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<h1>Weighted Montgomery’s identities for higher order differentiable functions of two variables</h1>
<p class="authors">
<span class="author">Asif R. Khan\(^{1,2}\), Josip Pečarić\(^{2,3}\) Marjan Praljak\(^{2,4}\)</span>
</p>
<p class="date">April 4, 2013.</p>
</div>
<p>\(^1\)Department of Mathematical Sciences, University of Karachi, University Road, Karachi, Pakistan, e-mail: <span class="tt">asif\(\_ \)rizkhan@yahoo.com</span> </p>
<p>\(^2\)Abdus Salam School of Mathematical Sciences, GC University, Lahore, Pakistan </p>
<p>\(^3\)Faculty Of Textile Technology, University of Zagreb, Prilaz baruna Filipovića 28a, Zagreb, Croatia, e-mail: <span class="tt">pecaric@mahazu.hazu.hr</span> </p>
<p>\(^4\)Faculty Of Food Technology and Biotechnology, University of Zagreb, Pierottijeva 6, Zagreb, Croatia, e-mail: <span class="tt">mpraljak@pbf.hr</span> </p>

<div class="abstract"><p> 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. </p>
<p><b class="bf">MSC.</b> 39B22, 26D15, 26D99 </p>
<p><b class="bf">Keywords.</b> Montgomery’s identities, Ostrowski-type inequalities, Grüss-type inequalities. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> 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 <span class="cite">
	[
	<a href="#S. Mitrinovic" >15</a>
	]
</span>, (see also <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span>), </p>
<div class="equation" id="introeq3">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(p(x,t)\) is the Peano kernel </p>
<div class="displaymath" id="a0000000003">
  \[  p(x,t)= \left\{  \begin{array}{rl} \tfrac {t-a}{b-a}, &  a\le t\le x, \\ \tfrac {t-b}{b-a}, &  x{\lt} t\le b. \end{array} \right.  \]
</div>
<p> 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 </p>
<div class="displaymath" id="a0000000004">
  \[  W (t )= \left\{  \begin{array}{ll} 0,&  t {\lt} a,\\ \int _a^t w(x) {\rm d}x, &  t\in [a, b], \\ 1, &  t {\gt} b. \end{array} \right.  \]
</div>
<p> The following identity (Pečarić <span class="cite">
	[
	<a href="#chebysev" >17</a>
	]
</span>) is a generalization of Montgomery’s identity, </p>
<div class="displaymath" id="a0000000005">
  \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*}
</div>
<p> where the weighted Peano kernel is </p>
<div class="displaymath" id="a0000000006">
  \[  p_w(x,t)= \left\{  \begin{array}{ll} W(t), &  a\le t\le x, \\ W(t)-1, &  x{\lt} t\le b. \end{array} \right.  \]
</div>
<p> In <span class="cite">
	[
	<a href="#Dragomir" >3</a>
	, 
	<a href="#S. Dragomir" >8</a>
	]
</span> 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: </p>
<div class="displaymath" id="a0000000007">
  \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*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000008">
  \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*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000009">
  \[  q(x,s)= \left\{  \begin{array}{rl} s-a, &  a\le s\le x, \\ s-b, &  x{\lt} s\le b, \end{array} \right.  \]
</div>
<div class="displaymath" id="a0000000010">
  \[  r(y,t)= \left\{  \begin{array}{rl} t-c, &  a\le t\le x, \\ t-d, &  x{\lt} t\le b. \end{array} \right.  \]
</div>
<p> We can also find weighted Montgomery’s identities for functions of two variables in <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span>. These identities may be summarized as: <div class="thm_thmwrapper " id="th01">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1.1</span>
  </div>
  <div class="thm_thmcontent">
  <p> Let \(p:[a,b]\times [c,d]\to \mathbb {R}\) be an integrable function and \(P(x,y)\) is defined as </p>
<div class="equation" id="P(0,0)">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> 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 \((x,y)\in [a,b]\times [c,d]\) </p>
<div class="displaymath" id="id01">
  \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}
</div>
<p> where </p>
<div class="displaymath" id="a0000000011">
  \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*}
</div>

  </div>
</div> <div class="thm_thmwrapper " id="th02">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1.2</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 \((x,y)\in [a,b]\times [c,d]\), we have </p>
<div class="displaymath" id="id02">
  \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}
</div>
<p> where \(p(.,.),\, P(a,c),\, \bar{P}(x,s,y,t)\) are as in Theorem <span class="rm"><a href="#th01">1.1</a></span>. </p>

  </div>
</div> <div class="thm_thmwrapper " id="th03">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1.3</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 \((x,y)\in [a,b]\times [c,d]\), we have </p>
<div class="displaymath" id="a0000000012">
  \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*}
</div>
<p> where \(p(.,.),\, P(a,c),\, \hat{P}(x,s),\, \tilde{P}(y,t),\, \bar{P}(x,s,y,t)\) are as in Theorem <span class="rm"><a href="#th01">1.1</a></span> and </p>
<div class="displaymath" id="a0000000013">
  \[ \check{P}(x,s,y,t)=2\hat{P}(x,s)\tilde{P}(y,t)-P(a,c)\bar{P}(x,s,y,t). \]
</div>

  </div>
</div> </p>
<p>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 <span class="cite">
	[
	<a href="#ostrapp1" >10</a>
	]
</span>) and Probability Theory (see <span class="cite">
	[
	<a href="#ostrapp2" >7</a>
	, 
	<a href="#ostrapp3" >14</a>
	]
</span>). We can also obtain Special Means with the help of such inequalities (for example see <span class="cite">
	[
	<a href="#ostrapp4" >1</a>
	, 
	<a href="#ostrapp5" >2</a>
	]
</span>). Very famous Čebyšev’s inequality is also an special case of Ostrowski-type inequalities (see <span class="cite">
	[
	<a href="#J. Pecaric" >18</a>
	, 
	<a href="#pecaric" >19</a>
	]
</span>). Grüss-type inequalities have applications in Numerical Integration and other fields (for reference see <span class="cite">
	[
	<a href="#grussapp3" >4</a>
	, 
	<a href="#grussapp1" >5</a>
	, 
	<a href="#grussapp2" >6</a>
	]
</span>). </p>
<p>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 <span class="cite">
	[
	<a href="#Dragomir" >3</a>
	, 
	<a href="#S. Dragomir" >8</a>
	, 
	<a href="#S.S. Dragomir" >9</a>
	, 
	<a href="#G-L" >12</a>
	, 
	<a href="#Ostrowski" >16</a>
	, 
	<a href="#Montgomery" >20</a>
	]
</span> etc. </p>
<h1 id="a0000000014">2 Weighted Montgomery’s identities for higher order differentiable functions of two variables</h1>
<p>In start of this section, we define some notation to reduce our lengthy expressions as follows: </p>
<div class="displaymath" id="P(i,j)">
  \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}
</div>
<div class="displaymath" id="P(i,0)">
  \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}
</div>
<p> For our next theorem, we give a lemma from <span class="cite">
	[
	<a href="#asif" >13</a>
	]
</span> using our notations as follows. <div class="lem_thmwrapper " id="lemma1">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">2.1</span>
  </div>
  <div class="lem_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="3-1">
  \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}
</div>

  </div>
</div> We give generalizations of Theorems \(\ref{th01}\), \(\ref{th02}\) and \(\ref{th03}\) respectively as follows: <div class="thm_thmwrapper " id="th1">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.2</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="id1">
  \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}
</div>
<p> where </p>
<div class="displaymath" id="a0000000015">
  \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*}
</div>
<p> 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. <div class="proof_wrapper" id="a0000000016">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Using Lemma <a href="#lemma1">2.1</a> for \([a,x]\times [c,y]\), we obtain the equality </p>
<div class="displaymath" id="a0000000017">
  \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*}
</div>
<p> Similarly, for \([x,b]\times [c,y]\), we get </p>
<div class="displaymath" id="a0000000018">
  \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*}
</div>
<p> For \([a,x]\times [y,d]\), we have </p>
<div class="displaymath" id="a0000000019">
  \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*}
</div>
<p> Finally, for \([x,b]\times [y,d]\) we obtain </p>
<div class="displaymath" id="a0000000020">
  \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*}
</div>
<div class="displaymath" id="a0000000021">
  \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*}
</div>
<p> Adding the four expressions, we get our required result. <div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="thm_thmwrapper " id="th2">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.3</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="id2">
  \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}
</div>
<p> 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. <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> First, we find an expression for </p>
<div class="displaymath" id="a0000000024">
  \[ \int _a^b\hat{P}^{(N,j)}(x,s,y)f_{(N+1,j)}(s,y)\, {\rm d}s \]
</div>
<p> by using integration by parts as follows </p>
<div class="displaymath" id="a0000000025">
  \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*}
</div>
<div class="displaymath" id="a0000000026">
  \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*}
</div>
<p> continuing in this way, we finally get </p>
<div class="displaymath" id="A">
  \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}
</div>
<p> Similarly </p>
<div class="displaymath" id="B">
  \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}
</div>
<p> If we put all these values in \((\ref{id1})\), then after some cancelation and some rearrangements, we get our required identity. <div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="thm_thmwrapper " id="th3">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2.4</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="id3">
  \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}
</div>
<div class="displaymath" id="a0000000028">
  \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}
</div>
<p> 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 \((\ref{P(0,0)})\). <div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Summing \((\ref{A})\) for \(j=0,\ldots ,M\) and \((\ref{B})\) for \(i=0,\ldots ,N\), we get respectively </p>
<div class="displaymath" id="C">
  \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}
</div>
<p> and </p>
<div class="displaymath" id="D">
  \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}
</div>
<p> for all \((x,y)\in [a,b]\times [c,d].\) </p>
<p>Formula \((\ref{C})\) applied for partial derivatives \(f_{(i,M+1)}\) for \(i=0,1,\ldots ,N\), gives </p>
<div class="displaymath" id="E">
  \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}
</div>
<p> Formula \((\ref{D})\) applied for partial derivatives \(f_{(N+1,j)}\) for \(j=0,1,\ldots ,M\), gives </p>
<div class="displaymath" id="F1">
  \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}
</div>
<p> Substituting \((\ref{E})\) and \((\ref{F1})\) into \((\ref{id1})\), we get </p>
<div class="displaymath" id="a0000000030">
  \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*}
</div>
<p> After some rearrangements and using Fubini’s Theorem, we get our required result. <div class="proof_wrapper" id="a0000000031">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000032">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">2.5</span>
  </div>
  <div class="rem_thmcontent">
  <p>For \(N=M=0\), Theorems <a href="#th01">1.1</a>, <a href="#th02">1.2</a> and <a href="#th03">1.3</a> become special cases of Theorem <a href="#th1">2.2</a>, <a href="#th2">2.3</a> and <a href="#th3">2.4</a> respectively (see also <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span>). </p>
<p>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: </p>
<div class="displaymath" id="a0000000033">
  \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*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000034">
  \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*}
</div>
<div class="displaymath" id="a0000000035">
  \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*}
</div>
<div class="displaymath" id="a0000000036">
  \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*}
</div>
<p> Particularly, if \(p(.,.)=1\) in identities \((\ref{id1})\), \((\ref{id2})\) and \((\ref{id3})\), then the expressions look like </p>
<div class="displaymath" id="a0000000037">
  \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*}
</div>
<div class="displaymath" id="a0000000038">
  \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*}
</div>

  </div>
</div> </p>
<h1 id="a0000000039">3 Ostrowski-type inequalities for double weighted integrals for higher order differentiable functions</h1>
<p> The following well known Ostrowski’s inequality is extracted from <span class="cite">
	[
	<a href="#Ostrowski" >16</a>
	]
</span>. </p>
<div class="equation" id="a0000000040">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> 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 <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span> 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. <div class="thm_thmwrapper " id="ost1">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3.1</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="a0000000041">
  \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*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000042">
  \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*}
</div>
<p> provided that </p>
<div class="displaymath" id="a0000000043">
  \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*}
</div>
<p> provided that </p>
<div class="displaymath" id="a0000000044">
  \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*}
</div>
<p> provided that </p>
<div class="displaymath" id="a0000000045">
  \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*}
</div>
<p> 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. <div class="proof_wrapper" id="a0000000046">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Identity \((\ref{id1})\) may be written as </p>
<div class="displaymath" id="a0000000047">
  \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*}
</div>
<p> Now, taking absolute value and applying Hölder’s inequality for double integrals, we easily obtain our required inequality. <div class="proof_wrapper" id="a0000000048">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000049">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">3.2</span>
  </div>
  <div class="rem_thmcontent">
  <p>For \(N=M=0\), Theorem 4 of <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span> becomes special case of Theorem <a href="#ost1">3.1</a> and we also retrieve results of <span class="cite">
	[
	<a href="#S.S. Dragomir" >9</a>
	]
</span> by simply putting \(p(.,.)=1\).<span class="qed">â–¡</span></p>

  </div>
</div> <div class="thm_thmwrapper " id="ost2">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3.3</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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., </p>
<div class="displaymath" id="a0000000050">
  \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*}
</div>
<p> \(1/p+1/q=1.\) Then, it follows that </p>
<div class="displaymath" id="a0000000051">
  \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*}
</div>
<p> for all \((x,y)\in [a,b]\times [c,d]\). <div class="proof_wrapper" id="a0000000052">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Identity \((\ref{id2})\) may be written as </p>
<div class="displaymath" id="a0000000053">
  \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*}
</div>
<p> Now taking absolute value and applying Hölder’s inequality for double integrals, we easily obtain our required inequality. <div class="proof_wrapper" id="a0000000054">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000055">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">3.4</span>
  </div>
  <div class="rem_thmcontent">
  <p>For \(N=M=0\), Theorem 5 of <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span> becomes special case of Theorem <a href="#ost2">3.3</a> and we also retrieve results of <span class="cite">
	[
	<a href="#Dragomir" >3</a>
	]
</span> and <span class="cite">
	[
	<a href="#S. Dragomir" >8</a>
	]
</span> by simply putting \(p(.,.)=1\).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000056">4 Grüss-type inequalities for double weighted integrals for higher order differentiable functions</h1>
<p> A celebrated integral inequality proved by Grüss <span class="cite">
	[
	<a href="#Gruss" >11</a>
	]
</span> in \(1935\), can be stated as follows (see <span class="cite">
	[
	<a href="#S. Mitrinovic" >15</a>
	, 
	p. 296
	]
</span>), </p>
<div class="displaymath" id="a0000000057">
  \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*}
</div>
<p> provided that \(f\) and \(g\) are two integrable functions on \([a, b]\) and satisfy the conditions </p>
<div class="displaymath" id="a0000000058">
  \[ m \le f (x) \le M,\quad {n \le g(x) \le N}, \]
</div>
<p> for all \(x \in [a, b]\), where \(m, M , n, N\) are real constants. </p>
<p>In <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span> 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. </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x000001AAA479BD70>">
  \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}
</div>
<div class="displaymath" id="a0000000059">
  \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}
</div>
<p> 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. </p>
<p>Now, we are ready to present our main results of this section by using notations defined above, which are as follows. <div class="thm_thmwrapper " id="gruss1">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">4.1</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="a0000000060">
  \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*}
</div>
<p> <div class="proof_wrapper" id="a0000000061">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> From \((\ref{id1})\), we have the following identities: </p>
<div class="displaymath" id="a">
  \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}
</div>
<div class="displaymath" id="b">
  \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}
</div>
<p> 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 </p>
<div class="displaymath" id="c">
  \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}
</div>
<p> Integrating \((\ref{c})\) over \([a,b]\times [c,d]\), we get </p>
<div class="displaymath" id="a0000000062">
  \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*}
</div>
<p> It may be written as </p>
<div class="displaymath" id="d">
  \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}
</div>
<p> Using \((\ref{A^{(i,j)}})\),\(\ldots \),\((\ref{CNM})\) we have the following inequalities </p>
<div class="displaymath" id="a0000000063">
  \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*}
</div>
<p> \(\forall (x,y)\in [a,b]\times [c,d]\). Taking absolute value on both sides in (<a href="#d">28</a>) and using all these inequalities in it, we get our required result. <div class="proof_wrapper" id="a0000000064">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="thm_thmwrapper " id="gruss2">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">4.2</span>
  </div>
  <div class="thm_thmcontent">
  <p> 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 </p>
<div class="displaymath" id="a0000000065">
  \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*}
</div>
<p> <div class="proof_wrapper" id="a0000000066">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> From \((\ref{id2})\), we have the following identities: </p>
<div class="displaymath" id="f">
  \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}
</div>
<p> 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 </p>
<div class="displaymath" id="h">
  \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}
</div>
<p> Integrating \((\ref{h})\) over \([a,b]\times [c,d]\), we get </p>
<div class="displaymath" id="i">
  \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}
</div>
<div class="equation" id="j">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> From \((\ref{i})\) and \((\ref{j})\), we obtain our required inequality. <div class="proof_wrapper" id="a0000000067">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> <div class="rem_thmwrapper " id="a0000000068">
  <div class="rem_thmheading">
    <span class="rem_thmcaption">
    Remark
    </span>
    <span class="rem_thmlabel">4.3</span>
  </div>
  <div class="rem_thmcontent">
  <p>For \(N=M=0\) Theorems 6 and 7 of <span class="cite">
	[
	<a href="#Montgomery" >20</a>
	]
</span> become special cases of Theorems <a href="#gruss1">4.1</a> and <a href="#gruss2">4.2</a> respectively and we also retrieve results of <span class="cite">
	[
	<a href="#Ostrowski" >16</a>
	]
</span> by simply putting \(p(.,.)=1\). For \(N=M=0\), we can also find similar results in <span class="cite">
	[
	<a href="#G-L" >12</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="acknowledgement_thmwrapper " id="a0000000069">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>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. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
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  <dd><p><i class="sc">J. Pečarić</i>, <i class="it">Some further remarks on the Ostrowski’s generalization of Čebyšev’s inequality</i>, J. Math. Anal. Appl., <b class="bfseries">123</b> (1987), pp. 18–33. </p>
</dd>
  <dt><a name="Montgomery">20</a></dt>
  <dd><p><i class="sc">J. Pečarić</i> and <i class="sc">A. Vukelić</i>, <i class="it">Montgomery’s identities for functions of two variables</i>, J. Math. Anal. Appl., <b class="bf">332</b> (2007), pp. 617–630. </p>
</dd>
</dl>


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