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A new interpretation of Jensen's inequality and geometric properties of ϕmeans
Journal of Inequalities and Applications volume 2011, Article number: 48 (2011)
Abstract
We introduce a mean of a realvalued measurable function f on a probability space induced by a strictly monotone function φ. Such a mean is called a φmean of f and written by M_{ φ } (f). We first give a new interpretation of Jensen's inequality by φmean. Next, as an application, we consider some geometric properties of M_{ φ } (f), for example, refinement, strictly monotone increasing (continuous) φmean path, convexity, etc.
Mathematics Subject Classification (2000): Primary 26E60; Secondary 26B25, 26B05.
1. Introduction
We are interested in means of realvalued measurable functions induced by strictly monotone functions. These means are somewhat different from continuously differentiable means, i.e., C^{1}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 realvalued 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 realvalued functions defined on I. Let {C}_{sm}^{+}\left(I\right) (resp. {C}_{sm}^{}\left(I\right)) be the set of all φ ∈ C(I) which is strictly monotone increasing (resp. decreasing) on I. Then {C}_{sm}^{+}\left(I\right) (resp. {C}_{sm}^{}\left(I\right)) is a positive (resp. negative) cone of C(I). Put {C}_{sm}\left(I\right)={C}_{sm}^{+}\left(I\right)\cup {C}_{sm}^{}\left(I\right). Then C_{ sm } (I) denotes the set of all strictly monotone continuous functions on I.
Let C_{ sm },_{ f } (I) be the set of all φ ∈ C_{ sm } (I) with φ ∘ f ∈ L^{1} (Ω, μ). Let φ be an arbitrary function of C_{ sm },_{ 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 M_{ φ } (f) ∈ I such that \int \left(\phi \circ f\right)\mathsf{\text{d}}\mu =\phi \left({M}_{\phi}\left(f\right)\right). Since φ is onetoone, it follows that
We call M_{ φ } (f) a φquasiarithmetic mean of f with respect to μ (or simply, φmean of f). A φmean of f has the following invariant property:
for each a, b ∈ ℝ with a ≠ 0.
Assume that μ(Ω\{ω_{1}, ..., ω_{ n } }) = 0 for some ω_{1}, ..., ω_{ n } ∈ Ω, f is a positive measurable function on Ω and I = ℝ. Then M_{ φ } (f) will denote a weighted arithmetic mean, a weighted geometric mean, a weighted harmonic mean, etc. of {f(ω_{1}), ..., f(ω_{ n } )} if φ(x) = x, φ(x) = log x, \phi \left(x\right)=\frac{1}{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 φmean function: ∇ _{ φ } → M_{ φ } (f) is orderpreserving as a new interpretation of Jensen's inequality (see Theorem 1). We next see that there is a strictly monotone increasing φmean (continuous) path between two φmeans (see Theorem 2). We next see that the φmean function is strictly concave (or convex) on a suitable convex subset of C_{ sm },_{ f } (I) (see Theorem 3). We also observe a certain boundedness of φmeans, more precisely,
under some conditions (see Theorem 4).
In Section 4, we treat a special φmean in which φ is a C^{2}functions with no stationary points.
In Section 5, we will give a new refinement of the geometricarithmetic 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 realvalued 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 λ _{c,φ} defined by
{\lambda}_{c,\phi}\left(x\right)=\frac{\phi \left(x\right)\phi \left(c\right)}{xc}\phantom{\rule{1em}{0ex}}\left(x\in I\backslash \left\{c\right\}\right)
is monotone increasing (resp. strictly monotone increasing) on I\{c}.

(iii)
For any c ∈ I°, there is a real constant m_{ c } ∈ ℝ such that
\phi \left(x\right)\phi \left(c\right){m}_{c}\left(xc\right)\ge 0\phantom{\rule{1em}{0ex}}\left(resp.>0\right)
for all x ∈ I\{c}, i.e., the line through (c, φ(c)) having slope m_{ c } is always below or on (resp. below) the graph of φ.
Here I° denotes the interior of I.
For φ, ψ ∈ C_{ sm } (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 \phi ,\psi \in {C}_{sm}^{+}\left(I\right). Then, the following three assertions are pairwise equivalent:

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

(ii)
For any c ∈ I°, there is a real constant m_{ c } ∈ ℝ such that
\psi \left(x\right)\psi \left(c\right){m}_{c}\left(\phi \left(x\right)\phi \left(c\right)\right)\ge 0\phantom{\rule{1em}{0ex}}\left(resp.>0\right)
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
If λ_{ c,φ,ψ } is monotone increasing (resp. strictly monotone increasing) on I\{c}, then λ_{ c,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 m_{ c } ∈ 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 m_{ c } ∈ ℝ which is independent of u such that
\psi \left(x\right)\psi \left(c\right){m}_{c}\left(\phi \left(x\right)\phi \left(c\right)\right)\ge 0\phantom{\rule{1em}{0ex}}\left(resp.>0\right),
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
If ψ ∘ φ^{1} is convex (resp. strictly convex) on φ(I), then by (2) and Lemma 11, λ_{ c,φ,ψ } ∘ φ^{1} and hence λ_{ c,φ,ψ } is monotone increasing (resp. strictly monotone increasing) on I\{c}. □
For each φ ∈ C_{ sm } (I), t ∈ [0, 1] and x, y ∈ I, put
This can be regarded as a φmean of {x, y} with respect to a probability measure which represents a weighted arithmetic mean (1t) x + ty.
For each φ ∈ C_{ sm } (I), denote by ∇ _{ φ } a three variable realvalued function:
on (0, 1) × {(x, y) ∈ I^{2} : x ≠ y}. For each φ, ψ ∈ C_{ sm } (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 ∈ C_{ sm } (I). Then

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

(ii)
If \psi \in {C}_{sm}^{+}\left(I\right), then ∇ _{ φ } ≤ ∇ _{ ψ } (resp. ∇ _{ φ } < ∇_{ ψ }) holds if and only if ψ ∘ φ ^{1} is convex (resp. strictly convex) on φ(I).

(iii)
If \psi \in {C}_{sm}^{}\left(I\right), 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) = aφ(x) + b for all x ∈ I. Since ψ is nonconstant, 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
\psi \left({\phi}^{1}\left(\left(1t\right)\phi \left(x\right)+t\phi \left(y\right)\right)\right)\le \left(1t\right)\psi \left(x\right)+t\psi \left(y\right)
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 } < ∇_{ ψ }) 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\phi ,\psi \in {C}_{sm}^{+}\left(I\right)(or{C}_{sm}^{}\left(I\right)) with ∇ _{ φ } < ∇_{ ψ }. For each s ∈ [0, 1], define ξ_{ s } = (1  s) φ + sψ. Then

(i)
Each ξ_{ s } belongs to {C}_{sm}^{+}\left(I\right) (resp. {C}_{sm}^{}\left(I\right) ) when \phi ,\psi \in {C}_{sm}^{+}\left(I\right) (resp. {C}_{sm}^{}\left(I\right) ).

(ii)
For each t ∈ (0, 1) and x, y ∈ I with x ≠ y, a function s\to x{\nabla}_{t,{\xi}_{s}}y is strictly monotone increasing on [0, 1].
Proof. (i) Straightforward.

(ii)
Assume \phi ,\psi \in {C}_{sm}^{+}\left(I\right) with ∇ _{ φ } < ∇_{ ψ }. Take t ∈ (0, 1) and x, y ∈ I with x ≠ y arbitrarily. To show that a function s\to x{\nabla}_{t,{\xi}_{s}}y is strictly monotone increasing on [0, 1], let 0 ≤ s _{1} < s _{2} ≤ 1. Take c ∈ I ^{∘} arbitrarily. Since ∇ _{ φ } < ∇ _{ ψ } holds, it follows from Lemmas 2 and 3 that λ_{ c,φ,ψ } is strictly monotone increasing on I\{c}. Moreover, we have
\begin{array}{lll}\hfill {\lambda}_{c,{\xi}_{{s}_{1}},{\xi}_{{s}_{2}}}\left(x\right)& =\frac{{\xi}_{{s}_{2}}\left(x\right){\xi}_{{s}_{2}}\left(c\right)}{{\xi}_{{s}_{1}}\left(x\right){\xi}_{{s}_{1}}\left(c\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\frac{{s}_{2}\left(\psi \left(x\right)\psi \left(c\right)\right)+\left(1{s}_{2}\right)\left(\phi \left(x\right)\phi \left(c\right)\right)}{{s}_{1}\left(\psi \left(x\right)\psi \left(c\right)\right)+\left(1{s}_{1}\right)\left(\phi \left(x\right)\phi \left(c\right)\right)}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\frac{{s}_{2}{\lambda}_{c,\phi ,\psi}\left(x\right)+1{s}_{2}}{{s}_{1}{\lambda}_{c,\phi ,\psi}\left(x\right)+1{s}_{1}}.\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}
for each x ∈ I\{c}. Therefore, we have
and
for each x ∈ I\{c}. If s_{1} = 0, then it is trivial by (3) that {\lambda}_{c,{\xi}_{{s}_{1}},{\xi}_{{s}_{2}}} is strictly monotone increasing on I\{c}. If s_{1} ≠ 0, then
for all x ∈ I\{c}. So, by (4), {\lambda}_{c,{\xi}_{{s}_{1}},{\xi}_{{s}_{2}}} is also strictly monotone increasing on I\{c}. Hence we see that {\nabla}_{{\xi}_{{s}_{1}}}<{\nabla}_{{\xi}_{{s}_{2}}} holds by (i), Lemmas 2 and 3. This implies that x{\nabla}_{t,{\xi}_{{s}_{1}}}y<x{\nabla}_{t,{\xi}_{{s}_{2}}}y. Then a function s\to x{\nabla}_{t,{\xi}_{s}}y is strictly monotone increasing on [0, 1], as required.
For the case of \phi ,\psi \in {C}_{sm}^{}\left(I\right), since \phi ,\psi \in {C}_{sm}^{+}\left(I\right), it follows from the above discussion that a function s\to x{\nabla}_{t,{\xi}_{s}}y is strictly monotone increasing on [0, 1]. However, by Lemma 3(i), x{\nabla}_{t,{\xi}_{s}}y=x{\nabla}_{t,{\xi}_{s}}y, 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) + tψ(y)  ((1  t)φ + tψ)(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 λ_{z,ψ}is monotone increasing (resp. decreasing). Therefore, we have
so that (1  t) φ(x) + tψ(y)  ((1  t)φ + tψ)(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 realvalued integrable function on Ω such that g(ω) ∈ I for almost all ω ∈ Ω and δ ∘ g ∈ L^{1} (Ω, μ). Then \delta \left(\int gd\mu \right)=\int \left(\delta \circ g\right)d\mu if and only if g is a constant function.
Proof. We first consider the strictly convex case. Put c=\int g\mathsf{\text{d}}\mu. 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 m_{ c } ∈ ℝ such that
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
Now assume that \delta \left(\int g\mathsf{\text{d}}\mu \right)=\int \left(\delta \circ g\right)\mathsf{\text{d}}\mu. Then (6) implies that
for almost all ω ∈ Ω. If μ({g ≠ c}) > 0, then we can find ω_{ c } ∈ Ω such that δ(g(ω_{ c } )) = m_{ c } (g(ω_{ c } )  c) + δ(c) and g(ω_{ c } ) ≠ 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 \delta \left(\int g\mathsf{\text{d}}\mu \right)=\int \left(\delta \circ g\right)\mathsf{\text{d}}\mu.
For the strictly concave case, since δ is strictly convex on I, it follows from the above discussion that \delta \left(\int g\mathsf{\text{d}}\mu \right)=\int \left(\delta \circ g\right)\mathsf{\text{d}}\mu iff g is a constant function on Ω. However, since \delta \left(\int g\mathsf{\text{d}}\mu \right)=\int \left(\delta \circ g\right)\mathsf{\text{d}}\mu iff \delta \left(\int g\mathsf{\text{d}}\mu \right)=\int \left(\delta \circ g\right)\mathsf{\text{d}}\mu, we obtain the desired result. □
Lemma 7. Suppose that f is nonconstant and φ, ψ ∈ C_{ sm,f } (I). Then

(i)
If either ψ ∘ φ ^{1} is convex (resp. strictly convex) on φ(I) and \psi \in {C}_{sm}^{+}\left(I\right) or ψ ∘ φ ^{1} is concave (resp. strictly concave) on φ(I) and \psi \in {C}_{sm}^{}\left(I\right), then
{M}_{\phi}\left(f\right)\le {M}_{\psi}\left(f\right)\phantom{\rule{1em}{0ex}}\left(resp.\phantom{\rule{0.3em}{0ex}}{M}_{\phi}\left(f\right)<{M}_{\psi}\left(f\right)\right)
holds.

(ii)
If either ψ ∘ φ ^{1} is convex (resp. strictly convex) on φ(I) and \psi \in {C}_{sm}^{}\left(I\right) or ψ ∘ φ ^{1} is concave (resp. strictly concave) on φ(I) and \psi \in {C}_{sm}^{+}\left(I\right), then
{M}_{\phi}\left(f\right)\ge {M}_{\psi}\left(f\right)\phantom{\rule{1em}{0ex}}\left(resp.\phantom{\rule{0.3em}{0ex}}{M}_{\phi}\left(f\right)>{M}_{\psi}\left(f\right)\right)
holds.
Proof. (i) Put δ = ψ ∘ φ^{1} and g = φ ∘ f. Assume that g is convex on φ(I) and \psi \in {C}_{sm}^{+}\left(I\right). Since g and δ ∘ g are integrable functions on Ω, we have
by Jensen's inequality. This means M_{ φ } (f) ≤ M_{ ψ } (f) because ψ is monotone increasing on I.
Next assume that g is concave on φ(I) and \psi \in {C}_{sm}^{}\left(I\right). Then
by Jensen's inequality. This also means M_{ φ } (f) ≤ M_{ ψ } (f) because ψ is monotone decreasing on I.
For the strict case, since g is a nonconstant 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 φmean. Next, as an application, we consider some geometric properties of φmeans of a realvalued measurable function f on Ω.
The first result asserts that a φmean function: ∇ _{ φ } → M_{ φ } (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 nonconstant and φ, ψ ∈ C_{ sm,f } (I). Then

(i)
If ∇ _{ φ } ≤ ∇ _{ ψ } holds, then M_{ φ } (f) ≤ M_{ ψ } (f).

(ii)
If ∇ _{ φ } < ∇ _{ ψ } holds, then M_{ φ } (f) < M_{ ψ } (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 M_{ φ } (f) ≤ M_{ ψ } (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 M_{ φ } (f) ≤ M_{ ψ } (f) by Lemma 7(i).

(ii)
Similarly. □
Let φ, ψ ∈ C_{ sm,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 + tψ is also an element of C_{ sm,f } (I) [cf. Lemma 4(i)]. The next result asserts that there is a strictly monotone increasing φmean (continuous) path between two φmeans.
Theorem 2. Suppose that f is nonconstant and φ, ψ ∈ C_{ sm,f } (I) with ∇ _{ φ } < ∇_{ ψ }.

(i)
If \phi ,\psi \in {C}_{sm}^{+}\left(I\right) [or {C}_{sm}^{}\left(I\right) ], then a function: s → M _{(1s)φ+sψ}(f) is strictly monotone increasing on [0, 1].

(ii)
If \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) [resp. {C}_{sm}^{}\left(I\right)] and ψ(x)  φ(x) ≥ 0 (resp. ≤ 0) for all x ∈ I, then a function: s → M _{(1s)φ+sψ}(f) is strictly monotone increasing and continuous on [0.1].
Proof. (i) Suppose that \phi ,\psi \in {C}_{sm}^{+}\left(I\right)[or{C}_{sm}^{}\left(I\right)]. For each s ∈ [0, 1], define ξ_{s} = (1  s)φ + s_{ ψ } . Let 0 ≤ s_{1} < s_{2} ≤ 1. Then, we must show that {M}_{{\xi}_{{s}_{1}}}\left(f\right)<{M}_{{\xi}_{{s}_{2}}}\left(f\right). By Lemma 4(ii), a function s\to x{\nabla}_{t,{\xi}_{s}}y is strictly monotone increasing on [0, 1] for each t ∈ (0, 1) and x, y ∈ I with x ≠ y, and hence we see that {\nabla}_{{\xi}_{{s}_{1}}}<{\nabla}_{{\xi}_{{s}_{2}}} holds. Therefore, we have from Theorem 1(ii) that {M}_{{\xi}_{{s}_{1}}}\left(f\right)<{M}_{{\xi}_{{s}_{2}}}\left(f\right), as required.

(ii)
Suppose that \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) and φ(x) ≤ ψ(x) for all x ∈ I. Since ψ = φ + (ψ φ), it follows that \psi \in {C}_{sm}^{+}\left(I\right). For each s ∈ [0, 1], put α_{ s } = M_{(1s)φ+ sψ}(f). Then, we must show that a function s → α_{ s } is continuous on [0, 1]. To do this, take 0 ≤ s < t ≤ 1 arbitrarily. By (i), we have α_{ s } < α_{ t } . Note that
\left(1t\right)\phi \left({\alpha}_{t}\right)+t\psi \left({\alpha}_{t}\right)=\left(1t\right)\int \left(\phi \circ f\right)\mathsf{\text{d}}\mu +t\int \left(\psi \circ f\right)\mathsf{\text{d}}\mu
and
Therefore, we have
Since \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) 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 → α_{ s } is continuous on [0, 1], as required.
For the case that \phi ,\psi \phi \in {C}_{sm}^{}\left(I\right) and φ(x) ≥ ψ(x) for all x ∈ I, a similar argument above implies that a function s → α_{ s } is also continuous on [0, 1]. □
The next result asserts that the φmean function is strictly concave (or convex) on a suitable convex subset of C_{ sm,f } (I).
Theorem 3. Suppose that f is nonconstant and φ, ψ ∈ C_{ sm,f } (I) with ∇ _{ φ } < ∇ _{ ψ }. Then

(i)
If \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) (resp. {C}_{sm}^{}\left(I\right) ) and ψ is convex (resp. concave) on I, then
\left(1t\right){M}_{\phi}\left(f\right)+t{M}_{\psi}\left(f\right)<{M}_{\left(1t\right)\phi +t\psi}\left(f\right)
holds for all t ∈ (0, 1).

(ii)
If \psi ,\phi \psi \in {C}_{sm}^{}\left(I\right) (resp. {C}_{sm}^{+}\left(I\right) ) and ψ is convex (resp. concave) on I, then
\left(1t\right){M}_{\phi}\left(f\right)+t{M}_{\psi}\left(f\right)>{M}_{\left(1t\right)\phi +t\psi}\left(f\right)
holds for all t ∈ (0, 1).
Proof. (i) Suppose that \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right)[resp. {C}_{sm}^{}\left(I\right)] and ψ is convex [resp. concave] on I. Since ψ = φ + (ψ φ), it follows from hypothesis that \psi \in {C}_{sm}^{+}\left(I\right) [resp. {C}_{sm}^{}\left(I\right)]. Put x = M_{ φ } (f) and y = M_{ ψ } (f), and so x < y by Theorem 1(ii). Also, we have from definition that
Let 0 < t < 1 and put u = M_{(1t)φ+tψ}(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 + tψ is strictly increasing (resp. decreasing), it follows that z < u, that is,
This means that (1  t)M_{ φ } (f) + tM_{ ψ } (f) < M_{(1t)φ+tψ}(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 φmeans.
Theorem 4. Suppose that f is nonconstant and φ, ψ ∈ C_{ sm,f } (I) with ∇ _{ φ } < ∇_{ ψ }.

(i)
If \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) [or {C}_{sm}^{}\left(I\right) ], then a function: s → M _{(1s)φ+sψ}(f) is strictly monotone increasing on [0, ∞) and
\underset{s\to \infty}{lim}{M}_{\left(1s\right)\phi +s\psi}\left(f\right)={M}_{\psi \phi}\left(f\right). 
(ii)
If \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right) [resp. {C}_{sm}^{}\left(I\right) ] and ψ(x)  φ(x) ≥ 0 (resp. ≤ 0) for all x ∈ I, then a function: s → M _{(1s)φ+sψ}(f) is strictly monotone increasing and continuous on [0, ∞).
Proof. (i) Suppose that \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right). For each s ≥ 1, put ξ_{ s } = (1  s)φ + sψ. Since ξ_{ s } = φ + s(ψ  φ), it follows from hypothesis that each ξ_{ s } is in {C}_{sm}^{+}\left(I\right), and then ξ_{ s } ∈ C_{ sm,f } (I). Since ψ = φ + (ψ φ), it follows from hypothesis that ψ is also in {C}_{sm}^{+}\left(I\right). Then by Lemmas 2 and 3, we have that λ_{ c,φ,ψ } is strictly monotone increasing on I\{c} for any c ∈ I°. Let 1 ≤ s_{1} < s_{2} < ∞ 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), {\lambda}_{c,{\xi}_{{s}_{1}},{\xi}_{{s}_{2}}}is also strictly monotone increasing on I\{c}. Then by Lemmas 2 and 3, we conclude that {\nabla}_{{\xi}_{{s}_{1}}}<{\nabla}_{{\xi}_{{s}_{2}}}. Therefore, we have from Theorem 1(ii) that {M}_{{\xi}_{{s}_{1}}}\left(f\right)<{M}_{{\xi}_{{s}_{2}}}\left(f\right) and then a function: s → M_{(1s)φ+sψ}(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 {\lambda}_{c,{\xi}_{s},\psi \phi} is strictly monotone increasing on I\{c} for each s ≥ 1 and c ∈ I°. Then by Lemmas 2 and 3, we conclude that {\nabla}_{{\xi}_{s}}<{\nabla}_{\psi \phi} for each s ≥ 1. Therefore, we have from Theorem 1(ii) that {M}_{{\xi}_{s}}\left(f\right)<{M}_{\psi \phi}\left(f\right) for each s ≥ 1.
Now take s ≥ 1 arbitrarily and put {\alpha}_{s}={M}_{{\xi}_{s}}\left(f\right) and α = M_{ψφ}(f), so α_{ s } < α. Since \phi ,\psi \phi \in {C}_{sm}^{+}\left(I\right), it follows that φ(α_{ s } ) < φ(α) and (ψ φ)(α_{ s } ) < (ψ φ)(α). By definition, we have
Also since {\xi}_{s}=s\left(\frac{1}{s}\phi +\psi \phi \right), it follows from an invariant property of φmean that {\alpha}_{s}={M}_{\frac{1}{s}\phi +\psi \phi}\left(f\right) 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. φmeans by C ^{2}functions
In this section, we treat a special φmean in which φ is a C^{2}functions with no stationary points. For each realvalued measurable function f on Ω, let {C}_{sm*,f}^{2}\left(I\right) be the set of all C^{2}functions φ in C_{ sm,f } (I) with no stationary points, that is, φ'(t) ≠ 0 for all t ∈ I.
Lemma 8. Let\phi ,\psi \in {C}_{sm*,f}^{2}\left(I\right). Then

(i)
The following two statements are equivalent:

(a)
ψ ∘ φ ^{1} is convex (resp. concave) on φ(I).

(b)
\left(\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)}\frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}\right){\psi}^{\prime}\left(x\right)\ge 0 (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)
\left(\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)}\frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}\right){\psi}^{\prime}\left(x\right)>0 (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 C^{1}functions on I. Then,

(i)
If φ'(x) < ψ'(x) for all x ∈ I° and ψ' is monotone increasing on I, then
\left(\left(1t\right)\phi +t\psi \right)\left(\left(1t\right)x+ty\right)<\left(1t\right)\phi \left(x\right)+t\psi \left(y\right)
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
\left(\left(1t\right)\phi +t\psi \right)\left(\left(1t\right)x+ty\right)>\left(1t\right)\phi \left(x\right)+t\psi \left(y\right)
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 + tψ)(z) < (1  t)φ(x) + tψ(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) + tψ(y)  ((1  t)φ + tψ)(z) > 0 from the preceding inequalities. Therefore, we obtain the desired inequality.

(ii)
Similarly. □
Corollary 1. Suppose that f is nonconstant and\phi ,\psi \in {C}_{sm*,f}^{2}\left(I\right). Then

(i)
If \frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}\le \frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)} for all x ∈ I°, then M_{ φ } (f) ≤ M_{ ψ } (f).

(ii)
If \frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}<\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)} for all x ∈ I° then M_{ φ } (f) < M_{ ψ } (f).
Proof. (i) Suppose that \frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}\le \frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)} 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 M_{ φ } (f) ≤ M_{ ψ } (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 M_{ φ } (f) ≤ M_{ ψ } (f) by Theorem 1(i).

(ii)
Similarly. □
Remark. Let (Ω, μ) be a probability space, 0 < p < q < ∞ and let f be a nonconstant realvalued function in L^{q} (Ω, μ). Then the wellknown inequality: f _{ p } < f _{ q } follows immediately from Corollary 1 (ii), by considering a family {φ_{ r } : r > 0} of functions on ℝ_{+}, where φ_{ r } (x) = x^{r} (x > 0).
Let \phi ,\psi \in {C}_{sm*,f}^{2}\left(I\right) 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 \left(1t\right)\phi +t\psi \in {C}_{sm*,f}^{2}\left(I\right). In this case, we have the following
Corollary 2. Suppose that f is nonconstant and\phi ,\psi \in {C}_{sm*,f}^{2}\left(I\right). If\frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}<\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)}and φ'(x)ψ'(x) > 0 for all x ∈ I°, then a function: s → M_{(1s)φ+sψ}(f) is strictly increasing on [0, 1].
Proof. Suppose that \frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}<\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)} and φ'(x)ψ'(x) > 0 for all x ∈ I°. We define ξ(s, x) = (1  s)φ(x) + sψ(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 M_{ φ } (f) < M_{(1s)φ+sψ}(f) for all s ∈ (0, 1). Similarly, we can see that M_{(1s)φ+sψ}(f) < M_{ ψ } (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_{(1s)φ+sψ}(f) is strictly increasing on (0, 1) and hence [0, 1]. □
Corollary 3. Suppose that f is nonconstant and that\phi ,\psi \in {C}_{sm*,f}^{2}\left(I\right)is such that\frac{{\phi}^{\u2033}\left(x\right)}{{\phi}^{\prime}\left(x\right)}<\frac{{\psi}^{\u2033}\left(x\right)}{{\psi}^{\prime}\left(x\right)}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
\left(1t\right){M}_{\phi}\left(f\right)+t{M}_{\psi}\left(f\right)<{M}_{\left(1t\right)\phi +t\psi}\left(f\right)
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
\left(1t\right){M}_{\phi}\left(f\right)+t{M}_{\psi}\left(f\right)>{M}_{\left(1t\right)\phi +t\psi}\left(f\right)
holds for all t ∈ (0, 1).
Proof. (i) Suppose that 0 < φ' < ψ' and ψ" ≥ 0 on I°. Put x = M_{ φ } (f) and y = M_{ ψ } (f), and so x < y by Corollary 1(ii). Take t ∈ ℝ with 0 < t < 1 arbitrarily. By hypothesis, (1  t)φ + tψ is strictly monotone increasing on I. Put u = M_{(1t)φ+tψ}(f). As observe in the proof of Theorem 3(i), we have
Put z = (1  t)x + ty. Then, by (10) and Lemma 9(i), we have
and then z < u, that is, (1  t)M_{ φ } (f) + tM_{ ψ } (f) < M_{(1t)φ+tψ}(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 \phi \left(x\right)=\frac{1}{x} and ψ(x) = x for each x ∈ I. Of course, these functions belong to C_{ sm } (I). The harmonicarithmetic mean inequality asserts that ∇ _{ φ } < ∇_{ ψ }. Take a nonconstant positive measurable function f on a probability space (Ω, μ) such that φ ∘ f and ψ ∘ f are in L ^{1}(Ω, μ). Then, we have from Theorem 1(ii) that M_{ φ } (f) < M_{ ψ } (f). Observe that this inequality means
1<\left(\int \frac{1}{f}\mathsf{\text{d}}\mu \right)\left(\int f\mathsf{\text{d}}\mu \right).
This is a special case of Jensen's inequality (or Schwarz's inequality). We note that if 0 < m ≤ f ≤ M, then \left(\int \frac{1}{f}\mathsf{\text{d}}\mu \right)\left(\int f\mathsf{\text{d}}\mu \right)\le \frac{{\left(m+M\right)}^{2}}{4mM}. The right side of this inequality is called a Kantorovich constant (cf. [5–7]).

(ii)
A similar consideration for the geometricarithmetic mean inequality yields that
\int logf\mathsf{\text{d}}\mu <log\int f\mathsf{\text{d}}\mu .
This is also a special case of Jensen's inequality. We note that if 0 < m ≤ f ≤ M, then log\int f\mathsf{\text{d}}\mu \int logf\mathsf{\text{d}}\mu \le {h}^{\frac{1}{h1}}{\left(elog{h}^{\frac{1}{h1}}\right)}^{1}, where h=\frac{M}{m}. 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 realvalued continuous function on ℝ_{+}. Denote by exp_{[t]}the inverse function of log_{[t]}. Let x _{1}, ..., x_{ n } > 0 and p _{1}, ..., p_{ n } > 0 with {\sum}_{k=1}^{n}{p}_{k}=1. Then Theorem 2(i) (or Corollary 2) implies that t\to {exp}_{\left[t\right]}\left({\sum}_{k=1}^{n}{p}_{k}{log}_{\left[t\right]}{x}_{k}\right) is strictly monotone increasing on [0, 1]. Note that {exp}_{\left[0\right]}\left({\sum}_{k=1}^{n}{p}_{k}{log}_{\left[0\right]}{x}_{k}\right)={\prod}_{k=1}^{n}{x}_{k}^{{p}_{k}} and {exp}_{\left[1\right]}\left({\sum}_{k=1}^{n}{p}_{k}{log}_{\left[1\right]}{x}_{k}\right)={\sum}_{k=1}^{n}{p}_{k}{x}_{k}. Therefore, we obtain that
\prod _{k=1}^{n}{x}_{k}^{{p}_{k}}\le {exp}_{\left[t\right]}\left(\sum _{k=1}^{n}{p}_{k}{log}_{\left[t\right]}{x}_{k}\right)\le \sum _{k=1}^{n}{p}_{k}{x}_{k}\left(0\le t\le 1\right).
This is a new refinement of geometricarithmetic mean inequality (cf. [9]).
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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 GrantinAid for Scientific Research, Japan Society for the Promotion of Science.
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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.
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Nakasuji, Y., Kumahara, K. & Takahasi, SE. A new interpretation of Jensen's inequality and geometric properties of ϕmeans. J Inequal Appl 2011, 48 (2011). https://doi.org/10.1186/1029242X201148
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DOI: https://doi.org/10.1186/1029242X201148