A reconsideration of Jensen’s inequality and its applications

Abstract

A finite form of Jensen’s inequality for a continuous convex function from a topological abelian semigroup to another topological ordered abelian semigroup is obtained under some assumption. As an application, a refinement of a mean inequality is also obtained.

MSC:39B62, 26B25, 26A51.

1 Introduction

This is the inheritance of the idea of [[1], Theorem 1] which gives a new interpretation of Jensen’s inequality by φ-mean. The finite form of Jensen’s inequality proved by Jensen in 1906 asserts that if $t_1,\dots ,t_n$ are positive numbers with ${\sum }_{i=1}^n{t}_i=1$ and f is a continuous convex (resp. concave) function on a real interval I, then

$$f\left(\sum _{i=1}^n{t}_i{x}_i\right)\le \sum _{i=1}^n{t}_{i}f(x_i)(\text{resp. }f\left(\sum _{i=1}^n{t}_i{x}_i\right)\ge \sum _{i=1}^n{t}_{i}f(x_i))$$

holds for all $x_1,\dots ,x_n\in I$.

We first introduce a concept called ‘$(\ast ,\circ )$-convex (or concave)’ for a continuous function from a topological abelian semigroup $(I,\ast )$ to another topological ordered abelian semigroup $(J,\circ )$ and give an interesting example of such a function (see Remark 1). Our purpose of this paper is to give a finite form of Jensen’s inequality for such a function under some assumption (see Theorem 1). Also, as an application, we give a refinement of a mean inequality (see Theorem 2).

2 Terminology and main theorem

Let I be a topological space, and let ∗ be a topological abelian semigroup operation on I. For any $x\in I$ and $n\in \mathbf{N}$ with $n\ge 1$, define the n th power $x^{{(n)}_{\ast }}$ of x recursively by $x^{{(1)}_{\ast }}=x$ and $x^{{(n+1)}_{\ast }}=x^{{(n)}_{\ast }}\ast x$ for $n\ge 1$. We assume that

(${\mathrm{♯}}_1$) any n th-power function $x\mapsto x^{{(n)}_{\ast }}$ is a bijection of I onto itself.

By the assumption (${\mathrm{♯}}_1$), for each $x\in I$ and $n\in \mathbf{N}$, there exists a unique element a of I such that $a^{{(n)}_{\ast }}=x$. Denote by $x^{{(1/n)}_{\ast }}$ such an element a. Moreover, we define

$$x^{{(m/n)}_{\ast }}={\left(x^{{(1/n)}_{\ast }}\right)}^{{(m)}_{\ast }}$$

for each $m,n\in \mathbf{N}$. Then we can easily see that this definition is well defined, that is,

$$\frac{m}{n}=\frac{m^{\prime }}{n^{\prime }}\Rightarrow {\left(x^{{(1/n)}_{\ast }}\right)}^{{(m)}_{\ast }}={\left(x^{{(1/n^{\prime })}_{\ast }}\right)}^{{(m^{\prime })}_{\ast }}(\mathrm{\forall }x\in I).$$

In this case, we can easily show that the following power laws:

$$x^{{(p+q)}_{\ast }}=x^{{(p)}_{\ast }}\ast x^{{(q)}_{\ast }},x^{{(pq)}_{\ast }}={\left(x^{{(p)}_{\ast }}\right)}^{{(q)}_{\ast }}\text{and}{(x\ast y)}^{{(p)}_{\ast }}=x^{{(p)}_{\ast }}\ast y^{{(p)}_{\ast }}$$

(1)

for all $p,q\in {\mathbf{Q}}_{+}$ and $x,y\in I$. Here ${\mathbf{Q}}_{+}$ denotes the set of all positive rational numbers. Moreover, we assume that

(${\mathrm{♯}}_2$) for each $x\in I$, the function $p\mapsto x^{{(p)}_{\ast }}$ is continuous on ${\mathbf{Q}}_{+}$ and it has a continuous extension to ${\mathbf{R}}_{+}$, say $t\mapsto x^{{(t)}_{\ast }}$.

Here ${\mathbf{R}}_{+}$ denotes the set of all positive real numbers. Therefore power laws (1) hold for all $p,q\in {\mathbf{R}}_{+}$. Denote by ${\mathcal{A}}_{+}(I)$ the set of all topological abelian semigroup operations on I satisfying the assumptions (${\mathrm{♯}}_1$) and (${\mathrm{♯}}_2$). Our assumption (${\mathrm{♯}}_1$) leads to the following important concept called ‘mean’. For each $x,y\in I$, put

$$M_{\ast }(x,y)={(x\ast y)}^{{(1/2)}_{\ast }}.$$

We call $M_{\ast }(x,y)$ the mean of x and y with respect to the operation ∗.

Moreover, let J be a topological ordered space with relation ≤, and denote by ${\mathcal{A}}_{+}^0(J,\le )$ the set of all operations $\circ \in {\mathcal{A}}_{+}(J)$ such that

(${\mathrm{♭}}_1$) $a\le b\iff a\circ c\le b\circ c$ ($a,b,c\in J$)

and

(${\mathrm{♭}}_2$) $a\le b\Rightarrow a^{{(t)}_{\circ }}\le b^{{(t)}_{\circ }}$ ($a,b\in J$, $t\in {\mathbf{R}}_{+}$).

Let $C(I,J)$ be the set of all continuous functions from I to J. Take $\ast \in {\mathcal{A}}_{+}(I)$, $\circ \in {\mathcal{A}}_{+}^0(J,\le )$ and $f\in C(I,J)$ arbitrarily. If f satisfies

$$f(M_{\ast }(x,y))\le M_{\circ }(f(x),f(y))(\text{resp. }f(M_{\ast }(x,y))\ge M_{\circ }(f(x),f(y)))$$

for all $x,y\in I$, then we say that f is said to be $(\ast ,\circ )$-convex (resp. concave).

The following theorem states a finite form of Jensen’s inequality for a $(\ast ,\circ )$-convex (or concave) function.

Theorem 1 Let $\ast \in {\mathcal{A}}_{+}(I)$ and $\circ \in {\mathcal{A}}_{+}^0(J,\le )$. If $f\in C(I,J)$ is $(\ast ,\circ )$-convex, then

$$f(x_1^{{(t_1)}_{\ast }}\ast \cdots \ast x_n^{{(t_n)}_{\ast }})\le f{(x_1)}^{{(t_1)}_{\circ }}\circ \cdots \circ f{(x_n)}^{{(t_n)}_{\circ }}$$

holds for all $n\in \mathbf{N}$, $x_1,\dots ,x_n\in I$ and $t_1,\dots ,t_n\in {\mathbf{R}}_{+}$ with $t_1+\cdots +t_n=1$.

If f is $(\ast ,\circ )$-concave, then the inequality above is reversed.

Remark 1 Let ${\mathbf{R}}_{+}^2$ be the product space of ${\mathbf{R}}_{+}$ with ordinary topology. Let ∗ be the operation on ${\mathbf{R}}_{+}^2$ defined by

$$(a,b)\ast (c,d)=(ac,ad+bc)$$

for each $(a,b),(c,d)\in {\mathbf{R}}_{+}^2$. Then ∗ is a topological abelian semigroup operation on ${\mathbf{R}}_{+}^2$ (cf. [[2], p.157-160]). In fact, $({\mathbf{R}}_{+}^2,\ast )$ is topologically isomorphic to an abelian subsemigroup of the semigroup of all $2\times 2$ matrices with usual product under the following mapping:

$$(a,b)\mapsto \left(\begin{array}{cc}a& b\\ 0& a\end{array}\right).$$

Note that

$${(a,b)}^{{(n)}_{\ast }}=(a^n,n{a}^{n-1}b)\text{and}{(a,b)}^{{(1/n)}_{\ast }}=(a^{1/n},\frac{b}{n{a}^{1-1/n}})$$

for all $n\in \mathbf{N}$. Then a simple calculation implies that

$${(a,b)}^{{(m/n)}_{\ast }}=(a^{m/n},\frac{m}{n}{a}^{\frac{m}{n}-1}b)$$

for each $m,n\in \mathbf{N}$. Therefore we see that $\ast \in {\mathcal{A}}_{+}({\mathbf{R}}_{+}^2)$. In this case, we obtain from a simple calculation that

$$M_{\ast }((a,b),(c,d))=(\sqrt{ac},\frac{ad+bc}{2\sqrt{ac}})$$

for each $(a,b),(c,d)\in {\mathbf{R}}_{+}^2$.

Now let ⋅ be the ordinary multiplication on ${\mathbf{R}}_{+}$. Since ${\mathbf{R}}_{+}$ becomes a topological ordered space with the ordinary topology and the ordinary order ≤, we have that $\cdot \in {\mathcal{A}}_{+}^0({\mathbf{R}}_{+},\le )$ and $M_{\cdot }(x,y)=\sqrt{xy}$ for each $x,y\in {\mathbf{R}}_{+}$. Let α and β be real numbers and put

$$f_{\alpha ,\beta }(a,b)=a^{\alpha }{b}^{\beta }$$

for each $(a,b)\in {\mathbf{R}}_{+}^2$. Then $f_{\alpha ,\beta }$ is a continuous function from ${\mathbf{R}}_{+}^2$ to ${\mathbf{R}}_{+}$ such that

$$f_{\alpha ,\beta }\left(M_{\ast }((a,b),(c,d))\right)={(\sqrt{ac})}^{\alpha -\beta }{\left(\frac{ad+bc}{2}\right)}^{\beta }$$

and

$$M_{\cdot }(f_{\alpha ,\beta }(a,b),f_{\alpha ,\beta }(c,d))={(\sqrt{ac})}^{\alpha }{(\sqrt{bd})}^{\beta }$$

for each $(a,b),(c,d)\in {\mathbf{R}}_{+}^2$. Therefore we can easily see that if $\beta \le 0$ (resp. $\beta \ge 0$), then $f_{\alpha ,\beta }$ is $(\ast ,\cdot )$-convex (resp. $(\ast ,\cdot )$-concave).

Remark 2 Let E be a nontrivial real interval with the ordinary topology and the ordinary order ≤. In this case, Craigen and Pales [3] showed that if ∘ is a continuous cancellative semigroup operation on E, then there exists a continuous order-preserving bijection φ of E onto another (necessarily unbounded) real interval such that

$$a\circ b={\phi }^{-1}(\phi (a)+\phi (b))$$

for all $a,b\in E$ (cf. Aczel [4, 5]). Therefore we can easily see that if $\phi (E)={\mathbf{R}}_{+}$, then all continuous cancellative semigroup operations on E are in ${\mathcal{A}}_{+}^0(E,\le )$.

Remark 3 It is clear that a direct product of ${\mathbf{R}}_{+}$ admits a semigroup operation in ${\mathcal{A}}_{+}^0$ that is given as the product of semigroup operations on each ${\mathbf{R}}_{+}$. However, the semigroup operation on ${\mathbf{R}}_{+}^2$ described in Remark 1 does not satisfy properties (${\mathrm{♭}}_1$) and (${\mathrm{♭}}_2$). So, it would be of interest to give an example of an ordered abelian semigroup with a semigroup operation in ${\mathcal{A}}_{+}^0$, which is not isomorphic, as a topological semigroup, to the direct product of the topological semigroup ${\mathbf{R}}_{+}$.

3 Lemmas and proof of Theorem 1

Throughout this section, let I and J be as in Section 2 and suppose that $\ast \in {\mathcal{A}}_{+}(I)$, $\circ \in {\mathcal{A}}_{+}^0(J,\le )$ and that $f\in C(I,J)$ is $(\ast ,\circ )$-convex.

Remark 4 If $f\in C(I,J)$ is $(\ast ,\circ )$-concave, then all inequalities in this section are reversed.

Lemma 1 The inequality

$$f(x_1^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^n)}_{\ast }})\le f{(x_1)}^{{(1/2)}_{\circ }}\circ \cdots \circ f{(x_{n-1})}^{{(1/2^{n-1})}_{\circ }}\circ {\left[f\left(x_n^{{(1/2)}_{\ast }}\right)\right]}^{{(1/2^{n-1})}_{\circ }}$$

holds for all $n\in \mathbf{N}$ and $x_1,\dots ,x_n\in I$.

Proof Let $n\in \mathbf{N}$ and $x_1,\dots ,x_n\in I$. Since f is $(\ast ,\circ )$-convex, it follows that

$$\begin{array}{c}f(x_1^{{(1/2)}_{\ast }}\ast x_2^{{(1/2^2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^n)}_{\ast }})\hfill \\ =f\left({(x_1\ast x_2^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-1})}_{\ast }})}^{{(1/2)}_{\ast }}\right)\hfill \\ \le {(f(x_1)\circ f(x_2^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-1})}_{\ast }}))}^{{(1/2)}_{\circ }}\hfill \\ =f{(x_1)}^{{(1/2)}_{\circ }}\circ {\left[f(x_2^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-1})}_{\ast }})\right]}^{{(1/2)}_{\circ }}\hfill \\ =f{(x_1)}^{{(1/2)}_{\circ }}\circ {\left[f\left({(x_2\ast x_3^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-2})}_{\ast }})}^{{(1/2)}_{\ast }}\right)\right]}^{{(1/2)}_{\circ }}\hfill \\ \le f{(x_1)}^{{(1/2)}_{\circ }}\circ {\left[{(f(x_2)\circ f(x_3^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-2})}_{\ast }}))}^{{(1/2)}_{\circ }}\right]}^{{(1/2)}_{\circ }}\hfill \\ =f{(x_1)}^{{(1/2)}_{\circ }}\circ f{(x_2)}^{{(1/2^2)}_{\circ }}\circ {\left[f(x_3^{{(1/2)}_{\ast }}\ast \cdots \ast x_n^{{(1/2^{n-2})}_{\ast }})\right]}^{{(1/2^2)}_{\circ }}\hfill \\ \cdots \hfill \\ \le f{(x_1)}^{{(1/2)}_{\circ }}\circ f{(x_2)}^{{(1/2^2)}_{\circ }}\circ \cdots \circ f{(x_{n-1})}^{{(1/2^{n-1})}_{\circ }}\circ {\left[f\left(x_n^{{(1/2)}_{\ast }}\right)\right]}^{{(1/2^{n-1})}_{\circ }}.\hfill \end{array}$$

Therefore we obtain the desired inequality. □

Let p and q be two nonnegative rational numbers. We define $x^{{(p)}_{\ast }}\ast y^{{(q)}_{\ast }}=y^{{(q)}_{\ast }}$ if $p=0$, $q>0$ and $x^{{(p)}_{\ast }}\ast y^{{(q)}_{\ast }}=x^{{(p)}_{\ast }}$ if $q=0$, $p>0$. We also define in the same way as for ∘. Applying these notations, we show the following lemma.

Lemma 2 The inequality $f(x^{{(p)}_{\ast }}\ast y^{{(1-p)}_{\ast }})\le f{(x)}^{{(p)}_{\circ }}\circ f{(y)}^{{(1-p)}_{\circ }}$ holds for all $x,y\in I$ and $p\in {\mathbf{R}}_{+}$ with $0<p<1$.

Proof Let $x,y\in I$ and $p\in {\mathbf{R}}_{+}$ with $0<p<1$. By the binary system, we have the expansion $p={\sum }_{i=1}^{\mathrm{\infty }}{p}_i/2^i$, where $p_i\in \{0,1\}$ ($i=1,2,\dots$). Since $p_i\in \{0,1\}$ ($i=1,2,\dots$), it follows that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}{\left[f(x^{{(p_1)}_{\ast }}\ast y^{{(1-p_1)}_{\ast }})\right]}^{{(1/2)}_{\circ }}\circ \cdots \circ {\left[f(x^{{(p_{n-1})}_{\ast }}\ast y^{{(1-p_{n-1})}_{\ast }})\right]}^{{(1/2^{n-1})}_{\circ }}\\ =\underset{n\to \mathrm{\infty }}{lim}[f{(x)}^{{(p_1/2)}_{\circ }}\circ f{(y)}^{{((1-p_1)/2)}_{\circ }}]\circ \cdots \\ \circ [f{(x)}^{{(p_n/2^{n-1})}_{\circ }}\circ f{(y)}^{{((1-p_{n-1})/2^{n-1})}_{\circ }}]\\ =\underset{n\to \mathrm{\infty }}{lim}f{(x)}^{{({\sum }_{i=1}^{n-1}{p}_i/2^i)}_{\circ }}\circ f{(y)}^{{({\sum }_{i=1}^{n-1}(1-p_i)/2^i)}_{\circ }}\\ =f{(x)}^{{(p)}_{\circ }}\circ f{(y)}^{{(1-p)}_{\circ }}.\end{array}$$

So, putting

$$a_n={\left[f(x^{{(p_1)}_{\ast }}\ast y^{{(1-p_1)}_{\ast }})\right]}^{{(1/2)}_{\circ }}\circ \cdots \circ {\left[f(x^{{(p_{n-1})}_{\ast }}\ast y^{{(1-p_{n-1})}_{\ast }})\right]}^{{(1/2^{n-1})}_{\circ }}$$

for each $n\in \mathbf{N}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=f{(x)}^{{(p)}_{\circ }}\circ f{(y)}^{{(1-p)}_{\circ }}.$$

(2)

Put

$${\mathbf{N}}_1=\{n\in \mathbf{N}:f\left({(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2)}_{\ast }}\right)=f\left(x^{{(1/2)}_{\ast }}\right)\}$$

and

$${\mathbf{N}}_0=\{n\in \mathbf{N}:f\left({(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2)}_{\ast }}\right)=f\left(y^{{(1/2)}_{\ast }}\right)\}.$$

Since

$$f\left({(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2)}_{\ast }}\right)=\{\begin{array}{cc}f(x^{{(1/2)}_{\ast }})\hfill & (p_n=1),\hfill \\ f(y^{{(1/2)}_{\ast }})\hfill & (p_n=0)\hfill \end{array}$$

for each $n\in \mathbf{N}$, it follows that $\mathbf{N}={\mathbf{N}}_1\cup {\mathbf{N}}_0$. Then either ${\mathbf{N}}_1$ or ${\mathbf{N}}_0$ are infinite. Put

$$e_n={\left[f\left({(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2)}_{\ast }}\right)\right]}^{{(1/2^{n-1})}_{\circ }}$$

for each $n\in \mathbf{N}$. If ${\mathbf{N}}_1$ is infinite, then we have

$$\underset{n\in {\mathbf{N}}_1}{lim}{e}_n\circ f\left(x^{{(1/2)}_{\ast }}\right)=\underset{n\in {\mathbf{N}}_1}{lim}{\left(f\left(x^{{(1/2)}_{\ast }}\right)\right)}^{(1+1/2^{n-1})\circ }=f\left(x^{{(1/2)}_{\ast }}\right).$$

(3)

Therefore it follows from (2) and (3) that

$$f{(x)}^{{(p)}_{\circ }}\circ f{(y)}^{{(1-p)}_{\circ }}\circ f\left(x^{{(1/2)}_{\ast }}\right)=\underset{n\in {\mathbf{N}}_1}{lim}{a}_n\circ e_n\circ f\left(x^{{(1/2)}_{\ast }}\right).$$

(4)

Also we have from Lemma 1 that

$$\begin{array}{r}f(x^{{({\sum }_{i=1}^n{p}_i/2^i)}_{\ast }}\ast y^{{({\sum }_{i=1}^n(1-p_i)/2^i)}_{\ast }})\\ =f({(x^{{(p_1)}_{\ast }}\ast y^{{(1-p_1)}_{\ast }})}^{{(1/2)}_{\ast }}\ast \cdots \ast {(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2^n)}_{\ast }})\\ \le {\left[f(x^{{(p_1)}_{\ast }}\ast y^{{(1-p_1)}_{\ast }})\right]}^{{(1/2)}_{\circ }}\circ \cdots \circ {\left[f(x^{{(p_{n-1})}_{\ast }}\ast y^{{(1-p_{n-1})}_{\ast }})\right]}^{{(1/2^{n-1})}_{\circ }}\\ \circ {\left[f\left({(x^{{(p_n)}_{\ast }}\ast y^{{(1-p_n)}_{\ast }})}^{{(1/2)}_{\ast }}\right)\right]}^{{(1/2^{n-1})}_{\circ }}\\ =a_n\circ e_n\end{array}$$

for all $n\in {\mathbf{N}}_1$. Then it follows from (${\mathrm{♭}}_1$) that

$$f(x^{{({\sum }_{i=1}^n{p}_i/2^i)}_{\ast }}\ast y^{{({\sum }_{i=1}^n(1-p_i)/2^i)}_{\ast }})\circ f\left(x^{{(1/2)}_{\ast }}\right)\le a_n\circ e_n\circ f\left(x^{{(1/2)}_{\ast }}\right)$$

(5)

for all $n\in {\mathbf{N}}_1$. Letting $n\in {\mathbf{N}}_1\to \mathrm{\infty }$ in (5), we obtain from (4) that

$$f(x^{{(p)}_{\ast }}\ast y^{{(1-p)}_{\ast }})\circ f\left(x^{{(1/2)}_{\ast }}\right)\le f{(x)}^{{(p)}_{\circ }}\circ f{(y)}^{{(1-p)}_{\circ }}\circ f\left(x^{{(1/2)}_{\ast }}\right).$$

(6)

Canceling $f(x^{{(1/2)}_{\ast }})$ in (6) by (${\mathrm{♭}}_1$), we obtain the desired inequality. Similarly, the desired inequality is obtained in case that ${\mathbf{N}}_0$ is infinite. □

Lemma 3 The inequality

$$f\left({(x_1\ast \cdots \ast x_n)}^{{(1/n)}_{\ast }}\right)\le {(f(x_1)\circ \cdots \circ f(x_n))}^{{(1/n)}_{\circ }}$$

holds for all $n\in \mathbf{N}$ and $x_1,\dots ,x_n\in I$.

Proof It is clear that the lemma holds for $n=1$. Suppose the lemma holds for $n=k$. Let $x_1,\dots ,x_k,x_{k+1}\in I$ and put

$$a={(f(x_1)\circ \cdots \circ f(x_k))}^{{(1/k)}_{\circ }}\text{and}x={(x_1\ast \cdots \ast x_k)}^{{(1/k)}_{\ast }}.$$

Then $a\ge f(x)$ by hypothesis, and hence $a^{{(k/k+1)}_{\circ }}\ge f{(x)}^{{(k/k+1)}_{\circ }}$ by (${\mathrm{♭}}_2$). It follows from (${\mathrm{♭}}_1$) that

$$a^{{(k/k+1)}_{\circ }}\circ f{(x_{k+1})}^{{(1/k+1)}_{\circ }}\ge f{(x)}^{{(k/k+1)}_{\circ }}\circ f{(x_{k+1})}^{{(1/k+1)}_{\circ }}.$$

Therefore we have from Lemma 2 that

$$\begin{array}{rl}{(f(x_1)\circ \cdots \circ f(x_{k+1}))}^{{(1/k+1)}_{\circ }}& ={(a^{{(k)}_{\circ }}\circ f(x_{k+1}))}^{{(1/k+1)}_{\circ }}\\ =a^{{(k/k+1)}_{\circ }}\circ f{(x_{k+1})}^{{(1/k+1)}_{\circ }}\\ \ge f{(x)}^{{(k/k+1)}_{\circ }}\circ f{(x_{k+1})}^{{(1/k+1)}_{\circ }}\\ \ge f(x^{{(k/k+1)}_{\ast }}\ast x_{k+1}^{{(1/k+1)}_{\ast }})\\ =f\left({(x_1\ast \cdots \ast x_k\ast x_{k+1})}^{{(1/k+1)}_{\ast }}\right).\end{array}$$

In other words, the lemma holds for $n=k+1$. Then, by mathematical induction, the lemma holds for all $n\in \mathbf{N}$. □

Lemma 4 The inequality

$$f(x_1^{{(p_1)}_{\ast }}\ast \cdots \ast x_n^{{(p_n)}_{\ast }})\le f{(x_1)}^{{(p_1)}_{\circ }}\circ \cdots \circ f{(x_n)}^{{(p_n)}_{\circ }}$$

holds for all $x_1,\dots ,x_n\in I$, $n\in \mathbf{N}$ and $p_1,\dots ,p_n\in {\mathbf{Q}}_{+}$ with $p_1+\cdots +p_n=1$.

Proof This result follows directly from Lemma 3. □

We are now in a position to prove Theorem 1.

Proof Let $n\in \mathbf{N}$, $x_1,\dots ,x_n\in I$ and $t_1,\dots ,t_n\in {\mathbf{R}}_{+}$ with $t_1+\cdots +t_n=1$. For each $1\le i\le n$, choose a sequence ${\{p_{ik}\}}_{k=1}^{\mathrm{\infty }}$ in ${\mathbf{Q}}_{+}$ which converges to $t_i$. Put $q_k=p_{1k}+\cdots +p_{nk}$ for each $k\in \mathbf{N}$. Then we have from Lemma 4 that

$$f(x_1^{{(p_{1k}/q_k)}_{\ast }}\ast \cdots \ast x_n^{{(p_{nk}/q_k)}_{\ast }})\le f{(x_1)}^{{(p_{1k}/q_k)}_{\circ }}\circ \cdots \circ f{(x_n)}^{{(p_{nk}/q_k)}_{\circ }}.$$

(7)

Hence, after taking the limit with respect to k in (7), we obtain the desired inequality:

$$f(x_1^{{(t_1)}_{\ast }}\ast \cdots \ast x_n^{{(t_n)}_{\ast }})\le f{(x_1)}^{{(t_1)}_{\circ }}\circ \cdots \circ f{(x_n)}^{{(t_n)}_{\circ }}.$$

Of course, if f is $(\ast ,\circ )$-concave, the above inequality is reversed, as stated in Remark 4. This completes the proof of Theorem 1. □

4 Applications

Let K be a topological ordered space with order ≤, and let ∘, ∗ and ⋄ be three operations in ${\mathcal{A}}_{+}^0(K,\le )$ which have the following properties:

$$(a\circ b)\ast c=(a\ast c)\circ (b\ast c)(\mathrm{\forall }a,b,c\in K),$$

(8)

$$\mathrm{\exists }e\in K:e\ast x=x(\mathrm{\forall }x\in K),$$

(9)

$$a\diamond b=a\circ b\circ (a\ast b)(\mathrm{\forall }a,b\in K).$$

(10)

In this case, it is clear that an element e in (9) is unique.

Lemma 5 The equality $x^{{(t)}_{\diamond }}\circ e={(x\circ e)}^{{(t)}_{\ast }}$ holds for each $x\in K$ and $t\in {\mathbf{R}}_{+}$.

Proof Take $x\in K$ arbitrarily. By (8) and (9), we have $(a\diamond b)\circ e=(a\circ e)\ast (b\circ e)$ for all $a,b\in K$. By mathematical induction, we have

$$(x_1\diamond \cdots \diamond x_k)\circ e=(x_1\circ e)\ast \cdots \ast (x_k\circ e)$$

(11)

for all $k\in \mathbf{N}$ and $x_1,\dots ,x_k\in K$. In particular, we have

$$x^{{(k)}_{\diamond }}\circ e={(x\circ e)}^{{(k)}_{\ast }}$$

(12)

for all $k\in \mathbf{N}$. Then we have

$$\begin{array}{rl}{(x\circ e)}^{{(n/m)}_{\ast }}& ={\left({(x\circ e)}^{{(n)}_{\ast }}\right)}^{{(1/m)}_{\ast }}\\ ={(x^{{(n)}_{\diamond }}\circ e)}^{{(1/m)}_{\ast }}\text{by (12)}\\ ={({\left(x^{{(n/m)}_{\diamond }}\right)}^{{(m)}_{\diamond }}\circ e)}^{{(1/m)}_{\ast }}\\ ={\left({(x^{{(n/m)}_{\diamond }}\circ e)}^{{(m)}_{\ast }}\right)}^{{(1/m)}_{\ast }}\text{by (12)}\\ =x^{{(n/m)}_{\diamond }}\circ e\end{array}$$

for all $n,m\in \mathbf{N}$. Therefore $x^{{(p)}_{\diamond }}\circ e={(x\circ e)}^{{(p)}_{\ast }}$ holds for all $p\in {\mathbf{Q}}_{+}$. Take $t\in {\mathbf{R}}_{+}$ arbitrarily, and choose a sequence $\{p_n\}$ in ${\mathbf{Q}}_{+}$ which converges to t. Then

$$x^{{(t)}_{\diamond }}\circ e=\underset{n\to \mathrm{\infty }}{lim}{x}^{{(p_n)}_{\diamond }}\circ e=\underset{n\to \mathrm{\infty }}{lim}{(x\circ e)}^{{(p_n)}_{\ast }}={(x\circ e)}^{{(t)}_{\ast }}$$

holds and so the proof is complete. □

Lemma 6 Suppose that $M_{\ast }(a,b)\le M_{\circ }(a,b)$ holds for all $a,b\in K$. Then $M_{\ast }(a,b)\le M_{\diamond }(a,b)\le M_{\circ }(a,b)$ holds for each $a,b\in K$.

Proof Let $a,b\in K$. We first show that $M_{\ast }(a,b)\le M_{\diamond }(a,b)$. Since

$${\left({(a\ast b)}^{{(1/2)}_{\ast }}\right)}^{{(2)}_{\circ }}\le a\circ b,$$

it follows from Lemma 5 that

$$\begin{array}{rl}{({(a\ast b)}^{{(1/2)}_{\ast }}\circ e)}^{{(2)}_{\ast }}& =(a\ast b)\circ {\left({(a\ast b)}^{{(1/2)}_{\ast }}\right)}^{{(2)}_{\circ }}\circ e\\ \le (a\ast b)\circ (a\circ b)\circ e\\ =(a\diamond b)\circ e\\ ={\left({(a\diamond b)}^{{(1/2)}_{\diamond }}\right)}^{{(2)}_{\diamond }}\circ e\\ ={({(a\diamond b)}^{{(1/2)}_{\diamond }}\circ e)}^{{(2)}_{\ast }}.\end{array}$$

Therefore we obtain from (${\mathrm{♭}}_2$) for ∗ that

$${(a\ast b)}^{{(1/2)}_{\ast }}\circ e\le {(a\diamond b)}^{{(1/2)}_{\diamond }}\circ e.$$

Canceling e in the above inequality, we obtain the desired inequality.

We next show that $M_{\diamond }(a,b)\le M_{\circ }(a,b)$. Since

$$a\ast b\le {\left({(a\circ b)}^{{(1/2)}_{\circ }}\right)}^{{(2)}_{\ast }},$$

it follows from Lemma 5 that

$$\begin{array}{rl}{({(a\diamond b)}^{{(1/2)}_{\diamond }}\circ e)}^{{(2)}_{\ast }}& ={\left({(a\diamond b)}^{{(1/2)}_{\diamond }}\right)}^{{(2)}_{\diamond }}\circ e\\ =(a\diamond b)\circ e\\ =(a\ast b)\circ a\circ b\circ e\\ \le {\left({(a\circ b)}^{{(1/2)}_{\circ }}\right)}^{{(2)}_{\ast }}\circ a\circ b\circ e\\ ={({(a\circ b)}^{{(1/2)}_{\circ }}\circ e)}^{{(2)}_{\ast }}.\end{array}$$

Therefore we obtain from (${\mathrm{♭}}_2$) for ∗ that

$${(a\diamond b)}^{{(1/2)}_{\diamond }}\circ e\le {(a\circ b)}^{{(1/2)}_{\circ }}\circ e.$$

Canceling e in the above inequality, we obtain the desired inequality and so the proof is complete. □

The following result is a refinement of the mean inequality.

Theorem 2 Let K be a topological ordered space with order ≤ and $\ast ,\circ ,\diamond \in {\mathcal{A}}_{+}^0(K,\le )$ satisfying (8), (9) and (10). If $M_{\ast }(x,y)\le M_{\circ }(x,y)$ holds for all $x,y\in K$, then

$$x_1^{{(t_1)}_{\ast }}\ast \cdots \ast x_n^{{(t_n)}_{\ast }}\le x_1^{{(t_1)}_{\diamond }}\diamond \cdots \diamond x_n^{{(t_n)}_{\diamond }}\le x_1^{{(t_1)}_{\circ }}\circ \cdots \circ x_n^{{(t_n)}_{\circ }}$$

holds for all $n\in \mathbf{N}$, $x_1,\dots ,x_n\in K$ and $t_1,\dots ,t_n\in {\mathbf{R}}_{+}$ with $t_1+\cdots +t_n=1$.

Proof This follows immediately from Theorem 1 and Lemma 6. □

5 Examples

Throughout this section, let ${\mathbf{R}}_{+}$ be an ordinary topological ordered space.

Example 1 Put $x{\circ }_{t}y={(x^t+y^t)}^{1/t}$ for each $x,y\in {\mathbf{R}}_{+}$ and $t\in \mathbf{R}\mathrm{\setminus }\{0\}$. Then each ${\circ }_t$ is a topological abelian semigroup operation on ${\mathbf{R}}_{+}$ such that $x^{{(n)}_{{\circ }_t}}=n^{1/t}x$ and $x^{{(1/n)}_{{\circ }_t}}={(1/n)}^{1/t}x$ for all $n\in \mathbf{N}$ and $x\in {\mathbf{R}}_{+}$. Also, since

$$x^{{(n/m)}_{{\circ }_t}}={\left(x^{{(1/m)}_{{\circ }_t}}\right)}^{{(n)}_{{\circ }_t}}=n^{1/t}{(1/m)}^{1/t}x={(n/m)}^{1/t}x$$

for each $m,n\in \mathbf{N}$, $t\in \mathbf{R}\mathrm{\setminus }\{0\}$ and $x\in {\mathbf{R}}_{+}$, it follows that ${\circ }_t\in {\mathcal{A}}_{+}^0({\mathbf{R}}_{+},\le )$ for each $t\in \mathbf{R}\mathrm{\setminus }\{0\}$. Let $x_1,\dots ,x_n,{\alpha }_1,\dots ,{\alpha }_n\in {\mathbf{R}}_{+}$ with ${\alpha }_1+\cdots +{\alpha }_n=1$. Since $x^{{(\alpha )}_{{\circ }_t}}={\alpha }^{1/t}x$ for each $x,\alpha \in {\mathbf{R}}_{+}$ and $t\in \mathbf{R}\mathrm{\setminus }\{0\}$, we have that

$$x_1^{{({\alpha }_1)}_{{\circ }_t}}{\circ }_t\cdots {\circ }_t{x}_n^{{({\alpha }_n)}_{{\circ }_t}}={({\alpha }_1{x}_1^t+\cdots +{\alpha }_n{x}_n^t)}^{1/t}$$

(13)

for all $t\in \mathbf{R}\mathrm{\setminus }\{0\}$. Note that

$$M_{{\circ }_s}(x,y)={\left(\frac{x^s+y^s}{2}\right)}^{1/s}\le {\left(\frac{x^t+y^t}{2}\right)}^{1/t}=M_{{\circ }_t}(x,y)$$

for all $x,y\in {\mathbf{R}}_{+}$ and $s,t\in \mathbf{R}\mathrm{\setminus }\{0\}$ with $s\le t$. Therefore Theorem 1 implies the following well-known inequality:

$${({\alpha }_1{x}_1^s+\cdots +{\alpha }_n{x}_n^s)}^{1/s}\le {({\alpha }_1{x}_1^t+\cdots +{\alpha }_n{x}_n^t)}^{1/t},$$

where $s,t\in \mathbf{R}\mathrm{\setminus }\{0\}$ with $s\le t$.

Example 2 Let t be a positive real number, and let ${\circ }_t$ be the operation on ${\mathbf{R}}_{+}$ defined in Example 1. Then we have ${\circ }_t\in {\mathcal{A}}_{+}^0({\mathbf{R}}_{+},\le )$. Also, if $x\ast y=xy$ ($x,y\in {\mathbf{R}}_{+}$), then $\ast \in {\mathcal{A}}_{+}^0({\mathbf{R}}_{+},\le )$ and $x^{{(\alpha )}_{\ast }}=x^{\alpha }$ for all $x,\alpha \in {\mathbf{R}}_{+}$. Hence we have

$$M_{\ast }(x,y)=\sqrt{xy}\le {\left(\frac{x^t+y^t}{2}\right)}^{1/t}=M_{{\circ }_t}(x,y)$$

for each $x,y\in {\mathbf{R}}_{+}$.

Let $n\in \mathbf{N}$ and $x_1,\dots ,x_n,{\alpha }_1,\dots ,{\alpha }_n\in {\mathbf{R}}_{+}$ with ${\alpha }_1+\cdots +{\alpha }_n=1$. Then it is clear that

$$x_1^{{({\alpha }_1)}_{\ast }}\ast \cdots \ast x_n^{{({\alpha }_n)}_{\ast }}=x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}.$$

(14)

If $x\diamond y=x{\circ }_{t}y{\circ }_t(x\ast y)$ ($x,y\in {\mathbf{R}}_{+}$), then $\diamond \in {\mathcal{A}}_{+}^0({\mathbf{R}}_{+},\le )$ and three operations ∗, ∘ and ⋄ on ${\mathbf{R}}_{+}$ satisfy (8), (9) and (10). If $x,\alpha \in {\mathbf{R}}_{+}$, we have from Lemma 5 that

$$x^{{(\alpha )}_{\diamond }}{\circ }_{t}1={(x{\circ }_{t}1)}^{{(\alpha )}_{\ast }}={(x{\circ }_{t}1)}^{\alpha }={(x^t+1)}^{\alpha /t}.$$

(15)

Then we have from (11) and (15) that

$$\begin{array}{rl}{x}_1^{{({\alpha }_1)}_{\diamond }}\diamond \cdots \diamond x_n^{{({\alpha }_n)}_{\diamond }}{\circ }_{t}1& =\left(x_1^{{({\alpha }_1)}_{\diamond }}{\circ }_{t}1\right)\ast \cdots \ast \left(x_n^{{({\alpha }_n)}_{\diamond }}{\circ }_{t}1\right)\\ =\left(x_1^{{({\alpha }_1)}_{\diamond }}{\circ }_{t}1\right)\cdots \left(x_n^{{({\alpha }_n)}_{\diamond }}{\circ }_{t}1\right)\\ ={(x_1^t+1)}^{{\alpha }_1/t}\cdots {(x_n^t+1)}^{{\alpha }_n/t},\end{array}$$

and hence

$$x_1^{{({\alpha }_1)}_{\diamond }}\diamond \cdots \diamond x_n^{{({\alpha }_n)}_{\diamond }}={({(x_1^t+1)}^{{\alpha }_1}\cdots {(x_n^t+1)}^{{\alpha }_n}-1)}^{1/t}.$$

(16)

Then we obtain from (13), (14), (16) and Theorem 2 that

$$x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}\le {({(x_1^t+1)}^{{\alpha }_1}\cdots {(x_n^t+1)}^{{\alpha }_n}-1)}^{1/t}\le {({\alpha }_1{x}_1^t+\cdots +{\alpha }_n{x}_n^t)}^{1/t}.$$

This is a refinement of the geometric-arithmetic mean inequality.