COMPARING THE INTEGRAL MEANS FOR FUNCTIONS OF TWO ...

Electronic Journal of Mathematical Analysis and Applications Vol. 4(2) July 2016, pp. 46-55. ISSN: 2090-729(online) http://fcag-egypt.com/Journals/EJMAA/ ————————————————————————————————

COMPARING THE INTEGRAL MEANS FOR FUNCTIONS OF TWO VARIABLES WITH BOUNDED VARIATION ¨ HUSEYIN BUDAK, MEHMET ZEKI SARIKAYA

Abstract. In this paper, we obtain an inequality for difference of the integral means using functions of two variables with bounded variation. An application to probability density functions is also given.

1. Introduction Let f : [a, b] → R be a differentiable mapping on (a, b) whoose derivative f ′ : (a, b) → R is bounded on (a, b) , i.e. ∥f ′ ∥∞ := sup |f ′ (t)| < ∞. Then we have the t∈(a,b)

inequality [ ) ] ( ∫b a+b 2 x − 1 1 2 f (x) − (b − a) ∥f ′ ∥∞ , f (t)dt ≤ + 2 b − a 4 (b − a)

(1)

a

for all x ∈ [a, b][23]. The constant 14 is the best possible. This inequality is well known in the literature as the Ostrowski inequality. In [15], Dragomir proved following Ostrowski type inequalities related functions of bounded variation: Theorem 1. Let f : [a, b] → R be a mapping of bounded variation on [a, b] . Then b ∫ ] ∨ [ b f (t)dt − (b − a) f (x) ≤ 1 (b − a) + x − a + b (f ) 2 2 a a

holds for all x ∈ [a, b] . The constant

1 2

is the best possible.

Hwang and Dragomir gave the following inequality in [21].

2010 Mathematics Subject Classification. 26D15, 26B30, 26D10, 41A55. Key words and phrases. Ostrowski type inequalities, Bounded variation, Riemann-Stieltjes integral, Probability density functions. Submitted July 1, 2015. 46

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Theorem 2. Let f : [a, b] → R be a mapping with bounded variation on [a, b]. Then, for a ≤ x < y ≤ b, we have the inequality ∫b ∫y 1 1 f (s)ds − f (s)ds (2) (b − a) (y − x) a

x

x−a∨ (f ) + max b−a a x

≤

{

x−a b−y , b−a b−a

}∨ y x

b−y ∨ (f ) b−a y b

(f ) +

The inequality (2) is sharp. Definition 1. (Vitali-Lebesque-Fr´echet-de la Vall´ee Poussin)[12]. We introduce the notation ∆11 f (xi , yj ) = f (xi+1 , yj+1 ) − f (xi+1 , yj ) − f (xi , yj+1 ) + f (xi , yj ), then, the function f (x, y) is said tobe of bounded variation if the sum m−1 , n−1 ∑

|∆11 f (xi , yj )|

i=0 , j=0

is bounded for all nets. Therefore, one can define the consept of total variation of a function of two variables, as follows: ∑ Let f be of bounded variation on Q = [a, b] × [c, d], and let (P ) denote the n ∑ m ∑ sum |∆11 f (xi , yj )| corresponding to the partition P of Q. The number i=1 j=1

∨

(f ) :=

d ∨ b ∨ c

Q

(f ) := sup

{∑

} (P ) : P ∈ P (Q) ,

a

is called the total variation of f on Q. Here P ([a, b]) denotes the family of partitions of [a, b] . In [20], Jawarneh and Noorani gave the following Lemmas for double RiemannStieltjes integral: Lemma 1. (Integrating by parts) If f ∈ RS(α) on Q, then α ∈ RS(f ) on Q, and we have ∫d ∫b ∫d ∫b f (t, s)dt ds α(t, s) + α(t, s)dt ds f (t, s) (3) c

=

a

c

a

f (b, d)α(b, d) − f (b, c)α(b, c) − f (a, d)α(a, d) + f (a, c)α(a, c).

Lemma 2. Assume that g ∈ RS(α) on Q and α is of bounded variation on Q, then d b ∫ ∫ ∨ g(x, y)dx dy α(x, y) ≤ sup |g(x, y)| (α) . (4) (x,y)∈Q Q c

a

In [20], Jawarneh and Noorani proved the following Ostrowski type inequality or functions of two variables with bounded variation:

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Theorem 3. Let f : Q →→ R be mapping of bounded variation on Q. Then for all (x, y) ∈ Q, we have inequality ∫d ∫b (b − a) (d − c) f (x, y) − f (t, s)dtds (5) c a ] [ 1 a + b ≤ (b − a) + x − 2 2 ] [ 1 c + d ∨ × (d − c) + y − (f ) 2 2 Q ∨ where (f ) denotes the total (double) variation of f on Q. Q

For more information and recent developments on integral inequalities for mappings of bounded variation (single variable and two variables), please refer to ([1][11], [13]-[22],[24]-[29]). In this paper, we compare the integral means using functions of two variables with bounded variation. A application to probability density functions is also given. 2. Main Results First, we give the following notations to simplify presentetion of some intervals. Q1

= [a, x1 ] × [c, y1 ] , Q2 = [a, x1 ] × [y1 , y2 ] , Q3 = [a, x1 ] × [y2 , d] ,

Q4

= [x1 , x2 ] × [c, y1 ] , Q5 = [x1 , x2 ] × [y1 , y2 ] , Q6 = [x1 , x2 ] × [y2 , d] ,

Q7

= [x2 , b] × [c, y1 ] , Q8 = [x1 , x2 ] × [y1 , y2 ] , Q9 = [x2 , b] × [y2 , d] .

Theorem 4. If the function f : Q = [a, b] × [c, d] → R is of bounded variation on Q, then, for a ≤ x1 < x2 ≤ b and c ≤ y1 < y2 ≤ d, we have the inequality ∫b ∫d ∫b ∫y2 1 1 f (t, s)dsdt − f (t, s)dsdt (6) (b − a) (d − c) (b − a) (y2 − y1 ) a

−

≤

c

1 (x2 − x1 ) (d − c)

a y1

∫x2 ∫d f (t, s)dsdt + x1 c

1 (x2 − x1 ) (y2 − y1 )

∫x2 ∫y2 x1 y 1

f (t, s)dsdt

] x2 +x1 a+b [(b − a) − (x2 − x1 )] [(d − c) − (y2 − y1 )] 1 2 − 2 + (b − a) (d − c) 2 (b − a) − (x2 − x1 ) ] b d [ y2 +y1 ∨∨ d+c − 1 2 2 + (f ). × 2 (d − c) − (y2 − y1 ) a c

where

b ∨ d ∨ a c

[

(f ) denotes the total (double) variation of f on Q.

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Proof. First, we define the mappings Kx1 ,x2 (t) and Ly1 ,y2 (s) by  a−t if t ∈ [a, x1 ]  b−a ,     t−x1 a−t Kx1 ,x2 (t) = x2 −x1 + b−a , if t ∈ (x1 , x2 ]      b−t if t ∈ (x2 , b] b−a , and

      Ly1 ,y2 (s) =

    

if s ∈ [c, y1 ]

c−s d−c , s−y1 y2 −y1

+

c−s d−c ,

if s ∈ (y1 , y2 ] . if s ∈ (y2 , d]

d−s d−c ,

Using the kernels Kx1 ,x2 (t) and Ly1 ,y2 (s), we have ∫b ∫d Kx1 ,x2 (t)Ly1 ,y2 (s)ds dt f (t, s) a

∫x1 ∫y1 = a

(7)

c

c

a−t c−s ds dt f (t, s) + b−ad−c

∫x1 ∫d + a y2

+ x1 y 1

∫x2 ∫d [ x1 y 2

a y1

a−td−s ds dt f (t, s) + b−ad−c

∫x2 ∫y2 [

+

∫x1 ∫y2

t − x1 a−t + x2 − x1 b−a

][

[ ] c−s a − t s − y1 + ds dt f (t, s) b − a y2 − y1 d−c

∫x2 ∫y1 [ x1 c

] a−t c−s t − x1 + ds dt f (t, s) x2 − x1 b−a d−c

] s − y1 c−s + ds dt f (t, s) y2 − y1 d−c

] ∫b ∫y1 t − x1 a−t d−s b−t c−s + ds dt f (t, s) + ds dt f (t, s) x2 − x1 b−a d−c b−ad−c x2 c

∫b ∫y2 +

[ ] ∫b ∫d c−s b−t d−s b − t s − y1 + ds dt f (t, s) + ds dt f (t, s) b − a y2 − y1 d−c b−ad−c x2 y 2

x2 y 1

= K1 + K2 + ... + K9 . Integrating the by parts, we have ∫x1 ∫y1 K1

= a

=

c

a−t c−s ds dt f (t, s) b−ad−c

(8)

  ∫x1 ∫y1 1 (a − x1 ) (c − y1 ) f (x1 , y1 ) − f (t, s)dsdt . (b − a) (d − c) a

c

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Similarly, we have ∫x1 ∫y2 K2

= a y1

=

[ ] a − t s − y1 c−s + ds dt f (t, s) b − a y2 − y1 d−c

a − x1 1 f (x1 , y2 ) + b−a (b − a) (y2 − y1 ) +

(9)

∫x1 ∫y2 f (t, s)dsdt a y1

1 [(a − x1 ) (c − y2 ) f (x1 , y2 ) (b − a) (d − c) ∫x1 ∫y2

− (a − x1 ) (c − y1 ) f (x1 , y1 ) −



f (t, s)dsdt ,

a y1

∫x1 ∫d K3

= a y2

=

a−td−s ds dt f (t, s) b−ad−c

(10)

  ∫x1 ∫d 1 − (a − x1 ) (d − y2 ) f (x1 , y2 ) − f (t, s)dsdt , (b − a) (d − c) a y2

∫x2 ∫y1 [ K4

= x1 c

=

] t − x1 a−t c−s + ds dt f (t, s) x2 − x1 b−a d−c

c − y1 1 f (x2 , y1 ) + d−c (x2 − x1 ) (d − c) +

(11)

∫x2 ∫y1 f (t, s)dsdt x1 c

1 [(a − x2 ) (c − y1 ) f (x2 , y1 ) (b − a) (d − c) ∫x2 ∫y1

− (a − x1 ) (c − y1 ) f (x1 , y1 ) −



f (t, s)dsdt ,

x1 c

(12) ∫x2 ∫y2 [ K5

= x1 y 1

a−t t − x1 + x2 − x1 b−a

][

]

s − y1 c−s + ds dt f (t, s) y2 − y1 d−c

1 = f (x2 , y2 ) − (x2 − x1 ) (y2 − y1 )

∫x2 ∫y2 f (t, s)dsdt x1 y 1

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∫x2 ∫y2

c − y2 c − y1 1 + f (x2 , y2 ) − f (x2 , y1 ) + d−c d−c (x2 − x1 ) (d − c)

f (t, s)dsdt x1 y 1

a − x2 a − x1 1 + f (x2 , y2 ) − f (x1 , y2 ) + b−a b−a (b − a) (y2 − y1 ) +

51

∫x2 ∫y2 f (t, s)dsdt x1 y 1

1 [(a − x2 ) (c − y2 ) f (x2 , y2 ) − (a − x1 ) (c − y2 ) f (x1 , y2 ) (b − a) (d − c) ∫x2 ∫y2

− (a − x2 ) (c − y1 ) f (x2 , y1 ) + (a − x1 ) (c − y1 ) f (x1 , y1 ) −



f (t, s)dsdt ,

x1 y 1

∫x2 ∫d [ K6

= x1 y2

] t − x1 a−t d−s + ds dt f (t, s) x2 − x1 b−a d−c

d − y2 1 = − f (x2 , y2 ) + d−c (x2 − x1 ) (d − c) +

(13)

∫x2 ∫d f (t, s)dsdt x1 y 2

1 [− (a − x2 ) (d − y2 ) f (x2 , y2 ) (b − a) (d − c) ∫x2 ∫d

+ (a − x1 ) (d − y2 ) f (x1 , y2 ) −

 f (t, s)dsdt ,

x1 y 2

∫b ∫y1 K7

= x2 c

=

b−t c−s ds dt f (t, s) b−ad−c

(14)

  ∫b ∫y1 1 − (b − x2 ) (c − y1 ) f (x2 , y1 ) − f (t, s)dsdt , (b − a) (d − c) x2 c

∫b ∫y2 K8

= x2 y 1

[ ] c−s b − t s − y1 + ds dt f (t, s) b − a y2 − y1 d−c

b − x2 1 = − f (x2 , y2 ) + b−a (b − a) (y2 − y1 )

(15)

∫b ∫y2 f (t, s)dsdt x2 y 1

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H. BUDAK, M.Z. SARIKAYA

+

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1 [− (b − x2 ) (c − y2 ) f (x2 , y2 ) (b − a) (d − c) ∫b ∫y2

+ (b − x2 ) (c − y1 ) f (x2 , y1 ) −

 f (t, s)dsdt

x2 y1

and ∫b ∫d K9

= x2 y2

b−t d−s ds dt f (t, s) b−ad−c

(16)

 =

1 (b − x2 ) (d − y2 ) f (x2 , y2 ) − (b − a) (d − c)

∫b ∫d

 f (t, s)dsdt .

x2 y 2

If we add the equalities (8)-(16) in (7), then we have ∫b ∫d Kx1 ,x2 (t)Ly1 ,y2 (s)ds dt f (t, s) a

c

1 (b − a) (y2 − y1 )

=

1 − (b − a) (d − c)

∫b ∫y2 a y1

∫b ∫d a

c

1 f (t, s)dsdt + (x2 − x1 ) (d − c)

∫x2 ∫d f (t, s)dsdt x1 c

1 f (t, s)dsdt − (x2 − x1 ) (y2 − y1 )

∫x2 ∫y2 f (t, s)dsdt. x1 y 1

On the other hand, using Lemma 2, we get b d ∫ ∫ K (t)L (s)d d f (t, s) x1 ,x2 y1 ,y2 s t a c x y x y ∫ 1 ∫ 2 ∫ 1 ∫ 1 [ ] c−s a − t s − y1 a−t c−s ds dt f (t, s) + + ds dt f (t, s) ≤ b−ad−c b − a y2 − y1 d−c a

c

a y1

x d x y ∫ 1 ∫ ∫ 2 ∫ 1 [ ] d − s t − x a − t c − s a − t 1 + ds dt f (t, s) + + ds dt f (t, s) b−ad−c x2 − x1 b−a d−c a y2

x1 c

x y ∫ 2 ∫ 2 [ ][ ] a − t s − y c − s t − x 1 1 + + ds dt f (t, s) + x2 − x1 b − a y2 − y1 d−c x1 y 1

x d b y ∫ 2 ∫ [ ∫ ∫ 1 ] a − t d − s c − s t − x b − t 1 + + ds dt f (t, s) + ds dt f (t, s) x − x b − a d − c b − a d − c 2 1 x1 y 2

x2 c

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53

b d b y ∫ ∫ ∫ ∫ 2 [ ] s − y c − s d − s b − t b − t 1 + ds dt f (t, s) + ds dt f (t, s) + b − a y2 − y1 d−c b−ad−c x2 y 1

≤

x1 − a y1 − c ∨ x1 − a (f ) + max b−a d−c b−a

{

Q1

+

y1 − c d − y2 , d−c d−c

}∨

(f )

Q2

{ } x1 − a d − y 2 ∨ x1 − a b − x1 y1 − c ∨ (f ) + max , (f ) b−a d−c b−a b−a d−c {

+ max { + max

+

x2 y 2

Q3

x1 − a b − x1 , b−a b−a x1 − a b − x1 , b−a b−a

b − x1 max b−a {

{

}

{ max

}

Q4

(f )

Q5

b − x1 y1 − c ∨ d − y2 ∨ (f ) + (f ) d−c b−a d−c Q6

y1 − c d − y2 , d−c d−c

x1 − a b − x1 , b−a b−a

y1 − c d − y2 , d−c d−c

}∨

}

}∨

(f ) +

Q8

Q7

b − x1 d − y2 ∨ (f ) b−a d−c Q9

{

}∨ b ∨ d

y1 − c d − y2 , (f ) d−c d−c a c ] [ x2 +x1 a+b [(b − a) − (x2 − x1 )] [(d − c) − (y2 − y1 )] 1 2 − 2 = + (b − a) (d − c) 2 (b − a) − (x2 − x1 ) ] b d [ y2 +y1 ∨∨ d+c 1 2 − 2 + × (f ) 2 (d − c) − (y2 − y1 ) a c

≤ max

max



which completes the proof.

2.1. Applications for Probability Density Functions. Let X and Y be two continuous random variables with f : [a, b]×[c, d] → R+ is a joint probability density ∫t ∫s function and F : [a, b] × [c, d] → R+ , F (t, s) = f (t, s)dsdt is its cumulative a c

distribution function. Proposition 1. Let f and F be as above. Then we have ∫b ∫s ∫t ∫d t − a s − c (t − a) (s − c) F (t, s) − f (ξ, η)dηdξ − f (ξ, η)dηdξ + b − a d − c (b − a) (d − c) a

c

a

(b − t) (t − a) (d − s) (s − c) ∨ ∨ (f ). (b − a) (d − c) a c b

≤

c

d

Proof. Taking x1 = a, x2 = t, y1 = c, y2 = s and using the fact that

∫b ∫d a c

1, we have the required result.

f (ξ, η)dηdξ = 

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[25] K-L. Tseng, Improvements of some inequalites of Ostrowski type and their applications, Taiwan. J. Math. 12 (9) (2008) 2427–2441. [26] K-L. Tseng, S-R. Hwang, and S. S. Dragomir, Generalizations of weighted Ostrowski type inequalities for mappings of bounded variation and applications, Computers and Mathematics with Applications, 55(2008), 1785-1793. [27] K-L. Tseng,S-R. Hwang, G-S. Yang, and Y-M. Chou, Improvements of the Ostrowski integral inequality for mappings of bounded variation I, Applied Mathematics and Computation 217 (2010) 2348–2355. [28] K-L. Tseng,S-R. Hwang, G-S. Yang, and Y-M. Chou, Weighted Ostrowski integral inequality for mappings of bounded variation, Taiwanese J. of Math., Vol. 15, No. 2, pp. 573-585, April 2011. [29] K-L. Tseng, Improvements of the Ostrowski integral inequality for mappings of bounded variation II, Applied Mathematics and Computation 218 (2012) 5841–5847. ¨seyin BUDAK, Department of Mathematics, Faculty of Science and Arts, Du ¨zce Hu ¨zce-TURKEY University, Du E-mail address: [email protected] Mehmet Zeki SARIKAYA, Department of Mathematics, Faculty of Science and Arts, ¨zce University, Du ¨zce-TURKEY Du E-mail address: [email protected]