Weighted Grüss Type Inequalities for Double ... - Semantic Scholar
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), pp. 1-25
International Journal of Pure and Applied Sciences and Technology ISSN 2229 - 6107 Available online at www.ijopaasat.in Research Paper
Weighted Grüss Type Inequalities for Double Integrals on Time Scales Nazir Ahmad Mir1,*, Roman Ullah2 and Zarin Khan3 1
Department of Mathematics, Preston University, Islamabad, Pakistan
2
Department of Mathematics, Comsats Institute of Information Technology, Islamabad, Pakistan
3
Department of Mathematics, Princess Naura University, Saudi Arabia
* Corresponding author, e-mail: (nazirahmad.mir@gmail) (Received: 21-4-11/ Accepted: 7-12-11)
Abstract: We prove some weighted Grüss type inequalities for double integrals on time scales and unify the corresponding continuous and discrete versions, which are the generalizations of the results proved earlier in the literature
Keywords: Weight function, Grüss type inequality, Double integral, Time scale.
1. Introduction One of the most important integral inequalities proved by Gerhard Grüss 8 in 1935 is the Grüss Integral Inequality, which gives estimation for the integral of a product in terms of product of integrals and is defined as:
b
b
b
1 fx gx dx 1 fx dx 1 gx dx 1 M mN n, 1.1 4 ba a ba a ba a (1.1) provided that f and g are two integrable functions on a, b and satisfy the condition m fx M, n gx N, for all x a, b, where m, M, n, N are given real constants. Pachpatte inequalities involving two independent variables have been established by Umut
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
2
Mutlu Ozkan and Huseyin Yildrim 20 for double integrals on time scales. The aim of this paper is to establish the weighted form of these inequalities. We also apply our results for the continuous and discrete cases.
2. Time Scale Essentials A time scale T is an arbitrary non-empty closed subset of the set of real numbers. Some important examples of time scales are, , , , and . If T is a time scale, then for t T , we define the forward and backward jump operators respectively as t inf s T s t and t sup s T s t . A point t is said to be right-scattered if t t and is left-scattered if t t. A point that is at the same time right-scattered as well as left-scattered is called an isolated point. The point t is called right-dense if t t . If t t, then t is said to be left-dense. If t t t, then the point t is called dense. A function f : T is called rd-continuous (denoted by C rd ) if it is continuous at each lim fs ft right-dense point or maximal point of T , and its left-sided limit st exists at left dense points of T . 0, satisfying t t t, Let t T , then two functions , T t t t are called graininess functions. If T has a left-scattered maximal point t, then TK T t, otherwise TK T . A function g T is defined as g t gt for all t T . Let fT be a function on time scale, then for s TK , we define f s to be the number, if one exists, such that for all 0 there is a neighbourhood N of s such that for all u N,
|f s fu f ss u| |s u|. In this case, f s is called the delta derivative of fs at s. f is said to be delta differentiable on T , if f is differentiable at each s T . A function F : T is said to be derivative for all f : T if, F t ft t TK , and in this case, we define the -integrable of f as b
ftt Fb Fa, a
for each a, b T . Let T1 , T2 be two time scales. Let i , i and i be the forward jump operator, the backward jump operator and the delta differentiation, respectively on Ti , for i 1,2. Let a, b and c, d are the half-closed a, b T1 , c, d T2 , with a b, c d. bounded intervals in T1 and T2 respectively. Let us introduce a "rectangle" in T1 T2 by R a, b c, d t 1 , t 2 : t 1 a, b, t 2 c, d.
Let f be a real valued function on T1 T2 . This function f is said to be rd-continuous in
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3
t 2 if a 1 T1 , then function fa 1 , t 2 is rd-continuous on T2 , and this function f is said to be rd-continuous in t 1 if a 2 T2 , then ft 1 , a 2 is rd-continuous on T1 . CC rd denotes the set of functions ft 1 , t 2 on T1 T2 having the properties: i f is rd-continuous in t 1 and t 2 , ii if x 1 , x 2 T1 T2 with x 1 right dense and x 2 right dense, then f is continuous at x 1 , x 2 , iii if x 1 and x 2 are left dense limits, then the limit of ft 1 , t 2 exists as t 1 , t 2 approaches to x 1 , x 2 along any path in the region:
RLL x 1 , x 2 t 1 , t 2 : t 1 a, x 1 T1 , c, x 2 T2 . 1 Let CC rd denote the set of all functions in CC rd for which both the 1 partial derivative and 2 partial derivative exist and are in CC rd . In 3, Bohner has defined the norm as fx, 2 y f 1 x , y f sup sup |fx, y| sup x 2 x 1 x,ya,b c,d x,yR x,yR 1 where f CC rd a, b c, d, . Let , be rd-continuous, a, b, c T and , , then
b
b
b
a
a
i t tt tt tt, a b
a
ii tt tt, a
b
b
c
b
a b
a
c
iii tt tt tt, b
iv t tt b a ttt; a b
a b
v t tt b a ttt; a a
a
vi tt 0. a
The weight function a, b 0, is a non-negative integrable function and b
tdt . a
3. Main Results Here, we give some notations used to simplify the results. b
d
a
c
K (u, y) u ( x, v) v,
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Bx, y x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t, R
B1 x, y x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R
x, y fx, t t, 2 st, 2 s 2 s 1 t, 1 t R
B2 x, y x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R
x, y gy, s 1 t, s 1 t, s 2 s 1 t, 2 s R
B3 x, y x, y fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s
x, y fx, tgy, s R
2 t, st, s s t, 2 1 1 t 2 s
H1 x h 2 x, a h 2 x, b,
H2 y h 2 c, y h 2 d, y, M1 x, y |x, y|x, yx, y |x, y|x, yx, y, M2 x, y |x, y|
2 t, st, s 1 t 2 s
|x, y|
Fx, y
d
b
c
a
2 t, st, s , 1 t 2 s
x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s,
4
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Gx, y
d
b
c
a
5
x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s, 1 for , CC rd a, b c, d , , CC 1rd such that a, b c, d 0, and fa, b a, b , gc, d c, d are given by
t
u, yu, fx, t
if t a, x ,
a t
u, yu,
if t x, b
b
and s
x, v v, gy, s
if s c, y,
c s
x, v v,
if s y, d.
d
In order to obtain our main results, we need the following lemmas: Lemma 3.1 1 Let a, b T1 , c, d T2 , , CC rd and the weight function a, b c, d 0, such that CC 1rd , then for all x, y R a, b c, d t 1 , t 2 : t 1 a, b, t 2 c, d, we have i |B1 x, y| M1 x, yd cH1 x, ii |B2 x, y| M1 x, yb aH2 y, iii |B3 x, y| M2 x, yH1 xH2 y. Proof: i Consider B1 x, y ( x, y ) f ( x, t ) ( (t , 2 ( s )) (t , 2 ( s ))) 2 s1t 1t R ( x, y ) f ( x, t ) ( (t , 2 ( s )) (t , 2 ( s ))) 2 s1t 1t R ( x, t ) f ( x, t ) | ( (t , 2 ( s )) (t , 2 ( s ))) | 2 s1t 1t R ( x, y )
f x, t | t ( (t , R
2
( s )) (t , 2 ( s ))) | 2 s1t
1
(3.1) Using the definition of norm, we have
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t, st, s x, yx, y, 2 2 1 t
sup x,yR
and
sup x,yR
t, st, s x, yx, y. 2 2 1 t
So, the inequality 3. 1 becomes
|B1 x, y| |x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y d
x t
b t
c
aa
xb
u, yu 1 t u, yu 1 t 2 s |x, y|x, yx, y |x, y|x, yx, yd ch 2 x, a h 2 x, b M 1 x, yd cH1 x . ii Similarly, |B2 x, y| x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R |x, y| |gy, s| R
t, s t, s s t 1 1 2 1 2 s
|x, y| |gy, s| R
t, s t, s s t 1 1 2 1 2 s
|x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y |gy, s| 2 s 1 t R
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|x, y|x, yx, y |x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y b
y s
d s
a
c c
yd
x, v v 2 s x, v v 2 s 1 t |x, y|x, yx, y |x, y|x, yx, yb ah 2 c, y h 2 d, y M1 x, yb aH2 y. iii By following the same steps as above, we have |B3 x, y| x, y fx, t gy, s R
x, y fx, t gy, s R
|x, y| |fx, t ||gy, s| R
2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s
|x, y| |fx, t ||gy, s| R
2 t, st, s 1 t 2 s
|x, y|
|fx, t ||gy, s| 2 s 1 t R
2 t, st, s 1 t 2 s
|x, y|
2 t, st, s s t 2 1 1 t 2 s
|x, y|
2 t, st, s 1 t 2 s
|x, y|
2 t, st, s 1 t 2 s
|fx, t ||gy, s| 2 s 1 t R
|fx, t ||gy, s| 2 s 1 t R
M 2 x, y |fx, t ||gy, s| 2 s 1 t R b
d
a
c
M 2 x, y |fx, t | |gy, s| 2 s d
1 t
b
M 2 x, y |gy, s| 2 s |fx, t | 1 t c
M 2 x, y
a y s
d s
c c
y d
x, v v 2 s x, v v 2 s
x t
b t
a a
x b
u, yu 1 t u, yu 1 t M 2 x, yh 2 c, y h 2 d, yh 1 x, a h 1 x, b M 2 x, yH1 x H2 y.
Hence
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|B3 x, y| M2 x, yH1 x H2 y. Lemma 3.2 1 Let a, b T1 , c, d T2 , , CC rd a, b c, d, , and the weight function a, b c, d 0, such that b
d
a
c
CC 1rd and u, yu 0, x, v v 0. Then for all x, y R a, b c, d, we have
x, yx, y 1 1 t, 2 s 1 t, 2 s 2 s 1 t K R gy, s 1 t, s 1 t, s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R fx, tgy, s R
2 t, st, s s t . 2 1 1 t 2 s
Proof We use the following identity which can be easily proved by integration by parts
uu
zz
1
uu
ku, zzz z
1
uu
3.2
(3.2) where k, , is given by z
uu, if z , u, ku, z
z
uu, if z u,
and C rd , , . Indeed, we have
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u z
uuzz z
z
u
uuzz| u
u
zzz
u
uuuu zzz,
and z
zzzz z u z
uuzz| u
zzz u
u
u
uuuu zzz. Adding the above two identities, we have
uuuu zzz ku, zzz z this implies that uu
1
uu
zzz
1
uu
ku, zzz z,
which is the desired identity 3. 2. Now, we write the identity 3. 2 for the map . , y. , y,
x, yx, y
y c, d to obtain:
b
1 t, y 1 t, y 1 t
1 b
u, yu
a
a
1 b
u, yu
b
fx, t t, yt, y 1 t, t a
3.3
1
a
for all x, y R. Also, if we write 3. 2 for the map t, . t, . , we get
(3.3)
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t, yt, y
10
d
t, 2 st, 2 s 2 s
1 d
x, v v
c
c
d
1 d
x, v v
gy, s t, st, s 2 s. s
3.4
2
c
c
(3.4) The formula 3. 2 applied for the partial derivative t, yt, y 1 t
t, . t, . 1t
will produce
d
1 d
x, v v
t, 2 st, 2 s 2 s t 1
c
c
d
gy, s
1 d
x, v v
c
2 t, st, s 2 s. 1 t 2 s
3.5
(3.5)
c
Putting 3. 4 and 3. 5 in 3. 3 , we have x, yx, y
b
1
u, yu
a
b
d
x, v v
a
d
1 t , 2 s 1 t , 2 s 2 s
1
c
c
d
1 d
x, v v
gy, s 1 t , s 1 t , s 2 s 1 t s 2
c
c
b
fx, t
1 b
u, yu
a
d
x, v v
a
1
d
t, 2 st, 2 s 2 s t c
1
c
d
gy, s
1 d
x, v v
c
2 t, st, s 2 s 1 t 2 s
1 t
c
1 t , 2 s 1 t , 2 s 2 s 1 t
1 b
d
a
c
u, yu x, v v
R
gy, s 1 t , s 1 t , s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R fx, t gy, s R
2 t, st, s 2 s 1 t 1 t 2 s
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1 ( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s)) 2 s1t K R
g ( y, s ) ( 1 (t ), s) ( 1 (t ), s) 2 s1t R
f ( x, t ) R
(t , 2 ( s)) (t , 2 ( s)) 2 s1t 1t
2 f ( x, t ) (t , s) (t , s) 2 s1t . R 1t 2 s Hence, the desired result. Lemma 3.3 Let a, b, x T1 with a x b and c, d, y T2 with c y d and the rectangle R a, b c, d be the union of four disjoint rectangles of forms R1 a, x c, y , R2 a, x y, d , R3 x, b c, y and R4 x, b y, d. 1 Let , CC rd a, b c, d, and a, b c, d 0, be the weight function, then
2 f ( x, t ) g ( y , s ) ((t , s) (t , s)) 2 s1t R 1t 2 s d
b
c
a
K ( x, y ) ( x, y ) ( x, v) v (( 1 (t ), y ) ( 1 (t ), y ))1t b
d
a
c
(u, y ) u (( x, 2 ( s)) ( x, 2 ( s))) 2 s (( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s))) 2 s1t. R
Proof 1 Let , CC rd a, b c, d, , then for the rectangle R1 , we have
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2 fx, tgy, s t, st, s 2 s 1 t
1 t 2 s
R1 x y
t
ac
a
s
u, yu x, v v c
x t
y
s
aa
c
c
x t
s
2 t, st, s s t 2 1 1 t 2 s
t, st, s 2 s 1 t u, yu x, v v t s 2
1
2
y u, yu x, v v t, st, s| c 1 t aa c y
t, 2 st, 2 s 2 s 1 t c 1 t y
x
t
x, v v u, yu t, yt, y 1 t 1 t c a a y
x t
u, yu t, 2 st, 2 s 1 t 2 s 1 t c aa y
x
x
a
a
x, v v u, yux, yx, y 1 t, y 1 t, y 1 t c y
x
u, yut, 2 st, 2 s c
a
x
1 t, 2 s 1 t, 2 s 1 t 2 s a
y
x
x, vv u, yux, yx, y c
a y
x
x, vv 1 t, y 1 t, y 1 t c x
a y
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.6
R1
Also, by similar computation for the rectangles R2 identities:
and R3 ,
we have the respective
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13
2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R2 d
x
y
a
x,vv u,yux,yx,y d
x
y
a
x
d
a
y
x,vv 1 t,y 1 t,y 1 t u,yu x, 2 sx, 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t 3.7 R2
(3.7) and 2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R3 y
b
x,vv u,yux,yx,y c
x y
b
x,vv 1 t,y 1 t,y 1 t c
x
b
y
x
c
u,yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.8
R3
(3.8) Now, finally for the rectangle R4 , we have
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2 fx, tgy, s t, st, s 2 s 1 t
1 t 2 s
R4 bd
t
s
x y
b
d
u, yu x, v v b t
s
xb
d
u, yu x, v v
2 t, st, s s t 2 1 1 t 2 s t, st, s| d y 1 t
d
t, 2 st, 2 s 2 s 1 t 1 t y d
b t
x, v v u, yu t, yt, y 1 t 1 t y x b
d
b t
u, yu t, 2 st, 2 s 1 t 2 s 1 t y x b
d
t
b
y
b
x
x, v v u, yut, yt, y| bx 1 t, y 1 t, y 1 t d
t
y
b
u, yut, 2 st, 2 s| bx b
1 t, 2 s 1 t, 2 s 1 t 2 s x
d
b
y
x
x,vv u,yux,yx,y y
b
x,vv 1 t,y 1 t,y 1 t c
x
b
d
x
y
u,yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.9
R4
(3.9) Adding the identities 3. 6,
3. 7,
3. 8 and 3. 9, we have
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2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R y
x
d
x
c
a
y
a
y
b
d
b
x
y
x
x,vv u,yu x,vv u,yu x,vv u,yu x,vv u,yu x,yx,y c y
d
x
y
a
x,vv x,vv 1 t,y 1 t,y 1 t
c y
d
b
y
x
x,vv x,vv 1 t,y 1 t,y 1 t c x
y
a
c
d
u,yu x, 2 sx, 2 s 2 s x, 2 sx, 2 s 2 s b
y
x
c
y d
u,yu x, 2 sx, 2 s 2 s x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R1
1 t, 2 s 1 t, 2 s 2 s 1 t R2
1 t, 2 s 1 t, 2 s 2 s 1 t R3
1 t, 2 s 1 t, 2 s 2 s 1 t R4
y
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y
16
d
x
y
a
x, v v x, v v u, yu
c y
d
b
x, v v x, v v c
u, yu x, yx, y
y d
x
c
a
x
x, v v 1 t, y 1 t, y 1 t
d
b
c
x
x, v v 1 t, y 1 t, y 1 t x
d
a
c
u, yu x, 2 sx, 2 s 2 s
b
d
x
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R d
b
x, v v u, yux, yx, y c
a d
b
x, v v 1 t, y 1 t, y 1 t c
a
b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R
d
b
c
a
Kx, yx, y x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t. R
Theorem 3.1 Let a, b T1 ,
c, d T2 ,
, CC 1rd a, b c, d, and let b
a, b c, d 0, be the weight function such that CC 1rd and , then
u, yu 0 a
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1 ( x, y ) ( x, y ) ( x, y ) 2 y1 x K R
1 ( x, y ) 2 y1 x ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x 2 R 2K R
1 ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x ( x, y ) 2 y1 x R 2K 2 R
1 [ M 1 ( x, y ) ((d c) H1 ( x) (b a) H 2 ( y )) 2K 2 R M 2 ( x, y ) H1 ( x) H 2 ( y )] 2 y 1 x.
(3.10) Proof: From the identity of Lemma 3. 2 , we have the following identities:
K ( x , y ) ( x, y )
( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s)) 2 s1t R
g ( y, s) ( 1 (t ), s) ( 1 (t ), s) 2 s1t R
f ( x, t ) R
(t , 2 ( s)) (t , 2 ( s)) 2 s1t 1t
2 f ( x, t ) g ( y , s ) (t , s ) (t , s) 2 s1t. R 1t 2 s (3.11) and
Kx, yx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
gy, s 1 t, s 1 t, s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R 2 fx, tgy, s t, st, s 2 s 1 t. 1 t 2 s R
3.12
(3.12) Multiplying 3. 11 with x, y and 3. 12 with x, y and then adding, we have
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2Kx, yx, yx, y x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R x, y fx, tgy, s R
x, y fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
B1 x, y B2 x, y B3 x, y.
3.13 (3.13)
Integrating 3. 13 on R , we get
x, y x, y x, y 2 y1 x R
1 x, y 2 y1 x 1 x , 2 y 1 x , 2 y 2 y1 x R 2 K R x, y 2 y1 x 1 x , 2 y 1 x , 2 y 2 y1 x R R 1 B1 x, y B2 x, y B3 x, y 2 y1 x. 2K R
This implies that
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1 ( x, y ) ( x, y ) ( x, y ) 2 y1 x K R
1 ( x, y ) 2 y1 x ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x 2 R R 2K
1 ( x, y ) 2 y1 ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x R 2K 2 R 1 [ B1 ( x, y ) B2 ( x, y ) B3 ( x, y )] 2 y 1 x 2K 2 R 1 [ M 1 ( x, y ) ((d c) H1 ( x) (b a) H 2 ( y )) 2K 2 R M 2 ( x, y ) H1 ( x) H 2 ( y )] 2 y1 x
.
(3.14) This completes the proof. Corollary 3.1 (Continuous case) Let T1 T2 , then our delta integral is the usual Riemann integral from calculus. This leads us to state the following inequality:
1 K
( x, y) ( x, y) ( x, y) d
2
yd1 x
R
1 2K 2
( x, y) d
2
yd1 x ( x, y ) ( x, y ) d 2 yd1 x
R
R
1 2K 2
( x, y) ( x, y) d
1 2K 2
[M ( x, y) ((d c) H ( x) (b a) H
2
yd1 x x, y d 2 yd1 x
R
R
1
1
2
( y ))
R
M 2 ( x, y ) H1 ( x) H 2 ( y )] d 2 yd1 x, b
d
a
c
where K (u, y) du ( x, v) dv. Corollary 3.2 (Discrete case) Let T1 T2 , a 0, b m, x, y u x , u y . Then
c 0,
d n,
x i,
y j and
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
1 m n 1 (ai , b j ) (ci , d j ) ( pi , q j ) K1 i 1 j 1 2 K12 1 2 K12
m
n
M
3
1 2 K12
m
20
m
n
m
n
(ai , b j ) (ci , d j ) ( pi , q j ) i 1 j 1
i 1 j 1
n
m
n
(ai , b j ) ( pi , q j ) (ci , d j ) i 1 j 1
i 1 j 1
( x, y ) ((d c) H1 ( x) (b a) H 2 ( y ))
i 1 j 1
M 4 ( x, y ) H1 ( x ) H 2 ( y ) , m
n
i 1
j 1
where K1 ( pi , y) ( x, q j ),
M3 |v x , v y |u x , u y w x , w y |u x , u y |v x , v y w x , w y and M4 |v x , v y | t s u t , u s w t , w s |u x , u y | t s v t , v s w t , w s . Theorem 3.2 Let a, b T1 ,
c, d T2 ,
, CC 1rd a, b, c, d, and let b
: a, b c, d 0, be the weight function such that CC 1rd and
u, yu 0, a
d
x, v v 0, c
then
1 x, yx, yx,y y x 2 1 K R 1 1 x,y 2 y 1 x 2 K R
1 x, y x, y y x 1 2 1 2 2 1 K R
1 1 x, y 2 y 1 x 2 K R
1 x, y x, y y x 1 2 1 2 2 1 K R
1 2 x,yFx, y x, yGx, y 2 y 1 x 2K R 1 2 M2 x,yH1 xH2 y 2 y 1 x. 2K R
3.15 (3.15)
Proof From identity 3. 6 in lemma 3.3, we have the following identities:
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
21
K x, yx, y d
b
x, v v 1 t, y 1 t, y 1 t c
a b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s
Fx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t. 2 1 1 t 2 s
3.16 (3.16)
and Kx, yx, y Gx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t. 2 1 1 t 2 s
3.17
(3.17) For x, y R, multiplying 3. 16 with gx, y and 3. 17 with fx, y and then adding, we have
2Kx,yx,yx,y x,yFx,y x,yGx,y x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R
x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R
x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R 2 x,y fx,tgy,s t,st,s 2 s 1 t 1 t 2 s R
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
22
x, yFx, y x, yGx, y
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
B3 x, y.
3.18 (3.18)
Integrating 3. 18 on R, we get
x, yx, yx, y 2 y 1 x R
1 x, yFx, y x, yGx, y 2 y 1 x 2K 1 x, y 2 y 1 x 1 x , 2 y 1 x , 2 y 2 y 1 x 2K R R x, y 2 y 1 x 1 x , 2 y 1 x , 2 y 2 y 1 x R
R
1 B3 x, y 2 y 1 x. 2K R
3.19 (3.19)
From the identity 3. 19 and the inequality iii in lemma 3.1, we have
1 x, yx, yx, y y x 2 1 K R 1 2
1 x, y y x 2 1 K R
1 x , y x , y y x 1 2 1 2 2 1 K R
1 2
1 x, y y x 2 1 K R
1 x , y x , y y x 1 2 1 2 2 1 K R
1 2 x, yFx, y x, yGx, y 2 y 1 x 2K R
1 |A x, y| y x 3 2 1 2K 2 R
1 M x, yH x H y y x. 2 1 2 2 1 2K 2 R
3.20
(3.20) This completes the proof. Corollary 3.3 (Continuous case) Let T1 T2 , then our delta integral is the usual Riemann integral from calculus. This leads us to state the following inequality:
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
23
1 ( x, y ) ( x, y ) ( x, y ) d 2 yd1 x K R 1 1 1 ( x, y ) d 2 yd1 x 1 ( x), 2 ( y ) 1 ( x), 2 ( y ) d 2 yd1 x 2 K R K R 1 1 1 ( x, y ) d 2 yd1 x 1 ( x), 2 ( y ) 1 ( x), 2 ( y ) d 2 yd1 x 2 K R K R 1 ( x, y ) F ( x, y ) ( x, y ) G ( x, y ) d 2 yd1 x 2K 2 R 1 M 2 ( x, y ) H1 ( x) H 2 ( y ) d 2 yd1 x, 2K 2 R
b
d
a
c
where K (u, y ) du ( x, v) dv,
F 1 x, y
d
b
b
d
c
a
a
c
d
b
b
d
c
a
a
c
x, v dv t, yt, yd 1 t u, ydu x, sx, sd 2 s
and
G 1 x, y
x, v dv t, yt, yd 1 t u, ydu x, sx, sd 2 s.
Corollary 3.4 (Discrete case) Let T1 T2 , a 0, b m, x, y u x , u y . Then
c 0,
d n,
x i,
y j and
1 m n (ai , b j ) (ci , d j ) ( pi , q j ) K1 i 1 j 1 1 m n 1 1 m n (ai , b j ) ( c , d ) ( p , q ) i j i j K i 1 j 1 2 K1 i 1 j 1 1 1 m n 1 1 m n (ci , d j ) ( a , b ) ( p , q ) i j i j K i 1 j 1 2 K1 i 1 j 1 1
1 m n (ci , d j ) F2 (ai , b j ) G2 2 K12 i 1 j 1
1 m n M 4 ( x, y) H1 ( x) H 2 ( y), 2 K12 i 1 j 1
where n
m
n
m
j 1
i 1
j 1
i 1
F2 (w, w j ) (ui , u) ( wi , w) ( wi , w) (u, u j ) (w, w j ) and
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
n
m
n
m
j 1
i 1
j 1
i 1
24
G2 (w, w j ) (vi , v) (wi , w) ( wi , w) (v, v j ) ( w, w j ). Remark 3.1 If we put the weight function ( x, y ) 1 in both the inequalities (3.10) and (3.15), we will obtain the results proved in [20].
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
[16] [17] [18] [19] [20] [21]
R. Agarwal, M. Bohner and A. Peterson, Inequalities on time scales: a survey, Inequal. Appl., 4 (4) (2001), 535-557. Farooq Ahmad, Arif Rafiq and Nazir Ahmad Mir, Weighted ostrowski-grüss type inequality for differentiable mappings, Glob. J. Pure Appl. Math., 2(2) (2006), 147-154. M. Bohner and A. Peterson, Dynamic Equations on Time Scales, An Introduction with Applications, Boston, MA. Birkhäser Boston Inc., (2001). M. Bohner and T. Mathews, The gruss inequality on time scales, Commun. Math. Anal., 3 (1) (2007), 1-8. M. Bohner and G. Sh. Guseinov, Double integral calculus of variations on time scales, Comput. Math. Appl., 54 (2007), 45-57. S. S. Dragomir and Young-Ho Kim, On certian new integral inequalities and their applications, J. Ineq. Pure Appl. Math., 3 (4) (2002), Article 65. S. S. Dragomir and A. Sofo, An integral inequality for twice differentiable mappings and applications, Tamkang Journal of Mathematics, 31(4) (2000), 257-266. G. Grüss, Uber das maximum des absoluten betrages von, Math. Z., 39 (1935), 215-226. S. Hilger, Ein ma kettenkül mit anwendung auf zentrumsmannigfaltigkeiten, PhD Thesis, Universität Würzburg, 1988. Wenjun Liu and Quôc-Anh Nĝo, An ostrowski-grüss type inequality on time scales, Comput. Math. Appl., 58 (2009), 1207-1210. W. J. Liu, Q. A. Nĝo and W. B. Chen, Ostrowski type inequalities on time scales for double integrals, Acta Appl. Math., 110(1) (2010), 477-497. N. A. Mir and A. Rafiq, A note on ostrowski like inequalities in L 1 (a, b) spaces, Gen. Math., 14(1) (2006), 23-30. N. A. Mir, A. Rafiq and F. Ahmad, Weighted ostrowski type inequality for differentiable mappings whose derivative belong to L ( a, b), , Gen. Math., 14(3) (2006), 27-38. Nazir Ahmad Mir, Arif Rafiq and Farooq Ahmad, Generalization of integral inequalities for functions whose modulus of nth derivatives are convex, Gen. Math., 14(4) (2007), 67-92. Nazir Ahmad Mir, Arif Rafiq and Muhammad Rizwan, Ostrowski-grüss cebysew type inequalities for functions whose modulus for second derivatives are convex, Gen. Math., 16(3) (2008), 111-134. A. Ostrowski, Uber die absolutabweichung einer differentienbaren Funktionen von ihren integralmittelwert, Comment. Math. Helv., 10 (1938), 226-227. B. G. Pachpatte, On grüss type inequality for double integrals, J. Math. Anal. Appl., 267 (2002), 454-459. B. G. Pachpatte, New inequalities of ostrowski and grüss type for triple integrals, Tamkang J. Math., 2 (2009), 117-127. U. M. Özkan and H. Yildirim, Ostrowski type inequality for double integrals on time scales, Acta Appl. Math., 110(1)(2010), 283-288. Umut Mutlu Özkan and Huseyin Yildrim, Grüss type inequalities for double integrals on time scales, Comput. Math. Appl., 57 (2009), 436-444. A. Rafiq, N. A. Mir and Fiza Zafar, A generalized ostrowski-grüss type inequality for twice differentiable mappings in euclidean norm, Gen. Math., 16(3) (2008), 51-72 .
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[23]
25
Arif Rafiq, Nazir Ahmad Mir and Fiza Zafar, A generalized ostrowski type inequality for a random variable whose probability density function belongs to L a, b . Demonstratio Math., 41(3) (2008), 723-732 . A. Sofo, Double integral inequalities based on multi-branch peano kernels, Math. Ineq. & Appl., 5(3) (2002), 491-504.
International Journal of Pure and Applied Sciences and Technology ISSN 2229 - 6107 Available online at www.ijopaasat.in Research Paper
Weighted Grüss Type Inequalities for Double Integrals on Time Scales Nazir Ahmad Mir1,*, Roman Ullah2 and Zarin Khan3 1
Department of Mathematics, Preston University, Islamabad, Pakistan
2
Department of Mathematics, Comsats Institute of Information Technology, Islamabad, Pakistan
3
Department of Mathematics, Princess Naura University, Saudi Arabia
* Corresponding author, e-mail: (nazirahmad.mir@gmail) (Received: 21-4-11/ Accepted: 7-12-11)
Abstract: We prove some weighted Grüss type inequalities for double integrals on time scales and unify the corresponding continuous and discrete versions, which are the generalizations of the results proved earlier in the literature
Keywords: Weight function, Grüss type inequality, Double integral, Time scale.
1. Introduction One of the most important integral inequalities proved by Gerhard Grüss 8 in 1935 is the Grüss Integral Inequality, which gives estimation for the integral of a product in terms of product of integrals and is defined as:
b
b
b
1 fx gx dx 1 fx dx 1 gx dx 1 M mN n, 1.1 4 ba a ba a ba a (1.1) provided that f and g are two integrable functions on a, b and satisfy the condition m fx M, n gx N, for all x a, b, where m, M, n, N are given real constants. Pachpatte inequalities involving two independent variables have been established by Umut
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
2
Mutlu Ozkan and Huseyin Yildrim 20 for double integrals on time scales. The aim of this paper is to establish the weighted form of these inequalities. We also apply our results for the continuous and discrete cases.
2. Time Scale Essentials A time scale T is an arbitrary non-empty closed subset of the set of real numbers. Some important examples of time scales are, , , , and . If T is a time scale, then for t T , we define the forward and backward jump operators respectively as t inf s T s t and t sup s T s t . A point t is said to be right-scattered if t t and is left-scattered if t t. A point that is at the same time right-scattered as well as left-scattered is called an isolated point. The point t is called right-dense if t t . If t t, then t is said to be left-dense. If t t t, then the point t is called dense. A function f : T is called rd-continuous (denoted by C rd ) if it is continuous at each lim fs ft right-dense point or maximal point of T , and its left-sided limit st exists at left dense points of T . 0, satisfying t t t, Let t T , then two functions , T t t t are called graininess functions. If T has a left-scattered maximal point t, then TK T t, otherwise TK T . A function g T is defined as g t gt for all t T . Let fT be a function on time scale, then for s TK , we define f s to be the number, if one exists, such that for all 0 there is a neighbourhood N of s such that for all u N,
|f s fu f ss u| |s u|. In this case, f s is called the delta derivative of fs at s. f is said to be delta differentiable on T , if f is differentiable at each s T . A function F : T is said to be derivative for all f : T if, F t ft t TK , and in this case, we define the -integrable of f as b
ftt Fb Fa, a
for each a, b T . Let T1 , T2 be two time scales. Let i , i and i be the forward jump operator, the backward jump operator and the delta differentiation, respectively on Ti , for i 1,2. Let a, b and c, d are the half-closed a, b T1 , c, d T2 , with a b, c d. bounded intervals in T1 and T2 respectively. Let us introduce a "rectangle" in T1 T2 by R a, b c, d t 1 , t 2 : t 1 a, b, t 2 c, d.
Let f be a real valued function on T1 T2 . This function f is said to be rd-continuous in
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
3
t 2 if a 1 T1 , then function fa 1 , t 2 is rd-continuous on T2 , and this function f is said to be rd-continuous in t 1 if a 2 T2 , then ft 1 , a 2 is rd-continuous on T1 . CC rd denotes the set of functions ft 1 , t 2 on T1 T2 having the properties: i f is rd-continuous in t 1 and t 2 , ii if x 1 , x 2 T1 T2 with x 1 right dense and x 2 right dense, then f is continuous at x 1 , x 2 , iii if x 1 and x 2 are left dense limits, then the limit of ft 1 , t 2 exists as t 1 , t 2 approaches to x 1 , x 2 along any path in the region:
RLL x 1 , x 2 t 1 , t 2 : t 1 a, x 1 T1 , c, x 2 T2 . 1 Let CC rd denote the set of all functions in CC rd for which both the 1 partial derivative and 2 partial derivative exist and are in CC rd . In 3, Bohner has defined the norm as fx, 2 y f 1 x , y f sup sup |fx, y| sup x 2 x 1 x,ya,b c,d x,yR x,yR 1 where f CC rd a, b c, d, . Let , be rd-continuous, a, b, c T and , , then
b
b
b
a
a
i t tt tt tt, a b
a
ii tt tt, a
b
b
c
b
a b
a
c
iii tt tt tt, b
iv t tt b a ttt; a b
a b
v t tt b a ttt; a a
a
vi tt 0. a
The weight function a, b 0, is a non-negative integrable function and b
tdt . a
3. Main Results Here, we give some notations used to simplify the results. b
d
a
c
K (u, y) u ( x, v) v,
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
Bx, y x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t, R
B1 x, y x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R
x, y fx, t t, 2 st, 2 s 2 s 1 t, 1 t R
B2 x, y x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R
x, y gy, s 1 t, s 1 t, s 2 s 1 t, 2 s R
B3 x, y x, y fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s
x, y fx, tgy, s R
2 t, st, s s t, 2 1 1 t 2 s
H1 x h 2 x, a h 2 x, b,
H2 y h 2 c, y h 2 d, y, M1 x, y |x, y|x, yx, y |x, y|x, yx, y, M2 x, y |x, y|
2 t, st, s 1 t 2 s
|x, y|
Fx, y
d
b
c
a
2 t, st, s , 1 t 2 s
x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s,
4
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
Gx, y
d
b
c
a
5
x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s, 1 for , CC rd a, b c, d , , CC 1rd such that a, b c, d 0, and fa, b a, b , gc, d c, d are given by
t
u, yu, fx, t
if t a, x ,
a t
u, yu,
if t x, b
b
and s
x, v v, gy, s
if s c, y,
c s
x, v v,
if s y, d.
d
In order to obtain our main results, we need the following lemmas: Lemma 3.1 1 Let a, b T1 , c, d T2 , , CC rd and the weight function a, b c, d 0, such that CC 1rd , then for all x, y R a, b c, d t 1 , t 2 : t 1 a, b, t 2 c, d, we have i |B1 x, y| M1 x, yd cH1 x, ii |B2 x, y| M1 x, yb aH2 y, iii |B3 x, y| M2 x, yH1 xH2 y. Proof: i Consider B1 x, y ( x, y ) f ( x, t ) ( (t , 2 ( s )) (t , 2 ( s ))) 2 s1t 1t R ( x, y ) f ( x, t ) ( (t , 2 ( s )) (t , 2 ( s ))) 2 s1t 1t R ( x, t ) f ( x, t ) | ( (t , 2 ( s )) (t , 2 ( s ))) | 2 s1t 1t R ( x, y )
f x, t | t ( (t , R
2
( s )) (t , 2 ( s ))) | 2 s1t
1
(3.1) Using the definition of norm, we have
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6
t, st, s x, yx, y, 2 2 1 t
sup x,yR
and
sup x,yR
t, st, s x, yx, y. 2 2 1 t
So, the inequality 3. 1 becomes
|B1 x, y| |x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y |fx, t| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y d
x t
b t
c
aa
xb
u, yu 1 t u, yu 1 t 2 s |x, y|x, yx, y |x, y|x, yx, yd ch 2 x, a h 2 x, b M 1 x, yd cH1 x . ii Similarly, |B2 x, y| x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R |x, y| |gy, s| R
t, s t, s s t 1 1 2 1 2 s
|x, y| |gy, s| R
t, s t, s s t 1 1 2 1 2 s
|x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y |gy, s| 2 s 1 t R
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7
|x, y|x, yx, y |x, y|x, yx, y |gy, s| 2 s 1 t R
|x, y|x, yx, y |x, y|x, yx, y b
y s
d s
a
c c
yd
x, v v 2 s x, v v 2 s 1 t |x, y|x, yx, y |x, y|x, yx, yb ah 2 c, y h 2 d, y M1 x, yb aH2 y. iii By following the same steps as above, we have |B3 x, y| x, y fx, t gy, s R
x, y fx, t gy, s R
|x, y| |fx, t ||gy, s| R
2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s
|x, y| |fx, t ||gy, s| R
2 t, st, s 1 t 2 s
|x, y|
|fx, t ||gy, s| 2 s 1 t R
2 t, st, s 1 t 2 s
|x, y|
2 t, st, s s t 2 1 1 t 2 s
|x, y|
2 t, st, s 1 t 2 s
|x, y|
2 t, st, s 1 t 2 s
|fx, t ||gy, s| 2 s 1 t R
|fx, t ||gy, s| 2 s 1 t R
M 2 x, y |fx, t ||gy, s| 2 s 1 t R b
d
a
c
M 2 x, y |fx, t | |gy, s| 2 s d
1 t
b
M 2 x, y |gy, s| 2 s |fx, t | 1 t c
M 2 x, y
a y s
d s
c c
y d
x, v v 2 s x, v v 2 s
x t
b t
a a
x b
u, yu 1 t u, yu 1 t M 2 x, yh 2 c, y h 2 d, yh 1 x, a h 1 x, b M 2 x, yH1 x H2 y.
Hence
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8
|B3 x, y| M2 x, yH1 x H2 y. Lemma 3.2 1 Let a, b T1 , c, d T2 , , CC rd a, b c, d, , and the weight function a, b c, d 0, such that b
d
a
c
CC 1rd and u, yu 0, x, v v 0. Then for all x, y R a, b c, d, we have
x, yx, y 1 1 t, 2 s 1 t, 2 s 2 s 1 t K R gy, s 1 t, s 1 t, s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R fx, tgy, s R
2 t, st, s s t . 2 1 1 t 2 s
Proof We use the following identity which can be easily proved by integration by parts
uu
zz
1
uu
ku, zzz z
1
uu
3.2
(3.2) where k, , is given by z
uu, if z , u, ku, z
z
uu, if z u,
and C rd , , . Indeed, we have
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9
u z
uuzz z
z
u
uuzz| u
u
zzz
u
uuuu zzz,
and z
zzzz z u z
uuzz| u
zzz u
u
u
uuuu zzz. Adding the above two identities, we have
uuuu zzz ku, zzz z this implies that uu
1
uu
zzz
1
uu
ku, zzz z,
which is the desired identity 3. 2. Now, we write the identity 3. 2 for the map . , y. , y,
x, yx, y
y c, d to obtain:
b
1 t, y 1 t, y 1 t
1 b
u, yu
a
a
1 b
u, yu
b
fx, t t, yt, y 1 t, t a
3.3
1
a
for all x, y R. Also, if we write 3. 2 for the map t, . t, . , we get
(3.3)
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t, yt, y
10
d
t, 2 st, 2 s 2 s
1 d
x, v v
c
c
d
1 d
x, v v
gy, s t, st, s 2 s. s
3.4
2
c
c
(3.4) The formula 3. 2 applied for the partial derivative t, yt, y 1 t
t, . t, . 1t
will produce
d
1 d
x, v v
t, 2 st, 2 s 2 s t 1
c
c
d
gy, s
1 d
x, v v
c
2 t, st, s 2 s. 1 t 2 s
3.5
(3.5)
c
Putting 3. 4 and 3. 5 in 3. 3 , we have x, yx, y
b
1
u, yu
a
b
d
x, v v
a
d
1 t , 2 s 1 t , 2 s 2 s
1
c
c
d
1 d
x, v v
gy, s 1 t , s 1 t , s 2 s 1 t s 2
c
c
b
fx, t
1 b
u, yu
a
d
x, v v
a
1
d
t, 2 st, 2 s 2 s t c
1
c
d
gy, s
1 d
x, v v
c
2 t, st, s 2 s 1 t 2 s
1 t
c
1 t , 2 s 1 t , 2 s 2 s 1 t
1 b
d
a
c
u, yu x, v v
R
gy, s 1 t , s 1 t , s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R fx, t gy, s R
2 t, st, s 2 s 1 t 1 t 2 s
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
11
1 ( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s)) 2 s1t K R
g ( y, s ) ( 1 (t ), s) ( 1 (t ), s) 2 s1t R
f ( x, t ) R
(t , 2 ( s)) (t , 2 ( s)) 2 s1t 1t
2 f ( x, t ) (t , s) (t , s) 2 s1t . R 1t 2 s Hence, the desired result. Lemma 3.3 Let a, b, x T1 with a x b and c, d, y T2 with c y d and the rectangle R a, b c, d be the union of four disjoint rectangles of forms R1 a, x c, y , R2 a, x y, d , R3 x, b c, y and R4 x, b y, d. 1 Let , CC rd a, b c, d, and a, b c, d 0, be the weight function, then
2 f ( x, t ) g ( y , s ) ((t , s) (t , s)) 2 s1t R 1t 2 s d
b
c
a
K ( x, y ) ( x, y ) ( x, v) v (( 1 (t ), y ) ( 1 (t ), y ))1t b
d
a
c
(u, y ) u (( x, 2 ( s)) ( x, 2 ( s))) 2 s (( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s))) 2 s1t. R
Proof 1 Let , CC rd a, b c, d, , then for the rectangle R1 , we have
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
12
2 fx, tgy, s t, st, s 2 s 1 t
1 t 2 s
R1 x y
t
ac
a
s
u, yu x, v v c
x t
y
s
aa
c
c
x t
s
2 t, st, s s t 2 1 1 t 2 s
t, st, s 2 s 1 t u, yu x, v v t s 2
1
2
y u, yu x, v v t, st, s| c 1 t aa c y
t, 2 st, 2 s 2 s 1 t c 1 t y
x
t
x, v v u, yu t, yt, y 1 t 1 t c a a y
x t
u, yu t, 2 st, 2 s 1 t 2 s 1 t c aa y
x
x
a
a
x, v v u, yux, yx, y 1 t, y 1 t, y 1 t c y
x
u, yut, 2 st, 2 s c
a
x
1 t, 2 s 1 t, 2 s 1 t 2 s a
y
x
x, vv u, yux, yx, y c
a y
x
x, vv 1 t, y 1 t, y 1 t c x
a y
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.6
R1
Also, by similar computation for the rectangles R2 identities:
and R3 ,
we have the respective
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
13
2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R2 d
x
y
a
x,vv u,yux,yx,y d
x
y
a
x
d
a
y
x,vv 1 t,y 1 t,y 1 t u,yu x, 2 sx, 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t 3.7 R2
(3.7) and 2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R3 y
b
x,vv u,yux,yx,y c
x y
b
x,vv 1 t,y 1 t,y 1 t c
x
b
y
x
c
u,yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.8
R3
(3.8) Now, finally for the rectangle R4 , we have
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
14
2 fx, tgy, s t, st, s 2 s 1 t
1 t 2 s
R4 bd
t
s
x y
b
d
u, yu x, v v b t
s
xb
d
u, yu x, v v
2 t, st, s s t 2 1 1 t 2 s t, st, s| d y 1 t
d
t, 2 st, 2 s 2 s 1 t 1 t y d
b t
x, v v u, yu t, yt, y 1 t 1 t y x b
d
b t
u, yu t, 2 st, 2 s 1 t 2 s 1 t y x b
d
t
b
y
b
x
x, v v u, yut, yt, y| bx 1 t, y 1 t, y 1 t d
t
y
b
u, yut, 2 st, 2 s| bx b
1 t, 2 s 1 t, 2 s 1 t 2 s x
d
b
y
x
x,vv u,yux,yx,y y
b
x,vv 1 t,y 1 t,y 1 t c
x
b
d
x
y
u,yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t.
3.9
R4
(3.9) Adding the identities 3. 6,
3. 7,
3. 8 and 3. 9, we have
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
15
2 fx,tgy,s t,st,s 2 s 1 t
1 t 2 s
R y
x
d
x
c
a
y
a
y
b
d
b
x
y
x
x,vv u,yu x,vv u,yu x,vv u,yu x,vv u,yu x,yx,y c y
d
x
y
a
x,vv x,vv 1 t,y 1 t,y 1 t
c y
d
b
y
x
x,vv x,vv 1 t,y 1 t,y 1 t c x
y
a
c
d
u,yu x, 2 sx, 2 s 2 s x, 2 sx, 2 s 2 s b
y
x
c
y d
u,yu x, 2 sx, 2 s 2 s x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R1
1 t, 2 s 1 t, 2 s 2 s 1 t R2
1 t, 2 s 1 t, 2 s 2 s 1 t R3
1 t, 2 s 1 t, 2 s 2 s 1 t R4
y
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
y
16
d
x
y
a
x, v v x, v v u, yu
c y
d
b
x, v v x, v v c
u, yu x, yx, y
y d
x
c
a
x
x, v v 1 t, y 1 t, y 1 t
d
b
c
x
x, v v 1 t, y 1 t, y 1 t x
d
a
c
u, yu x, 2 sx, 2 s 2 s
b
d
x
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R d
b
x, v v u, yux, yx, y c
a d
b
x, v v 1 t, y 1 t, y 1 t c
a
b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R
d
b
c
a
Kx, yx, y x, v v 1 t, y 1 t, y 1 t b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t. R
Theorem 3.1 Let a, b T1 ,
c, d T2 ,
, CC 1rd a, b c, d, and let b
a, b c, d 0, be the weight function such that CC 1rd and , then
u, yu 0 a
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
17
1 ( x, y ) ( x, y ) ( x, y ) 2 y1 x K R
1 ( x, y ) 2 y1 x ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x 2 R 2K R
1 ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x ( x, y ) 2 y1 x R 2K 2 R
1 [ M 1 ( x, y ) ((d c) H1 ( x) (b a) H 2 ( y )) 2K 2 R M 2 ( x, y ) H1 ( x) H 2 ( y )] 2 y 1 x.
(3.10) Proof: From the identity of Lemma 3. 2 , we have the following identities:
K ( x , y ) ( x, y )
( 1 (t ), 2 ( s)) ( 1 (t ), 2 ( s)) 2 s1t R
g ( y, s) ( 1 (t ), s) ( 1 (t ), s) 2 s1t R
f ( x, t ) R
(t , 2 ( s)) (t , 2 ( s)) 2 s1t 1t
2 f ( x, t ) g ( y , s ) (t , s ) (t , s) 2 s1t. R 1t 2 s (3.11) and
Kx, yx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
gy, s 1 t, s 1 t, s 2 s 1 t R
fx, t t, 2 st, 2 s 2 s 1 t 1 t R 2 fx, tgy, s t, st, s 2 s 1 t. 1 t 2 s R
3.12
(3.12) Multiplying 3. 11 with x, y and 3. 12 with x, y and then adding, we have
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
18
2Kx, yx, yx, y x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y gy, s 1 t, s 1 t, s 2 s 1 t 2 s R x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R x, y fx, t t, 2 st, 2 s 2 s 1 t 1 t R x, y fx, tgy, s R
x, y fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s 2 t, st, s s t 2 1 1 t 2 s
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
B1 x, y B2 x, y B3 x, y.
3.13 (3.13)
Integrating 3. 13 on R , we get
x, y x, y x, y 2 y1 x R
1 x, y 2 y1 x 1 x , 2 y 1 x , 2 y 2 y1 x R 2 K R x, y 2 y1 x 1 x , 2 y 1 x , 2 y 2 y1 x R R 1 B1 x, y B2 x, y B3 x, y 2 y1 x. 2K R
This implies that
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19
1 ( x, y ) ( x, y ) ( x, y ) 2 y1 x K R
1 ( x, y ) 2 y1 x ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x 2 R R 2K
1 ( x, y ) 2 y1 ( 1 ( x), 2 ( y )) ( 1 ( x), 2 ( y )) 2 y1 x R 2K 2 R 1 [ B1 ( x, y ) B2 ( x, y ) B3 ( x, y )] 2 y 1 x 2K 2 R 1 [ M 1 ( x, y ) ((d c) H1 ( x) (b a) H 2 ( y )) 2K 2 R M 2 ( x, y ) H1 ( x) H 2 ( y )] 2 y1 x
.
(3.14) This completes the proof. Corollary 3.1 (Continuous case) Let T1 T2 , then our delta integral is the usual Riemann integral from calculus. This leads us to state the following inequality:
1 K
( x, y) ( x, y) ( x, y) d
2
yd1 x
R
1 2K 2
( x, y) d
2
yd1 x ( x, y ) ( x, y ) d 2 yd1 x
R
R
1 2K 2
( x, y) ( x, y) d
1 2K 2
[M ( x, y) ((d c) H ( x) (b a) H
2
yd1 x x, y d 2 yd1 x
R
R
1
1
2
( y ))
R
M 2 ( x, y ) H1 ( x) H 2 ( y )] d 2 yd1 x, b
d
a
c
where K (u, y) du ( x, v) dv. Corollary 3.2 (Discrete case) Let T1 T2 , a 0, b m, x, y u x , u y . Then
c 0,
d n,
x i,
y j and
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
1 m n 1 (ai , b j ) (ci , d j ) ( pi , q j ) K1 i 1 j 1 2 K12 1 2 K12
m
n
M
3
1 2 K12
m
20
m
n
m
n
(ai , b j ) (ci , d j ) ( pi , q j ) i 1 j 1
i 1 j 1
n
m
n
(ai , b j ) ( pi , q j ) (ci , d j ) i 1 j 1
i 1 j 1
( x, y ) ((d c) H1 ( x) (b a) H 2 ( y ))
i 1 j 1
M 4 ( x, y ) H1 ( x ) H 2 ( y ) , m
n
i 1
j 1
where K1 ( pi , y) ( x, q j ),
M3 |v x , v y |u x , u y w x , w y |u x , u y |v x , v y w x , w y and M4 |v x , v y | t s u t , u s w t , w s |u x , u y | t s v t , v s w t , w s . Theorem 3.2 Let a, b T1 ,
c, d T2 ,
, CC 1rd a, b, c, d, and let b
: a, b c, d 0, be the weight function such that CC 1rd and
u, yu 0, a
d
x, v v 0, c
then
1 x, yx, yx,y y x 2 1 K R 1 1 x,y 2 y 1 x 2 K R
1 x, y x, y y x 1 2 1 2 2 1 K R
1 1 x, y 2 y 1 x 2 K R
1 x, y x, y y x 1 2 1 2 2 1 K R
1 2 x,yFx, y x, yGx, y 2 y 1 x 2K R 1 2 M2 x,yH1 xH2 y 2 y 1 x. 2K R
3.15 (3.15)
Proof From identity 3. 6 in lemma 3.3, we have the following identities:
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
21
K x, yx, y d
b
x, v v 1 t, y 1 t, y 1 t c
a b
d
a
c
u, yu x, 2 sx, 2 s 2 s 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t 2 1 1 t 2 s
Fx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t. 2 1 1 t 2 s
3.16 (3.16)
and Kx, yx, y Gx, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
fx, tgy, s R
2 t, st, s s t. 2 1 1 t 2 s
3.17
(3.17) For x, y R, multiplying 3. 16 with gx, y and 3. 17 with fx, y and then adding, we have
2Kx,yx,yx,y x,yFx,y x,yGx,y x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R
x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R
x,y 1 x, 2 y 1 x, 2 y 2 y 1 x R 2 x,y fx,tgy,s t,st,s 2 s 1 t 1 t 2 s R
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
22
x, yFx, y x, yGx, y
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
x, y 1 t, 2 s 1 t, 2 s 2 s 1 t R
B3 x, y.
3.18 (3.18)
Integrating 3. 18 on R, we get
x, yx, yx, y 2 y 1 x R
1 x, yFx, y x, yGx, y 2 y 1 x 2K 1 x, y 2 y 1 x 1 x , 2 y 1 x , 2 y 2 y 1 x 2K R R x, y 2 y 1 x 1 x , 2 y 1 x , 2 y 2 y 1 x R
R
1 B3 x, y 2 y 1 x. 2K R
3.19 (3.19)
From the identity 3. 19 and the inequality iii in lemma 3.1, we have
1 x, yx, yx, y y x 2 1 K R 1 2
1 x, y y x 2 1 K R
1 x , y x , y y x 1 2 1 2 2 1 K R
1 2
1 x, y y x 2 1 K R
1 x , y x , y y x 1 2 1 2 2 1 K R
1 2 x, yFx, y x, yGx, y 2 y 1 x 2K R
1 |A x, y| y x 3 2 1 2K 2 R
1 M x, yH x H y y x. 2 1 2 2 1 2K 2 R
3.20
(3.20) This completes the proof. Corollary 3.3 (Continuous case) Let T1 T2 , then our delta integral is the usual Riemann integral from calculus. This leads us to state the following inequality:
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
23
1 ( x, y ) ( x, y ) ( x, y ) d 2 yd1 x K R 1 1 1 ( x, y ) d 2 yd1 x 1 ( x), 2 ( y ) 1 ( x), 2 ( y ) d 2 yd1 x 2 K R K R 1 1 1 ( x, y ) d 2 yd1 x 1 ( x), 2 ( y ) 1 ( x), 2 ( y ) d 2 yd1 x 2 K R K R 1 ( x, y ) F ( x, y ) ( x, y ) G ( x, y ) d 2 yd1 x 2K 2 R 1 M 2 ( x, y ) H1 ( x) H 2 ( y ) d 2 yd1 x, 2K 2 R
b
d
a
c
where K (u, y ) du ( x, v) dv,
F 1 x, y
d
b
b
d
c
a
a
c
d
b
b
d
c
a
a
c
x, v dv t, yt, yd 1 t u, ydu x, sx, sd 2 s
and
G 1 x, y
x, v dv t, yt, yd 1 t u, ydu x, sx, sd 2 s.
Corollary 3.4 (Discrete case) Let T1 T2 , a 0, b m, x, y u x , u y . Then
c 0,
d n,
x i,
y j and
1 m n (ai , b j ) (ci , d j ) ( pi , q j ) K1 i 1 j 1 1 m n 1 1 m n (ai , b j ) ( c , d ) ( p , q ) i j i j K i 1 j 1 2 K1 i 1 j 1 1 1 m n 1 1 m n (ci , d j ) ( a , b ) ( p , q ) i j i j K i 1 j 1 2 K1 i 1 j 1 1
1 m n (ci , d j ) F2 (ai , b j ) G2 2 K12 i 1 j 1
1 m n M 4 ( x, y) H1 ( x) H 2 ( y), 2 K12 i 1 j 1
where n
m
n
m
j 1
i 1
j 1
i 1
F2 (w, w j ) (ui , u) ( wi , w) ( wi , w) (u, u j ) (w, w j ) and
Int. J. Pure Appl. Sci. Technol., 8(2) (2012), 1-25.
n
m
n
m
j 1
i 1
j 1
i 1
24
G2 (w, w j ) (vi , v) (wi , w) ( wi , w) (v, v j ) ( w, w j ). Remark 3.1 If we put the weight function ( x, y ) 1 in both the inequalities (3.10) and (3.15), we will obtain the results proved in [20].
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