)(I), we conclude that

that is,

For the decreasing case, replacing φ and ψ by -φ and -ψ, respectively, apply the above discussion for the increasing case.

(ii) Refer to the Proof of Theorem 2-(ii). □

### 4. φ-means by C2-functions

In this section, we treat a special φ-mean in which φ is a C2-functions with no stationary points. For each real-valued measurable function f on Ω, let be the set of all C2-functions φ in Csm,f(I) with no stationary points, that is, φ'(t) ≠ 0 for all t I.

Lemma 8. Let . Then

(i) The following two statements are equivalent:

(a) ψ φ-1 is convex (resp. concave) on φ(I).

(b) (resp. ≤ 0) for all x I°.

(ii) The following two statements are equivalent:

(c) ψ φ-1 is strictly convex (resp. strictly concave) on φ(I).

(d) (resp. < 0) for all x I°.

Proof. (i) Define τ(u) = ψ((φ-1(u)) for each u φ(I). Then a simple calculation yields that

for all u ∈ (φ(I))°, where x = φ-1(u). This equation implies that (a) and (b) are equivalent.

(ii) Similarly. □

Lemma 9. Let φ and ψ be C1-functions on I. Then,

(i) If φ'(x) < ψ'(x) for all x I° and ψ' is monotone increasing on I, then

holds for all x, y I with x < y and t ∈ (0, 1).

(ii) If φ'(x) > ψ'(x) for all x I° and ψ' is monotone decreasing on I, then

holds for all x, y I with x < y and t ∈ (0, 1).

Proof. (i) Suppose that φ'(x) < ψ'(x) for all x I° and ψ' is monotone increasing on I. Let x, y I with x < y and t ∈ (0, 1). Put z = (1 - t)x + ty. Then, we must show that ((1 - t)φ + )(z) < (1 - t)φ(x) + (y). By the mean value theorem, we have

for some θ, θ'∈ (0, 1) because z + (y- z)θ x + (z- x)θ' and hence ψ'(z + (y- z)θ) ≥ ψ'( x + (z- x)θ') by hypothesis. Since x + (z- x)θ' ∈ I°, it follows from hypothesis that

and so (1 - t)φ(x) + (y) - ((1 - t)φ + )(z) > 0 from the preceding inequalities. Therefore, we obtain the desired inequality.

(ii) Similarly. □

Corollary 1. Suppose that f is non-constant and . Then

(i) If for all x I°, then Mφ(f) ≤ Mψ(f).

(ii) If for all x I° then Mφ(f) < Mψ(f).

Proof. (i) Suppose that for all x I°. If ψ is monotone increasing on I, then ψ'(x) > 0 for all x I° and hence ψ φ-1 is convex on φ(I) by Lemma 8-(i). Therefore, by Lemma 3-(ii), ∇φ ≤ ∇ψ holds and then Mφ(f) ≤ Mψ(f) by Theorem 1-(i). If ψ is monotone decreasing on I, then ψ'(x) < 0 for all x I° and hence ψ φ-1 is concave on φ(I) by Lemma 8-(i). Therefore, by Lemma 3-(iii), ∇φ ≤ ∇ψ also holds and then Mφ(f) ≤ Mψ(f) by Theorem 1-(i).

(ii) Similarly. □

Remark. Let (Ω, μ) be a probability space, 0 < p < q < ∞ and let f be a non-constant real-valued function in Lq(Ω, μ). Then the well-known inequality: ||f||p < ||f||q follows immediately from Corollary 1 (ii), by considering a family {φr : r > 0} of functions on ℝ+, where φr(x) = xr (x > 0).

Let and let t ∈ (0, 1). Then, we can easily see that if either both φ and ψ are monotone increasing on I or both φ and ψ are monotone decreasing on I, then . In this case, we have the following

Corollary 2. Suppose that f is non-constant and . If and φ'(x)ψ'(x) > 0 for all x I°, then a function: s M(1-s)φ+(f) is strictly increasing on [0, 1].

Proof. Suppose that and φ'(x)ψ'(x) > 0 for all x I°. We define ξ(s, x) = (1 - s)φ(x) + (x) for each s ∈ (0, 1). We can easily see that

for each s ∈ (0, 1) and x I°. Then, we have from Corollary 1-(ii) that Mφ(f) < M(1-s)φ+(f) for all s ∈ (0, 1). Similarly, we can see that M(1-s)φ+(f) < Mψ(f) for all s ∈ (0, 1). Now put

for each s ∈ (0, 1) and x I°. Then a simple calculation implies that

for each s ∈ (0, 1) and x I°. Therefore, for a fixed x I°, a function: s A(s, x) is strictly increasing on (0, 1). Therefore, Corollary 1-(ii) implies that a function: s M(1-s)φ+(f) is strictly increasing on (0, 1) and hence [0, 1]. □

Corollary 3. Suppose that f is non-constant and that is such that for for all x I°. Then

(i) If either 0 < φ' < ψ' and ψ" ≥ 0 on I° or ψ' < φ' < 0 and ψ" ≤ 0 on I°, then

holds for all t ∈ (0, 1).

(ii) If either φ' < ψ' < 0 and ψ" ≥ 0 on I° or 0 < ψ' < φ' and ψ" ≤ 0 on I°, then

holds for all t ∈ (0, 1).

Proof. (i) Suppose that 0 < φ' < ψ' and ψ" ≥ 0 on I°. Put x = Mφ(f) and y = Mψ(f), and so x < y by Corollary 1-(ii). Take t ∈ ℝ with 0 < t < 1 arbitrarily. By hypothesis, (1 - t)φ + is strictly monotone increasing on I. Put u = M(1-t)φ+(f). As observe in the proof of Theorem 3-(i), we have

(10)

Put z = (1 - t)x + ty. Then, by (10) and Lemma 9-(i), we have

and then z < u, that is, (1 - t)Mφ(f) + tMψ(f) < M(1-t)φ+(f).

In the case of ψ' < φ' < 0 and ψ ≤ 0 on I°, we apply Lemma 9-(ii).

(2) Similarly. □

### 5. Remarks

(i) Let I = ℝ+. Put and ψ(x) = x for each x I. Of course, these functions belong to Csm(I). The harmonic-arithmetic mean inequality asserts that ∇φ < ∇ψ. Take a non-constant positive measurable function f on a probability space (Ω, μ) such that φ f and ψ f are in L1(Ω, μ). Then, we have from Theorem 1-(ii) that Mφ(f) < Mψ(f). Observe that this inequality means

This is a special case of Jensen's inequality (or Schwarz's inequality). We note that if 0 < m f M, then . The right side of this inequality is called a Kantorovich constant (cf. [5-7]).

(ii) A similar consideration for the geometric-arithmetic mean inequality yields that

This is also a special case of Jensen's inequality. We note that if 0 < m f M, then , where . The right side of this inequality is called Specht's ratio (cf. [8]).

(iii) For each t ∈ [0, 1], put log[t] x = (1 - t)log x + tx(x > 0). Then log[t] is a strictly monotone increasing real-valued continuous function on ℝ+. Denote by exp[t] the inverse function of log[t]. Let x1, ..., xn > 0 and p1, ..., pn > 0 with . Then Theorem 2-(i) (or Corollary 2) implies that is strictly monotone increasing on [0, 1]. Note that and . Therefore, we obtain that

This is a new refinement of geometric-arithmetic mean inequality (cf. [9]).

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

YN carried out the design of the study and performed the analysis. KK conceived of the study, and participated in its design and coordination. ST participated in the sequence alignment and drafted the manuscript. All authors read and approved the final manuscript.

### Acknowledgements

The authors are grateful to the referee, for the careful reading of the paper and for the helpful suggestions and comments. Also, we would like to thank Professor Masatoshi Fujii for his helpful informations of ∇ and Specht's ratio. S.-E. Takahasi is partially supported by Grant-in-Aid for Scientific Research, Japan Society for the Promotion of Science.

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