A generalized form of Grüss type inequality and other integral inequalities

Abstract

The aim of this presentation is to show several integral inequalities. Among these inequalities we have the inequality ${var}_h(f)\le ({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)$, where ${var}_h(f)$ denotes the h-variance of f, which is a bounded function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, and ${\gamma }_1$, ${\mathrm{\Gamma }}_1$ are two constants. This inequality is important because it proves a generalized form of the Grüss type inequality. This improvement is given by the inequality

$$0\le \frac{{[{cov}_h(f,g){cov}_h(f,q)-{cov}_h(g,q){var}_h(f)]}^2}{{var}_h(f){var}_h(q)-{[{cov}_h(f,g)]}^2}\le {var}_h(f){var}_h(g)-{[{cov}_h(f,g)]}^2.$$

Using the integral arithmetic mean and h-integral arithmetic mean for a Riemann-integrable function f we can also rewrite several integral inequalities. In addition, we will give a generalization of inequality of Grüss for normalized isotonic linear functionals.

MSC:26D15, 26D10.

1 Introduction

In 1935, Grüss [1] proved the following integral inequality which gives an approximation for the integral of a product of two functions in terms of the product of integrals of the two functions:

Let f and g be two bounded functions defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$ and ${\gamma }_2\le g(x)\le {\mathrm{\Gamma }}_2$ , where ${\gamma }_1$ , $y_2$ , ${\mathrm{\Gamma }}_1$ , ${\mathrm{\Gamma }}_2$ are four constants. Then we have

$$|\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx-\frac{1}{b-a}{\int }_a^{b}f(x)dx\frac{1}{b-a}{\int }_a^{b}g(x)dx|\le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2)$$

and the inequality is sharp, in the sense that the constant $1/4$ cannot be replaced by a smaller one.

It is well known that an important resource for studying inequalities is [2–4]. In [5], Peng and Miao established a form of inequality of Grüss type for functions whose first and second derivatives are absolutely continuous and the third derivative is bound. Also, in [6] Dragomir presented several integral inequalities of Grüss type, and in [7] he showed some Grüss type inequalities in inner product spaces and applications for the integral. Another improvement of the Grüss inequality was obtained by Mercer in [8]. Moreover, in [9], a Grüss type inequality was used in order to obtain some sharp Ostrowski-Grüss type inequalities by Liu. Kechriniotis and Delibasis showed in [10] several refinements of inequality of Grüss in inner product spaces using a Kurepa’s results for Gramians. New generalizations of the inequality of Gruss were presented in [11] using Riemann-Liouville fractional integrals. Cerone and Dragomir studied in [12] some refinements of Grüss’ inequality. Florea and Niculescu in [13] treated the problem of estimating the deviation of the values of a function from its mean value.

The estimation of the deviation of a function from its mean value is characterized in terms of random variables.

We denote by $\mathcal{R}([a,b])$ the space of Riemann-integrable functions on the interval $[a,b]$, and by $C^0([a,b])$ the space of real-valued continuous functions on the interval $[a,b]$.

The integral arithmetic mean for a Riemann-integrable function $f:[a,b]\to \mathbf{R}$ is the number

$$M_1[f]=\frac{1}{b-a}{\int }_a^{b}f(x)dx.$$

If $f,h\in \mathcal{R}([a,b])$ and ${\int }_a^{b}h(x)dx>0$, then a generalization for the integral arithmetic mean is the number

$$M_h[f]=\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}$$

called the h-integral arithmetic mean for a Riemann-integrable function f.

We find the following property of the h-integral arithmetic mean for a Riemann-integrable function f:

$$M_h[f\pm k]=M_h[f]\pm k,$$

where k is a real constant.

If the function f is a Riemann-integrable function, we denote by

$$var(f)=M_1\left[{(f-M_1(f))}^2\right]$$

the variance of f.

The expression for the variance can be expanded thus:

$$var(f)=\frac{1}{b-a}{\int }_a^b{(f(x)-\frac{1}{b-a}{\int }_a^{b}f(t)dt)}^{2}dx.$$

In the same way we defined the h-variance of a Riemann-integrable function f by

$${var}_h(f)=M_h\left[{(f-M_h(f))}^2\right].$$

The expression for the h-variance can be expanded thus:

$${var}_h(f)=\frac{1}{{\int }_a^{b}h(x)dx}{\int }_a^b{(f(x)-\frac{{\int }_a^{b}f(t)h(t)dt}{{\int }_a^{b}h(t)dt})}^{2}h(x)dx.$$

It is easy to see another form of the h-variance, given by the following:

$${var}_h(f)=M_h\left[f^2\right]-M_h^2[f].$$

We note the following property of the h-variance of an integrable function f:

$${var}_h(f\pm k)={var}_h(f),$$

where k is a constant.

In [14], Aldaz showed a refinement of the AM-GM inequality and used in the proof that

$$\frac{1-{\int }_a^b{f}^{\frac{1}{2}}(x)dx}{{({\int }_a^{b}f(x)dx)}^{\frac{1}{2}}}$$

is a measure of the dispersion of $f^{\frac{1}{2}}$ about its mean value, which is, in fact, comparable to the variance,

$$Var\left(\frac{f^{\frac{1}{2}}(x)}{{\parallel f^{\frac{1}{2}}(x)\parallel }_2}\right),$$

where ${\parallel f(x)\parallel }_2=\sqrt{{\int }_a^b{f}^2(x)dx}$.

The covariance is a measure of how much two Riemann-integrable functions change together and is defined as

$$cov(f,g)=M_1\left[(f-M_1[f])(g-M_1[g])\right],$$

and it is equivalent to the form

$$\begin{array}{rl}cov(f,g)& =M_1[fg]-M_1[f]M_1[g]\\ =\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx-\frac{1}{b-a}{\int }_a^{b}f(x)dx\frac{1}{b-a}{\int }_a^{b}g(x)dx.\end{array}$$

In fact the covariance is the Chebyshev functional attached to functions f and g. In [15] is written as $T(f,g)$. The properties of the Chebyshev functional have been studied by Elezović, Marangunić and Pec̆arić in their paper, [16]. For other generalizations of the Grüss’ inequality, see [17, 18].

The h-covariance is a measure of how much two integrable functions change together and is defined as

$${cov}_h(f,g)=M_h\left[(f-M_h[f])(g-M_h[g])\right],$$

and it is equivalent to the form

$$\begin{array}{rl}{cov}_h(f,g)& =M_h[fg]-M_h[f]M_h[g]\\ =\frac{{\int }_a^{b}f(x)g(x)h(x)dx}{{\int }_a^{b}h(x)dx}-\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}\cdot \frac{{\int }_a^{b}g(x)h(x)dx}{{\int }_a^{b}h(x)dx}.\end{array}$$

In [18], Pečarić used the generalization of the Chebyshev functional notion attached of functions f and g to the Chebyshev h-functional attached of functions f and g defined by $T(f,g;h)$. Here, Pečarić showed some generalizations of the inequality of Grüss by the Chebyshev h-functional. It is easy to see that, in terms of the covariance, this can be written as $T(f,g;h)={cov}_h(f,g)$.

In terms of covariance the inequality of Grüss becomes

$$|cov(f,g)|\le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2),$$

(1)

and in terms of Chebyshev functional the inequality of Grüss becomes

$$|T(f,g)|\le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2).$$

(2)

Let $x_1,x_2,\dots ,x_n$ be real numbers, assume $m\le x_i\le M$ for all $i=\overline{1,n}$ and the average $\overline{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i$, and X a discrete random variable given by

$$X={\left(\begin{array}{c}{x}_i\\ \frac{1}{n}\end{array}\right)}_{i=\overline{1,n}}.$$

In 1935, Popoviciu [19] proved the following inequality:

$$Var(X)=\frac{1}{n}\sum _{i=1}^n{(x_i-\overline{x})}^2\le \frac{1}{4}{(M-m)}^2.$$

(3)

Bhatia and Davis showed in [20] that the following inequality holds:

$$Var(X)=\frac{1}{n}\sum _{i=1}^n{(x_i-\overline{x})}^2\le (M-\overline{x})(\overline{x}-m).$$

(4)

The inequality of Bhatia and Davis represents an improvement of Popoviciu’s inequality, because ${(M-m)}^2\ge 4(M-\overline{x})(\overline{x}-m)$.

If there is additional information about the mean values of the two functions in the inequality of Grüss then Zitikis argued in his paper, [21], that the inequality can be sharpened and he gave also a probabilistic interpretation for it.

2 Main results

We will present in this paper the integral version of inequalities (3) and (4). Therefore we have the following inequalities.

Lemma 1 Let f be a Riemann-integrable function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, where ${\gamma }_1$ and ${\mathrm{\Gamma }}_1$ are two constants. Then we have

$${var}_h(f)\le \frac{1}{4}{({\mathrm{\Gamma }}_1-{\gamma }_1)}^2,$$

(5)

where $h:[a,b]\to [0,\mathrm{\infty })$ is a Riemann-integrable function with ${\int }_a^{b}h(x)dx>0$.

Proof Since ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, we obtain the following inequality:

$$|f(x)-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2}|\le \frac{{\mathrm{\Gamma }}_1-{\gamma }_1}{2}.$$

But it is easy to see that

$$\begin{array}{rl}{var}_h(f)& ={var}_h(f-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2})\\ =M_h\left[{(f-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2})}^2\right]-M_h^2[f-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2}]\\ \le M_h\left[{(f-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2})}^2\right]=\frac{1}{{\int }_a^{b}h(t)dt}{\int }_a^b{(f-\frac{{\mathrm{\Gamma }}_1+{\gamma }_1}{2})}^{2}h(t)dt\\ \le \frac{1}{{\int }_a^{b}h(t)dt}{\int }_a^b{\left(\frac{{\mathrm{\Gamma }}_1-{\gamma }_1}{2}\right)}^{2}h(t)dt=\frac{{({\mathrm{\Gamma }}_1-{\gamma }_1)}^2}{4}.\end{array}$$

 □

Lemma 2 Let f be a Riemann-integrable function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, where ${\gamma }_1$ and ${\mathrm{\Gamma }}_1$ are two constants and a Riemann-integrable function $h:[a,b]\to [0,\mathrm{\infty })$ with ${\int }_a^{b}h(x)dx>0$. Then we have the following relations:

$${var}_h(f)\le ({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)$$

or

$${var}_h(f)\le ({\mathrm{\Gamma }}_1-\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx})(\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}-{\gamma }_1).$$

(6)

Proof It is easy to see that ${var}_h(f)=\frac{1}{{\int }_a^{b}h(x)dx}{\int }_a^b{(f(x)-\frac{{\int }_a^{b}f(t)h(t)dt}{{\int }_a^{b}h(t)dt})}^{2}h(x)dx$ can be rewritten thus:

$${var}_h(f)=\frac{{\int }_a^b{f}^2(x)h(x)dx}{{\int }_a^{b}h(x)dx}-{\left[\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}\right]}^2.$$

Next, using the idea of Dragomir [6], we will make several calculations, namely

$$\begin{array}{r}\frac{{\int }_a^b({\mathrm{\Gamma }}_1-f(x))(f(x)-{\gamma }_1)h(x)dx}{{\int }_a^{b}h(x)dx}\\ =-{\mathrm{\Gamma }}_1{\gamma }_1+({\mathrm{\Gamma }}_1+{\gamma }_1)\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}-\frac{{\int }_a^b{f}^2(x)h(x)dx}{{\int }_a^{b}h(x)dx}\\ =-{\mathrm{\Gamma }}_1{\gamma }_1+({\mathrm{\Gamma }}_1+{\gamma }_1)\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}-{var}_h(f)-{\left[\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}\right]}^2\\ =({\mathrm{\Gamma }}_1-\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx})(\frac{{\int }_a^{b}f(x)h(x)dx}{{\int }_a^{b}h(x)dx}-{\gamma }_1)-{var}_h(f).\end{array}$$

Since f is a bounded function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, and the function $h:[a,b]\to [0,\mathrm{\infty })$, it follows that $({\mathrm{\Gamma }}_1-f(x))(f(x)-{\gamma }_1)\ge 0$. Therefore, it is easy to see that $\frac{{\int }_a^b({\mathrm{\Gamma }}_1-f(x))(f(x)-{\gamma }_1)h(x)dx}{{\int }_a^{b}h(x)dx}\ge 0$. This inequality proved the inequality of the statement. □

Next we show several relations between h-variance and h-covariance.

Lemma 3 If $f,g\in \mathcal{R}([a,b])$, and a Riemann-integrable function $h:[a,b]\to [0,\mathrm{\infty })$ with ${\int }_a^{b}h(x)dx>0$, then we have the following equality:

$${var}_h(af+bg)=a^2{var}_h(f)+b^2{var}_h(g)+2ab{cov}_h(f,g),$$

(7)

where a and b are real numbers.

Proof From the expression of the variance, we have

$$\begin{array}{rl}{var}_h(af+bg)=& M_h\left[{(af+bg)}^2\right]-M_h^2[af+bg]\\ =& M_h[a^2{f}^2+2abfg+b^2{g}^2]-{(a{M}_h[f]+b{M}_h[g])}^2\\ =& a^2{M}_h\left[f^2\right]+2ab{M}_h[fg]+b^2{M}_h\left[g^2\right]-a^2{M}_h^2[f]\\ -2ab{M}_h[f]M_h[g]-b^2{M}_h^2[g]\\ =& a^2{var}_h(f)+b^2{var}_h(g)+2ab{cov}_h(f,g).\end{array}$$

Therefore, we deduce relationship (1). □

Proposition 4 If $f,g\in \mathcal{R}([a,b])$, then we have the following equality:

$${var}_h(f+g)={var}_h(f)+{var}_h(g)+2cov(f,g)$$

(8)

and

$${var}_h(f-g)={var}_h(f)+{var}_h(g)-2cov(f,g).$$

(9)

Proof If we take $a=b=1$ and $a=1$, $b=-1$ in relation (7), then we obtain equalities (8) and (9). □

Remark 1 From relations (8) and (9), we find the parallelogram law in terms of h-variance, namely

$${var}_h(f+g)+{var}_h(f-g)=2{var}_h(f)+2{var}_h(g).$$

(10)

Lemma 5 If $f,g,p,q\in \mathcal{R}([a,b])$, then we have the following equality:

$$\begin{array}{r}{cov}_h(af+bg,cp+dq)\\ =ac{cov}_h(f,p)+ad{cov}_h(f,q)+bc{cov}_h(g,p)+bd{cov}_h(g,q)\end{array}$$

(11)

where a, b, c, and d are real numbers.

Proof From the expression of the covariance, we have

$$\begin{array}{rl}{cov}_h(af+bg,cp+dq)=& M_h[(af+bg)(cp+dq)]-M_h[af+bg]M_h[cp+dq]\\ =& M_h[acfp+adfq+bcgp+bdgq]\\ -(a{M}_h[f]+b{M}_h[g])(c{M}_h[p]+d{M}_h[q])\\ =& ac(M_h[fp]-M_h[f]M_h[p])+ad(M_h[fq]-M_h[f]M_h[q])\\ +bc(M_h[gp]-M_h[g]M_h[p])+bd(M_h[gq]-M_h[g]M_h[q])\\ =& ac{cov}_h(f,p)+ad{cov}_h(f,q)+bc{cov}_h(g,p)+bd{cov}_h(g,q).\end{array}$$

 □

We can prove an inequality for integrable functions similar to the inequality of Cauchy-Schwarz for random variables given by the following.

Theorem 6 If $f,g,h\in \mathcal{R}([a,b])$, then we have the inequality

$$|{cov}_h(f,g){|}^2\le {var}_h(f){var}_h(g)\mathit{\text{or}}|{cov}_h(f,g)|\le \sqrt{{var}_h(f){var}_h(g)}.$$

(12)

Proof If ${var}_h(g)=0$, then relation (12) is true. If ${var}_h(g)\ne 0$, then we calculate the h-variance for the function:

$$w=f-\frac{{cov}_h(f,g)}{{var}_h(g)}g\text{thus}{var}_h(w)={var}_h(f)-\frac{{cov}_h^2(f,g)}{{var}_h(g)}\ge 0,$$

because ${var}_h(w)\ge 0$. □

Proposition 7 Let f and g be two Riemann-integrable functions defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$ and ${\gamma }_2\le g(x)\le {\mathrm{\Gamma }}_2$, where ${\gamma }_1$, ${\gamma }_2$, ${\mathrm{\Gamma }}_1$, ${\mathrm{\Gamma }}_2$ are four constants, and we have a Riemann-integrable h function, $h:[a,b]\to [0,\mathrm{\infty })$ with ${\int }_a^{b}h(x)dx>0$. Then we have

$$\begin{array}{rl}|{cov}_h(f,g)|& =|T(f,g;h)|\\ \le \sqrt{({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)({\mathrm{\Gamma }}_2-M_h[g])(M_h[g]-{\gamma }_2)}\\ \le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2).\end{array}$$

(13)

Proof Using Theorem 6 and Lemma 2, we deduce the statement. □

Remark 2 Inequality (13) is a refinement of inequality of Grüss, because

$$\begin{array}{rl}|{cov}_h(f,g)|& =|T(f,g;h)|=|M_h[fg]-M_h[f]M_h[g]|\\ \le \sqrt{({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)({\mathrm{\Gamma }}_2-M_h[g])(M_h[g]-{\gamma }_2)}\\ \le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2).\end{array}$$

Remark 3 In the proof of Lemma 2 we found the equality in terms of random variables, given by

$$cov({\mathrm{\Gamma }}_1-f,f-{\gamma }_1)=({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)-{var}_h(f).$$

(14)

Remark 4 (a) Let f be a Riemann-integrable function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, where ${\gamma }_1$, ${\mathrm{\Gamma }}_1$ are two constants. Then we have

$$var(f)\le \frac{1}{4}{({\mathrm{\Gamma }}_1-{\gamma }_1)}^2.$$

(15)

(b) Let f be a Riemann-integrable function defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$, where ${\gamma }_1$, ${\mathrm{\Gamma }}_1$ are two constants. Then we have

$$var(f)\le ({\mathrm{\Gamma }}_1-M_h[f])(M_h[f]-{\gamma }_1)$$

or

$$var(f)\le ({\mathrm{\Gamma }}_1-\frac{1}{b-a}{\int }_a^{b}f(x)dx)(\frac{1}{b-a}{\int }_a^{b}f(x)dx-{\gamma }_1).$$

(16)

(c) Let f and g be two Riemann-integrable functions defined on $[a,b]$ with ${\gamma }_1\le f(x)\le {\mathrm{\Gamma }}_1$ and ${\gamma }_2\le g(x)\le {\mathrm{\Gamma }}_2$ where ${\gamma }_1$, ${\gamma }_2$, ${\mathrm{\Gamma }}_1$, ${\mathrm{\Gamma }}_2$ are four constants. Then we have

$$\begin{array}{r}|\frac{1}{b-a}{\int }_a^{b}f(x)g(x)dx-\frac{1}{b-a}{\int }_a^{b}f(x)dx\frac{1}{b-a}{\int }_a^{b}g(x)dx|\\ \le \sqrt{({\mathrm{\Gamma }}_1-\frac{1}{b-a}{\int }_a^{b}f(x)dx)(\frac{1}{b-a}{\int }_a^{b}f(x)dx-{\gamma }_1)}\\ \times \sqrt{({\mathrm{\Gamma }}_2-\frac{1}{b-a}{\int }_a^{b}g(x)dx)(\frac{1}{b-a}{\int }_a^{b}g(x)dx-{\gamma }_2)}\\ \le \frac{1}{4}({\mathrm{\Gamma }}_1-{\gamma }_1)({\mathrm{\Gamma }}_2-{\gamma }_2).\end{array}$$

(17)

Theorem 8 If $f,g,q\in \mathcal{R}([a,b])$, with $f\ne kq$ and $var(f)\ne 0$, then we have the inequality

$$0\le \frac{{[{cov}_h(f,g){cov}_h(f,q)-{cov}_h(g,q){var}_h(f)]}^2}{{var}_h(f){var}_h(q)-{[{cov}_h(f,g)]}^2}\le {var}_h(f){var}_h(g)-{[{cov}_h(f,g)]}^2.$$

(18)

Proof For the integrable functions f, g and q, with $var(f)\ne 0$, we take the following integrable function:

$$w=\frac{{cov}_h(f,g)+\lambda {cov}_h(f,q)}{{var}_h(f)}f-g-\lambda q.$$

We calculate the variance of the function w, thus:

$${var}_h(w)={var}_h((\frac{{cov}_h(f,g)}{{var}_h(f)}f-g)+\lambda (\frac{{cov}_h(f,q)}{{var}_h(f)}f-q))$$

and applying Lemma 3, we have

$$\begin{array}{rl}{var}_h(w)=& {var}_h(\frac{{cov}_h(f,g)}{{var}_h(f)}f-g)+{\lambda }^2{var}_h(\frac{{cov}_h(f,q)}{{var}_h(f)}f-q)\\ +2\lambda {cov}_h(\frac{{cov}_h(f,g)}{{var}_h(f)}f-g,\frac{{cov}_h(f,q)}{{var}_h(f)}f-q)\\ =& {var}_h(g)-\frac{{[{cov}_h(f,g)]}^2}{{var}_h(f)}+{\lambda }^2({var}_h(q)-\frac{{[{cov}_h(f,q)]}^2}{{var}_h(f)})\\ +2\lambda {cov}_h(\frac{{cov}_h(f,g)}{{var}_h(f)}f-g,\frac{{cov}_h(f,q)}{{var}_h(f)}f-q).\end{array}$$

Using Lemma 2, we deduce the following inequality:

$$\begin{array}{r}{cov}_h(\frac{{cov}_h(f,g)}{{var}_h(f)}f-g,\frac{{cov}_h(f,q)}{{var}_h(f)}f-q)\\ =\frac{{cov}_h(f,g){cov}_h(f,q)}{{var}_h(f){var}_h(f)}{cov}_h(f,f)\\ -\frac{{cov}_h(f,g){cov}_h(f,q)}{{var}_h(f)}-\frac{{cov}_h(f,q){cov}_h(f,g)}{{var}_h(f)}+{cov}_h(g,q)\\ ={cov}_h(g,q)-\frac{{cov}_h(f,g){cov}_h(f,q)}{{var}_h(f)}.\end{array}$$

Returning to calculation of the function w, we have

$$\begin{array}{rl}{var}_h(w)=& {var}_h(g)-\frac{{[{cov}_h(f,g)]}^2}{{var}_h(f)}+{\lambda }^2({var}_h(q)-\frac{{[{cov}_h(f,q)]}^2}{{var}_h(f)})\\ +2\lambda ({cov}_h(g,q)-\frac{{cov}_h(f,g){cov}_h(f,q)}{{var}_h(f)}).\end{array}$$

Therefore, we deduce the equality

$$\begin{array}{rl}{var}_h(f){var}_h(w)=& {var}_h(f){var}_h(g)-{[{cov}_h(f,g)]}^2+{\lambda }^2({var}_h(f){var}_h(q)-{[{cov}_h(f,q)]}^2)\\ +2\lambda ({var}_h(f){cov}_h(g,q)-{cov}_h(f,g){cov}_h(f,q)).\end{array}$$

Since ${var}_h(f){var}_h(w)\ge 0$, it follows that

$$\begin{array}{r}{\lambda }^2({var}_h(f){var}_h(q)-{[{cov}_h(f,q)]}^2)+2\lambda ({var}_h(f){cov}_h(g,q)-{cov}_h(f,g){cov}_h(f,q))\\ +{var}_h(f){var}_h(g)-{[{cov}_h(f,g)]}^2\ge 0\end{array}$$

for every $\lambda \in \mathbf{R}$.

This implies that

$$\begin{array}{r}({var}_h(f){var}_h(q)-{[{cov}_h(f,q)]}^2)({var}_h(f){var}_h(g)-{[{cov}_h(f,g)]}^2)\\ \ge {({var}_h(f){cov}_h(g,q)-{cov}_h(f,g){cov}_h(f,q))}^2.\end{array}$$

(19)

Taking into account that ${var}_h(f){var}_h(q)-{[{cov}_h(f,q)]}^2\ne 0$, because $f\ne kq$ and dividing by ${var}_h(f){var}_h(q)-{[{cov}_h(f,q)]}^2$, we obtain the inequality of the statement. □

Remark 5 Let f, g and q be three integrable functions, with ${var}_h(g)\ne 0$ and ${var}_h(q)\ne 0$. If we take the following function:

$$w=f-\frac{{cov}_h(f,g)}{{var}_h(g)}g-\lambda q,$$

then we have the inequality

$$\begin{array}{rl}0& \le \frac{{[{cov}_h(f,q){cov}_h(g,q)-{cov}_h(f,q){var}_h(g)]}^2}{{var}_h(g){var}_h(q)}\\ \le {var}_h(f){var}_h(g)-|{cov}_h(f,g){|}^2.\end{array}$$

(20)

Lemma 9 Let f, g be two Riemann-integrable functions defined on $[a,b]$. Then we have

$$M_h\left[f^2\right]M_h\left[g^2\right]\ge M_h^2[fg].$$

(21)

Proof If $M_h[f^2]=0$, then relation (21) is true. Now we consider that $M_h[f^2]\ne 0$.

$$M_h\left[{(\lambda f-g)}^2\right]\ge 0,$$

for all $\lambda \in \mathbf{R}$, so, we have

$$M_h\left[{(\lambda f-g)}^2\right]={\lambda }^2{M}_h\left[f^2\right]-2\lambda M_h[fg]+M_h\left[g^2\right],$$

which means that

$${\lambda }^2{M}_h\left[f^2\right]-2\lambda M_h[fg]+M_h\left[f^2\right]\ge 0,$$

for all $\lambda \in \mathbf{R}$, which implies

$$M_h\left[f^2\right]M_h\left[g^2\right]\ge M_h^2[fg].$$

Lemma 9 represents generalized integral variant of inequality of Cauchy. □

Now we compare inequalities (21) and (12) to see which is stronger.

Theorem 10 Let f, g be two Riemann-integrable functions defined on $[a,b]$. Then we have

$$M_h\left[f^2\right]M_h\left[g^2\right]-M_h^2[fg]\ge [{var}_h(f){var}_h(g)-{cov}_h^2(f,g)]\ge 0.$$

(22)

Proof We calculate the difference of the terms which appear in inequality (12) and (21), thus:

$$\begin{array}{r}{M}_h\left[f^2\right]M_h\left[g^2\right]-M_h^2[fg]-[{var}_h(f){var}_h(g)-{cov}_h^2(f,g)]\\ =M_h\left[f^2\right]M_h^2[g]-2M_h[fg]M_h[f]M_h[g]+M_h\left[g^2\right]M_h^2[f].\end{array}$$

But, applying the inequality between the arithmetic mean and the geometric mean and Lemma 9, we deduce the relation

$$\begin{array}{rl}{M}_h\left[f^2\right]M_h^2[g]+M_h\left[g^2\right]M_h^2[f]& \ge 2\sqrt{M_h\left[f^2\right]M_h\left[g^2\right]}{M}_h[f]M_h[g]\\ \ge 2M_h[fg]M_h[f]M_h[g].\end{array}$$

From this, we obtain the inequality

$$M_h\left[f^2\right]M_h^2[g]-2M_h[fg]M_h[f]M_h[g]+M_h\left[g^2\right]M_h^2[f]\ge 0.$$

Consequently the statement is true. □

3 A refinement of Grüss’ inequality for normalized isotonic linear functionals

There are many directions in which the inequality of Grüss [1] has been generalized.

Using the notion of normalized isotonic linear functional which appears in the paper [22], we will give a generalization of inequality of Grüss which is related to a theorem of Andrica and Badea (1988), [23].

Let E be a nonempty set, L a linear class of real-valued functions and $g:E\to \mathbf{R}$ having the properties:

• (L1) $f,g?L$ imply $(af+ßg)?L$ for all $a,ß?\mathbf{R}$,

• (L2) $\mathbf{1}?L$, i.e. if $f_0(t)=1$, $(\mathrm{?})t?E$, then $f_0?L$.

An isotonic linear functional (in [24] is called positive definite functional) $A:L\to \mathbf{R}$ is a functional satisfying:

• (A1) $A(af+ßg)=aA(f)+ßA(g)$, for all $f,g?L$ and $a,ß?\mathbf{R}$.

• (A2) If $f?L$ and $f=0$, then $A(f)=0$.

• (A3) The mapping A is said to be normalized if $A(\mathbf{1})=1$.

Theorem 11 Let $f\in L$ be such that $f^2\in L$ and assume that there exist real numbers ${\gamma }_1$ and ${\mathrm{\Gamma }}_1$ so that ${\gamma }_1\le f\le {\mathrm{\Gamma }}_1$.

Then for any normalized isotonic linear functional $A:L\to \mathbf{R}$ one has the inequality

$$A\left(f^2\right)-{[A(f)]}^2\le ({\mathrm{\Gamma }}_1-A(f))(A(f)-{\gamma }_1).$$

(23)

Proof Taking into account the hypothesis we have by using (L1) and (L2) that $(f-{\gamma }_1\cdot \mathbf{1})({\mathrm{\Gamma }}_1\cdot \mathbf{1}-f)\ge 0$ and thus by (A2), we see that

$$A[(f-{\gamma }_1\cdot \mathbf{1})({\mathrm{\Gamma }}_1\cdot \mathbf{1}-f)]\ge 0.$$

Using also (A1) and (A3) we have,

$$A[f{\mathrm{\Gamma }}_1\cdot \mathbf{1}-{\gamma }_1{\mathrm{\Gamma }}_1\cdot \mathbf{1}-f^2+{\gamma }_{1}f\cdot \mathbf{1}]\ge 0$$

or

$$A(f){\mathrm{\Gamma }}_1-{\gamma }_1{\mathrm{\Gamma }}_1\cdot A(\mathbf{1})-A\left(f^2\right)+{\gamma }_{1}A(f)\ge 0$$

or

$$({\mathrm{\Gamma }}_1+{\gamma }_1)A(f)-{\gamma }_1{\mathrm{\Gamma }}_1\cdot A(\mathbf{1})\ge A\left(f^2\right).$$

Therefore

$$({\mathrm{\Gamma }}_1+{\gamma }_1)A(f)-{\gamma }_1{\mathrm{\Gamma }}_1\cdot A(\mathbf{1})-{[A(f)]}^2\ge A\left(f^2\right)-{[A(f)]}^2$$

and from this we obtain

$$({\mathrm{\Gamma }}_1-A(f))(A(f)-{\gamma }_1)\ge A\left(f^2\right)-{[A(f)]}^2.$$

From the inequality of Cauchy-Schwarz for a normalized isotonic linear functional, [22], a counterpart of the CBS inequality, we obtain for $f,g,f^2,g^2\in L$ where $f,g:E\to \mathbf{R}$ and $A:L\to \mathbf{R}$ is any normalized isotonic linear functional

$${[A(fg)]}^2\le A\left(f^2\right)A\left(g^2\right).$$

(24)

Now if we consider this inequality and previous theorem we deduce as in Lemma 2 the following result. □

Theorem 12 Let $f,g,fg\in L$ such that $f^2,g^2\in L$ and ${\gamma }_1\le f\le {\mathrm{\Gamma }}_1$, ${\gamma }_2\le g\le {\mathrm{\Gamma }}_2$ where ${\gamma }_1$, ${\gamma }_2$, ${\mathrm{\Gamma }}_1$, ${\mathrm{\Gamma }}_2$ are given real numbers. Then for any normalized linear isotonic functional $A:L\to \mathbf{R}$ one has the inequality

$$A(fg)-A(f)A(g)\le \sqrt{({\mathrm{\Gamma }}_1-A(f))(A(f)-{\gamma }_1)({\mathrm{\Gamma }}_2-A(g))(A(g)-{\gamma }_2)}.$$

(25)

Proof Because we can write

$$\begin{array}{rl}A\left[(f-A(f)\cdot \mathbf{1})(g-A(g)\cdot \mathbf{1})\right]& =A\left[(f-A(f)\cdot \mathbf{1})g\right]-A[(f-A(f)\cdot \mathbf{1})A(g)]\\ =A(fg)-A(f)A(g)\end{array}$$

we can apply the CBS-inequality for a normalized linear isotonic functional, if ${(f-A(f)\cdot \mathbf{1})}^2,{(g-A(g)\cdot \mathbf{1})}^2,(f-A(f)\cdot \mathbf{1})(g-A(g)\cdot \mathbf{1})\in L$, by Theorem 11 we have

$$\begin{array}{rl}A(fg)-A(f)A(g)& =A\left[(f-A(f)\cdot \mathbf{1})(g-A(g)\cdot \mathbf{1})\right]\\ \le \sqrt{A\left({(f-A(f)\cdot \mathbf{1})}^2\right)A\left({(g-A(g)\cdot \mathbf{1})}^2\right)}\\ \le \sqrt{({\mathrm{\Gamma }}_1-A(f))(A(f)-{\gamma }_1)({\mathrm{\Gamma }}_2-A(g))(A(g)-{\gamma }_2)}.\end{array}$$

 □

Remark 6 If we take in the first theorem $A=\frac{1}{b-a}{\int }_a^b$, $L=L(a,b)$ (the Lebesque space of integrable functions on $[a,b]$) and g satisfying the condition ${\gamma }_2\le g\le {\mathrm{\Gamma }}_2$ on the interval $[a,b]$, then we obtain inequality (17).

Theorem 13 Let $f,g,h\in L$ be such that $h\ge 0$, $fh,gh,fgh\in L$ and there exist real constants ${\gamma }_2$, ${\mathrm{\Gamma }}_2$ so that ${\gamma }_2\le g\le {\mathrm{\Gamma }}_2$. Then for any $B:L\to \mathbf{R}$ an isotonic linear functional so that $B(h)>0$ one has the inequality

$$\begin{array}{r}|\frac{B(fgh)}{B(h)}-\frac{B(fh)}{B(h)}\frac{B(gh)}{B(h)}|\\ \le \sqrt{({\mathrm{\Gamma }}_1-\frac{B(fh)}{B(h)})(\frac{B(fh)}{B(h)}-{\gamma }_1)({\mathrm{\Gamma }}_2-\frac{B(gh)}{B(h)})(\frac{B(gh)}{B(h)}-{\gamma }_2)}.\end{array}$$

(26)

Proof We use the normalized isotonic linear functional A on L defined by

$$A_h(f)=\frac{1}{B(h)}B(hf),$$

$A:L\to \mathbf{R}$. □

4 Properties of h-variance

• 1. If $f,g\in \mathcal{R}([a,b])$, then we have the following inequality:

  $$\sqrt{{var}_h(f+g)}\le \sqrt{{var}_h(f)}+\sqrt{{var}_h(g)}.$$

  (27)

Proof From equality (8), we have

$$\begin{array}{rl}{var}_h(f+g)& ={var}_h(f)+{var}_h(g)+2{cov}_h(f,g)\\ ={(\sqrt{{var}_h(f)}+\sqrt{{var}_h(g)})}^2-2(\sqrt{{var}_h(f){var}_h(g)}-{cov}_h(f,g)).\end{array}$$

Applying the inequality of Cauchy-Schwarz for integrable functions given by

$$|{cov}_h(f,g)|\le \sqrt{{var}_h(f){var}_h(g)},$$

it follows that

$${var}_h(f+g)\le {(\sqrt{{var}_h(f)}+\sqrt{{var}_h(g)})}^2,$$

which implies the inequality of the statement. □

• 2. If $f,g\in \mathcal{R}([a,b])$, then we have the following inequality:

  $$\sqrt{{var}_h(f-g)}\ge |\sqrt{{var}_h(f)}-\sqrt{{var}_h(g)}|.$$

  (28)

Proof From relation (9), we have

$$\begin{array}{rl}{var}_h(f-g)& ={var}_h(f)+{var}_h(g)-2{cov}_h(f,g)\\ ={(\sqrt{{var}_h(f)}-\sqrt{{var}_h(g)})}^2+2(\sqrt{{var}_h(f){var}_h(g)}-{cov}_h(f,g)).\end{array}$$

Applying the inequality of Cauchy-Schwarz for integrable functions, we obtain

$${var}_h(f-g)\ge {(\sqrt{{var}_h(f)}-\sqrt{{var}_h(g)})}^2,$$

which implies the inequality of the statement. □

• 3. The natural way to obtain these quantities is by introducing and using the standard inner product (also known as the dot product) on $C^0([a,b])$. The inner product of any two continuous f and g functions is defined by

  $${〈f,g〉}_h={cov}_h(f,g).$$

The result is always a real number. Therefore the set $C^0([a,b])$ can be organized as an Euclidean space.

The inner product of $C^0([a,b])$ with itself is always non-negative. This product allows us to define the ‘length’ of an integrable function f through

$${\parallel f\parallel }_h=\sqrt{{〈f,f〉}_h}=\sqrt{{cov}_h(f,f)}=\sqrt{{var}_h(f)}.$$

This length function satisfies the required properties of a norm and is called the Euclidean norm on $C^0([a,b])$.

Finally, one can use the norm to define a metric (or distance function) on $C^0([a,b])$ by

$$d_h(f,g)={\parallel f-g\parallel }_h=\sqrt{{var}_h(f-g)}.$$

This distance function is called the Euclidean metric. This formula expresses a special case of the Pythagorean theorem.

From relation (27), we have

$$\sqrt{{var}_h(f-q)}=\sqrt{{var}_h((f-g)+(g-q))}\le \sqrt{{var}_h(f-g)}+\sqrt{{var}_h(g-q)},$$

so we obtain the triangle inequality,

$$d_h(f,q)\le d_h(f,g)+d_h(g,q).$$

Remark 7 From [25], the analog of the arithmetic mean in the context of finite measure spaces $(X,\mathrm{\Sigma };\mu )$ is the integral arithmetic mean, which, for a μ-integrable function $f:X\to \mathbf{R}$ is the number

$$M_1(f;\mu )=\frac{1}{\mu (X)}{\int }_{X}fd\mu ,$$

where $M_1(f;\mu )$ represents in probability theory the conditional expectation of the random variable f.

A result which represents another estimate of Jensen’s inequality, in the sense of generalizations of the integral arithmetic mean and variance, can also be found in [25], page 53.

Lemma 14 ([25], 1.8) (Another estimate of Jensen’s inequality)

Let $(X,\mathrm{\Sigma };\mu )$ be a finite measure space and let $g:X\to \mathbf{R}$ be a μ-integrable function (or $g\in L^2(\mu )$). If f is a twice differentiable function given on an I interval that includes the image of g; and $m\le f^{″}\le M$, then

$$\frac{m}{2}var(g)\le M_1(f\circ g;\mu )-f(M_1(g;\mu ))\le \frac{M}{2}var(g),$$

where $var(g)=M_1({(g-M_1(g;\mu ))}^2;\mu )$ denotes the variance of g.

Izumio et al. showed in [26] some extensions of Grüss’ inequality and they also studied the integral-type Grüss’ inequalities for Lebesque space $L_{\mu }(\mathrm{\Omega })$ for a finite positive measure μ on Ω.

5 Applications

• 1. Taking into account the integral arithmetic mean and h-integral arithmetic mean for a Riemann-integrable function $f:[a,b]\to \mathbf{R}$ we can rewrite the following inequalities.

  • (a) In the case when $p\ge 0$ the integral form of the inequality from Theorem 2.4 (see [27]) was given by Theorem 2.5. Under the conditions of Theorem 2.5, the inequality becomes

    $$M_1\left[\frac{f^{m+1}}{g^p}\right]\ge \frac{M_1^{m+1}[f]}{M_1^p[g]}.$$

  • (b) In [28], Mortici gave a new refinement of Radon’s inequality. Using the integral form of the reverse of inequality from Theorem 2.5 (see [27]) we obtain, for $p\in (-1,0)$, $m\in (-1,0)$ and $m\le p$, if $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ with $g(x)>0$, $(\mathrm{\forall })x\in [a,b]$ a continuous function on $[a,b]$, the inequality

    $$M_1\left[\frac{f^{m+1}}{g^p}\right]\le \frac{M_1^{m+1}[f]}{M_1^p[g]}.$$

  • (c) From the integral form of the inequality from Consequence 1 (see [29]) we deduce if $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions, g a continuous function on $[a,b]$, $g(x)>0$, $(\mathrm{\forall })x\in [a,b]$ and $f(x)\in (0,1)$, $(\mathrm{\forall })x\in [a,b]$ the following inequality:

    $$M_1\left[\frac{f}{g^r(1-f)}\right]\ge \frac{1}{M_1^r[g]}\frac{M_1[f]}{1-M_1[f]},$$

where $r\in [0,1)$.

• 2. (a) Starting from Theorem 2.3, the inequalities (2.5) and (2.6), Theorem 2.7 and Theorem 2.9, given in [30], by using Theorem 12 (see [31]) and Theorem 8 (see [32]) we can obtain the next properties for certain Riemann-integrable functions. If $g(x)>0$ and if $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ with $m\le \frac{f(x)}{g(x)}\le M$, $(\mathrm{\forall })x\in [a,b]$ and $M<1$, then

  $$\begin{array}{rl}0& \le M_1\left[\frac{fg}{g-f}\right]-\frac{M_1^2[g]}{M_1[g]-M_1[f]}\\ \le M_1\left[\frac{f{g}^2}{{(g-f)}^2}\right]-\frac{M_1[f]M_1^2[g]}{M_1^2[g]-M_1^2[f]}-\frac{M_1[f]}{M_1[g]}\cdot (M_1\left[\frac{g^3}{{(g-f)}^2}\right]-\frac{M_1^3[g]}{{(M_1[g]-M_1[f])}^2})\end{array}$$

and

$$\begin{array}{rl}0& \le M_1\left[\frac{fg}{g-f}\right]-\frac{M_1^2[g]}{M_1[g]-M_1[f]}\\ \le \frac{1}{4}(M-m)(\frac{1}{{(1-M)}^2}-\frac{1}{{(1-m)}^2})M_1[g].\end{array}$$

• (b) If $g(x)>0$ and if $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ with $m\le \frac{f(x)}{g(x)}\le M$, $(\mathrm{\forall })x\in [a,b]$ and $M<1$, then we have

  $$0\le M_1\left[\frac{fg}{g-f}\right]-\frac{M_1[f]M_1[g]}{M_1[f-g]}\le (\frac{M}{1-M}+\frac{m}{1-m}-\frac{2(M+m)}{2-M+m})M_1[g]$$

and

$$\begin{array}{rl}0& \le M_1\left[\frac{fg}{g-f}\right]-\frac{M_1[f]M_1[g]}{M_1[f-g]}\\ \le M_1[g]\frac{(M+m)M_1[g]-M_1[f]}{M_1[f+g]-(M+m)M_1[g]}-2\frac{M+m}{2-(M+m)}{M}_1[g]+M_1\left[\frac{fg}{g-f}\right].\end{array}$$

• (c) Using the integral form of the inequality (2.5) and (2.6) from Theorem 2.3 (see [30]), under the conditions of Theorem 5 (see [33]) we find that, for every $n\ge 2$, $p\ge 1$, $g(x)>0$ and if $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two continuous functions on $[a,b]$ with $m={inf}_{[a,b]}\frac{f(x)}{g(x)}$, $M={sup}_{[a,b]}\frac{f(x)}{g(x)}$, we have

  $$0\le M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[g]}\le \frac{p}{4}(M-m)(M^{p-1}-m^{p-1})M_1[g].$$

If $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ then

$$\begin{array}{rl}0& \le M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[g]}\\ \le p(M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[g]}-\frac{M_1[f]}{M_1[g]}(M_1\left[\frac{f^{p-1}}{g^{p-2}}\right]-\frac{M_1^{p-1}[f]}{M_1^{p-2}[g]})).\end{array}$$

• (d) Now we will rewrite the integral form of the inequality (2.19) from Theorem 2.9 (see [30]) which is given in Theorem 6 (see [33]). If $p\ge 1$, f and g are two continuous functions $f,g:[a,b]\to {\mathbf{R}}_{+}$ on $[a,b]$, with $m={inf}_{[a,b]}\frac{f(x)}{g(x)}$, $M={sup}_{[a,b]}\frac{f(x)}{g(x)}$, then we have

  $$\begin{array}{rl}0& \le M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[q]}\\ \le \frac{{[(M+m)M_1[g]-M_1[f]]}^p}{M_1^{p-1}[f]}-\frac{{(M+m)}^p}{2^{p-1}}{M}_1[g]+M_1\left[\frac{f^p}{g^{p-1}}\right].\end{array}$$

• 3. (i) Under the previous conditions we have

  $$0\le M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[q]}\le [M^p+m^p-\frac{{(M+m)}^p}{2^{p-1}}]M_1[g].$$

• (ii) If $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ with $g(x)>0$, $(\mathrm{\forall })x\in [a,b]$, $p>1$ and $mg(x)\le f(x)\le Mg(x)$, $(\mathrm{\forall })x\in [a,b]$, then we have

  $$\begin{array}{rl}p(p-1)m^{p-2}(M_1\left[\frac{f^2}{g}\right]-\frac{M_1^2[f]}{M_1[g]})& \le M_1\left[\frac{f^p}{g^{p-1}}\right]-\frac{M_1^p[f]}{M_1^{p-1}[g]}\\ \le p(p-1)M^{p-2}(M_1\left[\frac{f^2}{g}\right]-\frac{M_1^2[f]}{M_1[g]}).\end{array}$$

• 4. The following two inequalities are rewritten here and have as a starting point an inequality from [34].

• (i) If $a,b\in \mathbf{R}$, $a<b$, $f:[a,b]\to {\mathbf{R}}_{+}$ is a convex and continuous function and $h,l:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$, and, in addition, if $g:[a,b]\to {\mathbf{R}}_{+}$ is integrable on $[a,b]$, then

  $$\begin{array}{r}\underset{x\in [a,b]}{sup}\left\{\frac{h(x)}{l(x)}\right\}[M_1[l\cdot (f\circ g)]-f(M_l[g])\cdot M_1[l]]\\ \ge [M_1[h\cdot (f\circ g)]-f(M_h[g])\cdot M_1[h]]\\ \ge \underset{x\in [a,b]}{inf}\left\{\frac{h(x)}{l(x)}\right\}[M_1[l\cdot (f\circ g)]-f(M_l[g])\cdot M_1[l]],\end{array}$$

or

$$\begin{array}{r}\underset{x\in [a,b]}{sup}\left\{\frac{h(x)}{l(x)}\right\}{M}_1[l][M_l[f\circ g]-f(M_l[g])]\\ \ge M_1[h][M_h[f\circ g]-f(M_h[g])]\\ \ge \underset{x\in [a,b]}{inf}\left\{\frac{h(x)}{l(x)}\right\}{M}_1[l][M_l[f\circ g]-f(M_l[g])].\end{array}$$

• (ii) If $a,b\in \mathbf{R}$, $a<b$, $f:[a,b]\to {\mathbf{R}}_{+}$ is a convex and continuous function, $p,q_1:[a,b]\to {\mathbf{R}}_{+}$ are two integrable functions on $[a,b]$ with $q\in (0,1)$, if $g:[a,b]\to {\mathbf{R}}_{+}$ is integrable on $[a,b]$ and $M\ge m\ge 0$ are such that $Mp(x)\ge q_1(x)\ge mp(x)$, $(\mathrm{\forall })x\in \mathbf{R}$, then

  $$\begin{array}{rl}{M}^{\frac{1}{q}}{\left(\frac{M_1[p]}{M_1[q_1]}\right)}^{\frac{1}{q}}[M_p[f\circ g]-f(M_p[g])]& \ge M_{q_1}[f\circ g]-f(M_{q_1}[g])\\ \ge m^{\frac{1}{q}}{\left(\frac{M_1[p]}{M_1[q_1]}\right)}^{\frac{1}{q}}[M_p[f\circ g]-f(M_p[g])].\end{array}$$

• 5. Starting from the inequality of Halliwell and Mercer (see [35]), we can establish the following result.

If $a,b\in \mathbf{R}$, $a<b$, $f,g:[a,b]\to {\mathbf{R}}_{+}$ are two continuous and strict positive functions on $[a,b]$ and $M_1[f]=M_1[g]$, then the following inequality holds:

$$\begin{array}{rl}M\left[\frac{g{(g-f)}^2}{g^2+{(max\{{sup}_{x\in [a,b]}f,{sup}_{x\in [a,b]}g\})}^2}\right]& \le M[fln\left(\frac{f}{g}\right)]\\ \le M\left[\frac{g{(g-f)}^2}{g^2+{(min\{{inf}_{x\in [a,b]}f,{inf}_{x\in [a,b]}g\})}^2}\right].\end{array}$$