On an inequality of Grüss type via variant of Pompeiu@s mean

Received 02/06/14
On an inequality of Grüss type via
variant of Pompeiu’s mean value
theorem
Mehmet Zeki SARIKAYA
Department of Mathematics, Faculty of Science and Arts,
Düzce University, Düzce-TURKEY, e-mail:
[email protected]
Abstract
The main of this paper is to establish an Grüss type inequality
by using a mean value theorem.
1
Introduction
In 1935, G. Grüss [4] proved the following inequality:
1
b
a
Zb
f (x)g(x)dx
1
b
a
a
Zb
f (x)dx
1
b
a
a
Zb
g(x)dx
1
(
4
')(
a
provided that f and g are two integrable function on [a; b] satisfying the
condition
'
f (x)
and
for all x 2 [a; b]:
g(x)
The constant 14 is best possible.
µ
In 1882, P. L. Cebyš
ev [2] gave the following inequality:
jT (f; g)j
1
(b
12
a)2 kf 0 k1 kg 0 k1 ;
where f; g : [a; b] ! R are absolutely continuous function, whose …rst
derivatives f 0 and g 0 are bounded,
0
10
1
Zb
Zb
Zb
1
1
1
T (f; g) =
f (x)g(x)dx @
f (x)dxA @
g(x)dxA
b a
b a
b a
a
a
a
and k:k1 denotes the norm in L1 [a; b] de…ned as kpk1 = ess sup jp(t)j :
t2[a;b]
1
);
For a di¤erentiable function f : [a; b] ! R; a b > 0, Pachpatte has in
[6] proved, using Pompeiu’s mean value theorem [9], the following Grüss
type inequality:
0 b
1
Z
Zb
Zb
Zb
Zb
1 @
f (t)dt: tg(t)dt + g(t)dt: tf (t)dtA
f (t)g(t)dt
2
b
a2
a
a
kf
`f 0 k1
Zb
1
2
jg(t)j
a
t
dt + kg
a+b
a
`g 0 k1
a
a
Zb
jf (t)j
1
2
t
dt
a+b
a
where `(t) = t; t 2 [a; b] :
In [7], Pecaric and Ungar proved a general estimate with the p-norm,
1 < p < 1, which will for p = 1 give the Pachpatte [6] result.
The interested reader is also referred to ([1], [3], [5]-[10]) for integral
inequalities by using Pompeiu’s mean value theorem.
In this paper, we establish some new integral inequalities similar to
that of the Grüss type integral inequality via Pompeiu’s mean value
theorem.
2
Main Results
First we give the following notations used to simplify the details of presentation
F (u; v) = uvfuv (u; v) ufu (u; v) vfv (u; v) + f (u; v)
G(u; v) = vguv (u; v) ugu (u; v) vgv (u; v) + g(u; v);
A(x; y; p)
= (b
(
+
+
+
1
a) p
1
(d
1
c) p
1
a2 q x 2 q
x2
+
(1 2q) (2 q)
(1
a2 q x 2 q
x2
+
(1 2q) (2 q)
(1
b2
(1
b2
(1
q
x2 q
x2
+
2q) (2 q)
(1
q
x2 q
x2
+
2q) (2 q)
(1
q
a1+q x1 2q
2q) (1 + q)
q
a1+q x1 2q
2q) (1 + q)
q
b1+q x1 2q
2q) (1 + q)
q
b1+q x1 2q
2q) (1 + q)
1
q
y2 q
y2
+
2q) (2 q)
(1
q
c1+q y 1 2q
2q) (1 + q)
d2 q y 2 q
y2
+
(1 2q) (2 q)
(1
q
d1+q y 1 2q
2q) (1 + q)
y2 q
y2
+
2q) (2 q)
(1
q
c1+q y 1 2q
2q) (1 + q)
d2 q y 2 q
y2
+
(1 2q) (2 q)
(1
q
d1+q y 1 2q
2q) (1 + q)
c2
(1
1
q
1
q
c2
(1
1
q
q
q
To prove our theorems, we need the following lemma:
2
1
q
1
q
1
q
1
q
)
;
Lemma 1 f : = [a; b] [c; d]! R be an absolutely continuous function
such that the partial derivative of order 2 exists for all (t; s) 2
with
0 < a < b; 0 < c < d: Then for any (t; s); (x; y) 2 ; we have
stf (x; y)
ytf (x; s)
xsf (t; y) + xyf (t; s) = xyst
Zx Zy
t
1 1
; ! R by
Proof. De…ne
: 1b ; a1
d c
function
is continuously di¤erentiable on
1 1
(x1 ; y1 ) ; (x2 ; y2 ) 2 1b ; a1
; ; we get
d c
(x1 ; y1 )
(x1 ; y2 )
Zx1 Zy1 2
@ (t; s)
=
dsdt
@t@s
=
x2 y 2
Zx1 Zy1
1 1
;
t s
f
1
fs
s
(x2 ; y1 ) +
1 1
;
t s
F (u; v)
dvdu
:
u2 v 2
s
(t; s) := tsf 1t ; 1s . The
1 1
; , and for all
d c
1 1
;
b a
(x2 ; y2 )
1
ft
t
1 1
;
t s
+
1
fts
ts
1 1
;
t s
dsdt:
x2 y 2
Using the change of the variable in last integrals with u =
we get
(x1 ; y1 )
1
x1
=
(x2 ; y1 ) +
[f (u; v)
vfv (u; v)
and v = 1s ;
(1)
(x2 ; y2 )
1
y1
Z Z
1
x2
(x1 ; y2 )
1
t
ufu (u; v) + uvfuv (u; v)]
dvdu
:
u2 v 2
1
y2
Denote x1 = x1 ; x2 = 1t ; y1 = y1 and y2 = 1s : Then for all (x; y) ; (t; s) 2
[a; b] [c; d] from (1), we have
1
1
f (x; y)
f (x; s)
xy
xs
Z t Zs
dvdu
=
F (u; v) 2 2
uv
x
1
1
f (t; y) + f (t; s)
ty
ts
y
which gives (2) and completes the proof.
Theorem 2 f : ! R be an absolutely continuous function such that
the partial derivative of order 2 exists for all (t; s) 2
with 0 < a <
3
b; 0 < c < d: Then for p1 + 1q = 1 with 1
we have
Zb Zd
f (x; y)g(x; y)dydx
a
;
(2)
c
1
(d2
c2 )
1
(b2
a2 )
Zb Zd Zd
[f (x; s)g(x; y) + g(x; s)f (x; y)] ydsdydx
a
c
c
Zb
Zd
Zb
[f (t; y)g(x; y) + g(t; y)f (x; y)] xdtdydx
a
+
1 any (t; s); (x; y) 2
p; q
c
a
2
(b2
+
a2 ) (d2
c2 )
2 (b
(b2
a
c
Zb
Zd
@
c2 )
a
0
a2 ) (d2
1
p
c
1
p
a) (d
a2 ) (d2
+ kl1 guv
@
Zb Zd
0
2
(b2
0
c) @
kl1 fuv
c2 )
l2 gu
l3 gv + gkp
10 b d
1
Z Z
xyg(x; y)dydxA @
f (t; s)dsdtA
a
c
Zb
a
Zd
Zb
Zd
10
xyf (x; y)dydxA @
l2 fu
l3 fv + f kp
c
a
Zb
a
Zd
c
1
g(t; s)dsdtA
xy jg(x; y)j A(x; y; p)dydx
c
1
xy jf (x; y)j A(x; y; p)dydxA
where l1 (x; y) = xy, l2 (x; ) = x and l3 ( ; y) = y for all (x; y) 2
:
Proof. From Lemma 1, we have
stf (x; y)
stg(x; y)
ytf (x; s)
ytg(x; s)
Zx Zy
F (u; v)
t
s
Zx
Zy
G(u; v)
xsf (t; y) + xyf (t; s) = xyst
xsg(t; y) + xyg(t; s) = xyst
t
dvdu
;
u2 v 2
dvdu
:
u2 v 2
s
Multiplying these identities by g(x; y) and f (x; y) respectively, and adding
the results gives
2tsf (x; y)g(x; y) ty [f (x; s)g(x; y) + g(x; s)f (x; y)]
xs [f (t; y)g(x; y) + g(t; y)f (x; y)] + xy [f (t; s)g(x; y) + g(t; s)f (x; y)]
Zx Zy
Zx Zy
dvdu
dvdu
= xtysg(x; y)
F (u; v) 2 2 + xtysf (x; y)
G(u; v) 2 2
uv
uv
t
s
t
4
s
By integrating with respect to (t; s) on [a; b]
(b2
a2 ) (d2
2
c2 )
(b2
f (x; y)g(x; y)
a2 )
2
[c; d] ; we get
y
Zd
[f (x; s)g(x; y) + g(x; s)f (x; y)] ds
c
(d2
c2 )
2
Zb
x
[f (t; y)g(x; y) + g(t; y)f (x; y)] dt
a
+xy
Zb Zd
a
[f (t; s)g(x; y) + g(t; s)f (x; y)] dsdt
c
2 x y
3
Z Z
dvdu
= xyg(x; y)
st 4
F (u; v) 2 2 5 dsdt
uv
a c
t s
2 x y
3
Z Z
Zb Zd
dvdu
G(u; v) 2 2 5 dsdt
+xyf (x; y)
st 4
uv
Zb Zd
a
t
c
s
and therefore by integrating with respect to (x; y) on [a; b]
taking the modulus, we obtain
(b2
a2 ) (d2
2
c2 )
Zb Zd
a
(b2
a2 )
2
c2 )
2
(3)
f (x; y)g(x; y)dydx
c
Zb Zd Zd
[f (x; s)g(x; y) + g(x; s)f (x; y)] ydsdydx
Zb Zd Zb
[f (t; y)g(x; y) + g(t; y)f (x; y)] xdtdydx
a
(d2
a
c
c
c
a
0 b d
10 b d
1
Z Z
Z Z
+@
xyg(x; y)dydxA @
f (t; s)dsdtA
a
0
+@
c
Zb Zd
a
c
[c; d] ; and
a
10
xyf (x; y)dydxA @
c
Zb Zd
a
5
c
1
g(t; s)dsdtA
0
2 x y
3
1
Z Z
dvdu
xyg(x; y) @
st 4
F (u; v) 2 2 5 dsdtA dydx
uv
a c
a c
t s
0 b d 2 x y
3
1
Zb Zd
Z Z
Z Z
dvdu
+
xyf (x; y) @
st 4
G(u; v) 2 2 5 dsdtA dydx :
uv
Zb Zd
a
Zb Zd
c
a
c
t
s
For the …rst added in the right-hand side of (3), we have
0 b d 2 x y
3
1
Z Z
Z Z
Zb Zd
dvdu
F (u; v) 2 2 5 dsdtA dydx
st 4
xyg(x; y) @
uv
t s
a c
a c
2 x y
3
Zb Zd
Zb Zd
Z Z
dvdu
jxyg(x; y)j
st 4
F (u; v) 2 2 5 dsdt dydx
uv
a
c
a
c
t
(4)
s
and for the second factor under the integral on (x; y) (i.e the outher most
integral) in (4), we have
3
2 x y
Zb Zd
Z Z
dvdu
(5)
st 4
F (u; v) 2 2 5 dsdt
uv
=
a
Zb
a
c
Zd
c
Zx
t
t
y
Z
Zx
Zy
Zt
Zs
a
c
x
jF (u; v)j dvdu dsdt
s
F (u; v)
ts
dvdu dsdt
u2 v 2
y
+
Zx
Z t Zs
F (u; v)
ts
dvdu dsdt
u2 v 2
+
Zb Zy Z t Zs
F (u; v)
ts
dvdu dsdt
u2 v 2
x
c
x
y
Zb
Zd
Zt
Zs
F (u; v)
ts
dvdu dsdt:
u2 v 2
a
+
x
Zd
s
y
y
x
x
y
y
6
By using Hölder’s inequality, the sum in the last line (5) is
0
@
Zx Zy
a
c
0
@
1
Zx Zy
t
1 p1 0
jF (u; v)jp dvduA dsdtA @
s
Zx Zy
a
c
0
@
Zx Zy
t
s
q q
1
1 1q
t s dvdu A
dsdtA (6)
u2q v 2q
0
1
1 p1 0 x d 0 x s
1
1 1q
Zx Zd Zx Zs
Z Z Z Z q q
t s dvdu A
@
@
jF (u; v)jp dvduA dsdtA @
+@
dsdtA
u2q v 2q
0
a
t
y
y
a
t
y
y
1
1 p1 0 b y 0 t y
0
1
1 1q
Zb Zy Z t Zy
Z Z Z Z q q
t s dvdu A
@
@
jF (u; v)jp dvduA dsdtA @
+@
dsdtA
u2q v 2q
0
0
x
c
x
s
x
y
x
y
x
x
c
s
0
1
1 p1 0 b d 0 t s
1
1 1q
Zb Zd Z t Zs
Z Z Z Z q q
t s dvdu A
@
@
+@
jF (u; v))jp dvduA dsdtA @
dsdtA
u2q v 2q
x
y
x
y
0
0
1
1 p1
Zb Zd Zb Zd
@
@
jF (u; v)jp dvduA dsdtA
a
c
a
c
80
0
1
1 1q 0 x d 0 x s
1
1 1q
>
Z Z Z Z q q
< Zx Zy Zx Zy tq sq dvdu
t s dvdu A
@
@
A dsdtA + @
@
dsdtA
2q
2q
2q v 2q
>
u
v
u
:
a
0
+@
c
Zb Zy
x
c
t
s
a
y
t
y
x
y
x
y
0 t y
1
1
0 b d0 t s
1
1 1q 9
>
Z Z q q
Z Z Z Z q q
=
t
s
dvdu
t
s
dvdu
@
A dsdtA + @
@
A dsdtA
>
u2q v 2q
u2q v 2q
;
1
q
x
s
The …rst factor in (6) equals
0
@
= (b
Zb Zd
a
c
0
@
1
p
Zb Zd
a
a) (d
c
1
1 p1
jF (u; v)jp dvduA dsdtA
1
c) p kl1 fuv
l2 fu
l3 fv + f kp :
and by Lemma, the second factor equals A(x; y; p): Plugging into (5)
7
shows that for the …rst added on the right-hand side in (4), we get.
3
1
0 b d 2 x y
Z Z
Z Z
Zb Zd
dvdu
xyg(x; y) @
st 4
F (u; v) 2 2 5 dsdtA dydx
uv
c
a
a
Zb Zd
a
= (b
jxyg(x; y)j (b
c
t
1
s
1
c) p kl1 fuv
a) p (d
l3 fv + f kp :A(x; y; p)dydx
l2 fu
c
1
p
a) (d
1
p
c) kl1 fuv
l2 fu
l3 fv + f kp
Zb Zd
a
xy jg(x; y)j A(x; y; p)dydx:
c
An analogous inequality holds for the second added in (3), so putting
(b2 a2 )(d2 c2 )
these two inequalities into (3) and dividing
gives the re2
quired inequality (2), proving the theorem.
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µ
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ev, Sur less expressions approximatives des integrales
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