Weighted Montgomery’s identities for higher order differentiable functions of two variables

Asif R. Khan$^{1,2}$, Josip Pečarić$^{2,3}$ Marjan Praljak$^{2,4}$

April 4, 2013.

$^1$Department of Mathematical Sciences, University of Karachi, University Road, Karachi, Pakistan, e-mail: asif$\_ $rizkhan@yahoo.com

$^2$Abdus Salam School of Mathematical Sciences, GC University, Lahore, Pakistan

$^3$Faculty Of Textile Technology, University of Zagreb, Prilaz baruna Filipovića 28a, Zagreb, Croatia, e-mail: pecaric@mahazu.hazu.hr

$^4$Faculty Of Food Technology and Biotechnology, University of Zagreb, Pierottijeva 6, Zagreb, Croatia, e-mail: mpraljak@pbf.hr

We give weighted Montgomery’s identities for higher order differentiable functions of two variables and by using these identities we obtain generalized Ostrowski-type and Grüss-type inequalities for double weighted integrals of higher order differentiable functions of two independent variables.

MSC. 39B22, 26D15, 26D99

Keywords. Montgomery’s identities, Ostrowski-type inequalities, Grüss-type inequalities.

1 Introduction

Let $f:[a,b]\to \mathbb {R}$ be a differentiable function on $[a,b]$ and $f':[a,b]\to \mathbb {R}$ be an integrable function. Then, the Montgomery identity holds [ 15 ] , (see also [ 20 ] ),

\begin{equation} \label{introeq1} f(x)=\tfrac {1}{b-a}\int _a^bf(t)\, {\rm d}t+\int _a^bp(x,t)f'(t)\, {\rm d}t, \end{equation}

where $p(x,t)$ is the Peano kernel

\[ p(x,t)= \left\{ \begin{array}{rl} \tfrac {t-a}{b-a}, & a\le t\le x, \\ \tfrac {t-b}{b-a}, & x{\lt} t\le b. \end{array} \right. \]

Suppose now that $w : [a, b] \to [0,\infty )$ is some probability density function, i.e., it is a positive integrable function satisfying $ \int _a^b w(t ) {\rm d} t = 1$ and

\[ W (t )= \left\{ \begin{array}{ll} 0,& t {\lt} a,\\ \int _a^t w(x) {\rm d}x, & t\in [a, b], \\ 1, & t {\gt} b. \end{array} \right. \]

The following identity (Pečarić [ 17 ] ) is a generalization of Montgomery’s identity,

\begin{equation*} \label{introeq2} f(x)=\int _a^bw(t)f(t)\, {\rm d}t+\int _a^bp_w(x,t)f(t)\, {\rm d}t, \end{equation*}

where the weighted Peano kernel is

\[ p_w(x,t)= \left\{ \begin{array}{ll} W(t), & a\le t\le x, \\ W(t)-1, & x{\lt} t\le b. \end{array} \right. \]

In [ 3 , 8 ] the authors obtained two identities which generalize $(\ref{introeq1})$ for functions of two variables. In fact, for a function $f : [a, b]\times [c, d ]\to \mathbb {R}$ such that the partial derivatives $\tfrac {\partial f(s,t)}{\partial s}$, $\tfrac {\partial f(s,t)}{\partial t}$ and $\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}$ exist and are continuous on $[a, b]\times [c, d ]$ and for all $(x, y)\in [a, b]\times [c, d ]$ they obtained:

\begin{align*} \label{introeq3} & (d-c)(b-a)f(x,y)=\\ & \quad =-\int _a^b\int _c^df(s,t)\, {\rm d}t\, {\rm d}s+(d-c)\int _a^bf(s,y)\, {\rm d}s\\ & \quad \quad +(b-a)\int _c^df(x,t)\, {\rm d}t+\int _a^b\int _c^dq(x,s)r(y,t)\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}\, {\rm d}t\, {\rm d}s, \end{align*}

and

\begin{align*} & (d-c)(b-a)f(x,y)=\\ & \quad =\int _a^b\int _c^df(s,t)\, {\rm d}t\, {\rm d}s+\int _a^b\int _c^dq(x,s)\tfrac {\partial f(s,t)}{\partial s}\, {\rm d}t\, {\rm d}s\\ & \quad \quad +\int _a^b\int _c^dr(y,t)\tfrac {\partial f(s,t)}{\partial t}\, {\rm d}t\, {\rm d}s+\int _a^b\int _c^d q(x,s)r(y,t)\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}\, {\rm d}t\, {\rm d}s, \end{align*}

where

\[ q(x,s)= \left\{ \begin{array}{rl} s-a, & a\le s\le x, \\ s-b, & x{\lt} s\le b, \end{array} \right. \]

\[ r(y,t)= \left\{ \begin{array}{rl} t-c, & a\le t\le x, \\ t-d, & x{\lt} t\le b. \end{array} \right. \]

We can also find weighted Montgomery’s identities for functions of two variables in [ 20 ] . These identities may be summarized as:

Theorem 1.1

Let $p:[a,b]\times [c,d]\to \mathbb {R}$ be an integrable function and $P(x,y)$ is defined as

\begin{equation} \label{P(0,0)} P(x,y)=\int _x^b \int _y^d p(\xi ,\eta ) \, {\rm d}\eta \, {\rm d}\xi . \end{equation}

Let $f:[a,b]\times [c,d]\to \mathbb {R}$ have continuous partial derivatives $\tfrac {\partial f(s,t)}{\partial s}$, $\tfrac {\partial f(s,t)}{\partial t}$ and $\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}$ on $[a,b]\times [c,d]$. Then, for all $(x,y)\in [a,b]\times [c,d]$

\begin{align} \label{id01} & f(x,y)P(a,c)=\\ & =\int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s+\int _a^b\hat{P}(x,s)\tfrac {\partial f(s,y)}{\partial s}\, {\rm d}s \nonumber \\ & \quad +\int _c^d\tilde{P}(y,t)\tfrac {\partial f(x,t)}{\partial t}\, {\rm d}t- \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}\, {\rm d}t\, {\rm d}s,\nonumber \end{align}

where

\begin{align*} \hat{P}(x,s)& = \left\{ \begin{array}{ll} \displaystyle {\int _a^s \int _c^d p(\xi ,\eta ) {\rm d}\eta }\, {\rm d}\xi , & a\le s\le x, \\ -P(s,c), & x{\lt} s\le b, \end{array} \right.\\ \tilde{P}^{(i,M)}(x,y,t)& = \left\{ \begin{array}{ll} \displaystyle {\int _a^b \int _c^t p(\xi ,\eta ) {\rm d}\eta \, {\rm d}\xi }, & c\le t\le y, \\ -P(a,t), & y{\lt} t\le d \end{array} \right. {\text and}\\ \bar{P}(x,s,y,t)& = \left\{ \begin{array}{ll} \displaystyle {\int _a^s \int _c^t p(\xi ,\eta ) {\rm d}\eta \, {\rm d}\xi }, & a\le s\le x,\, c\le t\le y, \\ -\displaystyle {\int _s^b \int _c^t p(\xi ,\eta ) {\rm d}\eta \, {\rm d}\xi }, & x{\lt} s\le b,\, c\le t\le y, \\ -\displaystyle {\int _a^s \int _t^d p(\xi ,\eta ) {\rm d}\eta \, {\rm d}\xi }, & a\le s\le x,\, y{\lt} t\le d,\\ P(s,t), & x{\lt} s\le b,\, y{\lt} t\le d. \end{array} \right. \end{align*}

Theorem 1.2

\begin{align} \label{id02} & f(x,y)P(a,c)=\\ & =-\int _a^b \int _c^d p(s,t)f(s,t) \, {\rm d}t\, {\rm d}s+\int _a^b \int _c^d p(s,t) f(s,y)\, {\rm d}t\, {\rm d}s\nonumber \\ & \quad + \int _a^b \int _c^d p(s,t)f(x,t)\, {\rm d}t\, {\rm d}s+ \int _a^b \int _c^d \bar{P}(x,s,y,t)\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}\, {\rm d}t\, {\rm d}s,\nonumber \end{align}

where $p(.,.),\, P(a,c),\, \bar{P}(x,s,y,t)$ are as in Theorem 1.1.

Theorem 1.3

\begin{align*} f(x,y)[P(a,c)]^2& =P(a,c)\int _a^b \int _c^d p(s,t)f(s,t)\, {\rm d}t\, {\rm d}s\\ & \quad +\int _a^b\left(\int _a^b \int _c^d p(\xi ,t)\hat{P}(x,s) \tfrac {\partial f(s,t)}{\partial s}\, {\rm d}t\, {\rm d}s\right){\rm d}\xi \\ & \quad + \int _c^d\left(\int _a^b \int _c^d p(s,\eta )\tilde{P}(y,t)\tfrac {\partial f(s,t)}{\partial t}\, {\rm d}t\, {\rm d}s\right){\rm d}\eta \\ & \quad + \int _a^b \int _c^d \check{P}(x,s,y,t)\tfrac {\partial ^2 f(s,t)}{\partial s \partial t}\, {\rm d}t\, {\rm d}s, \end{align*}

where $p(.,.),\, P(a,c),\, \hat{P}(x,s),\, \tilde{P}(y,t),\, \bar{P}(x,s,y,t)$ are as in Theorem 1.1 and

\[ \check{P}(x,s,y,t)=2\hat{P}(x,s)\tilde{P}(y,t)-P(a,c)\bar{P}(x,s,y,t). \]

Montgomery’s identities have many applications and capture other well known and important identities and inequalities which includes Ostrowski-type, Čebyšev-type and Grüss-type inequalities, as we can see in this paper also. Ostrowski’s inequalities have many applications in field of Numerical Integration (for an extensive reference see [ 10 ] ) and Probability Theory (see [ 7 , 14 ] ). We can also obtain Special Means with the help of such inequalities (for example see [ 1 , 2 ] ). Very famous Čebyšev’s inequality is also an special case of Ostrowski-type inequalities (see [ 18 , 19 ] ). Grüss-type inequalities have applications in Numerical Integration and other fields (for reference see [ 4 , 5 , 6 ] ).

The structure of this paper based on four sections. In the second section, we give generalization of Montgomery’s identities using higher order differentiable functions of two variables. In the third and the fourth sections respectively, we obtain some generalized Ostrowski-type and Grüss-type inequalities for higher order differentiable functions of two independent variables by using identities proved in the second section. These identities and inequalities generalize many results given in [ 3 , 8 , 9 , 12 , 16 , 20 ] etc.

2 Weighted Montgomery’s identities for higher order differentiable functions of two variables

In start of this section, we define some notation to reduce our lengthy expressions as follows:

\begin{align} \label{P(i,j)} P^{(i,j)}_{(a,c)\to (b,d)}(x,y)& =\int _a^b \int _c^d p(\xi ,\eta ) \tfrac {(\xi -x)^i}{i!} \tfrac {(\eta -y)^j}{j!}\, {\rm d}\eta \, {\rm d}\xi ,\\ \label{P(0,j)} P^{(0,j)}_{(a,c)\to (b,d)}(y)& =\int _a^b \int _c^d p(\xi ,\eta ) \tfrac {(\eta -y)^j}{j!}\, {\rm d}\eta \, {\rm d}\xi , \end{align}

\begin{align} \label{P(i,0)} & P^{(i,0)}_{(a,c)\to (b,d)}(x)=\int _a^b \int _c^d p(\xi ,\eta ) \tfrac {(\xi -x)^i}{i!}\, {\rm d}\eta \, {\rm d}\xi ,\\ \label{R} & R(x,y;f)=\\ \nonumber & =-\sum _{i=1}^{N} \sum _{j=1}^{M} f_{(i,j)}(x,y)P^{(i,j)}_{(a,c)\to (b,d)}(x,y)\\ \nonumber & \quad -\sum _{j=1}^{M} f_{(0,j)}(x,y)P^{(0,j)}_{(a,c)\to (b,d)}(y)-\sum _{i=1}^{N} f_{(i,0)}(x,y)P^{(i,0)}_{(a,c)\to (b,d)}(x). \end{align}

For our next theorem, we give a lemma from [ 13 ] using our notations as follows.

Lemma 2.1

Let $p,f:[a,b]\times [c,d] \to \mathbb {R}$ be integrable functions and $f\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then we have

\begin{align} \label{3-1} \int _a^b \int _c^d p(x,y)f(x,y) {\rm d}y\, {\rm d}x =& \sum _{i=0}^{N} \sum _{j=0}^{M} P^{(i,j)}_{(a,c)\to (b,d)}(a,c)f_{(i,j)}(a,c)\\ \nonumber & +\sum _{j=0}^{M}\int _a^b P^{(N,j)}_{(x,c)\to (b,d)}(x,c) f_{(N+1,j)}(x,c) \, {\rm d}x \\ \nonumber & +\sum _{i=0}^{N} \int _c^d P^{(i,M)}_{(a,y)\to (b,d)}(a,y) f_{(i,M+1)}(a,y) \, {\rm d}y\\ \nonumber & +\int _a^b \int _c^d P^{(N,M)}_{(x,y)\to (b,d)}(x,y) f_{(N+1,M+1)}(x,y) \, {\rm d}y\, {\rm d}x. \end{align}

We give generalizations of Theorems $\ref{th01}$, $\ref{th02}$ and $\ref{th03}$ respectively as follows:

Theorem 2.2

Let $p,f:[a,b]\times [c,d] \to \mathbb {R}$ be integrable functions and $f\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then we have

\begin{align} \label{id1} f(x,y)P(a,c)& =R(x,y;f)+\int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s\\ & \quad +\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s\nonumber \\ & \quad +\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t\nonumber \\ & \quad - \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s,\nonumber \end{align}

where

\begin{align*} \hat{P}^{(N,j)}(x,s,y)& = \left\{ \begin{array}{rl} P^{(N,j)}_{(a,c)\to (s,d)}(s,y), & a\le s\le x, \\ -P^{(N,j)}_{(s,c)\to (b,d)}(s,y), & x{\lt} s\le b, \end{array} \right.\\ \tilde{P}^{(i,M)}(x,y,t)& = \left\{ \begin{array}{rl} P^{(i,M)}_{(a,c)\to (b,t)}(x,t), & c\le t\le y, \\ -P^{(i,M)}_{(a,t)\to (b,d)}(x,t), & y{\lt} t\le d, \end{array} \right.\ {\text and}\\ \bar{P}^{(N,M)}(x,s,y,t)& = \left\{ \begin{array}{rl} P^{(N,M)}_{(a,c)\to (s,t)}(s,t), & a\le s\le x,\, c\le t\le y, \\ -P^{(N,M)}_{(s,c)\to (b,t)}(s,t), & x{\lt} s\le b,\, c\le t\le y, \\ -P^{(N,M)}_{(a,t)\to (s,d)}(s,t), & a\le s\le x,\, y{\lt} t\le d,\\ P^{(N,M)}_{(s,t)\to (b,d)}(s,t), & x{\lt} s\le b,\, y{\lt} t\le d, \end{array} \right. \end{align*}

where $P^{(i,j)}_{(.,.)\to (.,.)}(.,.)$ for $i,j \in \{ N,M\} $ is defined in $(\ref{P(i,j)})$, and $P(a,c)$ and $R(x,y;f)$ are defined in $(\ref{P(0,0)})$ and $(\ref{R})$ respectively.

Proof â–¼

Using Lemma 2.1 for $[a,x]\times [c,y]$, we obtain the equality

\begin{eqnarray*} & & \int _a^x \int _c^y p(s,t)f(s,t) {\rm d}t\, {\rm d}s=\int _x^a \int _y^c p(s,t)f(s,t) {\rm d}t\, {\rm d}s=\\ & & = \sum _{i=0}^{N} \sum _{j=0}^{M} P^{(i,j)}_{(x,y)\to (a,c)}(x,y) f_{(i,j)}(x,y)\! +\! \sum _{j=0}^{M}\int _x^a P^{(N,j)}_{(s,y)\to (a,c)}(s,y) f_{(N+1,j)}(s,y)\, {\rm d}s \\ & & \quad + \sum _{i=0}^{N} \int _y^c P^{(i,M)}_{(x,t)\to (a,c)}(x,t) f_{(i,M+1)}(x,t)\, {\rm d}t\\ & & \quad +\int _x^a \int _y^c P^{(N,M)}_{(s,t)\to (a,c)}(s,t) f_{(N+1,M+1)}(s,t)\, {\rm d}t \, {\rm d}s\\ & & = \sum _{i=0}^{N} \sum _{j=0}^{M} f_{(i,j)}(x,y) \left[P^{(i,j)}_{(x,y)\to (b,d)}(x,y) P^{(i,j)}_{(x,c)\to (b,d)}(x,y)-P^{(i,j)}_{(a,y)\to (b,d)}(x,y)\right.\\ & & \left.\quad +P^{(i,j)}_{(a,c)\to (b,d)}(x,y)\right] - \sum _{j=0}^{M}\int _a^x f_{(N+1,j)}(s,y) \left[P^{(N,j)}_{(s,y)\to (b,d)}(s,y)\right.\\ & & \left.\quad -P^{(N,j)}_{(s,c)\to (b,d)}(s,y)-P^{(N,j)}_{(a,y)\to (b,d)}(s,y)+ P^{(N,j)}_{(a,c)\to (b,d)}(s,y)\right] \, {\rm d}s \\ & & \quad -\sum _{i=0}^{N} \int _c^y f_{(i,M+1)}(x,t)\left[P^{(i,M)}_{(x,t)\to (b,d)}(x,t)\! -\! P^{(i,M)}_{(x,c)\to (b,d)}(x,t)\! -\! P^{(i,M)}_{(a,t)\to (b,d)}(x,t)\right.\\ & & \quad +\left.P^{(i,M)}_{(a,c)\to (b,d)}(x,t)\right]\, {\rm d}t+\int _a^x \int _c^y f_{(N+1,M+1)}(s,t)\left[P^{(N,M)}_{(s,t)\to (b,d)}(s,t)\right.\\ & & \quad \left.-P^{(N,M)}_{(s,c)\to (b,d)}(s,t)-P^{(N,M)}_{(a,t)\to (b,d)}(s,t)+ P^{(N,M)}_{(a,c)\to (b,d)}(s,t)\right]\, {\rm d}t \, {\rm d}s. \end{eqnarray*}

Similarly, for $[x,b]\times [c,y]$, we get

\begin{eqnarray*} & & \int _x^b \int _c^y p(s,t)f(s,t) {\rm d}t\, {\rm d}s=-\int _x^b \int _y^c p(s,t)f(s,t) {\rm d}t\, {\rm d}s=\\ & & = -\sum _{i=0}^{N} \sum _{j=0}^{M} f_{(i,j)}(x,y)\left[P^{(i,j)}_{(x,y)\to (b,d)}(x,y) P^{(i,j)}_{(x,c)\to (b,d)}(x,y)\right] \\ & & \quad - \sum _{j=0}^{M}\int _x^b f_{(N+1,j)}(s,y)\left[ P^{(N,j)}_{(s,y)\to (b,d)}(s,y)-P^{(N,j)}_{(s,c)\to (b,d)}(s,y)\right]\, {\rm d}s \\ & & \quad + \sum _{i=0}^{N} \int _c^y f_{(i,M+1)}(x,t)\left[P^{(i,M)}_{(x,t)\to (b,d)}(x,t)-P^{(i,M)}_{(x,c)\to (b,d)}(x,t)\right]\, {\rm d}t\\ & & \quad + \int _x^b \int _c^y f_{(N+1,M+1)}(s,t)\left[P^{(N,M)}_{(s,t)\to (b,d)}(s,t)-P^{(N,M)}_{(s,c)\to (b,d)}(s,t)\right]\, {\rm d}t \, {\rm d}s. \end{eqnarray*}

For $[a,x]\times [y,d]$, we have

\begin{eqnarray*} & & \int _a^x \int _y^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s=-\int _x^a \int _y^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s=\\ & & = -\sum _{i=0}^{N} \sum _{j=0}^{M} f_{(i,j)}(x,y)\left[ P^{(i,j)}_{(x,y)\to (b,d)}(x,y)-P^{(i,j)}_{(a,y)\to (b,d)}(x,y)\right] \\ & & \quad + \sum _{j=0}^{M}\int _a^x f_{(N+1,j)}(s,y)\left[ P^{(N,j)}_{(s,y)\to (b,d)}(s,y)- P^{(N,j)}_{(a,y)\to (b,d)}(s,y)\right]\, {\rm d}s \\ & & \quad - \sum _{i=0}^{N} \int _y^d f_{(i,M+1)}(x,t)\left[P^{(i,M)}_{(x,t)\to (b,d)}(x,t)-P^{(i,M)}_{(a,t)\to (b,d)}(x,t)\right]\, {\rm d}t\\ & & \quad + \int _a^x \int _y^d f_{(N+1,M+1)}(s,t)\left[P^{(N,M)}_{(s,t)\to (b,d)}(s,t)-P^{(N,M)}_{(a,t)\to (b,d)}(s,t)\right]\, {\rm d}t \, {\rm d}s. \end{eqnarray*}

Finally, for $[x,b]\times [y,d]$ we obtain

\begin{align*} \int _x^b \int _y^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s& =\sum _{i=0}^{N} \sum _{j=0}^{M} f_{(i,j)}(x,y)P^{(i,j)}_{(x,y)\to (b,d)}(x,y)= \\ & \quad + \sum _{j=0}^{M}\int _x^b f_{(N+1,j)}(s,y)P^{(N,j)}_{(s,y)\to (b,d)}(s,y)\, {\rm d}s+ \end{align*}

\begin{align*} & \quad +\sum _{i=0}^{N} \int _y^d f_{(i,M+1)}(x,t)P^{(i,M)}_{(x,t)\to (b,d)}(x,t)\, {\rm d}t\\ & \quad + \int _x^b \int _y^d f_{(N+1,M+1)}(s,t)P^{(N,M)}_{(s,t)\to (b,d)}(s,t)\, {\rm d}t \, {\rm d}s. \end{align*}

Adding the four expressions, we get our required result.

Proof â–¼

Theorem 2.3

Let $p,f:[a,b]\times [c,d] \to \mathbb {R}$ be integrable functions and $f\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then we have

\begin{align} \label{id2} & f(x,y)P(a,c)=\\ & \quad =R(x,y;f)+\sum _{j=1}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -y)^j}{j!}f_{(0,j)}(s,y)\, {\rm d}\eta \, {\rm d}s\nonumber \\ & \quad \quad \! +\! \sum _{i=1}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(i,0)} (x,t)\, {\rm d}t \, {\rm d}\xi \! -\! \int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s\nonumber \\ & \quad \quad +\int _a^b\int _c^d p(s,t) f(s,y)\, {\rm d}t \, {\rm d}s+\int _a^b\int _c^d p(s,t)f(x,t)\, {\rm d}t\, {\rm d}s\nonumber \\ & \quad \quad +\int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s,\nonumber \end{align}

where $\bar{P}^{(N,M)}(x,s,y,t)$ is as in Theorem $\ref{th1}$, and $P(a,c)$ and $R(x,y,f)$ are defined in $(\ref{P(0,0)})$ and $(\ref{R})$ respectively.

Proof â–¼

First, we find an expression for

\[ \int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s \]

by using integration by parts as follows

\begin{align*} & \int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s=\\ & =\int _a^xP^{(N,j)}_{(a,c)\to (s,d)}(s,y) f_{(N+1,j)}(s,y)\, {\rm d}s -\int _x^bP^{(N,j)}_{(s,c)\to (b,d)}(s,y) f_{(N+1,j)}(s,y)\, {\rm d}s\\ & =\int _a^xP^{(N,j)}_{(a,c)\to (s,d)}(s,y) f_{(N+1,j)}(s,y)\, {\rm d}s +\int _x^bP^{(N,j)}_{(b,c)\to (s,d)}(s,y) f_{(N+1,j)}(s,y)\, {\rm d}s \end{align*}

\begin{align*} & =P^{(N,j)}_{(a,c)\to (x,d)}(x,y) f_{(N,j)}(x,y) +\int _a^xP^{(N-1,j)}_{(a,c)\to (s,d)}(s,y) f_{(N,j)}(s,y)\, {\rm d}s \\ & \quad +P^{(N,j)}_{(x,c)\to (b,d)}(x,y) f_{(N,j)}(x,y)+\int _x^bP^{(N-1,j)}_{(b,c)\to (s,d)}(s,y) f_{(N,j)}(s,y)\, {\rm d}s\\ & =P^{(N,j)}_{(a,c)\to (b,d)}(x,y) f_{(N,j)}(x,y) +\int _a^xP^{(N-1,j)}_{(a,c)\to (s,d)}(s,y) f_{(N,j)}(s,y)\, {\rm d}s \\ & \quad +\int _x^bP^{(N-1,j)}_{(b,c)\to (s,d)}(s,y) f_{(N,j)}(s,y)\, {\rm d}s\\ & =P^{(N,j)}_{(a,c)\to (b,d)}(x,y) f_{(N,j)}(x,y) +\int _a^bP^{(N-1,j)}_{(a,c)\to (s,d)}(s,y) f_{(N,j)}(s,y)\, {\rm d}s, \end{align*}

continuing in this way, we finally get

\begin{eqnarray} \label{A} & & \int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s=\\ \nonumber & & =\int _a^b \int _c^d p(\xi ,\eta ) \tfrac {(\eta -y)^j}{j!}\left[\sum _{k=0}^N\tfrac {(\xi -x)^k}{k!} f_{(k,j)}(x,y)\right]\, {\rm d}\eta \, {\rm d}\xi \\ \nonumber & & \quad -\int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -y)^j}{j!} f_{(0,j)}(s,y)\, {\rm d}\eta \, {\rm d}s. \end{eqnarray}

Similarly

\begin{eqnarray} \label{B} & & \int _c^d\tilde{P}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t=\\ \nonumber & & =\int _a^b \int _c^d p(\xi ,\eta ) \tfrac {(\xi -x)^i}{i!}\left[\sum _{l=0}^M\tfrac {(\eta -y)^l}{l!} f_{(i,l)}(x,y)\right]\, {\rm d}\eta \, {\rm d}\xi \\ \nonumber & & \quad -\int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!} f_{(i,0)}(x,t)\, {\rm d}\xi \, {\rm d}t. \end{eqnarray}

If we put all these values in $(\ref{id1})$, then after some cancelation and some rearrangements, we get our required identity.

Proof â–¼

Theorem 2.4

Let $f:[a,b]\times [c,d]\to \mathbb {R}$ be a function such that $f\in C^{(2N+1,2M+1)}([a,b]\times [c,d])$. Then for all $(x,y)\in [a,b]\times [c,d]$ we have

\begin{align} \label{id3} & f(x,y)[P(a,c)]^2=\\ & =P(a,c)R(x,y;f)+P(a,c)\int _a^b \int _c^d p(s,t)f(s,t){\rm d}t\, {\rm d}s\nonumber \\ & \nonumber \quad +\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)R(x,t;f_{(i,M+1)})\, {\rm d}t+ \end{align}

\begin{align} & \nonumber \quad +\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)R(s,y;f_{(N+1,j)})\, {\rm d}s\\ \nonumber & \quad +\sum _{i=0}^{N}\sum _{j=0}^{M}\int _a^b\int _a^b \int _c^d\hat{P}^{(N,j)}(x,s,y) p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(N+1+i,j)} (s,t)\, {\rm d}t \, {\rm d}s\, {\rm d}\xi \\ \nonumber & \quad +\sum _{i=0}^{N}\sum _{j=0}^{M} \int _c^d \int _a^b \int _c^d\tilde{P}^{(i,M)}(x,y,t) p(s,\eta ) \tfrac {(\eta -y)^j}{j!}f_{(i,M+1+j)} (s,t)\, {\rm d}t\, {\rm d}s\, {\rm d}\eta \\ \nonumber & \quad +2\int _a^b \int _c^d \sum _{i=0}^{N}\sum _{j=0}^{M}\hat{P}^{(N,j)}(x,s,y)\tilde{P}^{(i,M)}(x,y,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}t\, {\rm d}s\\ \nonumber & \quad -\int _a^b \int _c^d\bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s, \end{align}

where $\hat{P}^{(N,j)}(x,s,y)\; \tilde{P}^{(i,M)}(x,y,t),\; \bar{P}^{(N,M)}(x,s,y,t)$ are as in Theorem $\ref{th1}$ and $P(a,c)$ is defined in $(\ref{P(0,0)})$.

Proof â–¼

Summing $(\ref{A})$ for $j=0,\ldots ,M$ and $(\ref{B})$ for $i=0,\ldots ,N$, we get respectively

\begin{align} \label{C} f(x,y)P(a,c)& =R(x,y;f)+\sum _{j=0}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -y)^j}{j!}f_{(0,j)} (s,y)\, {\rm d}\eta \, {\rm d}s\\ \nonumber & \quad +\sum _{j=0}^M\int _a^b \hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s, \end{align}

and

\begin{align} \label{D} f(x,y)P(a,c)& =R(x,y;f)+\sum _{i=0}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(i,0)} (x,t)\, {\rm d}t \, {\rm d}\xi \\ \nonumber & \quad +\sum _{i=0}^N\int _c^d \tilde{P}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t, \end{align}

for all $(x,y)\in [a,b]\times [c,d].$

Formula $(\ref{C})$ applied for partial derivatives $f_{(i,M+1)}$ for $i=0,1,\ldots ,N$, gives

\begin{eqnarray} \label{E} & & f_{(i,M+1)}(x,t)P(a,c)=\\ \nonumber & & =R(x,t;f_{(i,M+1)})\\ \nonumber & & \quad +\sum _{j=0}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -t)^j}{j!}f_{(i,M+1+j)} (s,t)\, {\rm d}\eta \, {\rm d}s\\ \nonumber & & \quad +\sum _{j=0}^M\int _a^b \hat{P}^{(N,j)}(x,s,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}s. \end{eqnarray}

Formula $(\ref{D})$ applied for partial derivatives $f_{(N+1,j)}$ for $j=0,1,\ldots ,M$, gives

\begin{eqnarray} \label{F1} & & f_{(N+1,j)}(s,y)P(a,c)=\\ \nonumber & & =R(s,y;f_{(N+1,j)})\\ \nonumber & & \quad +\sum _{i=0}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -s)^i}{i!}f_{(N+1+i,j)} (s,t)\, {\rm d}t \, {\rm d}\xi \\ \nonumber & & \quad +\sum _{i=0}^N\int _c^d \tilde{P}^{(i,M)}(s,y,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}t. \end{eqnarray}

Substituting $(\ref{E})$ and $(\ref{F1})$ into $(\ref{id1})$, we get

\begin{eqnarray*} & & f(x,y)P(a,c)=\\ & & =R(x,y;f) +\int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s\\ & & \quad +\tfrac {1}{P(a,c)}\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)\bigg[R(s,y;f_{(N+1,j)})\\ & & \quad +\sum _{i=0}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -s)^i}{i!}f_{(N+1+i,j)} (s,t)\, {\rm d}t \, {\rm d}\xi \\ & & \quad +\sum _{i=0}^N\int _c^d \tilde{P}^{(i,M)}(s,y,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}t\bigg]\, {\rm d}s\\ & & \quad +\tfrac {1}{P(a,c)}\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)\bigg[R(x,t;f_{(i,M+1)})\\ & & \quad +\sum _{j=0}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -t)^j}{j!}f_{(i,M+1+j)} (s,t)\, {\rm d}\eta \, {\rm d}s\\ & & \quad +\sum _{j=0}^M\int _a^b \hat{P}^{(N,j)}(x,s,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}s\bigg]\, {\rm d}t\\ & & \quad - \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s. \end{eqnarray*}

After some rearrangements and using Fubini’s Theorem, we get our required result.

Proof â–¼

Remark 2.5

For $N=M=0$, Theorems 1.1, 1.2 and 1.3 become special cases of Theorem 2.2, 2.3 and 2.4 respectively (see also [ 20 ] ).

If $p(s,t)=q(s)r(t)$ in identities $(\ref{id1})$, $(\ref{id2})$ and $(\ref{id3})$, then we get respectively the following special cases:

\begin{eqnarray*} \nonumber & & f(x,y)P_{a\to b}(q)P_{c\to d}(r)=\\ & & =Q(x,y;f)+\int _a^b \int _c^d q(s)r(t)f(s,t) {\rm d}t\, {\rm d}s\\ & & \quad +\sum _{j=0}^{M}\int _a^b\hat{Q}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s\\ & & \quad +\sum _{i=0}^{N}\int _c^d\tilde{Q}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t\\ & & \quad -\int _a^b \int _c^d \bar{Q}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s,\\ & & f(x,y)P_{a\to b}(q)P_{c\to d}(r)=\\ & & =Q(x,y;f)+\sum _{j=1}^{M} \int _a^b q(s)f_{(0,j)} (s,y)\, {\rm d}s\, Q^{(j)}_{c\to d}(r,y)\\ & & \quad +\sum _{i=1}^{N} Q^{(i)}_{a\to b}(q,x)\int _c^d r(t) f_{(i,0)} (x,t)\, {\rm d}t-\int _a^b \int _c^d q(s)r(t)f(s,t) {\rm d}t\, {\rm d}s\\ \nonumber & & \quad +\int _a^b\int _c^d q(s)r(t) f(s,y)\, {\rm d}t \, {\rm d}s+\int _a^b\int _c^d q(s)r(t)f(x,t)\, {\rm d}t\, {\rm d}s\\ \nonumber & & \quad -\int _a^b \int _c^d \bar{Q}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s,\\ & & f(x,y)[P_{a\to b}(q)P_{c\to d}(r)]^2=\\ & & =P_{a\to b}(q)P_{c\to d}(r)Q(x,y;f)\\ & & \quad +\sum _{j=0}^{M}\int _a^b\hat{Q}^{(N,j)}(x,s,y)Q(s,y;f_{(N+1,j)})\, {\rm d}s\\ & & \quad +\sum _{i=0}^{N}\int _c^d\tilde{Q}^{(i,M)}(x,y,t)Q(x,t;f_{(i,M+1)})\, {\rm d}t \\ \nonumber & & \quad +P_{a\to b}(q)P_{c\to d}(r)\int _a^b \int _c^d q(s)r(t)f(s,t) {\rm d}t\, {\rm d}s\\ \nonumber & & \quad +\sum _{i=0}^{N}\sum _{j=0}^{M}Q^{(i)}_{a\to b}(q,x)\int _a^b \int _c^d\hat{Q}^{(N,j)}(x,s,y) r(t) f_{(N+1+i,j)} (s,t)\, {\rm d}t \, {\rm d}s\\ \nonumber & & \quad +\sum _{i=0}^{N}\sum _{j=0}^{M} Q^{(j)}_{c\to d}(r,y)\int _a^b \int _c^d\tilde{Q}^{(i,M)}(x,y,t) q(s)\, f_{(i,M+1+j)} (s,t)\, {\rm d}t\, {\rm d}s\, \\ \nonumber & & \quad +2\int _a^b \int _c^d \sum _{i=0}^{N}\sum _{j=0}^{M}\hat{Q}^{(N,j)}(x,s,y)\tilde{Q}^{(i,M)}(x,y,t)f_{(N+1+i,M+1+j)}(s,t)\, {\rm d}t\, {\rm d}s\\ & & \quad -\int _a^b \int _c^d\bar{Q}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s, \end{eqnarray*}

where

\begin{align*} & P_{a\to b}(q)=\int _a^bq(s)\, {\rm d}s,\quad Q^{(i)}_{a\to b}(q,x)=\int _a^b q(\xi )\tfrac {(\xi -x)^i}{i!}\, {\rm d}\xi ,\\ & Q^{(i,j)}_{(a,c)\to (b,d)}(x,y)=Q^{(i)}_{a\to b}(q,x)Q^{(j)}_{c\to d}(r,y), \\ & Q^{(0,j)}_{(a,c)\to (b,d)}(y)=P_{a\to b}(q)\; Q^{(j)}_{c\to d}(r,y),\\ & Q^{(i,0)}_{(a,c)\to (b,d)}(x)= Q^{(i)}_{a\to b}(q,x)\; P_{c\to d}(r),\end{align*}

\begin{align*} Q(x,y;f)& =-\sum _{i=1}^{N} \sum _{j=1}^{M} f_{(i,j)}(x,y)Q^{(i,j)}_{(a,c)\to (b,d)}(x,y)\\ & \quad -\sum _{j=1}^{M} f_{(0,j)}(x,y)Q^{(0,j)}_{(a,c)\to (b,d)}(y)-\sum _{i=1}^{N} f_{(i,0)}(x,y)Q^{(i,0)}_{(a,c)\to (b,d)}(x), \end{align*}

\begin{align*} \hat{Q}^{(N,j)}(x,s,y)& = \left\{ \begin{array}{rl} Q^{(N,j)}_{(a,c)\to (s,d)}(s,y), & a\le s\le x, \\ -Q^{(N,j)}_{(s,c)\to (b,d)}(s,y), & x{\lt} s\le b, \end{array} \right.\\ \tilde{Q}^{(i,M)}(x,y,t)& = \left\{ \begin{array}{rl} Q^{(i,M)}_{(a,c)\to (b,t)}(x,t), & c\le t\le y, \\ -Q^{(i,M)}_{(a,t)\to (b,d)}(x,t), & y{\lt} t\le d, \end{array} \right. \ {\rm and}\\ \bar{Q}^{(N,M)}(x,s,y,t)& = \left\{ \begin{array}{rl} Q^{(N,M)}_{(a,c)\to (s,t)}(s,t), & a\le s\le x,\, c\le t\le y, \\ -Q^{(N,M)}_{(s,c)\to (b,t)}(s,t), & x{\lt} s\le b,\, c\le t\le y, \\ -Q^{(N,M)}_{(a,t)\to (s,d)}(s,t), & a\le s\le x,\, y{\lt} t\le d,\\ Q^{(N,M)}_{(s,t)\to (b,d)}(s,t), & x{\lt} s\le b,\, y{\lt} t\le d. \end{array} \right. \end{align*}

Particularly, if $p(.,.)=1$ in identities $(\ref{id1})$, $(\ref{id2})$ and $(\ref{id3})$, then the expressions look like

\begin{align*} P_{a\to b}& =b-a, \quad Q^{(i)}_{a\to b}(x)=\tfrac {(b-x)^{i+1}-(a-x)^{i+1}}{(i+1)!},\\ Q(x,y;f)& =-\sum _{i=1}^{N} \sum _{j=1}^{M}\tfrac {(b-x)^{i+1}-(a-x)^{i+1}}{(i+1)!}\times \tfrac {(d-y)^{j+1}-(c-y)^{j+1}}{(j+1)!}f_{(i,j)}(x,y)\\ & \quad -(b-a)\sum _{j=1}^{M} \tfrac {(d-y)^{j+1}-(c-y)^{j+1}}{(j+1)!}\; f_{(0,j)}(x,y) \\ & \quad -(d-c)\sum _{i=1}^{N} \tfrac {(b-x)^{i+1}-(a-x)^{i+1}}{(i+1)!}\; f_{(i,0)}(x,y), \end{align*}

\begin{align*} \hat{Q}^{(N,j)}(x,s,y)& = \left\{ \begin{array}{rl} -\tfrac {(a-s)^{N+1}}{(N+1)!}\; \tfrac {(d-y)^{j+1}-(c-y)^{j+1}}{(j+1)!}, & a\le s\le x, \\ -\tfrac {(b-s)^{N+1}}{(N+1)!}\; \tfrac {(d-y)^{j+1}-(c-y)^{j+1}}{(j+1)!}, & x{\lt} s\le b, \end{array} \right.\\ \tilde{Q}^{(i,M)}(x,y,t)& = \left\{ \begin{array}{rl} -\tfrac {(c-t)^{M+1}}{(M+1)!}\; \tfrac {(b-x)^{i+1}-(a-x)^{i+1}}{(i+1)!}, & c\le t\le y, \\ -\tfrac {(d-t)^{M+1}}{(M+1)!}\; \tfrac {(b-x)^{i+1}-(a-x)^{i+1}}{(i+1)!}, & y{\lt} t\le d \end{array} \right.\ {\rm and}\\ \bar{Q}^{(N,M)}(x,s,y,t)& = \left\{ \begin{array}{rl} \tfrac {(a-s)^{N+1}}{(N+1)!}\; \tfrac {(c-t)^{M+1}}{(M+1)!}, & a\le s\le x,\, c\le t\le y, \\ \tfrac {(b-s)^{N+1}}{(N+1)!}\; \tfrac {(c-t)^{M+1}}{(M+1)!}, & x{\lt} s\le b,\, c\le t\le y, \\ \tfrac {(a-s)^{N+1}}{(N+1)!}\; \tfrac {(d-t)^{M+1}}{(M+1)!}, & a\le s\le x,\, y{\lt} t\le d,\\ \tfrac {(b-s)^{N+1}}{(N+1)!}\; \tfrac {(d-t)^{M+1}}{(M+1)!}, & x{\lt} s\le b,\, y{\lt} t\le d. \hfil \qed \end{array} \right. \end{align*}

3 Ostrowski-type inequalities for double weighted integrals for higher order differentiable functions

The following well known Ostrowski’s inequality is extracted from [ 16 ] .

\begin{equation} \left|f(x)-\tfrac {1}{b-a}\int _a^bf(t)\, {\rm d}t\right|\le \left[\tfrac {1}{4}+\tfrac {(x-\tfrac {a+b}{2})^2}{(b-a)^2}\right](b-a)M, \qquad {x\in [a,b]}, \end{equation}

where $f : [a, b]\to \mathbb {R}$ is a differentiable function such that $|f'(x)| \le M$ for every $x \in [a, b]$. This inequality undergoes many generalizations and in [ 20 ] Pečarić and Vukeli$\acute{c}$ provided additional ones for two independent variables using identities $(\ref{id01})$ and $(\ref{id02})$. By using identities $(\ref{id1})$ and $(\ref{id2})$, we can give generalized results of Ostrowski-type for higher order differentiable functions of two independent variables as follows.

Theorem 3.1

Let $f:[a,b]\times [c,d]\to \mathbb {R}$ be such that $f\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then $\forall \, (x,y)\in [a,b]\times [c,d]$ we have

\begin{eqnarray*} & & \left|f(x,y)-\tfrac {1}{P(a,c)}\int _a^b \int _c^dp(s,t)f(s,t)\, {\rm d}t\, {\rm d}s\right|\le \\ & & \le D(x,y)+\sum _{j=0}^M\hat{D}^{(0,j)}(x,y)+\sum _{i=0}^N\tilde{D}^{(i,0)}(x,y)+\bar{D}(x,y), \end{eqnarray*}

where

\begin{align*} D(x,y)& =\tfrac {1}{|P(a,c)|}|R(x,y;f)|,\\ \hat{D}^{(0,j)}(x,y)& =\tfrac {1}{|P(a,c)|}\bigg(\sum _{j=0}^{M}\int _a^b|\hat{P}^{(N,j)}(x,s,y)|^{\hat{q}_j}{\rm d}s\bigg)^{1/\hat{q}_j}.\| f_{(N+1,j)}\| _{\hat{p}_j},\\ \end{align*}

provided that

\begin{eqnarray*} & & f_{(N+1,j)}\in L_{\hat{p}_j}([a,b]\times [c,d]),\; 1/\hat{p}_j+1/\hat{q}_j=1,\\ & & \tilde{D}^{(i,0)}(x,y)=\tfrac {1}{|P(a,c)|}\bigg(\sum _{i=0}^{N}\int _c^d |\tilde{P}^{(i,M)}(x,y,t)|^{\tilde{q}_i}dt\bigg)^{1/\tilde{q}_i}.\| f_{(i,M+1)}\| _{\tilde{p}_i}, \end{eqnarray*}

provided that

\begin{eqnarray*} & & f_{(i,M+1)}\in L_{\tilde{p}_i}([a,b]\times [c,d]),\; 1/\tilde{p}_i+1/\tilde{q}_i=1,\\ & & \bar{D}(x,y)=\tfrac {1}{|P(a,c)|}\bigg( \int _a^b \int _c^d |\bar{P}^{(N,M)}(x,s,y,t)|^{\bar{q}}dt\, ds\bigg)^{1/\bar{q}}.\| f_{(N+1,M+1)}\| _{\bar{p}}, \end{eqnarray*}

provided that

\begin{eqnarray*} & & f_{(N+1,M+1)}\in L_{\bar{p}}([a,b]\times [c,d]),\quad 1/\bar{p}+1/\bar{q}=1, \end{eqnarray*}

where $\hat{P}^{(N,j)}(x,s,y)\; \tilde{P}^{(i,M)}(x,y,t),\; \bar{P}^{(N,M)}(x,s,y,t)$ are as in Theorem $\ref{th1}$ whereas $P(a,c)$ and $R(x,y,f)$ are defined in $(\ref{P(0,0)})$ and $(\ref{R})$ respectively.

Proof â–¼

Identity $(\ref{id1})$ may be written as

\begin{eqnarray*} & & f(x,y)-\tfrac {1}{P(a,c)}\int _a^b \int _c^dp(s,t)f(s,t)\, {\rm d}t\, {\rm d}s=\\ & & =\tfrac {1}{P(a,c)}\bigg[R(x,y;f)+\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s\\ & & \quad +\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t\\ & & \quad -\int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s\bigg]. \end{eqnarray*}

Now, taking absolute value and applying Hölder’s inequality for double integrals, we easily obtain our required inequality.

Proof â–¼

Remark 3.2

For $N=M=0$, Theorem 4 of [ 20 ] becomes special case of Theorem 3.1 and we also retrieve results of [ 9 ] by simply putting $p(.,.)=1$.â–¡

Theorem 3.3

Let $f:[a,b]\times [c,d]\to \mathbb {R}$ be a continuous function on $[a,b]\times [c,d]$ such that $f\in C^{(N+1,M+1)}((a,b)\times (c,d))$ and $|f_{(N+1,M+1)}|^p$ be an integrable function, i.e.,

\begin{equation*} \left\| f_{(N+1,M+1)}\right\| _{p}:=\bigg(\int _a^b \int _c^d|f_{(N+1,M+1)}(s,t)|^p\, {\rm d}t\, {\rm d}s\bigg)^{1/p}{\lt}\infty , \end{equation*}

$1/p+1/q=1.$ Then, it follows that

\begin{align*} & \bigg|\int _a^b\int _c^d p(s,t)f(x,t)\, {\rm d}t\, {\rm d}s-\bigg[R(x,y;f)\\ & \quad +\sum _{j=1}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -y)^j}{j!}f_{(0,j)} (s,y)\, {\rm d}\eta \, {\rm d}s\\ & \quad +\sum _{i=1}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(i,0)} (x,t)\, {\rm d}t \, {\rm d}\xi \\ & \quad +\int _a^b\int _c^d p(s,t) f(x,t)\, {\rm d}t \, {\rm d}s\\ & \quad +\int _a^b \int _c^d p(s,t)f(s,y) {\rm d}t\, {\rm d}s-f(x,y)P(a,c)\bigg]\bigg|\le \\ & \le \bigg(\int _a^b \int _c^d |\bar{P}^{(N,M)}(x,s,y,t)|\, {\rm d}t\, {\rm d}s\bigg)^{1/q}\| f_{(N+1,M+1)}\| _{p}. \end{align*}

for all $(x,y)\in [a,b]\times [c,d]$.

Proof â–¼

Identity $(\ref{id2})$ may be written as

\begin{eqnarray*} & & \int _a^b\int _c^d p(s,t)f(s,t)\, {\rm d}t\, {\rm d}s-\bigg[R(x,y;f)\\ & & \quad +\int _a^b\int _c^d p(s,t) f(s,y)\, {\rm d}t \, {\rm d}s+\int _a^b \int _c^d p(s,t)f(x,t) {\rm d}t\, {\rm d}s \\ & & \quad +\sum _{j=1}^{M} \int _a^b \int _c^d p(s,\eta ) \tfrac {(\eta -y)^j}{j!}f_{(0,j)} (s,y)\, {\rm d}\eta \, {\rm d}s\\ & & \quad +\sum _{i=1}^{N} \int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(i,0)} (x,t)\, {\rm d}t \, {\rm d}\xi -f(x,y)P(a,c)\bigg]=\\ & & =\int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s. \end{eqnarray*}

Now taking absolute value and applying Hölder’s inequality for double integrals, we easily obtain our required inequality.

Proof â–¼

Remark 3.4

For $N=M=0$, Theorem 5 of [ 20 ] becomes special case of Theorem 3.3 and we also retrieve results of [ 3 ] and [ 8 ] by simply putting $p(.,.)=1$.â–¡

4 Grüss-type inequalities for double weighted integrals for higher order differentiable functions

A celebrated integral inequality proved by Grüss [ 11 ] in $1935$, can be stated as follows (see [ 15 , p. 296 ] ),

\begin{eqnarray*} & & \left|\tfrac {1}{b-a}\int _a^bf(x)g(x)\, {\rm d}x-\left(\tfrac {1}{b-a}\int _a^bf(x)\, {\rm d}x\right)\left(\tfrac {1}{b-a}\int _a^bg(x)\, {\rm d}x\right)\right|\le \\ & & \quad \le \tfrac {1}{4}(M-m)(N-n) \end{eqnarray*}

provided that $f$ and $g$ are two integrable functions on $[a, b]$ and satisfy the conditions

\[ m \le f (x) \le M,\quad {n \le g(x) \le N}, \]

for all $x \in [a, b]$, where $m, M , n, N$ are real constants.

In [ 20 ] Pečarić and Vukeli$\acute{c}$ gave new Grüss-type inequalities for double weighted integrals by using identities $(\ref{id01})$ and $(\ref{id02})$. Now, we give more generalized results by using higher order differentiable functions of two independent variables but in order to simplify the details of the presentations we define the following notations.

\begin{align} \label{A^{(i,j)}} A^{(i,j)}(x,y)& =p(x,y)[f_{(i,j)}(x,y)g(x,y)+g_{(i,j)}(x,y)f(x,y)]\times \\ \nonumber & \quad \times P^{(i,j)}_{(a,c)\to (b,d)}(x,y),\\ A(x,y)& =p(x,y)\int _a^b\int _c^d p(s,t)[f(s,t)g(x,y)+g(s,t)f(x,y)]\, {\rm d}t\, {\rm d}s,\\ \hat{A}^{(N,j)}(x,y)& =p(x,y)\int _a^b[f_{(N+1,j)}(s,y)g(x,y)+g_{(N+1,j)}(s,y)f(x,y)]\times \\ \nonumber & \quad \times \hat{P}^{(N,j)}(x,s,y)\, {\rm d}s,\\ \tilde{A}^{(i,M)}(x,y)& =p(x,y)\int _c^d[f_{(i,M+1)}(x,t)g(x,y)+g_{(i,M+1)}(x,t)f(x,y)]\times \\ \nonumber & \quad \times \tilde{P}^{(i,M)}(x,y,t)\, {\rm d}t,\\ \label{BIJ} \bar{A}^{(N,M)}(x,y)& =p(x,y)\int _a^b\int _c^d[f_{(N+1,M+1)}(s,t)g(x,y)\\ \nonumber & \quad +g_{(N+1,M+1)}(s,t)f(x,y)]\bar{P}^{(N,M)}(x,s,y,t)\, {\rm d}t\, {\rm d}s,\\ B^{(i,j)}(x,y)& =|p(x,y)g(x,y)|\; \| f_{(i,j)}(x,y)\| _\infty +|p(x,y)f(x,y)|\times \\ \nonumber & \quad \times \| g_{(i,j)}(x,y)\| _\infty , \end{align}

\begin{align} C^{(i,j)}(x,y)& =\tfrac {(\max \{ b-x,x-a\} )^{i+1}}{(i+1)!}\; \tfrac {(\max \{ d-y,y-c\} )^{j+1}}{(j+1)!}\, \times \\ \nonumber & \quad \times \int _a^b \int _c^d |p(\xi ,\eta )| \, {\rm d}\eta \, {\rm d}\xi ,\\ C^{(0,j)}(y)& =(b-a)\tfrac {(\max \{ d-y,y-c\} )^{j+1}}{(j+1)!}\, \int _a^b \int _c^d |p(\xi ,\eta )| \, {\rm d}\eta \, {\rm d}\xi ,\\ C^{(i,0)}(x)& =(d-c)\tfrac {(\max \{ b-x,x-a\} )^{i+1}}{(i+1)!}\, \int _a^b \int _c^d |p(\xi ,\eta )| \, {\rm d}\eta \, {\rm d}\xi ,\\ \hat{C}^{(N,j)}(x,y)& =\int _a^b|\hat{P}^{(N,j)}(x,s,y)|\, {\rm d}s,\\ \tilde{C}^{(i,M)}(x,y)& =\int _c^d|\tilde{P}^{(i,M)}(x,y,t)|\, {\rm d}t,\\ \label{CNM} \bar{C}^{(N,M)}(x,y)& =\int _a^b\int _c^d|\bar{P}^{(N,M)}(x,s,y,t)|\, {\rm d}t\, {\rm d}s,\\ \label{F} F(x,y)& =R(x,y;f)+\int _a^b\int _c^d p(s,t) f(s,y)\, {\rm d}t \, {\rm d}s \\ \nonumber & \quad +\int _a^b\int _c^d p(s,t)f(x,t)\, {\rm d}t\, {\rm d}s\\ \nonumber & \quad +\sum _{j=1}^{M} \int _a^b \int _c^d p(s,\eta )\tfrac {(\eta -y)^j}{j!}f_{(0,j)} (s,y)\, {\rm d}\eta \, {\rm d}s \\ & \quad +\sum _{i=1}^{N}\int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}f_{(i,0)} (x,t)\, {\rm d}t\, {\rm d}\xi , \nonumber \\ \label{G} G(x,y)& =R(x,y;g)+\int _a^b\int _c^d p(s,t)g(s,y)\, {\rm d}t \, {\rm d}s\\ \nonumber & \quad +\int _a^b\int _c^d p(s,t)g(x,t)\, {\rm d}t\, {\rm d}s\\ \nonumber & \quad +\sum _{j=1}^{M} \int _a^b \int _c^d p(s,\eta )\tfrac {(\eta -y)^j}{j!}g_{(0,j)} (s,y)\, {\rm d}\eta \, {\rm d}s \\ \nonumber & \quad + \sum _{i=1}^{N}\int _a^b \int _c^d p(\xi ,t) \tfrac {(\xi -x)^i}{i!}g_{(i,0)} (x,t)\, {\rm d}t\, {\rm d}\xi , \end{align}

Now, we are ready to present our main results of this section by using notations defined above, which are as follows.

Theorem 4.1

Let $f,g:[a,b]\times [c,d]\to \mathbb {R}$ be two functions such that $f,\, g\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then

\begin{eqnarray*} & & \left|\tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)f(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right.\\ & & \quad \quad -\left(\tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)f(x,y)\, {\rm d}y\, {\rm d}x\right)\\ & & \quad \quad \times \left.\left(\tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right)\right|\le \\ & & \quad \le \tfrac {1}{2[P(a,c)]^2}\int _a^b\int _c^d\Bigg[\sum _{i=1}^{N} \sum _{j=1}^{M}B^{(i,j)}(x,y)C^{(i,j)}(x,y)\\ & & \quad \quad +\sum _{j=1}^{M}B^{(0,j)}(y)C^{(0,j)}(y)+\sum _{i=1}^{N}B^{(i,0)}(x)C^{(i,0)}(x)\\ & & \quad \quad +B^{(N+1,j)}(x,y)\hat{C}^{(N,j)}(x,y)+B^{(i,M+1)}(x,y)\tilde{C}^{(i,M)}(x,y)\\ & & \quad \quad +B^{(N+1,M+1)}(x,y)\bar{C}^{(N,M)}(x,y)\Bigg]\, {\rm d}y\, {\rm d}x. \end{eqnarray*}

Proof â–¼

From $(\ref{id1})$, we have the following identities:

\begin{align} \label{a} f(x,y)P(a,c)& =R(x,y;f)+\int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s\\ \nonumber & \quad +\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s\\ \nonumber & \quad +\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)f_{(i,M+1)}(x,t)\, {\rm d}t\\ \nonumber & \quad - \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s \end{align}

\begin{align} \label{b} g(x,y)P(a,c)& =R(x,y;g)+\int _a^b \int _c^d p(s,t)g(s,t) {\rm d}t\, {\rm d}s\\ \nonumber & \quad +\sum _{j=0}^{M}\int _a^b\hat{P}^{(N,j)}(x,s,y)g_{(N+1,j)}(s,y)\, {\rm d}s\\ \nonumber & \quad +\sum _{i=0}^{N}\int _c^d\tilde{P}^{(i,M)}(x,y,t)g_{(i,M+1)}(x,t)\, {\rm d}t\\ \nonumber & \quad - \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)g_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s \end{align}

for $(x,y)\in [a,b]\times [c,d]$. Multiplying $(\ref{a})$ by $p(x,y)g(x,y)$ and $(\ref{b})$ by $p(x,y)f(x,y)$ and adding the resulting identities, we obtain

\begin{eqnarray} \label{c} & & 2P(a,c)p(x,y)f(x,y)g(x,y)=-\sum _{i=1}^{N} \sum _{j=1}^{M}A^{(i,j)}(x,y)-\sum _{j=1}^{M}A^{(0,j)}(y)\\ \nonumber & & \quad -\sum _{i=1}^{N}A^{(i,0)}(x)+A(x,y)+\hat{A}^{(N,j)}(x,y)+\tilde{A}^{(i,M)}(x,y)-\bar{A}^{(N,M)}(x,y) \end{eqnarray}

Integrating $(\ref{c})$ over $[a,b]\times [c,d]$, we get

\begin{eqnarray*} \nonumber & & \int _a^b \int _c^dp(x,y)f(x,y)g(x,y)\, {\rm d}y\, {\rm d}x=\tfrac {1}{2P(a,c)}\int _a^b \int _c^d\bigg[-\sum _{i=1}^{N} \sum _{j=1}^{M}A^{(i,j)}(x,y)\\ & & \quad -\sum _{j=1}^{M}A^{(0,j)}(y)-\sum _{i=1}^{N}A^{(i,0)}(x+\left. A(x,y)+\hat{A}^{(N,j)}(x,y)+\tilde{A}^{(i,M)}(x,y)\right.\\ \nonumber & & \quad -\bar{A}^{(N,M)}(x,y)\bigg]\, {\rm d}y\, {\rm d}x \end{eqnarray*}

It may be written as

\begin{eqnarray} \label{d} & & \tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)f(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\\ \nonumber & & \quad -\left(\tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)f(x,y)\, {\rm d}y\, {\rm d}x\right)\\ \nonumber & & \quad \times \left(\tfrac {1}{P(a,c)}\int _a^b\int _c^dp(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right)\\ \nonumber & & =\tfrac {1}{2[P(a,c)]^2}\int _a^b \int _c^d\bigg[-\sum _{i=1}^{N} \sum _{j=1}^{M}A^{(i,j)}(x,y)-\sum _{j=1}^{M}A^{(0,j)}(y)\\ \nonumber & & \quad -\sum _{i=1}^{N}A^{(i,0)}(x)+\hat{A}^{(N,j)}(x,y)+\tilde{A}^{(i,M)}(x,y)-\bar{A}^{(N,M)}(x,y)\bigg]\, {\rm d}y\, {\rm d}x. \end{eqnarray}

Using $(\ref{A^{(i,j)}})$,$\ldots $,$(\ref{CNM})$ we have the following inequalities

\begin{eqnarray*} & & |A^{(i,j)}(x,y)|\le B^{(i,j)}(x,y)\; C^{(i,j)}(x,y),\\ & & |A^{(0,j)}(y)|\le B^{(0,j)}(y)\; C^{(0,j)}(y),\\ & & |A^{(i,0)}(x)|\le B^{(i,0)}(x)\; C^{(i,0)}(x),\\ & & |\hat{A}^{(N,j)}(x,y)|\le B^{(N+1,j)}(x,y)\; \hat{C}^{(N,j)}(x,y),\\ & & |\tilde{A}^{(i,M)}(x,y)|\le B^{(i,M+1)}(x,y)\; \tilde{C}^{(i,M)}(x,y),\\ & & |\bar{A}^{(N,M)}(x,y)|\le B^{(N+1,M+1)}(x,y)\; \bar{C}^{(N,M)}(x,y), \end{eqnarray*}

$\forall (x,y)\in [a,b]\times [c,d]$. Taking absolute value on both sides in (28) and using all these inequalities in it, we get our required result.

Proof â–¼

Theorem 4.2

Let $f,g:[a,b]\times [c,d]\to \mathbb {R}$ be two continuous functions on $[a,b]\times [c,d]$, such that $f,\, g\in C^{(N+1,M+1)}([a,b]\times [c,d])$. Then

\begin{eqnarray*} & & \left|\tfrac {1}{P(a,c)}\int _a^b \int _c^dp(x,y)f(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right.\\ & & \quad +\left(\tfrac {1}{P(a,c)}\int _a^b \int _c^dp(x,y)f(x,y)\, {\rm d}y\, {\rm d}x\right)\left(\tfrac {1}{P(a,c)}\int _a^b \int _c^dp(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right)\\ & & \quad \left.-\tfrac {1}{2[P(a,c)]^2}\int _a^b \int _c^dp(x,y)[g(x,y)F(x,y)+f(x,y)G(x,y)]\, {\rm d}y\, {\rm d}x\right|\le \\ & & \le \tfrac {1}{2[P(a,c)]^2}\int _a^b \int _c^d B^{(N+1,M+1)}(x,y)\; \bar{C}^{(N,M)}(x,y)\, {\rm d}y\, {\rm d}x \end{eqnarray*}

Proof â–¼

From $(\ref{id2})$, we have the following identities:

\begin{align} \label{f} f(x,y)P(a,c)& =F(x,y)-\int _a^b \int _c^d p(s,t)f(s,t) {\rm d}t\, {\rm d}s\\ \nonumber & \quad + \int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)f_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s,\\ \label{g} g(x,y)P(a,c)& =G(x,y)-\int _a^b \int _c^d p(s,t)g(s,t) {\rm d}t\, {\rm d}s \\ \nonumber & \quad +\int _a^b \int _c^d \bar{P}^{(N,M)}(x,s,y,t)g_{(N+1,M+1)}(s,t)\, {\rm d}t\, {\rm d}s, \end{align}

for $(x,y)\in [a,b]\times [c,d]$. Multiplying $(\ref{f})$ by $p(x,y)g(x,y)$ and $(\ref{g})$ by $p(x,y)f(x,y)$ and adding the resulting identities, we obtain

\begin{eqnarray} \label{h} & & 2P(a,c)p(x,y)f(x,y)g(x,y)=p(x,y)g(x,y)F(x,y)\\ \nonumber & & \quad +p(x,y)f(x,y)G(x,y)-A(x,y)+\bar{A}^{(N,M)}(x,y) \end{eqnarray}

Integrating $(\ref{h})$ over $[a,b]\times [c,d]$, we get

\begin{eqnarray} \label{i} & & \int _a^b \int _c^dp(x,y)f(x,y)g(x,y)\, {\rm d}y\, {\rm d}x=\\ \nonumber & & =\tfrac {1}{2P(a,c)}\int _a^b \int _c^dp(x,y)[g(x,y)F(x,y)+f(x,y)G(x,y)]\, {\rm d}y\, {\rm d}x\\ \nonumber & & \quad -\tfrac {1}{P(a,c)}\left(\int _a^b \int _c^dp(x,y)f(x,y)\, {\rm d}y\, {\rm d}x\right)\left(\int _a^b \int _c^dp(x,y)g(x,y)\, {\rm d}y\, {\rm d}x\right)\\ \nonumber & & \quad +\tfrac {1}{2P(a,c)}\int _a^b \int _c^d\bar{A}^{(N,M)}(x,y)\, {\rm d}y\, {\rm d}x \end{eqnarray}

\begin{equation} \label{j} |\bar{A}^{(N,M)}(x,y)|\le B^{(N+1,M+1)}(x,y)\; \bar{C}^{(N,M)}(x,y) \end{equation}

From $(\ref{i})$ and $(\ref{j})$, we obtain our required inequality.

Proof â–¼

Remark 4.3

For $N=M=0$ Theorems 6 and 7 of [ 20 ] become special cases of Theorems 4.1 and 4.2 respectively and we also retrieve results of [ 16 ] by simply putting $p(.,.)=1$. For $N=M=0$, we can also find similar results in [ 12 ] .â–¡

Acknowledgement

This research work is funded by Higher Education Commission Pakistan. The research of the second author was supported by the Croatian Ministry of Science, Education and Sports under the Research Grants 117-1170889-0888.

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