New Refinements of the Generalized Versions of Hölder’s Inequality

Abstract

In this paper, we present new refinements of the generalized versions of Hölder’s inequality. Such refinements are rare. Our results are novel and clearly illustrate the essence of some recent specific refinements. As for applications, we present new inequalities for integral power means, give a new refinement of the generalized Opial–Olech inequality, refine the well-known inequality between Rényi’s entropies with different parameters by utilizing the concept of “useful” information, and finally, we obtain a refinement of the Cauchy–Bunyakovsky–Schwarz inequality.

1. Introduction

Hölder’s inequality plays a very important role in different branches of modern mathematics, such as real and complex analysis, numerical analysis, probability theory, qualitative theory of differential equations, optimization, information theory, and machine learning, and their applications. Due to its importance, Hölder’s inequality has been extensively investigated and continues to be studied today. The motivation for this paper stems from two recently published refinements of Hölder’s inequality.

The first result can be found in [1]. This is an interesting and frequently cited result (see, without claiming to be exhaustive, [2,3,4,5]), which refines a special case of Hölder’s inequality (classical Lebesgue integral is used on a compact interval).

Theorem 1.

Let $p>1$ and $1/p+1/q=1$. If f and g are real functions defined on $[a,b]$ such that ${\left|f\right|}^p$ and ${\left|g\right|}^q$ are integrable functions on $[a,b]$, then

$${\displaystyle {\int }_a^b}\left|f\left(t\right)g\left(t\right)\right|dt\le {\displaystyle \frac{1}{b-a}}\left\{{\left({\displaystyle {\int }_a^b}\left(b-t\right){\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}\left(b-t\right){\left|g\left(t\right)\right|}^{q}dt\right)}^{1/q}\right.$$

$$\left.+{\left({\displaystyle {\int }_a^b}\left(t-a\right){\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}\left(t-a\right){\left|g\left(t\right)\right|}^{q}dt\right)}^{1/q}\right\}$$

$$\le {\left({\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}{\left|g\left(t\right)\right|}^{q}dt\right)}^{1/q}.$$

A recent paper [2] contains the next generalization of the previous result.

Theorem 2.

Let $p>1$ and $1/p+1/q=1$, and let $-\infty \le a<b\le \infty$. Let u, $v:\left[a,b\right]\to \left[0,\infty \right]$ be two weight functions and let $w:=u+v$. If f and g are real functions defined on $[a,b]$ such that ${u\left|f\right|}^p$, ${v\left|f\right|}^p$, ${v\left|g\right|}^q$ and ${v\left|g\right|}^q$ are integrable, then

$${\displaystyle {\int }_a^b}w\left(t\right)\left|f\left(t\right)g\left(t\right)\right|dt\le {\left({\displaystyle {\int }_a^b}u\left(t\right){\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}u\left(t\right){\left|g\left(t\right)\right|}^{q}dt\right)}^{1/q}$$

$$+{\left({\displaystyle {\int }_a^b}v\left(t\right){\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}{\left|v\left(t\right)g\left(t\right)\right|}^{q}dt\right)}^{1/q}$$

$$\le {\left({\displaystyle {\int }_a^b}w\left(t\right){\left|f\left(t\right)\right|}^{p}dt\right)}^{1/p}{\left({\displaystyle {\int }_a^b}w\left(t\right){\left|g\left(t\right)\right|}^{q}dt\right)}^{1/q}.$$

If $-\infty <a<b<\infty$, then Theorem 1 can be obtained from Theorem 2 by choosing the functions u, $v:\left[a,b\right]\to \left[0,\infty \right]$ as follows:

$$u\left(t\right):={\displaystyle \frac{b-t}{b-a}},v\left(t\right):={\displaystyle \frac{t-a}{b-a}}.$$

The previous two refinements are very special cases of Theorem 4.8 in [6], which contains a refinement of Hölder’s inequality formulated in measure spaces. This result and its proof clearly illustrate the essence of the construction of refinements given in the above theorems.

In this paper, we present refinements of generalized versions of Hölder’s inequality whose structure is similar to the refinements obtained in the first two theorems.

Before presenting the generalized forms of Hölder’s inequality, we introduce a few notations.

Let $\left(X,\mathcal{A},\mu \right)$ be an arbitrary measure space ($\mathcal{A}$ always means a $\sigma$-algebra; $\mu :\mathcal{A}\to \left[0,\infty \right]$ is $\sigma$-additive).

Let $p>0$. As usual, we say that $f\in L_p\left(X,\mathcal{A},\mu \right)$ if f is a complex-valued $\mathcal{A}$-measurable function defined $\mu$-a.e. on X such that ${\left|f\right|}^p$ is $\mu$-integrable.

If $p<0$, we will also use the notation $L_p\left(X,\mathcal{A},\mu \right)$: in this case $f\in L_p\left(X,\mathcal{A},\mu \right)$ means that f is a complex-valued $\mathcal{A}$-measurable function, it is defined and nonzero $\mu$-a.e. on X, and that ${\left|f\right|}^p$ is $\mu$-integrable.

A generalized form of Hölder’s inequality is as follows (see [7]).

Theorem 3.

Let $p_j>1$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$.

(a) (Integral Hölder inequality for positive exponent) Let $\left(X,\mathcal{A},\mu \right)$ be a measure space. If $f_j\in L_{p_j}\left(X,\mathcal{A},\mu \right)$ $\left(j=1,\dots ,k\right)$, then ${\displaystyle {\prod }_{j=1}^k}{f}_j\in L_1\left(X,\mathcal{A},\mu \right)$, and

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu \le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d\mu \right)}^{1/p_j}.$$

(1)

(b) (Discrete Hölder inequality for positive exponent) Let $\left({\alpha }_1,\dots ,{\alpha }_m\right)$ be a nonnegative m-tuples. If $\left(x_{j1},\dots ,x_{jm}\right)$ is a complex m-tuples $\left(j=1,\dots ,k\right)$, then

$${\displaystyle {\sum }_{i=1}^m}{\alpha }_i{\displaystyle {\prod }_{j=1}^k}\left|x_{ji}\right|\le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle {\sum }_{i=1}^m}{\alpha }_i{\left|x_{ji}\right|}^{p_j}\right)}^{1/p_j}.$$

(2)

An even more general version can be found in [8].

Another generalized version of the classical Hölder inequality is as follows (see [7,9]).

Theorem 4.

Let $0<p_1<1$, and let $p_j<0$ $\left(j=2,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$.

(a) (Integral Hölder inequality for negative exponent) Let $\left(X,\mathcal{A},\mu \right)$ be a measure space. Let $f_j\in L_p\left(X,\mathcal{A},\mu \right)$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d\mu >0$ $\left(j=2,\dots ,k\right)$. If ${\displaystyle {\prod }_{j=1}^k}{f}_j$ is μ-integrable, then the reverse inequality holds in (1).

(b) (Discrete Hölder inequality for negative exponent) Let $\left({\alpha }_1,\dots ,{\alpha }_m\right)$ be a nonnegative m-tuples. If $\left(x_{j1},\dots ,x_{jm}\right)$ is a complex m-tuples $\left(j=1,\dots ,k\right)$ such that ${\displaystyle {\sum }_{i=1}^m}{\alpha }_i{\left|x_{ji}\right|}^{p_j}>0$ $\left(j=2,\dots ,k\right)$, then the reverse inequality holds in (2).

Few papers deal with refinements of generalized versions of Hölder’s inequality (see [10,11,12,13]). Ref. [10] examines only the integral Hölder inequality for positive exponent; refs. [11,13] contain discrete refinements, while ref. [12] uses classical Lebesgue measure on a compact interval.

In this paper, building on the two known refinements mentioned above, we present a new method by which generalized integral Hölder inequalities can also be refined. The results are novel. The method used here differs from those used in the previously mentioned papers. The proofs rely on repeated applications of generalized Hölder inequalities together with suitable measure decompositions. These refinements contain Theorems 1 and 2. In some cases, we specify the explicit form of the discrete refinements that can be derived from the refinements obtained for generalized integral Hölder inequalities. We also provide the exact form of the resulting refinement in certain cases where the measure employed is absolutely continuous with respect to another measure.

As an application, we investigate inequalities involving power means, some of which are entirely new, while others generalize known inequalities. It is worth noting that the proofs are based almost exclusively on Hölder’s inequality; Jensen’s inequality is not used. As a further application of our results, we obtain a new refinement of a generalized version of the Opial–Olech inequality. In our next application, we refine the well-known inequality between Rényi’s entropies with different parameters by utilizing the concept of “useful” information. The result demonstrates the applicability of the refinements presented in the paper, as it naturally incorporates the quantity of “useful” information. Finally, we present a new refinement of the Cauchy–Bunyakovsky–Schwarz inequality. This not only illustrates the applicability of our refinements, but the idea itself and the method of its proof also stem from our results.

2. Preliminary Results

The first part of the following statement is well known (see, e.g., [7]), but the second part is rarely encountered, so for the sake of completeness, we prove it.

Lemma 1.

Let $\left(X,\mathcal{A},\mu \right)$ be a measure space with $0<\mu \left(X\right)<\infty$, and let f be a positive function from $L_1\left(X,\mathcal{A},\mu \right)$. Then,

(a)

$$\underset{p\to 0+}{\lim }{\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu \right)}^{1/p}=\exp \left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu \right).$$

(b) If f belongs to $L_{t_0}\left(X,\mathcal{A},\mu \right)$ for some $t_0<0$ too, then

$$\underset{p\to 0-}{\lim }{\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu \right)}^{1/p}=\exp \left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu \right).$$

Proof. 

(a) See [7].

(b) Since f is positive and $\mu \left(X\right)>0$, ${\displaystyle \underset{X}{\int }}fd\mu >0$.

(i) Let $a>0$. An elementary argument shows that the function

$$t\to {\displaystyle \frac{a^t-1}{t}},t\in \mathbb{R}\setminus \left\{0\right\}$$

is increasing on both $\left[-\infty ,0\right]$ and $\left[0,\infty \right]$, and

$$\underset{t\to 0}{\lim }{\displaystyle \frac{a^t-1}{t}}=\log \left(a\right).$$

(ii) If we replace t by

$$f\left(x\right)/{\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}fd\mu$$

in the inequality $\log \left(t\right)\le t-1$ $\left(t>0\right)$, we obtain

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu \le \log \left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}fd\mu \right).$$

(3)

According to $f\in L_{t_0}\left(X,\mathcal{A},\mu \right)$, $f\in L_p\left(X,\mathcal{A},\mu \right)$ for all $t_0<p<0$.

Applying the inequality $\log \left(t\right)\le t-1$ $\left(t>0\right)$ again, and then (3), we obtain

$${\displaystyle \frac{1}{p}}\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu -1\right)\le {\displaystyle \frac{1}{p}}\log \left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu \right)$$

$$\le {\displaystyle \frac{1}{p}}\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ f^{p}d\mu \right)={\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu ,t_0<p<0.$$

(4)

By (i) and the monotone convergence theorem,

$${\displaystyle \frac{1}{p}}\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu -1\right)={\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{\displaystyle \frac{f^p-1}{p}}d\mu \stackrel{p\to 0-}{\to }{\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu ,$$

and the result follows from this and (4).

The proof is complete. □

Remark 1.

Of course, we could have used Jensen’s inequality to prove inequality (3), but the approach described above is more elementary.

Petrov (see [14]) provided a stronger version of the integral Hölder inequality when $k=2$. His result can be generalized to any $k\ge 2$, as we show in the following assertion.

Theorem 5.

Let $p_j>1$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$. Let $\left(X,\mathcal{A},\mu \right)$ be a measure space. If $f_j\in L_{p_j}\left(X,\mathcal{A},\mu \right)$ $\left(j=1,\dots ,k\right)$, then ${\displaystyle {\prod }_{j=1}^k}{f}_j\in L_1\left(X,\mathcal{A},\mu \right)$, and

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu \le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{H}{\int }}{\left|f_j\right|}^{p_j}d\mu \right)}^{1/p_j},$$

where $H:=\left\{x\in X\mid {\displaystyle {\prod }_{j=1}^k}{f}_j\left(x\right)\ne 0\right\}$.

Proof. 

It is obvious that $H\in \mathcal{A}$. Since

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu ={\displaystyle \underset{H}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu ,$$

the result follows from Theorem 3 (a).

The proof is complete. □

Remark 2. 

(a) The result above is clearly sharper than Theorem 3 (a) and can be regarded as a refinement of it.

(b) A sharper version of Theorem 3 (b), analogous to Theorem 5, can also be formulated.

3. Main Results

The following result contains refinements of generalized integral Hölder inequalities.

Theorem 6.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let ${\mu }_i$ $\left(i=1,\dots ,n\right)$ be measures on $\mathcal{A}$, and let $\mu :={\displaystyle {\sum }_{i=1}^n}{\mu }_i$.

(a) Let $p_j>1$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$. If $f_j\in L_{p_j}\left(X,\mathcal{A},\mu \right)$ $\left(j=1,\dots ,k\right)$, then ${\displaystyle {\prod }_{j=1}^k}{f}_j\in L_1\left(X,\mathcal{A},\mu \right)$, and

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu \le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j}$$

$$\le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d\mu \right)}^{1/p_j}.$$

(5)

(b) Let $0<p_1<1$, and let $p_j<0$ $\left(j=2,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$. Let $f_j\in L_p\left(X,\mathcal{A},\mu \right)$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i>0$ $\left(j=2,\dots ,k,i=1,\dots ,n\right)$. If ${\displaystyle {\prod }_{j=1}^k}{f}_j$ is μ-integrable, then the reverse inequalities hold in (5).

(c) Let $n=2$, and let $0<p<1$ and $q<0$ such that $1/p+1/q=1$. Let $f\in L_p\left(X,\mathcal{A},\mu \right)$, and let $g\in L_q\left(X,\mathcal{A},\mu \right)$ such that ${\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d\mu >0$. If $fg$ is μ-integrable, then

$${\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{p}d\mu \right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d\mu \right)}^{1/q}\le \left({\left({\displaystyle \underset{X}{\int }}\left|fg\right|d{\mu }_1\right)}^p{\left({\displaystyle \frac{{\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d{\mu }_1}{{\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d\mu }}\right)}^{-p/q}\right.$$

$${\left.+{\left({\displaystyle \underset{X}{\int }}\left|fg\right|d{\mu }_2\right)}^p{\left({\displaystyle \frac{{\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d{\mu }_2}{{\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d\mu }}\right)}^{-p/q}\right)}^{1/p}\le {\displaystyle \underset{X}{\int }}\left|fg\right|d\mu .$$

(6)

Proof. 

(a) It is obvious that $L_p\left(X,\mathcal{A},\mu \right)={\displaystyle {\bigcap }_{i=1}^n}{L}_p\left(X,\mathcal{A},{\mu }_i\right)$ $\left(p\in \mathbb{R}\right)$.

By applying the integral Hölder inequality for positive exponent, we obtain

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu ={\displaystyle {\sum }_{i=1}^n}{\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d{\mu }_i$$

$$\le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j},$$

which is precisely the first inequality.

This follows from the discrete Hölder inequality for positive exponent by choosing ${\alpha }_i:=1$ $\left(i=1,\dots ,n\right)$ and

$$x_{ji}:={\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j},j=1,\dots ,k,i=1,\dots ,n.$$

(b) It can be proven similarly to (a), using the integral and discrete Hölder inequalities for negative exponents.

(c) For the sake of simpler notation, we assume that f is a nonnegative function from $L_p\left(X,\mathcal{A},\mu \right)$ and g is a nonnegative function from $L_q\left(X,\mathcal{A},\mu \right)$ such that ${\displaystyle \underset{X}{\int }}{g}^{q}d\mu >0$.

Since ${\displaystyle \underset{X}{\int }}{g}^{q}d\mu >0$ and $q<0$, we have that $g>0$ $\mu$-a.e. on X. Let $r:={\displaystyle \frac{1}{p}}$, and define $z:=g^{-p}$ and $w:={\left(fg\right)}^p$. If $s>1$ such that $1/r+1/s=1$, then $z^s=g^q$, and therefore $z\in L_s\left(X,\mathcal{A},\mu \right)$. Since $fg$ is $\mu$-integrable, $w\in L_r\left(X,\mathcal{A},\mu \right)$. Applying (5) with p replaced by r and q replaced by s we have

$${\displaystyle \underset{X}{\int }}{f}^{p}d\mu ={\displaystyle \underset{X}{\int }}zwd\mu \le {\left({\displaystyle \underset{X}{\int }}{w}^{r}d{\mu }_1\right)}^{1/r}{\left({\displaystyle \underset{X}{\int }}{z}^{s}d{\mu }_1\right)}^{1/s}$$

$$+{\left({\displaystyle \underset{X}{\int }}{w}^{r}d{\mu }_2\right)}^{1/r}{\left({\displaystyle \underset{X}{\int }}{z}^{s}d{\mu }_2\right)}^{1/s}\le {\left({\displaystyle \underset{X}{\int }}{w}^{r}d\mu \right)}^{1/r}{\left({\displaystyle \underset{X}{\int }}{z}^{s}d\mu \right)}^{1/s}.$$

According to $-sp=q$, we obtain

$${\displaystyle \underset{X}{\int }}{f}^{p}d\mu \le {\left({\displaystyle \underset{X}{\int }}fgd{\mu }_1\right)}^p{\left({\displaystyle \underset{X}{\int }}{g}^{q}d{\mu }_1\right)}^{-p/q}$$

$$+{\left({\displaystyle \underset{X}{\int }}fgd{\mu }_2\right)}^p{\left({\displaystyle \underset{X}{\int }}{g}^{q}d{\mu }_2\right)}^{-p/q}\le {\left({\displaystyle \underset{X}{\int }}fgd\mu \right)}^p{\left({\displaystyle \underset{X}{\int }}{g}^{q}d\mu \right)}^{-p/q}$$

and this implies the result.

The proof is complete. □

Remark 3.

(a) The method used and the results are novel.

It is significant in several respects that we express the refinements in general measure spaces: the appropriate discrete refinements can be derived from our results; moreover, we are not limited to using finite sums, and variants that use weight functions can also be easily derived.

Any refinement of the Hölder’s inequality (whether presented in this paper or already known) has its own justification. It is difficult to compare the various types of refinements, since each one uses a different method to obtain sharper inequalities. The usefulness and importance of the various refinements are demonstrated by the fact that they can be applied in different areas of mathematics.

(b) Theorem 2 (and hence Theorem 1) is contained in Theorem 6 (a): Let ${\mu }_1$ and ${\mu }_2$ be defined on the Lebesgue-measurable sets on $\left[a,b\right]$ by

$${\mu }_1\left(A\right):={\displaystyle \underset{A}{\int }}u\left(t\right)dt,{\mu }_2\left(A\right):={\displaystyle \underset{A}{\int }}v\left(t\right)dt,$$

respectively. In this case, ${\mu }_1$ and ${\mu }_2$ are absolutely continuous measures with respect to the Lebesgue measure.

Remark 4.

(a) Part (a) of Theorem 6 can be sharpened in two ways using Theorem 5: If $H:=\left\{x\in X\mid {\displaystyle {\prod }_{j=1}^k}{f}_j\left(x\right)\ne 0\right\}$, then

(i)

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu \le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{H}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j}$$

$$\le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{H}{\int }}{\left|f_j\right|}^{p_j}d\mu \right)}^{1/p_j}.$$

(ii)

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|d\mu \le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{H}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j}$$

$$\le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d{\mu }_i\right)}^{1/p_j}\le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}d\mu \right)}^{1/p_j}.$$

(b) Assume there exists a σ-finite measure ν on $\mathcal{A}$ such that ${\mu }_1,\dots ,{\mu }_n$ are absolutely continuous with respect to ν, and let $w_i$ be the Radon–Nikodým derivative of ${\mu }_i$ with respect to ν $\left(i=1,\dots ,n\right)$. Then, $w_i$ is an essentially unique nonnegative function from $L_1\left(X,\mathcal{A},\nu \right)$ $\left(i=1,\dots ,n\right)$. It is easy to think that μ is also absolutely continuous with respect to ν, and $w:={\displaystyle {\sum }_{i=1}^n}{w}_i$ is the Radon–Nikodým derivative of μ with respect to ν. In this case, (5) can be written in the following form

$${\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}\left|f_j\right|wd\nu \le {\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}{w}_{i}d\nu \right)}^{1/p_j}$$

$$\le {\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{\left|f_j\right|}^{p_j}wd\nu \right)}^{1/p_j}.$$

Of course, (6) can also be rewritten in a similar way.

The conditions for equality in the inequalities under consideration can be precisely formulated by applying the well-known conditions for equality in Hölder’s inequality. We illustrate this by analyzing Theorem 6 (a), taking $k=2$ for the sake of simplicity.

Proposition 1.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let ${\mu }_i$ $\left(i=1,\dots ,n\right)$ be measures on $\mathcal{A}$, and let $\mu :={\displaystyle {\sum }_{i=1}^n}{\mu }_i$. Let p, $q>1$ such that ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$. Assume $f\in L_p\left(X,\mathcal{A},\mu \right)$ and $g\in L_q\left(X,\mathcal{A},\mu \right)$. Then,

(a)

$${\displaystyle \underset{X}{\int }}\left|fg\right|d\mu ={\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{p}d{\mu }_i\right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d{\mu }_i\right)}^{1/q}$$

if and only if there are nonnegative numbers $A_i$ and $B_i$, which are not both zero $\left(i=1,\dots ,n\right)$, such that

$$A_i{\left|f\right|}^p=B_i{\left|g\right|}^q{\mu }_i\text{-}\mathrm{a}.\mathrm{e}.,i=1,\dots ,n.$$

(b)

$${\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{p}d{\mu }_i\right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d{\mu }_i\right)}^{1/q}={\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{p}d\mu \right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d\mu \right)}^{1/q}$$

if and only if there are nonnegative numbers A and B, which are not both zero, such that

$$A{\displaystyle \underset{X}{\int }}{\left|f\right|}^{p}d{\mu }_i=B{\displaystyle \underset{X}{\int }}{\left|g\right|}^{q}d{\mu }_i,i=1,\dots ,n.$$

Proof. 

They follow from the known equality conditions in the integral and discrete Hölder inequalities (see [7]). □

It’s more or less clear that the conditions described in parts (a) and (b) of the previous statement are independent, but the simple examples below clearly illustrate this fact.

For a set X, the family of all subsets of X is denoted by $P\left(X\right)$.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, and let a be a fixed point of X. The unit mass at a is denoted by ${\epsilon }_a$.

Example 1.

Let $X:=\left\{1,2,3\right\}$, $\mathcal{A}:=P\left(X\right)$, ${\mu }_1:={\epsilon }_1+{\epsilon }_2$ and ${\mu }_2:={\epsilon }_3$. Let $p=q=2$.

(a) If f, $g:X\to \mathbb{R}$ are defined by

$$f\left(1\right):=1,f\left(2\right):=0,f\left(3\right):=2$$

and

$$g\left(1\right):=1,g\left(2\right):=0,g\left(3\right):=1,$$

then

$${\displaystyle \underset{X}{\int }}\left|fg\right|d\mu =3={\displaystyle {\sum }_{i=1}^2}{\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{2}d{\mu }_i\right)}^{1/2}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{2}d{\mu }_i\right)}^{1/2}$$

$$<\sqrt{10}={\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{2}d\mu \right)}^{1/2}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{2}d\mu \right)}^{1/2}.$$

(b) If f, $g:X\to \mathbb{R}$ are defined by

$$f\left(1\right):=1,f\left(2\right):=1,f\left(3\right):=2$$

and

$$g\left(1\right):=1,g\left(2\right):=2,g\left(3\right):=\sqrt{10},$$

then

$${\displaystyle \underset{X}{\int }}\left|fg\right|d\mu =3+2\sqrt{10}<{\displaystyle {\sum }_{i=1}^2}{\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{2}d{\mu }_i\right)}^{1/2}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{2}d{\mu }_i\right)}^{1/2}$$

$$=3\sqrt{10}={\left({\displaystyle \underset{X}{\int }}{\left|f\right|}^{2}d\mu \right)}^{1/2}{\left({\displaystyle \underset{X}{\int }}{\left|g\right|}^{2}d\mu \right)}^{1/2}.$$

One of the main results of the paper [3] (Theorem 2.1 of [3]) is a Hölder-type inequality, the origin of which also goes back to Theorem 1. In the following statement, we obtain a general form of it.

Proposition 2.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let ${\mu }_i$ $\left(i=1,\dots ,n\right)$ be measures on $\mathcal{A}$, and let $\mu :={\displaystyle {\sum }_{i=1}^n}{\mu }_i$. Let p, $q>1$ such that $1/p+1/q=1$. If $f\in L_p\left(X,\mathcal{A},\mu \right)$, $g\in L_q\left(X,\mathcal{A},\mu \right)$ and $\left|f\right|{\left|g\right|}^q\in L_1\left(X,\mathcal{A},\mu \right)$, then

$${\displaystyle \underset{X}{\int }}\left|fg\right|d\mu \le {\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \underset{X}{\int }}\left|f\right|d{\mu }_i\right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}\left|f\right|{\left|g\right|}^{q}d{\mu }_i\right)}^{1/q}$$

$$\le {\left({\displaystyle \underset{X}{\int }}\left|f\right|d\mu \right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}\left|f\right|{\left|g\right|}^{q}d\mu \right)}^{1/q}.$$

Proof. 

Since $f\in L_p\left(X,\mathcal{A},\mu \right)$, f also belongs to $L_1\left(X,\mathcal{A},\mu \right)$.

We can follow the line of reasoning in the proof of part (a) of Theorem 6, applying first the integral and then the discrete Hölder inequality for positive exponents.

$${\displaystyle \underset{X}{\int }}\left|fg\right|d\mu ={\displaystyle {\sum }_{i=1}^n}{\displaystyle \underset{X}{\int }}{\left|f\right|}^{1/p}\left({\left|f\right|}^{1/q}\left|g\right|\right)d{\mu }_i$$

$$\le {\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \underset{X}{\int }}\left|f\right|d{\mu }_i\right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}\left|f\right|{\left|g\right|}^{q}d{\mu }_i\right)}^{1/q}$$

$$\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \underset{X}{\int }}\left|f\right|d{\mu }_i\right)}^{1/p}{\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \underset{X}{\int }}\left|f\right|{\left|g\right|}^{q}d{\mu }_i\right)}^{1/q}$$

$$={\left({\displaystyle \underset{X}{\int }}\left|f\right|d\mu \right)}^{1/p}{\left({\displaystyle \underset{X}{\int }}\left|f\right|{\left|g\right|}^{q}d\mu \right)}^{1/q}$$

The proof is complete. □

Remark 5.

Theorem 2.1 of [3] is contained in Proposition 2: Let ${\mu }_1$ and ${\mu }_2$ be defined on the Lebesgue-measurable sets on $\left[a,b\right]$ by

$${\mu }_1\left(A\right):={\displaystyle \underset{A}{\int }}{\displaystyle \frac{b-t}{b-a}}dt,{\mu }_2\left(A\right):={\displaystyle \underset{A}{\int }}{\displaystyle \frac{t-a}{b-a}}dt,$$

respectively. In this case, ${\mu }_1$ and ${\mu }_2$ are absolutely continuous measures with respect to the Lebesgue measure.

In the following two results, we give the discrete versions of Theorem 6 for $n=2$.

Corollary 1.

Let the index set I denote either $\{1,\dots ,m\}$ for some $m\ge 1$ or $\{1,2,\dots \}$. Let ${\left({\alpha }_i\right)}_{i\in I}$ and ${\left({\beta }_i\right)}_{i\in I}$ be two nonnegative sequences.

(a) Let p, $q>1$ such that $1/p+1/q=1$. If ${\left(x_i\right)}_{i\in I}$ and ${\left(y_i\right)}_{i\in I}$ are two complex sequences such that the series ${\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p\left({\alpha }_i+{\beta }_i\right)$ and ${\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)$ are absolutely convergent, then ${\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|\left({\alpha }_i+{\beta }_i\right)$ is absolutely convergent and

$${\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|\left({\alpha }_i+{\beta }_i\right)\le {\left({\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p{\alpha }_i\right)}^{1/p}{\left({\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\alpha }_i\right)}^{1/q}$$

$$+{\left({\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p{\beta }_i\right)}^{1/p}{\left({\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\beta }_i\right)}^{1/q}$$

$$\le {\left({\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p\left({\alpha }_i+{\beta }_i\right)\right)}^{1/p}{\left({\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)\right)}^{1/q}.$$

(7)

(b) Let $0<p<1$ and $q<0$ such that $1/p+1/q=1$. If ${\left(x_i\right)}_{i\in I}$ and ${\left(y_i\right)}_{i\in I}$ are two complex sequences such that ${\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\alpha }_i>0$ and ${\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\beta }_i>0$, then the reverse inequalities hold in (7).

(c) Let $0<p<1$ and $q<0$ such that $1/p+1/q=1$. If ${\left(x_i\right)}_{i\in I}$ and ${\left(y_i\right)}_{i\in I}$ are two complex sequences such that the series ${\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p\left({\alpha }_i+{\beta }_i\right)$, ${\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)$ and ${\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|\left({\alpha }_i+{\beta }_i\right)$ are absolutely convergent, and ${\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)>0$, then

$${\left({\displaystyle {\sum }_{i\in I}}{\left|x_i\right|}^p\left({\alpha }_i+{\beta }_i\right)\right)}^{1/p}{\left({\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)\right)}^{1/q}$$

$$\le \left({\left({\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|{\alpha }_i\right)}^p{\left({\displaystyle \frac{{\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\alpha }_i}{{\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)}}\right)}^{-p/q}\right.$$

$${\left.+{\left({\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|{\beta }_i\right)}^p{\left({\displaystyle \frac{{\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q{\beta }_i}{{\displaystyle {\sum }_{i\in I}}{\left|y_i\right|}^q\left({\alpha }_i+{\beta }_i\right)}}\right)}^{-p/q}\right)}^{1/p}\le {\displaystyle {\sum }_{i\in I}}\left|x_i{y}_i\right|\left({\alpha }_i+{\beta }_i\right).$$

Proof. 

We can apply Theorem 6 in the following special case: X is either $\{1,\dots ,m\}$ or $\{1,2,\dots \}$, $\mathcal{A}:=P\left(X\right)$, ${\mu }_1:={\displaystyle {\sum }_{i\in I}}{\alpha }_i{\epsilon }_i$, ${\mu }_2:={\displaystyle {\sum }_{i\in I}}{\beta }_i{\epsilon }_i$, $f\left(i\right):=x_i$ and $g\left(i\right):=y_i$ $\left(i\in I\right)$. □

Remark 6.

Of course, if I is a finite set, the convergence conditions for the sums can be omitted.

4. Applications

Let $\left(X,\mathcal{A},\mu \right)$ be a measure space with $0<\mu \left(X\right)<\infty$. Integral power means of order $p\in \mathbb{R}$ are defined as follows:

(i) If $p\ne 0$ and f is a nonnegative function from $L_p\left(X,\mathcal{A},\mu \right)$, then

$$M_p\left(f,\mu \right):={\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{p}d\mu \right)}^{1/p}.$$

(ii) If $p=0$ and f is a positive function from $L_1\left(X,\mathcal{A},\mu \right)$, then

$$M_0\left(f,\mu \right):=\exp \left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}\log \circ fd\mu \right).$$

First, we obtain inequalities for the integral power means.

The principal tool in the next assertions is Theorem 6.

Theorem 7.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let ${\mu }_i$ be measures on $\mathcal{A}$ with $0<{\mu }_i\left(X\right)<\infty$ $\left(i=1,\dots ,n\right)$, and let $\mu :={\displaystyle {\sum }_{i=1}^n}{\mu }_i$. Let $p_j>1$ $\left(j=1,\dots ,k\right)$ such that ${\displaystyle {\sum }_{j=1}^k}{\displaystyle \frac{1}{p_j}}=1$. Let $r_j\ne 0$ $\left(j=1,\dots ,k\right)$ such that $r:={\displaystyle {\sum }_{j=1}^k}{r}_j\ne 0$. Assume f is a positive function from $\left({\displaystyle {\bigcap }_{j=1}^k}{L}_{r_j{p}_j}\left(X,\mathcal{A},\mu \right)\right){\displaystyle \bigcap }{L}_r\left(X,\mathcal{A},\mu \right)$.

(a) If $r>0$, then

$$M_r\left(f,\mu \right)\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{\displaystyle {\prod }_{j=1}^k}{M}_{r_j{p}_j}^{r_j}\left(f,{\mu }_i\right)\right)}^{1/r}\le {\displaystyle {\prod }_{j=1}^k}{M}_{r_j{p}_j}^{r_j/r}\left(f,\mu \right).$$

(8)

(b) If $r<0$, then the reverse inequalities hold in (8).

Proof. 

(a) By Theorem 6 (a),

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu ={\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{\displaystyle {\prod }_{j=1}^k}{f}^{r_j}d\mu \le {\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle {\sum }_{i=1}^n}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{f}^{r_j{p}_j}d{\mu }_i\right)}^{1/p_j}$$

$$\le {\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle {\prod }_{j=1}^k}{\left({\displaystyle \underset{X}{\int }}{f}^{r_j{p}_j}d\mu \right)}^{1/p_j}.$$

Since the function $t\to t^{1/r}$ $\left(t\ge 0\right)$ is increasing,

$${\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu \right)}^{1/r}$$

$$\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{\displaystyle {\prod }_{j=1}^k}{\left({\left({\displaystyle \frac{1}{{\mu }_i\left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r_j{p}_j}d{\mu }_i\right)}^{1/r_j{p}_j}\right)}^{r_j}\right)}^{1/r}$$

$$\le {\displaystyle {\prod }_{j=1}^k}{\left({\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r_j{p}_j}d\mu \right)}^{1/r_j{p}_j}\right)}^{r_j/r}.$$

(b) It can be proved similarly to (a).

The proof is complete. □

Remark 7.

While Hölder’s inequality can naturally be applied to inequalities involving integral power means and their proofs, Jensen’s inequality is generally used instead. The previous result illustrates that new and interesting inequalities can be obtained for power means by applying only Hölder’s inequality. The method used is not the standard procedure, even when $k=2$.

The following result is related to the well-known inequality $M_r\left(f,\mu \right)\le M_s\left(f,\mu \right)$ $\left(r<s\right)$.

Theorem 8.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let ${\mu }_i$ be measures on $\mathcal{A}$ with $0<{\mu }_i\left(X\right)<\infty$ $\left(i=1,\dots ,n\right)$, and let $\mu :={\displaystyle {\sum }_{i=1}^n}{\mu }_i$.

(a) If $0<r<s$ and f is a nonnegative function from $L_s\left(X,\mathcal{A},\mu \right)$, then

$$M_r\left(f,\mu \right)\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{M}_s^r\left(f,{\mu }_i\right)\right)}^{1/r}\le M_s\left(f,\mu \right).$$

(b) If $r<s<0$ and f is a nonnegative function from $L_r\left(X,\mathcal{A},\mu \right)$, then

$$M_r\left(f,\mu \right)\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{M}_r^s\left(f,{\mu }_i\right)\right)}^{1/s}\le M_s\left(f,\mu \right).$$

(c) If $0=r<s$ and f is a positive function from $L_s\left(X,\mathcal{A},\mu \right)\cap L_1\left(X,\mathcal{A},\mu \right)$, then

$$M_0\left(f,\mu \right)\le {\displaystyle {\prod }_{i=1}^n}{\left(M_s\left(f,{\mu }_i\right)\right)}^{{\mu }_i\left(X\right)/\mu \left(X\right)}\le M_s\left(f,\mu \right).$$

(d) If $r<s=0$ and f is a positive function from $L_r\left(X,\mathcal{A},\mu \right)\cap L_1\left(X,\mathcal{A},\mu \right)$, then

$$M_r\left(f,\mu \right)\le {\displaystyle {\prod }_{i=1}^n}{\left(M_r\left(f,{\mu }_i\right)\right)}^{{\mu }_i\left(X\right)/\mu \left(X\right)}\le M_0\left(f,\mu \right).$$

(e) If $r<0<s$ and f is a positive function from $L_r\left(X,\mathcal{A},\mu \right)\cap L_s\left(X,\mathcal{A},\mu \right)\cap L_1\left(X,\mathcal{A},\mu \right)$, then

$$M_r\left(f,\mu \right)\le {\displaystyle {\prod }_{i=1}^n}{\left(M_r\left(f,{\mu }_i\right)\right)}^{{\mu }_i\left(X\right)/\mu \left(X\right)}\le M_0\left(f,\mu \right)$$

$$\le {\displaystyle {\prod }_{i=1}^n}{\left(M_s\left(f,{\mu }_i\right)\right)}^{{\mu }_i\left(X\right)/\mu \left(X\right)}\le M_s\left(f,\mu \right).$$

(f) If $r<0<s$ and f is a positive function from $L_r\left(X,\mathcal{A},\mu \right)\cap L_s\left(X,\mathcal{A},\mu \right)\cap L_1\left(X,\mathcal{A},\mu \right)$, then

$$M_r\left(f,\mu \right)\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{M}_s^r\left(f,{\mu }_i\right)\right)}^{1/r}$$

$$\le {\displaystyle {\prod }_{i=1}^n}{\left(M_s\left(f,{\mu }_i\right)\right)}^{{\mu }_i\left(X\right)/\mu \left(X\right)}\le M_s\left(f,\mu \right).$$

Proof. 

(a) Since $\mu$ is a finite measure, it is easy to think that $f\in L_s\left(X,\mathcal{A},\mu \right)$ implies $f\in L_r\left(X,\mathcal{A},\mu \right)$.

Let $p:={\displaystyle \frac{s}{r}}$ and $q:={\displaystyle \frac{s}{s-r}}$. Then, p, $q>1$ such that $1/p+1/q=1$. By Theorem 6 (a),

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu ={\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^r·1d\mu$$

$$\le {\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \underset{X}{\int }}{f}^{rp}d{\mu }_i\right)}^{1/p}{\left({\mu }_i\left(X\right)\right)}^{1/q}\le {\displaystyle \frac{1}{\mu \left(X\right)}}{\left({\displaystyle \underset{X}{\int }}{f}^{rp}d\mu \right)}^{1/p}{\left(\mu \left(X\right)\right)}^{1/q},$$

that is,

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu \le {\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \frac{1}{{\mu }_i\left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_i\right)}^{1/p}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}\le {\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{1/p}.$$

It follows that

$${\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu \right)}^{1/r}\le {\left({\displaystyle {\sum }_{i=1}^n}{\left({\displaystyle \frac{1}{{\mu }_i\left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_i\right)}^{r/s}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}\right)}^{1/r}$$

$$\le {\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{1/s}.$$

(b) Since $\mu$ is a finite measure, it is easy to think that $f\in L_r\left(X,\mathcal{A},\mu \right)$ implies $f\in L_s\left(X,\mathcal{A},\mu \right)$.

Let $p:={\displaystyle \frac{r}{s}}$ and $q:={\displaystyle \frac{r}{r-s}}$. Then, p, $q>1$ such that $1/p+1/q=1$. We can continue in the same way as in (a).

(c) Let us apply (a) and Lemma 1.

(d) It follows from (b) and Lemma 1.

(e) We get it by using (c) and (d).

(f) Inequalities

$$M_r\left(f,\mu \right)\le {\left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{M}_s^r\left(f,{\mu }_i\right)\right)}^{1/r}\le M_s\left(f,\mu \right)$$

can be proven similarly to (a) by using Theorem 6 (b).

The third inequality comes from (c).

By applying the discrete Jensen inequality to the concave function log, we obtain

$$\log \left({\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}{\left({\displaystyle \frac{1}{{\mu }_i\left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_i\right)}^{r/s}\right)$$

$$\ge r{\displaystyle {\sum }_{i=1}^n}{\displaystyle \frac{{\mu }_i\left(X\right)}{\mu \left(X\right)}}\log {\left({\displaystyle \frac{1}{{\mu }_i\left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_i\right)}^{1/s},$$

which gives the second inequality, since $r<0$.

The proof is complete. □

Remark 8.

(a) Our results, on the one hand, extend inequalities presented in [4], and on the other hand, are new (for example (e) and (f)). In that paper $X:=\left[a,b\right]\subset \mathbb{R}$, $n=2$, and the measures μ, ${\mu }_1$ and ${\mu }_2$ are absolutely continuous with respect to the Lebesgue measure on the Borel subsets of $\left[a,b\right]$. It should also be noted that some of our results are special cases of the very general inequalities regarding quasi-arithmetic means obtained in [15]. However, on the one hand, power means are not discussed separately in that paper, and on the other hand, the proofs are based on the Jensen’s inequality.

(b) I could not find a single refinement similar to (f) in the literature.

The following inequality involving power means is unique because it arises from a refinement of Hölder’s inequality for exponents in the range $0<p<1$.

Theorem 9.

Let $\left(X,\mathcal{A},{\mu }_i\right)$ be measure spaces with $0<{\mu }_i\left(X\right)<\infty$ $\left(i=1,2\right)$, and let $\mu :={\mu }_1+{\mu }_2$. If $r<0<s$ and f is a positive function from $L_r\left(X,\mathcal{A},\mu \right)\cap L_s\left(X,\mathcal{A},\mu \right)\cap L_1\left(X,\mathcal{A},\mu \right)$, then

$${\displaystyle \frac{M_r\left(f,\mu \right)}{M_s\left(f,\mu \right)}}$$

$$\le {\left({\displaystyle \frac{{\mu }_1\left(X\right)}{\mu \left(X\right)}}{\left({\displaystyle \frac{M_r\left(f,{\mu }_1\right)}{M_s\left(f,{\mu }_1\right)}}\right)}^{{\displaystyle \frac{rs}{s-r}}}+{\displaystyle \frac{{\mu }_2\left(X\right)}{\mu \left(X\right)}}{\left({\displaystyle \frac{M_r\left(f,{\mu }_2\right)}{M_s\left(f,{\mu }_2\right)}}\right)}^{{\displaystyle \frac{rs}{s-r}}}\right)}^{{\displaystyle \frac{s-r}{rs}}}\le 1.$$

Proof. 

Let $p:={\displaystyle \frac{s}{s-r}}$ and $q:={\displaystyle \frac{s}{r}}$. Then, $0<p<1$ and $q<0$ such that $1/p+1/q=1$. By Theorem 6 (c),

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu ={\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^r·1d\mu$$

$$\ge {\displaystyle \frac{1}{\mu \left(X\right)}}\left({\left({\displaystyle \underset{X}{\int }}{f}^{r}d{\mu }_1\right)}^p{\left({\displaystyle \underset{X}{\int }}{f}^{rq}d{\mu }_1/{\displaystyle \underset{X}{\int }}{f}^{rq}d\mu \right)}^{-p/q}\right.$$

$${\left.+{\left({\displaystyle \underset{X}{\int }}{f}^{r}d{\mu }_2\right)}^p{\left({\displaystyle \underset{X}{\int }}{f}^{rq}d{\mu }_2/{\displaystyle \underset{X}{\int }}{f}^{rq}d\mu \right)}^{-p/q}\right)}^{1/p}$$

$$\ge {\displaystyle \frac{1}{\mu \left(X\right)}}{\left({\displaystyle \underset{X}{\int }}{f}^{rq}d\mu \right)}^{1/q}{\left(\mu \left(X\right)\right)}^{1/p}.$$

This implies that

$${\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{r}d\mu$$

$$\ge {\displaystyle \frac{1}{\mu \left(X\right)}}\left({\left({\displaystyle \underset{X}{\int }}{f}^{r}d{\mu }_1\right)}^{s/\left(s-r\right)}{\left({\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_1/{\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{-r/\left(s-r\right)}\right.$$

$${\left.+{\left({\displaystyle \underset{X}{\int }}{f}^{r}d{\mu }_2\right)}^{s/\left(s-r\right)}{\left({\displaystyle \underset{X}{\int }}{f}^{s}d{\mu }_2/{\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{-r/\left(s-r\right)}\right)}^{\left(s-r\right)/s}$$

$$\ge {\displaystyle \frac{1}{\mu \left(X\right)}}{\left({\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{r/s}{\left(\mu \left(X\right)\right)}^{\left(s-r\right)/s}={\left({\displaystyle \frac{1}{\mu \left(X\right)}}{\displaystyle \underset{X}{\int }}{f}^{s}d\mu \right)}^{r/s},$$

and the result follows from a simple calculation.

The proof is complete. □

Next, we present some discrete versions of the previous statements.

Corollary 2.

Let $\left({\alpha }_1,\dots ,{\alpha }_m\right)$ and $\left({\beta }_1,\dots ,{\beta }_m\right)$ be two nonnegative m-tuples such that $A:={\displaystyle {\sum }_{i=1}^m}{\alpha }_i>0$ and $B:={\displaystyle {\sum }_{i=1}^m}{\beta }_i>0$.

(a) If $0<r<s$ and $\left(x_1,\dots ,x_m\right)$ is a nonnegative m-tuples, then

$${\left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^r\left({\alpha }_i+{\beta }_i\right)\right)}^{1/r}$$

$$\le {\left({\left({\displaystyle \frac{1}{A}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\alpha }_i\right)}^{r/s}{\displaystyle \frac{A}{A+B}}+{\left({\displaystyle \frac{1}{B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\beta }_i\right)}^{r/s}{\displaystyle \frac{B}{A+B}}\right)}^{1/r}$$

$$\le {\left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s\left({\alpha }_i+{\beta }_i\right)\right)}^{1/s}.$$

(b) If $0=r<s$ and $\left(x_1,\dots ,x_m\right)$ is a positive m-tuples, then

$$\exp \left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}\log \left(x_i\right)\right)\le {\left({\left({\displaystyle \frac{1}{A}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\alpha }_i\right)}^{1/s}\right)}^{A/\left(A+B\right)}$$

$$·{\left({\left({\displaystyle \frac{1}{B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\beta }_i\right)}^{1/s}\right)}^{A/\left(A+B\right)}\le {\left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s\left({\alpha }_i+{\beta }_i\right)\right)}^{1/s}.$$

(c) If $r<0<s$ and $\left(x_1,\dots ,x_m\right)$ is a positive m-tuples, then

$${\left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^r\left({\alpha }_i+{\beta }_i\right)\right)}^{1/r}/{\left({\displaystyle \frac{1}{A+B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s\left({\alpha }_i+{\beta }_i\right)\right)}^{1/s}$$

$$\le \left({\displaystyle \frac{A}{A+B}}{\left({\left({\displaystyle \frac{1}{A}}{\displaystyle {\sum }_{i=1}^m}{x}_i^r{\alpha }_i\right)}^{1/r}/{\left({\displaystyle \frac{1}{A}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\alpha }_i\right)}^{1/s}\right)}^{{\displaystyle \frac{rs}{s-r}}}\right.$$

$${\left.+{\displaystyle \frac{B}{A+B}}{\left({\left({\displaystyle \frac{1}{B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^r{\beta }_i\right)}^{1/r}/{\left({\displaystyle \frac{1}{B}}{\displaystyle {\sum }_{i=1}^m}{x}_i^s{\beta }_i\right)}^{1/s}\right)}^{{\displaystyle \frac{rs}{s-r}}}\right)}^{{\displaystyle \frac{s-r}{rs}}}\le 1.$$

Proof. 

These are special cases of the following, in order: Theorem 8 (a), (c) and Theorem 9 (see the proof of Corollary 1). □

As another application of our main results, we present a refinement of the generalized Opial–Olech inequality, as follows:

Theorem 10

(see [16,17]). Let $f:\left[a,b\right]\to \mathbb{R}$ be absolutely continuous and $f\left(a\right)=0$.

(a) If $r>0$ and ${\left|f^{\prime }\right|}^{r+1}$ is Lebesgue-integrable, then

$${\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^r\left|f^{\prime }\left(t\right)\right|dt\le {\displaystyle \frac{{\left(b-a\right)}^r}{r+1}}{\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}dt.$$

(9)

(b) If either $r<-1$ and both ${\left|f\right|}^r\left|f^{\prime }\right|$ and ${\left|f^{\prime }\right|}^{r+1}$ are Lebesgue-integrable, or $-1<r<0$ and ${\left|f\right|}^r\left|f^{\prime }\right|$ is Lebesgue-integrable, the reverse inequality holds in (9).

The original versions are taken from [18] ($r=1$, $a=0$ and $f\left(b\right)=0$) and [19] ($r=1$ and $a=0$).

Let ${\mathcal{L}}_{\left[a,b\right]}$ be the $\sigma$-algebra of Lebesgue-measurable sets on $\left[a,b\right]$, and let $\lambda$ be the Lebesgue measure on ${\mathcal{L}}_{\left[a,b\right]}$.

Theorem 11.

Let the measures ${\mu }_1$ and ${\mu }_2$ be defined on ${\mathcal{L}}_{\left[a,b\right]}$ such that $\lambda ={\mu }_1+{\mu }_2$. Let $f:\left[a,b\right]\to \mathbb{R}$ be absolutely continuous and $f\left(a\right)=0$.

(a) If $r>0$ and ${\left|f^{\prime }\right|}^{r+1}$ is Lebesgue-integrable, then

$${\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^r\left|f^{\prime }\left(t\right)\right|dt\le {\displaystyle \frac{1}{r+1}}\left({\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}d{\mu }_1\left(t\right)\right)}^{1/\left(r+1\right)}{\mu }_1{\left(\left[a,b\right]\right)}^{1/s}\right.$$

$${\left.+{\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}d{\mu }_2\left(t\right)\right)}^{1/\left(r+1\right)}{\mu }_2{\left(\left[a,b\right]\right)}^{1/s}\right)}^{r+1}\le {\displaystyle \frac{{\left(b-a\right)}^r}{r+1}}{\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}dt,$$

(10)

where $s:={\displaystyle \frac{r+1}{r}}$.

(b) If either $r<-1$ and both ${\left|f\right|}^r\left|f^{\prime }\right|$ and ${\left|f^{\prime }\right|}^{r+1}$ are Lebesgue-integrable, or $-1<r<0$ and ${\left|f\right|}^r\left|f^{\prime }\right|$ is Lebesgue-integrable, then the reverse inequalities hold in (10).

Proof. 

Introducing the function

$$g:\left[a,b\right]\to \mathbb{R},g\left(x\right):={\displaystyle {\int }_a^x}\left|f^{\prime }\left(t\right)\right|dt,$$

we can easily obtain

$$g\left(x\right)\ge \left|f\left(x\right)\right|,x\in \left[a,b\right],$$

and

$$g^{\prime }=\left|f^{\prime }\right|,\lambda \text{-}\mathrm{a}.\mathrm{e}.\mathrm{on}\left[a,b\right].$$

It follows that for every $r>0$

$${\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^r\left|f^{\prime }\left(t\right)\right|dt\le {\displaystyle {\int }_a^b}{g}^r\left(t\right)g^{\prime }\left(t\right)dt={\displaystyle \frac{1}{r+1}}{\left({\displaystyle {\int }_a^b}\left|f^{\prime }\left(t\right)\right|dt\right)}^{r+1},$$

(11)

and the reverse inequality holds if either $r<-1$ or $-1<r<0$.

Let $p:=r+1$ and $q:=s$ for all $r\in \mathbb{R}\setminus \left\{-1,0\right\}$ (then ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$).

(a) Applying Theorem 6 (a), we have that

$${\displaystyle {\int }_a^b}\left|f^{\prime }\left(t\right)\right|·1dt\le {\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}d{\mu }_1\left(t\right)\right)}^{1/\left(r+1\right)}{\mu }_1{\left(\left[a,b\right]\right)}^{1/s}$$

$$+{\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}d{\mu }_2\left(t\right)\right)}^{1/\left(r+1\right)}{\mu }_2{\left(\left[a,b\right]\right)}^{1/s}$$

$$\le {\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}d\left(t\right)\right)}^{1/\left(r+1\right)}{\left(b-a\right)}^{1/s}.$$

(12)

The result follows from (11) and (12).

(b) If $-1<r<0$, then $0<p<1$, while if $r<-1$, then $0<q<1$. It follows from Theorem 6 (b) that the reverse inequalities hold in (12).

We can conclude the proof in the same way as in part (a).

The proof is complete. □

Remark 9.

(a) The main result of [17] (see Theorem 4 of [17]) is also a refinement of the inequality (9); both the refinement and its proof differ from those we have provided. Hölder’s inequality is not used at all in [17].

(b) A result for $r=1$ based on a similar line of reasoning but using a different refinement of Hölder’s inequality can be found in [20].

(c) Assume ${\mu }_1$ and ${\mu }_2$ are absolutely continuous with respect to λ, and let u and v be the Radon–Nikodým derivatives of ${\mu }_1$ and ${\mu }_2$ with respect to λ, respectively. Then, $u\left(x\right)+v\left(x\right)=1$ $\left(\lambda \text{-}\mathrm{a}.\mathrm{e}.\mathrm{on}\left[a,b\right]\right)$. In this case, (10) can be written in the following form

$${\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^r\left|f^{\prime }\left(t\right)\right|dt\le {\displaystyle \frac{1}{r+1}}\left({\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}u\left(t\right)dt\right)}^{1/\left(r+1\right)}{\left({\displaystyle {\int }_a^b}u\left(t\right)dt\right)}^{1/s}\right.$$

$${\left.+{\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}v\left(t\right)dt\right)}^{1/\left(r+1\right)}{\left({\displaystyle {\int }_a^b}v\left(t\right)dt\right)}^{1/s}\right)}^{r+1}$$

$$\le {\displaystyle \frac{{\left(b-a\right)}^r}{r+1}}{\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}dt.$$

(13)

(d) If, as in Theorem 1,

$$u\left(x\right):={\displaystyle \frac{b-x}{b-a}},v\left(x\right):={\displaystyle \frac{x-a}{b-a}},x\in \left[a,b\right],$$

then (13) gives

$${\displaystyle {\int }_a^b}{\left|f\left(t\right)\right|}^r\left|f^{\prime }\left(t\right)\right|dt\le {\displaystyle \frac{{\left(b-a\right)}^{r-1}}{2^r\left(r+1\right)}}\left({\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}\left(b-t\right)dt\right)}^{1/\left(r+1\right)}\right.$$

$${\left.+{\left({\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}\left(t-a\right)dt\right)}^{1/\left(r+1\right)}\right)}^{r+1}\le {\displaystyle \frac{{\left(b-a\right)}^r}{r+1}}{\displaystyle {\int }_a^b}{\left|f^{\prime }\left(t\right)\right|}^{r+1}dt.$$

Our next application is related to information theory.

Let $P:=\left(p_1,\dots ,p_m\right)$ be a probability distribution $(p_i\ge 0,i=1,\dots ,m,$ and ${\displaystyle {\sum }_{i=1}^m}{p}_i=1)$, $U:=\left(u_1,\dots ,u_m\right)$ be the utility distribution $\left(u_i>0,i=1,\dots ,m\right)$, $\alpha >0$ and $\alpha \ne 1$. Gurdial and F. Pessoa (see [21]) introduced the so called “useful” information of order $\alpha$:

$$H_{\alpha }\left(P,U\right):={\displaystyle \frac{1}{1-\alpha }}\log \left({\displaystyle {\sum }_{i=1}^m}{\displaystyle \frac{p_i^{\alpha }{u}_i}{{\displaystyle {\sum }_{j=1}^m}{p}_j{u}_j}}\right).$$

This is a generalization of Rényi’s [22] entropy of order $\alpha$ when $u_i=1$ for all i. It will be denoted by $H_{\alpha }\left(P\right)$.

If $p_i>0$ $\left(i=1,\dots ,m\right)$, then

$$\underset{\alpha \to 1}{\lim }{H}_{\alpha }\left(P\right)=-{\displaystyle {\sum }_{i=1}^m}{p}_i\log \left(p_i\right),$$

which is Shannon’s entropy.

It is known (see [22]) that Rényi’s entropies are strictly decreasing of their order parameter $\alpha$ if the base of the logarithm is greater than 1: for any two orders $0<\alpha \le \beta$, the corresponding Rényi entropies satisfy

$$H_{\alpha }\left(P\right)\ge H_{\beta }\left(P\right)$$

(14)

for any probability distribution P.

In the next statement, we refine inequality (14) by using the “useful” information.

Theorem 12.

Let $P:=\left(p_1,\dots ,p_m\right)$ be a probability distribution, and let $U:=\left(u_1,\dots ,u_m\right)$ and $V:=\left(v_1,\dots ,v_m\right)$ such that $u_i$, $v_i>0$ and $u_i+v_i=1$ $\left(i=1,\dots ,m\right)$. Suppose the base a of the logarithm is greater than 1.

(a) If either $0<\alpha <1<\beta$ or $1<\alpha <\beta$, then

$$H_{\alpha }\left(P\right)\ge {\displaystyle \frac{1}{1-\alpha }}\log \left(\left({\displaystyle {\sum }_{i=1}^m}{p}_i{u}_i\right)·{\exp }_a\left(\left(1-\alpha \right)H_{\beta }\left(P,U\right)\right)\right.$$

$$\left.+\left({\displaystyle {\sum }_{i=1}^m}{p}_i{v}_i\right)·{\exp }_a\left(\left(1-\alpha \right)H_{\beta }\left(P,V\right)\right)\right)\ge H_{\beta }\left(P\right).$$

(b) If $0<\alpha <\beta <1$, then

$$H_{\alpha }\left(P\right)\ge {\displaystyle \frac{1}{1-\beta }}\log \left(\left({\displaystyle {\sum }_{i=1}^m}{p}_i{u}_i\right)·{\exp }_a\left(\left(1-\beta \right)H_{\alpha }\left(P,U\right)\right)\right.$$

$$\left.+\left({\displaystyle {\sum }_{i=1}^m}{p}_i{v}_i\right)·{\exp }_a\left(\left(1-\beta \right)H_{\alpha }\left(P,V\right)\right)\right)\ge H_{\beta }\left(P\right).$$

Proof. 

(a) Let $\vartheta :={\displaystyle \frac{1-\alpha }{1-\beta }}$.

If $1<\alpha <\beta$, then $0<\vartheta <1$. Since

$${\displaystyle {\sum }_{i=1}^m}{p}_i^{\alpha }={\displaystyle {\sum }_{i=1}^m}{p}_i^{\vartheta \beta }·p_i^{1-\vartheta },$$

we can apply Corollary 1 (a) with $x_i:=p_i^{\vartheta \beta }$, $y_i:=p_i^{1-\vartheta }$, ${\alpha }_i:=u_i$, ${\beta }_i:=v_i$, $p:=1/\vartheta$ and $q:=1/\left(1-\vartheta \right)$, and obtain

$${\displaystyle {\sum }_{i=1}^m}{p}_i^{\alpha }\le {\left({\displaystyle {\sum }_{i=1}^m}{p}_i^{\beta }{u}_i\right)}^{\vartheta }{\left({\displaystyle {\sum }_{i=1}^m}{p}_i{u}_i\right)}^{1-\vartheta }$$

$$+{\left({\displaystyle {\sum }_{i=1}^m}{p}_i^{\beta }{v}_i\right)}^{\vartheta }{\left({\displaystyle {\sum }_{i=1}^m}{p}_i{v}_i\right)}^{1-\vartheta }\le {\left({\displaystyle {\sum }_{i=1}^m}{p}_i^{\beta }\right)}^{\vartheta }.$$

Taking the logarithm of the expressions, we get

$${\displaystyle \frac{1}{1-\alpha }}\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i^{\alpha }\right)$$

$$\ge {\displaystyle \frac{1}{1-\alpha }}\log \left({\exp }_a\left(\vartheta \log \left({\displaystyle {\sum }_{i=1}^m}{\displaystyle \frac{p_i^{\beta }{u}_i}{{\displaystyle {\sum }_{j=1}^m}{p}_j{u}_j}}\right)+\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i{u}_i\right)\right)\right.$$

$$\left.+{\exp }_a\left(\vartheta \log \left({\displaystyle {\sum }_{i=1}^m}{\displaystyle \frac{p_i^{\beta }{v}_i}{{\displaystyle {\sum }_{j=1}^m}{p}_j{v}_j}}\right)+\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i{v}_i\right)\right)\right)$$

$$\ge {\displaystyle \frac{1}{1-\beta }}\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i^{\beta }\right).$$

This implies that

$$H_{\alpha }\left(P\right)\ge {\displaystyle \frac{1}{1-\alpha }}\log \left({\exp }_a\left(\left(1-\alpha \right)H_{\beta }\left(P,U\right)+\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i{u}_i\right)\right)\right.$$

$$\left.+{\exp }_a\left(\left(1-\alpha \right)H_{\beta }\left(P,V\right)+\log \left({\displaystyle {\sum }_{i=1}^m}{p}_i{v}_i\right)\right)\right)\ge H_{\beta }\left(P\right),$$

and the result follows.

If $0<\alpha <1<\beta$, then $\vartheta <0$, and in this case, we can copy the proof of the first part by applying Corollary 1 (b) instead of Corollary 1 (a).

(b) Let $\vartheta :={\displaystyle \frac{1-\beta }{1-\alpha }}$. Then, $0<\vartheta <1$. Since

$${\displaystyle {\sum }_{i=1}^m}{p}_i^{\beta }={\displaystyle {\sum }_{i=1}^m}{p}_i^{\vartheta \alpha }·p_i^{1-\vartheta },$$

we can apply Corollary 1 (a) with $x_i:=p_i^{\vartheta \alpha }$, $y_i:=p_i^{1-\vartheta }$, ${\alpha }_i:=u_i$, ${\beta }_i:=v_i$, $p:=1/\vartheta$ and $q:=1/\left(1-\vartheta \right)$. The proof follows a similar procedure to that in the first part, we omit the details.

The proof is complete. □

Remark 10.

(a) The preceding result is not merely a refinement of an important information-theoretic inequality, but also clearly demonstrates the applicability of the refinements presented in this paper: on the one hand, the generalized Hölder inequality is employed, and on the other hand, the concept of “useful” information emerges naturally within this refinement.

(b) Hölder’s inequality and its refinements are often used to obtain inequalities in information theory (see, e.g., [23]).

Our final result is a refinement of the Cauchy–Bunyakovsky–Schwarz inequality that, on the one hand, utilizes the refinements obtained for Hölder’s inequality and, on the other hand, derives the idea behind its formulation from our results.

Theorem 13.

Let H be a linear space over $\mathbb{R}$ or $\mathbb{C}$ with inner products ${〈·,·〉}_i$ $\left(i=1,\dots ,m\right)$, and let $〈·,·〉:={\displaystyle {\sum }_{i=1}^m}{〈·,·〉}_i$. If $\left({\alpha }_1,\dots ,{\alpha }_m\right)$ and $\left({\beta }_1,\dots ,{\beta }_m\right)$ are two nonnegative m-tuples such that ${\alpha }_i+{\beta }_i=1$ $\left(i=1,\dots ,m\right)$, then

$$\left|〈x,y〉\right|\le {\left({\displaystyle {\sum }_{i=1}^m}{\alpha }_i{〈x,x〉}_i\right)}^{1/2}{\left({\displaystyle {\sum }_{i=1}^m}{\alpha }_i{〈y,y〉}_i\right)}^{1/2}$$

$$+{\left({\displaystyle {\sum }_{i=1}^m}{\beta }_i{〈x,x〉}_i\right)}^{1/2}{\left({\displaystyle {\sum }_{i=1}^m}{\beta }_i{〈y,y〉}_i\right)}^{1/2}$$

$$\le \sqrt{〈x,x〉}\sqrt{〈y,y〉},x,y\in H.$$

Proof. 

It is obvious that $〈·,·〉$ is also an inner product on H.

Applying the Cauchy–Bunyakovsky–Schwarz inequality first, then Corollary 1 (a), we obtain

$$\left|〈x,y〉\right|=\left|{\displaystyle {\sum }_{i=1}^m}{〈x,y〉}_i\right|\le {\displaystyle {\sum }_{i=1}^m}\left|{〈x,y〉}_i\right|\le {\displaystyle {\sum }_{i=1}^m}\sqrt{{〈x,x〉}_i}\sqrt{{〈y,y〉}_i}$$

$$\le {\left({\displaystyle {\sum }_{i=1}^m}{\alpha }_i{〈x,x〉}_i\right)}^{1/2}{\left({\displaystyle {\sum }_{i=1}^m}{\alpha }_i{〈y,y〉}_i\right)}^{1/2}$$

$$+{\left({\displaystyle {\sum }_{i=1}^m}{\beta }_i{〈x,x〉}_i\right)}^{1/2}{\left({\displaystyle {\sum }_{i=1}^m}{\beta }_i{〈y,y〉}_i\right)}^{1/2}$$

$$\le {\left({\displaystyle {\sum }_{i=1}^m}{〈x,x〉}_i\right)}^{1/2}{\left({\displaystyle {\sum }_{i=1}^m}{〈y,y〉}_i\right)}^{1/2}=\sqrt{〈x,x〉}\sqrt{〈y,y〉}.$$

The proof is complete. □

Remark 11.

The previous result includes part (a) of Theorem 6 for the case $p=q=2$.