Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
30 (2014), 43–52
www.emis.de/journals
ISSN 1786-0091
INEQUALITIES OF POMPEIU’S TYPE FOR ABSOLUTELY
CONTINUOUS FUNCTIONS WITH APPLICATIONS TO
OSTROWSKI’S INEQUALITY
S. S. DRAGOMIR
Abstract. In this paper, some new Pompeiu’s type inequalities for complexvalued absolutely continuous functions are provided. They are applied to
obtain some new Ostrowski type inequalities.
1. Introduction
In 1946, Pompeiu [6] derived a variant of Lagrange’s mean value theorem,
now known as Pompeiu’s mean value theorem (see also [8, p. 83]).
Theorem 1 (Pompeiu, 1946 [6]). For every real valued function f differentiable on an interval [a, b] not containing 0 and for all pairs x1 6= x2 in [a, b] ,
there exists a point ξ between x1 and x2 such that
(1.1)
x1 f (x2 ) − x2 f (x1 )
= f (ξ) − ξf 0 (ξ) .
x1 − x2
In 1938, A. Ostrowski [4] proved the following result in the estimating the
integral mean:
Theorem 2 (Ostrowski, 1938 [4]). Let f : [a, b] → R be continuous on [a, b]
and differentiable on (a, b) with |f 0 (t)| ≤ M < ∞ for all t ∈ (a, b). Then for
any x ∈ [a, b] , we have the inequality

!2 
Z b
a+b
x− 2
1
f (x) − 1
 M (b − a) .
f (t) dt ≤  +
(1.2)
b−a a
4
b−a
The constant 14 is best possible in the sense that it cannot be replaced by a
smaller quantity.
2010 Mathematics Subject Classification. 25D10.
Key words and phrases. Ostrowski inequality, Pompeiu’s mean inequality, Integral
inequalities.
43
44
S. S. DRAGOMIR
In order to provide another approximation of the integral mean, by making
use of the Pompeiu’s mean value theorem, the author proved the following
result:
Theorem 3 (Dragomir, 2005 [3]). Let f : [a, b] → R be continuous on [a, b]
and differentiable on (a, b) with [a, b] not containing 0. Then for any x ∈ [a, b] ,
we have the inequality
Z b
a + b f (x)
1
(1.3) ·
−
f (t) dt
2
x
b−a a
≤

b − a 1
+
|x|
4
x−
b−a
a+b
2
!2 
 kf − `f 0 k ,
∞
where ` (t) = t, t ∈ [a, b].
The constant 14 is sharp in the sense that it cannot be replaced by a smaller
constant.
In [7], E. C. Popa using a mean value theorem obtained a generalization of
(1.3) as follows:
Theorem 4 (Popa, 2007 [7]). Let f : [a, b] → R be continuous on [a, b] and
differentiable on (a, b). Assume that α ∈
/ [a, b]. Then for any x ∈ [a, b] , we
have the inequality
Z
a+b
α−x b
(1.4) − α f (x) +
f (t) dt
2
b−a a

!2 
a+b
x− 2
1
 (b − a) kf − `α f 0 k ,
≤ +
∞
4
b−a
where `α (t) = t − α, t ∈ [a, b].
In [5], J. Pečarić and S. Ungar have proved a general estimate with the
p-norm, 1 ≤ p ≤ ∞ which for p = ∞ give Dragomir’s result.
Theorem 5 (Pečarić and Ungar, 2006 [5]). Let f : [a, b] → R be continuous
on [a, b] and differentiable on (a, b) with 0 < a < b. Then for 1 ≤ p, q ≤ ∞
with p1 + 1q = 1 we have the inequality
(1.5)
Z b
a + b f (x)
1
f (t) dt ≤ P U (x, p) kf − `f 0 kp ,
2 · x − b−a
a
INEQUALITIES OF POMPEIU’S TYPE
45
for x ∈ [a, b] , where
P U (x, p) := (b − a)
1
−1
p
"
1/q
x2−q − a1+q x1−2q
a2−q − x2−q
+
(1 − 2q) (2 − q)
(1 − 2q) (1 + q)
2−q
1/q #
b
− x2−q
x2−q − b1+q x1−2q
.
+
+
(1 − 2q) (2 − q)
(1 − 2q) (1 + q)
In the cases (p, q) = (1, ∞) , (∞, 1) and (2, 2) the quantity P U (x, p) has to be
taken as the limit as p → 1, ∞ and 2, respectively.
For other inequalities in terms of the p-norm of the quantity f − `α f 0 , where
`α (t) = t − α, t ∈ [a, b] and α ∈
/ [a, b] see [2] and [1].
In this paper, some new Pompeiu’s type inequalities for complex-valued
absolutely continuous functions are provided. They are applied to obtain some
new Ostrowski type inequalities.
2. Pompeiu’s Type Inequalities
The following inequality is useful to derive some Ostrowski type inequalities.
Corollary 1 (Pompeiu’s Inequality). With the assumptions of Theorem 1 and
if kf − `f 0 k∞ = supt∈(a,b) |f (t) − tf 0 (t)| < ∞ where ` (t) = t, t ∈ [a, b] , then
(2.1)
|tf (x) − xf (t)| ≤ kf − `f 0 k∞ |x − t|
for any t, x ∈ [a, b].
The inequality (2.1) was stated by the author in [3].
We can generalize the above inequality (2.1) for the larger class of functions
that are absolutely continuous and complex-valued as well as for other norms
of the difference f − `f 0 .
Theorem 6. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] with b > a > 0. Then for any t, x ∈ [a, b] we have
(2.2) |tf (x) − xf (t)|


kf − `f 0 k∞ |x − t|



 1 kf − `f 0 k xq −
p tq−1
≤ 2q−1




kf − `f 0 k max{t,x} ,
1 min{t,x}
x
tq 1/q
q−1
if f − `f 0 ∈ L∞ [a, b] ,
if f − `f 0 ∈ Lp [a, b] , p > 1,
1
+ 1q = 1,
p
46
S. S. DRAGOMIR
or, equivalently
f (x) f (t) (2.3) −
x
t 
1 1
0

kf
−
`f
k

∞ t − x


 1 kf − `f 0 k 1 −
p t2q−1
≤ 2q−1




1
kf − `f 0 k
2 2 .
1 1/q
2q−1
x
if f − `f 0 ∈ L∞ [a, b] ,
if f − `f 0 ∈ Lp [a, b] , p > 1,
1
+ 1q = 1,
p
1 min{t ,x }
Proof. If f is absolutely continuous, then f /` is absolutely continuous on the
interval [a, b] that does not containing 0 and
0
Z x
f (x) f (t)
f (s)
ds =
−
s
x
t
t
for any t, x ∈ [a, b] with x 6= t.
Since
0
Z x
Z x 0
f (s)
f (s) s − f (s)
ds =
ds
s
s2
t
t
then we get the following identity
Z x 0
f (s) s − f (s)
(2.4)
tf (x) − xf (t) = xt
ds
s2
t
for any t, x ∈ [a, b].
We notice that the equality (2.4) was proved for the smaller class of differentiable real valued functions and in a different manner in [5].
Taking the modulus in (2.4) we have
(2.5) |tf (x) − xf (t)|
Z x 0
Z
f
(s)
s
−
f
(s)
= xt
ds ≤ xt 2
s
t
t
x
0
f (s) s − f (s) ds := I
s2
and utilizing Hölder’s integral inequality we deduce

R x 1 0

sup
|f
(s)
s
−
f
(s)|
ds ,
R s∈[t,x]([x,t])
t s2 R
1/p
1/q
(2.6) I ≤ xt tx |f 0 (s) s − f (s)|p ds tx s12q ds , p > 1, p1 +

R
 x 0
sups∈[t,x]([x,t]) 12 ,
|f
(s)
s
−
f
(s)|
ds
s
t

0

kf − `f k∞ |x − t| ,
1
xq
tq 1/q
≤ 2q−1
kf − `f 0 kp tq−1
− xq−1
,
p > 1, p1 + 1q = 1


kf − `f 0 k1 max{t,x}
,
min{t,x}
and the inequality (2.3) is proved.
1
q
= 1,
Remark 1. The first inequality in (2.2) also holds in the same form for 0 >
b > a.
INEQUALITIES OF POMPEIU’S TYPE
47
Remark 2. If we take in (2.2) x = A = A (a, b) := a+b
(the arithmetic mean)
2
√
and t = G = G (a, b) := ab (the geometric mean) then we get the simple
inequality for functions of means:
(2.7) |Gf (A) − Af (G)|

kf − `f 0 k∞ (A − G)



1/q

 1
(A2q−1 −G2q−1 )
0
kf
−
`f
k
p
A1/p G1/p
≤ 2q−1





A
kf − `f 0 k1 G
.
if f − `f 0 ∈ L∞ [a, b] ,
if f − `f 0 ∈ Lp [a, b] , p > 1,
1
+ 1q = 1,
p
3. Evaluating the Integral Mean
The following new result holds.
Theorem 7. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] with b > a > 0. Then for any x ∈ [a, b] we have
Z b
a + b f (x)
1
(3.1) ·
−
f (t) dt
2
x
b−a a

a+b 2 
 b−a 1 + x− 2
kf − `f 0 k∞
if f − `f 0 ∈ L∞ [a, b] ,

x
4
b−a




1
kf − `f 0 kp [Bq (a, b; x)]1/q if f − `f 0 ∈ Lp [a, b] , p > 1,
≤ (2q−1)x(b−a)1/q

1

+ 1q = 1,

p



 1 kf − `f 0 k ln x + b2 −x2 2 ,
b−a
where
(3.2)
1
a
2x
 q
x
q−2

− aq−2 − bq−2 )
 2−q (2x
Bq (a, b; x) = + xq−11(q+1) (bq+1 + aq+1 − 2xq+1 ) ,

 2 x2 b3 +a3 −2x3
x ln ab +
,
3x
q 6= 2
q = 2.
Proof. The first inequality can be proved in an identical way to the case of
differentiable functions from [3] by utilizing the first inequality in (2.2).
Utilising the second inequality in (2.2) we have
Z b
a + b
x
(3.3)
·
f
(x)
−
f
(t)
dt
2
b−a a
Z b
1
≤
|tf (x) − xf (t)| dt
b−a a
Z b q
q 1/q
x
t
1
0
kf − `f kp
≤
tq−1 − xq−1 dt.
(2q − 1) (b − a)
a
48
S. S. DRAGOMIR
Utilising Hölder’s integral inequality we have
1/q
Z b q
x
tq (3.4)
tq−1 − xq−1 dt
a
#q !1/q
Z b 1/p Z b " q
q 1/q
x
t
≤
dt
dt
tq−1 − xq−1 a
a
1/q
Z b q
x
tq 1/p
= (b − a)
.
tq−1 − xq−1 dt
a
For q 6= 2 we have
Z b q
q x
t
tq−1 − xq−1 dt
a
Z x q
Z b q
tq
xq
x
t
=
−
dt +
−
dt
tq−1 xq−1
xq−1 tq−1
a
x
Z x
Z x
Z b
Z b
1
dt
1
1
q
q
q
q
=x
− q−1
dt
t dt + q−1
t dt − x
q−1
q−1
x
x
a t
a
x
x t
1
xq
1
1
=
− 2−q − q−1
xq+1 − aq+1
2−q
2−q x
a
x (q + 1)
1
1
xq
1
q+1
q+1
+ q−1
b
−x
−
−
x (q + 1)
2 − q b2−q x2−q
1
xq
1
1
1
=
−
−
+
2 − q x2−q a2−q b2−q x2−q
1
+ q−1
bq+1 − xq+1 − xq+1 + aq+1
x (q + 1)
xq
1
=
2xq−2 − aq−2 − bq−2 + q−1
bq+1 + aq+1 − 2xq+1
2−q
x (q + 1)
= Bq (a, b; x) .
For q = 2 we have
Z b 2
Z x 2
Z b 2
2
2
2
x
t
x
t
x
t
− dt =
−
−
dt +
dt
t
x
t
x
x
t
a
a
x
Z x
Z b
Z
Z
dt 1 x 2
1
1 b 2
2
2
=x
−
t dt +
t dt − x
dt
t
x a
x x
a
x t
b
x 1 x3 − a 3 1 b 3 − x 3
+
− x2 ln
= x2 ln −
a x 3
x 3
x
2
3
3
3
x
1 b + a − 2x
= x2 ln
+
= B2 (a, b; x) .
ab x
3
Utilizing (3.3) and (3.4) we get the second inequality in (3.1).
INEQUALITIES OF POMPEIU’S TYPE
49
Utilising the third inequality in (2.2) we have
Z b
Z b
a + b
x
1
(3.5) · f (x) −
f (t) dt ≤
|tf (x) − xf (t)| dt
2
b−a a
b−a a
Z b
1
max {t, x}
0
≤
kf − `f k1
dt.
b−a
a min {t, x}
Since
Z
b
a
max {t, x}
dt =
min {t, x}
Z
a
x
x
dt +
t
Z
x
b
t
x 1 b 2 − x2
dt = x ln +
,
x
a x 2
then by (3.5) we have
Z b
Z b
a + b
x
1
f (t) dt ≤
|tf (x) − xf (t)| dt
2 · f (x) − b − a
b−a a
a
1
x 1 b2 − x2
0
≤
kf − `f k1 x ln +
,
b−a
a x 2
and the last part of (3.1) is thus proved.
Remark 3. If we take in (3.1) x = A = A (a, b) :=
a+b
,
2
then we get
Z b
1
(3.6) f (A) −
f (t) dt
b−a a

b−a

kf − `f 0 k∞
if f − `f 0 ∈ L∞ [a, b] ,

4A


1/q
1
0

if f − `f 0 ∈ Lp [a, b] , p > 1,
1/q kf − `f kp [Bq (a, b; A)]
(2q−1)A(b−a)
≤
1

+ 1q = 1,

p


 1 kf − `f 0 k ln A + 1 (b − a) a+3b A ,
b−a
1
a
2
4
where
Bq (a, b; A)
( q
2A
(Aq−2 − A (aq−2 , bq−2 )) +
2−q
=
A
2A2 ln G
+ 12 (b − a)2 ,
2
(q+1)Aq−1
(A (bq+1 , aq+1 ) − Aq+1 ) , q =
6 2
q = 2.
4. A Related Result
The following new result also holds.
50
S. S. DRAGOMIR
Theorem 8. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] with b > a > 0. Then for any x ∈ [a, b] we have
Z b
f (x)
1
f
(t)
(4.1) −
dt
x
b−a a t

a+b
−x
2
0

√x + 2

kf
−
`f
k
,
ln

∞
b−a
x
ab





1
≤
kf − `f 0 kp (Cq (a, b; x))1/q ,
(2q−1)(b−a)1/q







 1 kf − `f 0 k x2 +ab−2ax ,
1
b−a
x2 a
if f − `f 0 ∈ L∞ [a, b] ,
if f − `f 0 ∈ Lp [a, b] , p > 1,
1
p
+
1
q
=1
where
(4.2)
Cq (a, b; x) =
1
x2q−1
(b + a − 2x) +
a2−2q + b2−2q − 2x2−2q
,
2 (q − 1)
q > 1.
Proof. From the first inequality in (3.2) we have
Z b
Z b
f (x)
f (x) f (t) f
(t)
1
1
dt
(4.3)
dt ≤
−
x − b−a
t
b−a a x
t a
Z b
1 1 1
0
− dt.
≤ kf − `f k∞
b − a a t x
Since
Z x Z b
Z b
1 1
1
1
1
1
− dt =
−
dt +
−
dt
x t t x
x t
a
a
x
x x−a b−x
b
= ln −
+
− ln
a
x
x
x
2
x
a + b − 2x
= ln
+
ab
x
!
a+b
−
x
x
= 2 ln √ + 2
x
ab
for any x ∈ [a, b] , then we deduce from (4.3) the first inequality in (4.1).
From the second inequality in (3.2) we have
Z b
Z b
f (x)
f (x) f (t) 1
1
f
(t)
dt
(4.4) −
dt ≤
−
x
b−a a t
b−a a x
t 1/q
Z b
1
1
1
0
dt.
≤
kf − `f kp
−
(2q − 1) (b − a)
t2q−1 x2q−1 a
INEQUALITIES OF POMPEIU’S TYPE
51
Utilising Hölder’s integral inequality we have
Z
(4.5)
b
a
Since
1/q
1/q #q !1/q
Z b 1/p Z b "
1
1
1
1
dt
dt
t2q−1 − x2q−1 dt ≤
t2q−1 − x2q−1 a
a
1/q
Z b 1
1
1/p
= (b − a)
.
−
t2q−1 x2q−1 dt
a
Z b
1
1
t2q−1 − x2q−1 dt
a
Z x
Z b
1
1
1
1
=
−
dt +
−
dt
t2q−1 x2q−1
x2q−1 t2q−1
a
x
x2−2q − a2−2q
1
1
b2−2q − x2−2q
− 2q−1 (x − a) + 2q−1 (b − x) −
=
2 − 2q
x
x
2 − 2q
2−2q
2−2q
2−2q
1
2x
−a
−b
= 2q−1 (b + a − 2x) +
x
2 − 2q
2−2q
1
a
+ b2−2q − 2x2−2q
= 2q−1 (b + a − 2x) +
= Cq (a, b; x)
x
2 (q − 1)
then by (4.4) and (4.5) we get
Z b
f (x)
1
f (t) dt
x − b−a
t
a
≤
1
kf − `f 0 kp (b − a)1/p (Cq (a, b; x))1/q
(2q − 1) (b − a)
and the second inequality in (4.1) is proved.
From the third inequality in (3.2) we have
Z b
Z b
f (x) f (t) f (x)
f (t) 1
1
dt
(4.6) −
dt ≤
−
x
b−a a t
b−a a x
t Z b
1
1
0
≤
kf − `f k1
dt.
2
2
b−a
a min {t , x }
Since
Z
a
b
Z b
dt
dt
x−a b−x
+
=
+ 2
2
2
xa
x
a t
x x
2
x + ab − 2ax
,
=
x2 a
1
dt =
min {t2 , x2 }
Z
x
then by (4.6) we deduce the last part of (4.1).
52
S. S. DRAGOMIR
Remark 4. If we take in (4.1) x = A = A (a, b) := a+b
, then we get
2
Z b
f (A)
1
f
(t)
(4.7) −
dt
A
b−a a t

A
2

kf − `f 0 k∞ ln G
,
if f − `f 0 ∈ L∞ [a, b] ,

b−a





1/q
1
≤ (2q−1)(b−a)1/q kf − `f 0 kp (Cq (a, b; A)) , if f − `f 0 ∈ Lp [a, b] , p > 1,


1

+ 1q = 1,


p

1
kf − `f 0 k1 A+a
,
2
A2 a
where
Cq (a, b; A) =
A (a2−2q , b2−2q ) − A2−2q
,
q−1
q > 1.
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Received February 6, 2014.
Mathematics, College of Engineering & Science,
Victoria University, PO Box 14428,
Melbourne City, MC 8001,
Australia
and
School of Computational & Applied Mathematics,
University of the Witwatersrand,
Private Bag 3,
Johannesburg 2050,
South Africa
E-mail address: sever.dragomir@vu.edu.au
URL: http://rgmia.org/dragomir