Khayyam J. Math. 1 (2015), no. 1, 1–35
A SURVEY ON OSTROWSKI TYPE INEQUALITIES RELATED
TO POMPEIU’S MEAN VALUE THEOREM
SILVESTRU S. DRAGOMIR
Communicated by M.S. Moslehian
Abstract. In this paper we survey some recent results obtained by the author related to Pompeiu’s mean value theorem and inequality. Natural applications to Ostrowski type inequalities that play an important role in Numerical
Analysis, Approximation Theory, Probability Theory & Statistics, Information
Theory and other fields, are given as well.
1. Introduction
In 1946, Pompeiu [11] derived a variant of Lagrange’s mean value theorem, now
known as Pompeiu’s mean value theorem (see also [13, p. 83]).
Theorem 1.1 (Pompeiu, 1946 [11]). 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
x1 f (x2 ) − x2 f (x1 )
= f (ξ) − ξf 0 (ξ) .
x1 − x2
(1.1)
Following [13, p. 84 – 85], we will mention here a geometrical interpretation of
Pompeiu’s theorem.
The equation of the secant line joining the points (x1 , f (x1 )) and (x2 , f (x2 ))
is given by
f (x2 ) − f (x1 )
(x − x1 ) .
y = f (x1 ) +
x2 − x1
Date: Received: 18 June 2014; Accepted: 18 July 2014.
2010 Mathematics Subject Classification. Primary 26D10; Secondary 26D15.
Key words and phrases. Ostrowski inequality, Pompeiu’s mean inequality, integral inequalities, special means.
1
2
S.S. DRAGOMIR
This line intersects the y−axis at the point (0, y) , where y is
f (x2 ) − f (x1 )
(0 − x1 )
x2 − x1
x1 f (x2 ) − x2 f (x1 )
=
.
x1 − x2
The equation of the tangent line at the point (ξ, f (ξ)) is
y = f (x1 ) +
y = (x − ξ) f 0 (ξ) + f (ξ) .
The tangent line intersects the y−axis at the point (0, y) , where
y = −ξf 0 (ξ) + f (ξ) .
Hence, the geometric meaning of Pompeiu’s mean value theorem is that the tangent of the point (ξ, f (ξ)) intersects on the y−axis at the same point as the secant
line connecting the points (x1 , f (x1 )) and (x2 , f (x2 )) .
The following inequality is a simple consequence of Pompeiu’s mean value theorem.
Corollary 1.2 (Pompeiu’s Inequality). With the assumptions of Theorem 1.1
and if kf − `f 0 k∞ = supt∈(a,b) |f (t) − tf 0 (t)| < ∞ where ` (t) = t, t ∈ [a, b] , then
|tf (x) − xf (t)| ≤ kf − `f 0 k∞ |x − t|
(1.2)
for any t, x ∈ [a, b] .
The inequality (1.2) was obtained by the author in [3], see also [4].
In 1938, A. Ostrowski [9] proved the following result in the estimating the
integral mean:
Theorem 1.3 (Ostrowski, 1938 [9]). 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) .
(1.3)
f (t) dt ≤  +
b−a a
4
b−a
The constant
quantity.
1
4
is best possible in the sense that it cannot be replaced by a smaller
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 1.4 (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

!2 
Z b
a+b
a + b f (x)
x
−
1
b − a 1
2

f (t) dt ≤
+
(1.4)
2 · x − b−a
|x|
4
b−a
a
× kf − `f 0 k∞ ,
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
3
where ` (t) = t, t ∈ [a, b] .
The constant 14 is sharp in the sense that it cannot be replaced by a smaller
constant.
In [12], E. C. Popa using a mean value theorem obtained a generalization of
(1.4) as follows:
Theorem 1.5 (Popa, 2007 [12]). 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
−
α
f
(x)
+
f
(t)
dt
(1.5)
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 [10], J. Pečarić and S. Ungar have proved a general estimate with the
p-norm, 1 ≤ p ≤ ∞ which for p = ∞ give Dragomir’s result.
Theorem 1.6 (Pečarić & Ungar, 2006 [10]). 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
1
+ 1q = 1 we have the inequality
p
Z b
a + b f (x)
1
≤ P U (x, p) kf − `f 0 k ,
·
−
f
(t)
dt
p
2
x
b−a a
(1.6)
for x ∈ [a, b] , where
"
a2−q − x2−q
x2−q − a1+q x1−2q
P U (x, p) : = (b − a)
+
(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)
1
−1
p
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 we survey some recent results obtained by the author related to
Pompeiu’s inequality presented above. Natural applications to Ostrowski type
inequalities that play an important role in Numerical Analysis, Approximation
Theory, Probability Theory & Statistics, Information Theory and other fields are
given as well.
4
S.S. DRAGOMIR
2. Ostrowski Via a Generalized Pompeiu’s Inequality
2.1. Pompeiu’s Inequality for p-Norms. We can generalize the above inequality (1.2) 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 as follows:
Lemma 2.1 (Dragomir, 2013 [5]). 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
|tf (x) − xf (t)|

kf − `f 0 k∞ |x − t|









xq
1
kf − `f 0 kp tq−1
−
≤
2q−1








 kf − `f 0 k max{t,x}
1 min{t,x}
or, equivalently
f (x) f (t) x − t 
kf − `f 0 k∞ 1t − x1 








1
1
0
kf
−
`f
k
≤
p t2q−1 −
2q−1








 kf − `f 0 k
1
1 min{t2 ,x2 }
(2.1)
if f − `f 0 ∈ L∞ [a, b]
1/q
q−1
tq
x
if f − `f 0 ∈ Lp [a, b]
p > 1,
1
+ 1q = 1
p
(2.2)
if f − `f 0 ∈ L∞ [a, b]
x
1 1/q
2q−1
if f − `f 0 ∈ Lp [a, b]
p > 1,
1
+ 1q = 1
p
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 (s)
f (x) f (t)
ds =
−
s
x
t
t
for any t, x ∈ [a, b] with x 6= t.
Since
0
Z x 0
Z x
f (s)
f (s) s − f (s)
ds =
ds
s
s2
t
t
then we get the following identity
Z
tf (x) − xf (t) = xt
t
x
f 0 (s) s − f (s)
ds
s2
(2.3)
for any t, x ∈ [a, b] .
We notice that the equality (2.3) was proved for the smaller class of differentiable function and in a different manner in [10].
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
5
Taking the modulus in (2.3) we have
Z x 0
f
(s)
s
−
f
(s)
|tf (x) − xf (t)| = xt
ds
s2
t
Z x 0
f (s) s − f (s) ds := I
≤ xt s2
t
(2.4)
and utilizing Hölder’s integral inequality we deduce
R x 1 
0
ds
sup
|f
(s)
s
−
f
(s)|

s∈[t,x]([x,t])
t s2





 R
1/p R x
1/q
x
I ≤ xt t |f 0 (s) s − f (s)|p ds t s12q ds






 R x 0
|f (s) s − f (s)| ds sups∈[t,x]([x,t]) s12
t

kf − `f 0 k∞ |x − t|







xq
p > 1,
1
tq 1/q
kf − `f 0 kp tq−1
− xq−1
1
≤
2q−1
+ 1q = 1

p






kf − `f 0 k1 max{t,x}
min{t,x}
1
p
p > 1,
+ 1q = 1
and the inequality (2.2) is proved.
(2.5)
Remark 2.2. The first inequality in (2.1) also holds in the same form for 0 > b > a.
(the arithmetic mean) and
Remark 2.3. If we take in (2.1) x = A = A (a, b) := a+b
2
√
t = G = G (a, b) := ab (the geometric mean) then we get the simple inequality
for functions of means:
|Gf (A) − Af (G)|

0
if f − `f 0 ∈ L∞ [a, b]

 kf − `f k∞ (A − G)






if f − `f 0 ∈ Lp [a, b]

2q−1 −G2q−1 1/q
A
(
)
1
p > 1,
kf − `f 0 kp
≤
2q−1
A1/p G1/p

1

+ 1q = 1

p






A
kf − `f 0 k1 G
2.2. Evaluating the Integral Mean. The following new result holds.
(2.6)
6
S.S. DRAGOMIR
Theorem 2.4 (Dragomir, 2013 [5]). 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
f (t) dt
(2.7)
2 · x − b−a
a

a+b 2 x−

b−a 1

+ b−a2
kf − `f 0 k∞
if f − `f 0 ∈ L∞ [a, b]


x
4








if f − `f 0 ∈ Lp [a, b]
1
≤
,
p > 1,
kf − `f 0 kp [Bq (a, b; x)]1/q

(2q−1)x(b−a)1/q

1
1

+q =1

p







 1 kf − `f 0 k ln x + b2 −x2 2 ,
1
b−a
a
2x
where
Bq (a, b; x) =











xq
2−q
(2xq−2 − aq−2 − bq−2 )
q 6= 2
+ xq−11(q+1)
2
x2 ln xab +
(b
q+1
+a
b3 +a3 −2x3
,
3x
q+1
− 2x
q+1
),
(2.8)
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 (2.1).
Utilising the second inequality in (2.1) we have
Z b
a + b
x
f (t) dt
(2.9)
2 · f (x) − b − a
a
Z b
1
≤
|tf (x) − xf (t)| dt
b−a a
Z b q
q 1/q
x
1
t
0
≤
kf − `f kp
tq−1 − xq−1 dt
(2q − 1) (b − a)
a
Utilising Hölder’s integral inequality we have
1/q
Z b q
Z b q
q 1/q
q x
x
t
t
1/p
.
tq−1 − xq−1 dt ≤ (b − a)
tq−1 − xq−1 dt
a
a
For q 6= 2 we have
Z b q
q x
t
tq−1 − xq−1 dt
a
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) .
(2.10)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
7
For q = 2 we have
Z b 2
2
3
3
3
2
x
t
− dt = x2 ln x + 1 b + a − 2x = B2 (a, b; x) .
t
x
ab x
3
a
Utilizing (2.9) and (2.10) we get the second inequality in (2.7).
Utilising the third inequality in (2.1) we have
Z b
Z b
a + b
x
1
f (t) dt ≤
|tf (x) − xf (t)| dt
(2.11)
2 · f (x) − b − a
b−a a
a
Z b
max {t, x}
1
0
kf − `f k1
dt.
≤
b−a
a min {t, x}
Since
Z
a
b
max {t, x}
dt =
min {t, x}
Z
a
x
x
dt +
t
Z
x
b
t
x 1 b2 − x 2
dt = x ln +
,
x
a x 2
then by (2.11) we have
Z b
Z b
a + b
1
x
f (t) dt ≤
|tf (x) − xf (t)| dt
2 · f (x) − b − a
b−a a
a
1
x 1 b2 − x 2
0
≤
kf − `f k1 x ln +
,
b−a
a x 2
and the last part of (2.7) is thus proved.
Remark 2.5. If we take in (2.7) x = A = A (a, b) :=
a+b
,
2
then we get
Z b
1
f (A) −
f (t) dt
(2.12)
b−a a
 b−a
kf − `f 0 k∞
if f − `f 0 ∈ L∞ [a, b]

4A







if f − `f 0 ∈ Lp [a, b]

1/q
1
p > 1,
kf − `f 0 kp [Bq (a, b; A)]
≤
,
(2q−1)A(b−a)1/q

1
1

+
=
1

p
q





 1
kf − `f 0 k1 ln Aa + 12 (b − a) a+3b
A ,
b−a
4
where
Bq (a, b; A) =











2Aq
2−q
(Aq−2 − A (aq−2 , bq−2 ))
q 6= 2
2
q+1 q+1
+ (q+1)A
, a ) − Aq+1 ) ,
q−1 (A (b
A
2A2 ln G
+ 12 (b − a)2 ,
q=2
8
S.S. DRAGOMIR
2.3. A Related Result. The following new result also holds.
Theorem 2.6 (Dragomir, 2013 [5]). 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)
−
dt
(2.13)
x
b−a a t

a+b
−x
2
0

√x + 2
kf
−
`f
k
ln
if f − `f 0 ∈ L∞ [a, b]

∞
b−a
x

ab







if f − `f 0 ∈ Lp [a, b]
1/q
1
≤
,
p > 1,
kf − `f 0 kp (Cq (a, b; x))

(2q−1)(b−a)1/q
1
1


+q =1

p





2
 1
kf − `f 0 k1 x +ab−2ax
,
b−a
x2 a
where
Cq (a, b; x) =
1
x2q−1
a2−2q + b2−2q − 2x2−2q
(b + a − 2x) +
, q > 1.
2 (q − 1)
Proof. From the first inequality in (2.8) we have
Z b
Z b
f (x)
f (x) f (t) f
(t)
1
1
dt
dt ≤
−
x − b−a
t
b−a a x
t a
Z b
1 1 1
0
≤ kf − `f k∞
− dt.
b − a a t x
(2.14)
(2.15)
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
x2 a + b − 2x
= ln
+
ab
x
for any x ∈ [a, b] , then we deduce from (2.15) the first inequality in (2.13).
From the second inequality in (2.8) we have
Z b
f (x)
1
f
(t)
dt
(2.16)
x − b−a
t
a
Z b
f (x) f (t) 1
dt
≤
−
b−a a x
t 1/q
Z b
1
1
1 0
≤
kf − `f kp
t2q−1 − x2q−1 dt.
(2q − 1) (b − a)
a
Utilising Hölder’s integral inequality we have
1/q
1/q
Z b Z b
1
1
1 1 1/p
.
t2q−1 − x2q−1 dt ≤ (b − a)
t2q−1 − x2q−1 dt
a
a
(2.17)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
9
Since
Z b
1
1 t2q−1 − x2q−1 dt
a
1
a2−2q + b2−2q − 2x2−2q
= Cq (a, b; x)
= 2q−1 (b + a − 2x) +
x
2 (q − 1)
then by (2.16) and (2.17) we get
Z b
f (x)
f
(t)
1
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 (2.13) is proved.
From the third inequality in (2.8) we have
Z b
Z b
f (x)
f (x) f (t) 1
f (t) 1
dt ≤
x − b−a
x − t dt
t
b
−
a
a
a
Z b
1
1
0
≤
kf − `f k1
dt.
2
2
b−a
a min {t , x }
(2.18)
Since
Z
a
b
1
dt =
min {t2 , x2 }
Z
a
x
dt
+
t2
Z
x
b
dt
x2 + ab − 2ax
=
,
x2
x2 a
then by (2.18) we deduce the last part of (2.13).
Remark 2.7. If we take in (2.13) x = A = A (a, b) := a+b
, then we get
2
Z b
f (A)
f (t) 1
dt
A − b−a
t
a
 2
A
0
kf
−
`f
k
ln
if f − `f 0 ∈ L∞ [a, b]

∞
b−a
G







if f − `f 0 ∈ Lp [a, b]

1/q
1
0
p > 1,
≤
kf − `f kp (Cq (a, b; A))
(2q−1)(b−a)1/q

1

+ 1q = 1

p





 1
kf − `f 0 k1 A+a
,
2
A2 a
(2.19)
,
where
Cq (a, b; A) =
A (a2−2q , b2−2q ) − A2−2q
, q > 1.
q−1
3. Ostrowski Via Power Pompeiu’s Inequality
3.1. Power Pompeiu’s Inequality. We can generalize the above (1.2) inequality for the power function as follows.
10
S.S. DRAGOMIR
Lemma 3.1 (Dragomir, 2013 [6]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] with b > a > 0. If r ∈ R, r 6= 0, then for any
t, x ∈ [a, b] we have
|tr f (x) − xr f (t)|
 1
0
r
r
0

 |r| kf ` − rf k∞ |t − x | , if f ` − rf ∈ L∞ [a, b]







1
tr
xr
,
1−q(r+1)−r

− t1−q(r+1)−r


|1−q(r+1)|
x





for r 6= − p1

kf 0 ` − rf kp
≤



 r r

t x |ln x − ln t| , for r = − p1





if f 0 ` − rf ∈ Lp [a, b]






tr xr
 kf 0 ` − rf k
1 min{xr+1 ,tr+1 }
or, equivalently
f (x) f (t) xr − tr 
1
1
0
r − 1r , if f 0 ` − rf ∈ L∞ [a, b]

kf
`
−
rf
k
 |r|
∞ x
t




1



1
1
1−q(r+1) − 1−q(r+1)
, for r 6= − 1



|1−q(r+1)|
p
x
t


kf 0 ` − rf kp
≤

|ln x − ln t| , for r = − p1




 if f 0 ` − rf ∈ Lp [a, b]






1
 kf 0 ` − rf k
1 min{xr+1 ,tr+1 }
where p > 1, p1 +
1
q
(3.1)
(3.2)
= 1.
Proof. If f is absolutely continuous, then f / (·)r is absolutely continuous on the
interval [a, b] and
0
Z x
f (s)
f (x) f (t)
ds =
− r
r
s
xr
t
t
for any t, x ∈ [a, b] with x 6= t.
Since
0
Z x
Z x 0
Z x 0
f (s)
f (s) sr − rsr−1 f (s)
f (s) s − rf (s)
ds =
ds =
ds
r
2r
s
s
sr+1
t
t
t
then we get the following identity
r
r
r r
Z
t f (x) − x f (t) = x t
t
for any t, x ∈ [a, b] .
x
f 0 (s) s − rf (s)
ds
sr+1
(3.3)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
11
Taking the modulus in (3.3) we have
Z x 0
f (s) s − rf (s) |t f (x) − x f (t)| = x t ds
sr+1
t
Z x 0
|f (s) s − rf (s)| r r
ds := I
≤ xt sr+1
t
r
r
r r
and utilizing Hölder’s integral inequality we deduce
R x 1

0
sup
|f
(s)
s
−
rf
(s)|
ds
r+1

s∈[t,x]([x,t])
t s



 R
R
1/q
r r
1
x |f 0 (s) s − rf (s)|p ds1/p x q(r+1)
I ≤ xt
ds
t s
 t



1  R x 0
|f (s) s − rf (s)| ds sups∈[t,x]([x,t]) sr+1
t
 1
1
0
r − 1r kf
`
−
rf
k

∞ x

|r|
t




1



1
1
1−q(r+1) − 1−q(r+1)
,



x
t



 |1−q(r+1)|
1
r 6= − p
≤ xr tr
kf 0 ` − rf kp








|ln x − ln t| , r = − p1






1
 kf 0 ` − rf k
1 min{xr+1 ,tr+1 } .
where p > 1, p1 +
1
q
(3.4)
(3.5)
= 1, and the inequality (3.1) is proved.
3.2. Some Ostrowski Type Results. The following new result also holds.
Theorem 3.2 (Dragomir, 2013 [6]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] with b > a > 0. If r ∈ R, r 6= 0, and f 0 ` − rf ∈
L∞ [a, b] , then for any x ∈ [a, b] we have
r+1
Z b
r+1
b
−
a
r
f
(x)
−
x
f
(t)
dt
(3.6)
r+1
a
1
≤
kf 0 ` − rf k∞
|r|

2rxr+1 −xr (a+b)(r+1)+br+1 +ar+1

, if r > 0

r+1
×

 xr (a+b)(r+1)−2rxr+1 −br+1 −ar+1 , if r ∈ (−∞, 0) {−1} .
r+1
Also, for r = −1, we have
Z b
x
b
1
f (x) ln −
≤ 2 kf 0 ` + f k
√
f
(t)
dt
ln
+
∞
a x a
ab
for any x ∈ [a, b] , provided f 0 ` + f ∈ L∞ [a, b]
The constant 2 in (3.7) is best possible.
a+b
2
−x
x
!
(3.7)
12
S.S. DRAGOMIR
Proof. Utilising the first inequality in (3.1) for r 6= −1 we have
r+1
Z b
Z b
b
− ar+1
r
|tr f (x) − xr f (t)| dt
(3.8)
f (t) dt ≤
r + 1 f (x) − x
a
a
Z b
1
0
≤
|tr − xr | dt.
kf ` − rf k∞
|r|
a
Observe that
 Rx
Rb
Z b
 a (xr − tr ) dt + x (tr − xr ) dt, if r > 0
|tr − xr | dt =
Rb r
 Rx r
a
r
(t
−
x
)
dt
+
(x − tr ) dt, if r ∈ (−∞, 0) {−1} .
a
x
Then for r > 0 we have
Z x
Z b
2rxr+1 − xr (a + b) (r + 1) + br+1 + ar+1
r
r
(x − t ) dt +
(tr − xr ) dt =
r+1
a
x
and for r ∈ (−∞, 0) {−1} we have
Z b
Z x
2rxr+1 − xr (a + b) (r + 1) + br+1 + ar+1
r
r
(xr − tr ) dt = −
(t − x ) dt +
.
r+1
x
a
Making use of (3.8) we get (3.6).
Utilizing the inequality (3.1) for r = −1 we have
−1
t f (x) − x−1 f (t) ≤ kf 0 ` + f k t−1 − x−1 ∞
if f 0 ` + f ∈ L∞ [a, b].
Integrating this inequality, we have
Z b
Z b
−1
f (x) ln b − x−1
≤
t f (x) − x−1 f (t) dt
f
(t)
dt
a
a
a
Z b
−1
0
t − x−1 dt.
≤ kf ` + f k∞
(3.9)
a
Since
Z b
1 1
− dt = 2 ln √x +
x t ab
a
a+b
2
−x
x
!
,
then by (3.9) we get the desired inequality (3.7).
Now, assume that (3.7) holds with a constant C > 0, i.e.
Z b
x
f (x) ln b − x−1
f (t) dt ≤ C kf 0 ` + f k∞ ln √ +
a
ab
a
for any x ∈ [a, b] .
If we take in (3.10) f (t) = 1, t ∈ [a, b] , then we get
!
a+b
b b − a
−
x
x
ln −
≤ C ln √ + 2
a
x x
ab
for any for any x ∈ [a, b] .
a+b
2
−x
x
!
(3.10)
(3.11)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
13
Making x = a in (3.10) produces the inequality
b b − a
1
b
b
−
a
ln −
≤C
− ln
a
a 2a
2 a
which implies that C ≥ 2.
This proves the sharpness of the constant 2 in (3.7).
Remark 3.3. Consider the r-Logarithmic mean
r+1
1/r
b
− ar+1
Lr = Lr (a, b) :=
(r + 1) (b − a)
defined for r ∈ R {0, −1} and the Logarithmic mean, defined as
b−a
.
L = L (a, b) :=
ln b − ln a
If A = A (a, b) := a+b
, then from (3.6) we get for x = A the inequality
2
Z b
r
Lr (b − a) f (A) − Ar
f
(t)
dt
(3.12)
a
≤



2
kf 0 ` − rf k∞

|r|

A(br+1 ,ar+1 )−Ar+1
,
r+1
if r > 0
Ar+1 −A(br+1 ,ar+1 )
,
r+1
if r ∈ (−∞, 0) {−1} ,
while from (3.7) we get
Z b
−1
A
−1
L (b − a) f (A) − A
f (t) dt ≤ 2 kf 0 ` + f k∞ ln
G
(3.13)
a
The following related result holds.
Theorem 3.4 (Dragomir, 2013 [6]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] with b > a > 0. If r ∈ R, r 6= 0, then for any
x ∈ [a, b] we have
Z b
f (x)
f
(t)
(b
−
a)
−
dt
(3.14)
xr
tr
a
1
≤
kf 0 ` − rf k∞
|r|
 2x1−r −a1−r −b1−r
+

1−r
×
 a1−r +b1−r −2x1−r
+
1−r
1
xr
(b + a − 2x) , r ∈ (0, ∞) {1}
,
1
xr
(2x − a − b) , if r < 0.
Also, for r = 1, we have
Z b
f (x)
x
f (t) dt ≤ 2 kf 0 ` − f k∞ ln √ +
x (b − a) −
t
ab
a
for any x ∈ [a, b] , provided f 0 ` − f ∈ L∞ [a, b] .
The constant 2 is best possible in (3.15).
a+b
2
−x
x
!
(3.15)
14
S.S. DRAGOMIR
Proof. From the first inequality in (3.2) we have
1
f (x) f (t) 1
1 0
xr − tr ≤ |r| kf ` − rf k∞ xr − tr ,
for any t, x ∈ [a, b] , provided f 0 ` − rf ∈ L∞ [a, b] .
Integrating over t ∈ [a, b] we get
Z b
Z b
f (x)
f (x) f (t) f
(t)
dt ≤
xr (b − a) −
xr − tr dt
tr
a
a
Z b
1
1
1 0
kf ` − rf k∞
≤
xr − tr dt,
|r|
a
(3.16)
(3.17)
for r ∈ R, r 6= 0.
For r ∈ (0, ∞) {1} we have
Z b
1−r
1
− a1−r − b1−r
1
1
− dt = 2x
+ r (b + a − 2x)
xr tr 1−r
x
a
for any x ∈ [a, b] .
For r < 0, we also have
Z b
1−r
1
1
+ b1−r − 2x1−r
1
− dt = a
+ r (2x − a − b)
xr tr 1−r
x
a
for any x ∈ [a, b] .
For r = 1 we have
Z b
1 1
− dt = 2 ln √x +
x t ab
a
a+b
2
−x
x
!
for any x ∈ [a, b] , and the inequality (3.15) is obtained.
The sharpness of the constant 2 follows as in the proof of Theorem 3.2 and the
details are omitted.
Remark 3.5. If we take x = A in Theorem 3.4, then we we have
Z b
f (A)
f
(t)
dt
Ar (b − a) −
r
t
a
 A1−r −A a1−r ,b1−r
(
)

, r ∈ (0, ∞) {1}

1−r
2
0
≤
kf ` − rf k∞
,

|r|
 A(a1−r ,b1−r )−A1−r
, if r < 0.
1−r
Also, for r = 1, we have
Z b
f (A)
f (t) A
dt ≤ 2 kf 0 ` − f k∞ ln .
A (b − a) −
t
G
a
(3.18)
(3.19)
Remark 3.6. The interested reader may obtain other similar results in terms of
the p-norms kf 0 ` − rf kp with p ≥ 1. However, since some calculations are too
complicated, the details are not presented here.
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
15
4. Ostrowski Via an Exponential Pompeiu’s Inequality
4.1. An Exponential Pompeiu’s Inequality. We can provide some similar
results for complex-valued functions with the exponential instead of `.
Lemma 4.1 (Dragomir, 2013 [7]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] and α ∈ C with Re (α) 6= 0. Then for any t, x ∈ [a, b]
we have
f (x)
f (t) exp (αx) − exp (αt) 
|Re (α)| kf 0 − αf k∞




1
1

×
−

exp(tRe(α))
exp(xRe(α)) 







q 1/q |Re (α)|1/q kf 0 − αf kp
≤
1/q


1
1


×
−

exp(tqRe(α))
exp(xqRe(α))








 kf 0 − αf k
1
1 min{exp(tRe(α)),exp(xRe(α))} ,
if f 0 −αf
∈ L∞ [a, b] ,
if f 0 −αf
∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.1)
or, equivalently
|exp (αt) f (x) − f (t) exp (αx)|


|Re (α)| kf 0 − αf k∞



× |exp (xRe (α)) − exp (tRe (α))|









q 1/q |Re (α)|1/q kf 0 − αf kp
≤



× |exp (xqRe (α)) − exp (tqRe (α))|1/q









 kf 0 − αf k max {exp (tRe (α)) , exp (xRe (α))} .
1
if f 0 −αf
∈ L∞ [a, b] ,
if f 0 −αf
∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.2)
Proof. If f is absolutely continuous, then f / exp (α·) is absolutely continuous on
the interval [a, b] and
Z
t
x
f (s)
exp (αs)
for any t, x ∈ [a, b] with x 6= t.
0
ds =
f (x)
f (t)
−
exp (αx) exp (αt)
16
S.S. DRAGOMIR
Since
x
Z
t
f (s)
exp (αs)
0
Z
x
ds =
Zt x
=
t
f 0 (s) exp (αs) − αf (s) exp (αs)
ds
exp (2αs)
f 0 (s) − αf (s)
ds,
exp (αs)
then we get the following identity
f (t)
f (x)
−
=
exp (αx) exp (αt)
Z
t
x
f 0 (s) − αf (s)
ds
exp (αs)
for any t, x ∈ [a, b] with x 6= t.
Taking the modulus in (4.3) we have
Z x 0
f (x)
f
(s)
−
αf
(s)
f
(t)
=
−
ds
exp (αx) exp (αt) exp
(αs)
Zt x 0
|f (s) − αf (s)| ds := I
≤
|exp (αs)|
(4.3)
(4.4)
t
and utilizing Hölder’s integral inequality we deduce
R

x
1
0

sup
|f
(s)
−
αf
(s)|
ds
,

s∈[t,x]([x,t])
t |exp(αs)|





 R
1/q
1/p R x
x
p
1
0
I≤
|f (s) − αf (s)| ds
t |exp(αs)|q ds ,
t





o
n


1
 R x |f 0 (s) − αf (s)| ds sup
s∈[t,x]([x,t]) |exp(αs)| ,
t
R

x
1
0


 kf − αf k∞ t |exp(αs)| ds ,





R
1/q
x
1
0
≤
kf − αf kp t |exp(αs)|q ds ,





n
o


1
 kf 0 − αf k sup
s∈[t,x]([x,t]) |exp(αs)| .
1
(4.5)
Now, since α = Re (α) + iIm (α) and s ∈ [a, b], then
|exp (αs)| = exp (sRe (α)) .
We have
Z
t
x
1
1
1
ds = Re (α)
−
|exp (αs)|
exp (tRe (α)) exp (xRe (α))
and by (4.4) and (4.5) we get
f (x)
f
(t)
1
1
0
≤ kf − αf k |Re (α)| −
−
∞
exp (αx) exp (αt) exp (tRe (α)) exp (xRe (α)) and the first part of (4.1) is proved.
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
We have
Z x
t
17
1
1
1
ds = qRe (α)
−
|exp (αs)|q
exp (tqRe (α)) exp (xqRe (α))
and by (4.4) and (4.5) we get the second part of (4.1).
We have
1
1
sup
=
|exp (αs)|
min {exp (tRe (α)) , exp (xRe (α))}
s∈[t,x]([x,t])
and by (4.4) and (4.5) we get the last part of (4.1).
The inequality (4.2) follows by (4.1) on multiplying with |exp (αx) exp (αt)|
and performing the required calculation.
The following particular case is of interest.
Corollary 4.2. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] . Then for any t, x ∈ [a, b] we have
f (x)
f
(t)
−
exp (x) exp (t) 
1

1 0

kf
−
f
k
−

∞ exp(t)
exp(x) 









1/q
1
≤
1
1/q
0
q kf − f kp exp(tq) − exp(xq) 











1
 kf 0 − f k
1 min{exp(t),exp(x)} ,
if f 0 − f
∈ L∞ [a, b] ,
if f 0 − f
∈ Lp [a, b]
p > 1,
1
+ 1q = 1,
p
(4.6)
or, equivalently
|exp (t) f (x) − f (t) exp (x)|
≤




kf 0 − f k∞ |exp (x) − exp (t)|










q 1/q kf 0 − f kp |exp (xq) − exp (tq)|1/q












 kf 0 − f k max {exp (t) , exp (x)} .
1
if f 0 − f
∈ L∞ [a, b] ,
if f 0 − f
∈ Lp [a, b]
p > 1,
1
+ 1q = 1,
p
(4.7)
18
S.S. DRAGOMIR
Remark 4.3. If Re (α) = 0 then the inequality (4.5) becomes
R

x
1
0

sup
|f
(s)
−
iIm
(α)
f
(s)|
ds
,

s∈[t,x]([x,t])
t |exp(iIm(α)s)|





 R
1/q
R
1
x |f 0 (s) − iIm (α) f (s)|p ds1/p x
I≤
ds ,
t |exp(iIm(α)s)|q
t





n
o

 R x 0
1

sups∈[t,x]([x,t])
|f
(s)
−
iIm
(α)
f
(s)|
ds
,
|exp(iIm(α)s)|
t
R x 

kf 0 − iIm (α) f k∞ t ds ,
kf 0 − iIm (α) f k∞ |x − t| ,










R
1/q
x
0
=
≤
kf 0 − iIm (α) f kp |x − t|1/q ,
,
kf − iIm (α) f kp t ds










0
kf 0 − iIm (α) f k1 .
kf − iIm (α) f k1 ,
Therefore we have

kf 0 − iIm (α) f k∞ |x − t| ,





f
(x)
f
(t)
1/q
0
exp (iIm (α) x) − exp (iIm (α) t) ≤  kf − iIm (α) f kp |x − t| ,




kf 0 − iIm (α) f k1 ,
(4.8)
or, equivalently
|exp (iIm (α) t) f (x) − f (t) exp (iIm (α) x)|

kf 0 − iIm (α) f k∞ |x − t| ,





≤
kf 0 − iIm (α) f kp |x − t|1/q ,





kf 0 − iIm (α) f k1
(4.9)
for any t, x ∈ [a, b] .
In particular, we have

kf 0 − if k∞ |x − t| ,




f (x)
f (t) 
1/q
0
exp (ix) − exp (it) ≤  kf − if kp |x − t| ,




kf 0 − if k1 ,
(4.10)
or, equivalently

kf 0 − if k∞ |x − t| ,





|exp (it) f (x) − f (t) exp (ix)| ≤
kf 0 − if kp |x − t|1/q ,





kf 0 − if k1 ,
for any t, x ∈ [a, b] .
(4.11)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
19
4.2. Inequalities of Ostrowski Type. The following result holds:
Theorem 4.4 (Dragomir, 2013 [7]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] and α ∈ C with Re (α) > 0. Then for any x ∈ [a, b]
we have
Z b
exp
(αb)
−
exp
(αa)
f (x)
f
(t)
dt
−
exp
(αx)
α
a




|Re (α)| kf 0 − αf k∞ B1 (a, b, x, α)










q 1/q |Re (α)|1/q (b − a)1/p
≤


 × kf 0 − αf kp |Bq (a, b, x, α)|1/q









 kf 0 − αf k B (a, b, x, α)
1 ∞
if f 0 −αf
∈ L∞ [a, b] ,
if f 0 −αf
∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.12)
where
a+b
Bq (a, b, x, α) := 2 exp (xqRe (α)) x −
2
1
exp (bqRe (α)) + exp (aqRe (α))
− exp (xqRe (α))
+
qRe (α)
2
for q ≥ 1 and
exp (bRe (α)) − exp (xRe (α))
B∞ (a, b, x, α) := exp (xRe (α)) (x − a) +
.
Re (α)
Proof. Utilising the first inequality in (4.2) we have
Z b
Z b
f (x)
exp
(αt)
dt
−
exp
(αx)
f
(t)
dt
(4.13)
a
a
Z b
|exp (αt) f (x) − f (t) exp (αx)| dt
≤
a
Z b
0
≤ |Re (α)| kf − αf k∞
|exp (xRe (α)) − exp (tRe (α))| dt
a
for any x ∈ [a, b] .
Observe that, since Re (α) > 0, then
Z b
|exp (xRe (α)) − exp (tRe (α))| dt
a
a+b
= 2 exp (xRe (α)) x −
2
1
exp (bRe (α)) + exp (aRe (α))
+
− exp (xRe (α))
Re (α)
2
20
S.S. DRAGOMIR
for any x ∈ [a, b] .
Also
Z
b
Z
b
exp (αt) dt − exp (αx)
f (t) dt
Z b
exp (αb) − exp (αa)
= f (x)
− exp (αx)
f (t) dt
α
a
for any x ∈ [a, b] and by (4.13) we get the first inequality in (4.12).
Using the second inequality in (4.2) we have
Z b
Z b
f (x)
f (t) dt
(4.14)
exp (αt) dt − exp (αx)
a
a
Z b
|exp (αt) f (x) − f (t) exp (αx)| dt
≤
a
Z b
1/q
1/q
0
≤ q |Re (α)| kf − αf kp
|exp (xqRe (α)) − exp (tqRe (α))|1/q dt
f (x)
a
a
a
for any x ∈ [a, b] .
By Hölder’s integral inequality we also have
Z b
|exp (xqRe (α)) − exp (tqRe (α))|1/q dt
a
1/p
Z
≤ (b − a)
b
1/q
,
|exp (xqRe (α)) − exp (tqRe (α))| dt
a
for any x ∈ [a, b] .
Observe that, as above, we have
Z b
|exp (xqRe (α)) − exp (tqRe (α))| dt
a
a+b
= 2 exp (xqRe (α)) x −
2
1
exp (bqRe (α)) + exp (aqRe (α))
+
− exp (xqRe (α))
qRe (α)
2
= Bq (a, b, x, α)
for any x ∈ [a, b] and by (4.14) we get the second part of (4.12).
Using the third inequality in (4.2) we have
Z b
Z b
f (x)
exp (αt) dt − exp (αx)
f (t) dt
a
a
Z b
≤
|exp (αt) f (x) − f (t) exp (αx)| dt
a
Z b
0
≤ kf − αf k1
max {exp (tRe (α)) , exp (xRe (α))} dt
a
for any x ∈ [a, b] .
(4.15)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
21
Observe that,
Z
b
max {exp (tRe (α)) , exp (xRe (α))} dt
Z x
max {exp (tRe (α)) , exp (xRe (α))} dt
=
a
Z b
max {exp (tRe (α)) , exp (xRe (α))} dt
+
x
Z b
Z x
exp (xRe (α)) dt +
exp (tRe (α)) dt =
=
a
a
x
= exp (xRe (α)) (x − a) +
exp (bRe (α)) − exp (xRe (α))
Re (α)
and by (4.15) we get the third part of (4.12).
Remark 4.5. If Re (α) < 0, then a similar result may be stated. However the
details are left to the interested reader.
Corollary 4.6. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b]. Then for any x ∈ [a, b] we have
Z b
f (x) [exp (b) − exp (a)] − exp (x)
f
(t)
dt
a

kf 0 − f k∞ B1 (a, b, x)








 1/q
q (b − a)1/p kf 0 − f kp
≤

× |Bq (a, b, x)|1/q








kf 0 − f k1 B∞ (a, b, x)
if f 0 −f ∈ L∞ [a, b] ,
if f 0 −f ∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
where
Bq (a, b, x)
1 exp (bq) + exp (aq)
a+b
exp (xq) +
− exp (xq)
:= 2 x −
2
q
2
for q ≥ 1 and
B∞ (a, b, x) := (x − a) exp (x) + exp (b) − exp (x) .
(4.16)
22
S.S. DRAGOMIR
Remark 4.7. The midpoint case is as follows:
Z b
f a + b [exp (b) − exp (a)] − exp a + b
f
(t)
dt
2
2
a

kf 0 − f k∞ B1 (a, b)








 1/q
q (b − a)1/p kf 0 − f kp
≤

× |Bq (a, b)|1/q








kf 0 − f k1 B∞ (a, b)
if f 0 −f ∈ L∞ [a, b] ,
if f 0 −f ∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.17)
where
2
Bq (a, b, x) :=
q
exp (bq) + exp (aq)
− exp
2
a+b
q
2
for q ≥ 1 and
b−a
B∞ (a, b) :=
exp
2
a+b
2
+ exp (b) − exp
a+b
2
.
The case Re (α) = 0 is different and may be stated as follows.
Theorem 4.8 (Dragomir, 2013 [7]). Let f : [a, b] → C be an absolutely continuous
function on the interval [a, b] and α ∈ C with Re (α) = 0 and Im (α) 6= 0. Then
for any x ∈ [a, b] we have
Z b
exp
(iIm
(α)
b)
−
exp
(iIm
(α)
a)
f (x)
−
exp
(iIm
(α)
x)
f
(t)
dt
iIm (α)
a
≤

0
kf


−iIm (α)f2 k∞



x− a+b
1
2

(b − a)2
+


4
b−a







q
kf 0 − iIm (α) f kp
q+1
h
i
q+1
q+1
x−a
q
q
× b−x
+
(b
b−a
b−a














 kf 0 − iIm (α) f k (b − a) .
1
if f 0 −iIm (α) f
∈ L∞ [a, b] ,
− a)
q+1
q
if f 0 −iIm (α) f
∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.18)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
23
Proof. Utilizing the inequality (4.9) we have
Z b
exp
(iIm
(α)
b)
−
exp
(iIm
(α)
a)
f (x)
f
(t)
dt
−
exp
(iIm
(α)
x)
iIm (α)
a
(4.19)
b
Z
|exp (iIm (α) t) f (x) − f (t) exp (iIm (α) x)| dt
≤
a

Rb

kf 0 − iIm (α) f k∞ a |x − t| dt,





Rb
≤
kf 0 − iIm (α) f kp a |x − t|1/q dt,





 kf 0 − iIm (α) f k R b dt.
1 a
Since
b
Z
a

1
|x − t| dt =  +
4
a+b
2
x−
b−a
!2 
 (b − a)2
and
Z
a
b
q
|x − t|1/q dt =
q+1
"
b−x
b−a
q+1
q
+
x−a
b−a
#
q+1
q
(b − a)
q+1
q
then we get from (4.19) the desired result (4.18).
,
Corollary 4.9. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] . Then for any x ∈ [a, b] we have
Z b
f (x) exp (ib) − exp (ia) − exp (ix)
f
(t)
dt
i
a
≤

kf0 − if k∞



a+b 2 

x−
1

× 4 + b−a2
(b − a)2









q
kf 0 − if kp
q+1
h
q+1
q
× b−x
+
b−a

i

q+1
q+1

x−a

q
q

(b
−
a)

b−a








 kf 0 − if k (b − a) .
1
if f 0 −if
∈ L∞ [a, b] ,
if f 0 −if
∈ Lp [a, b]
p > 1,
1 1
+ = 1,
p q
(4.20)
24
S.S. DRAGOMIR
Remark 4.10. The midpoint case is as follows
Z b
f a + b exp (ib) − exp (ia) − exp i a + b
f
(t)
dt
2
i
2
(4.21)
a
≤



1
4


q
(q+1)21/q
kf 0 − if k∞ (b − a)2 , if f 0 −if ∈ L∞ [a, b] ,
kf 0 − if kp (b − a)
q+1
q
, if f 0 −if ∈ Lp [a, b] .
Similar inequalities may be stated if one uses (4.1) and integrates over t on
[a, b] . The details are left to the interested reader.
5. Ostrowski Via a Two Functions Pompeiu’s Inequality
5.1. A General Pompeiu’s Inequality. We start with the following generalization of Pompeiu’s inequality:
Theorem 5.1 (Dragomir, 2013 [8]). Let f, g : [a, b] → C be absolutely continuous
functions on the interval [a, b] with g (t) 6= 0 for all t ∈ [a, b] . Then for any
t, x ∈ [a, b] we have
f (x) f (t) g (x) − g (t) 
R
x 1
0
0

kf
g
−
f
g
k
ds
if f 0 g − f g 0 ∈ L∞ [a, b] ,

2
∞

t
|g(s)|







if f 0 g − f g 0 ∈ Lp [a, b]

1/q
R
x
1
p > 1,
≤
kf 0 g − f g 0 kp t |g(s)|2q ds

1

+ 1q = 1,

p





n
o


1
 kf 0 g − f g 0 k sups∈[t,x]([x,t])
1
|g(s)|2
(5.1)
or, equivalently
|g (t) f (x) − f (t) g (x)|

R
x 1
0
0

kf
g
−
f
g
k
|g
(t)
g
(x)|
ds
if f 0 g − f g 0 ∈ L∞ [a, b] ,

2
∞

t
|g(s)|







if f 0 g − f g 0 ∈ Lp [a, b]

R
1/q
x
1
p > 1,
≤
kf 0 g − f g 0 kp |g (t) g (x)| t |g(s)|2q ds

1

+ 1q = 1,

p





n
o


1
 kf 0 g − f g 0 k |g (t) g (x)| sups∈[t,x]([x,t])
.
1
|g(s)|2
(5.2)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
25
Proof. If f and g are absolutely continuous and g (t) 6= 0 for all t ∈ [a, b], then
f /g is absolutely continuous on the interval [a, b] and
0
Z x
f (s)
f (x) f (t)
−
ds =
g (s)
g (x)
g (t)
t
for any t, x ∈ [a, b] with x 6= t.
Since
0
Z x 0
Z x
f (s) g (s) − f (s) g 0 (s)
f (s)
ds =
ds,
g (s)
g 2 (s)
t
t
then we get the following identity
Z x 0
f (s) g (s) − f (s) g 0 (s)
f (x) f (t)
−
=
ds
g (x)
g (t)
g 2 (s)
t
(5.3)
for any t, x ∈ [a, b] .
Taking the modulus in (5.3) we have
Z
f (x) f (t) x f 0 (s) g (s) − f (s) g 0 (s) ds
g (x) − g (t) = g 2 (s)
t
Z x 0
|f (s) g (s) − f (s) g 0 (s)| ≤
ds := I
|g (s)|2
t
and utilizing Hölder’s integral inequality we deduce

R
x 1
0
0

sups∈[t,x]([x,t]) |f (s) g (s) − f (s) g (s)| t |g(s)|2 ds ,







 R
1/q
R
x |f 0 (s) g (s) − f (s) g 0 (s)|p ds1/p x 1 2q ds
I≤
t
t |g(s)|






n
o
R


1
 x |f 0 (s) g (s) − f (s) g 0 (s)| ds sups∈[t,x]([x,t])
,
t
|g(s)|2

R
x 1

kf 0 g − f g 0 k∞ t |g(s)|

2 ds ,







R
1/q
p > 1,
x 1
0
0
≤
kf g − f g kp t |g(s)|2q ds
1
1
+
= 1,

p
q





o
n


1
 kf 0 g − f g 0 k sups∈[t,x]([x,t])
1
|g(s)|2
and the inequality (5.1) is proved.
1
p
(5.4)
p > 1,
+ 1q = 1,
The following particular case extends Pompeiu’s inequality to other p-norms
than p = ∞ obtained in (5.2).
26
S.S. DRAGOMIR
Corollary 5.2. 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
f (x) f (t) x − t 
kf − `f 0 k∞ 1t − x1 








1
1
0
kf
−
`f
k
≤
p t2q−1 −
2q−1








 kf − `f 0 k
1
1 min{t2 ,x2 }
if f − `f 0 ∈ L∞ [a, b] ,
x
1 1/q
2q−1
if f − `f 0 ∈ Lp [a, b]
p > 1,
1
1
+
= 1,
p
q
(5.5)
or, equivalently
|tf (x) − xf (t)|

kf − `f 0 k∞ |x − t|







xq
1
kf − `f 0 kp tq−1
−
≤
2q−1







kf − `f 0 k1 max{t,x}
,
min{t,x}
if f − `f 0 ∈ L∞ [a, b] ,
1/q
q−1
tq
x
if f − `f 0 ∈ Lp [a, b]
p > 1, p1 + 1q = 1,
(5.6)
where ` (t) = t, t ∈ [a, b] .
The proof follows by (5.1) for g (t) = ` (t) = t, t ∈ [a, b] .
The general case for power functions is as follows.
Corollary 5.3. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] with b > a > 0. If r ∈ R, r 6= 0, then for any t, x ∈ [a, b] we have
f (x) f (t) xr − tr 
1
1
0
r − 1r , if f 0 ` − rf ∈ L∞ [a, b] ,

kf
`
−
rf
k

∞
|r|
x
t







kf 0 ` − rf kp 



1
1
1
1


 |1−q(r+1)| x1−q(r+1) − t1−q(r+1) , for r 6= − p
≤
×



|ln x − ln t| , for r = − p1





if f 0 ` − rf ∈ Lp [a, b] ,






1
 kf 0 ` − rf k
1 min{xr+1 ,tr+1 } ,
(5.7)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
27
or, equivalently
|tr f (x) − xr f (t)|

1
kf 0 ` − rf k∞ |tr − xr | , if f 0 ` − rf ∈ L∞ [a, b] ,


|r|







kf 0 ` − rf kp 



1
xr
1
tr


 |1−q(r+1)| x1−q(r+1)−r − t1−q(r+1)−r , for r 6= − p
≤
×

 r r

t x |ln x − ln t| , for r = − p1





if f 0 ` − rf ∈ Lp [a, b] ,






tr xr
 kf 0 ` − rf k
1 min{xr+1 ,tr+1 } ,
where p > 1, p1 +
1
q
(5.8)
= 1.
The proof follows by (5.1) for g (t) = tr , t ∈ [a, b] . The details for calculations
are omitted.
We have the following result for exponential.
Corollary 5.4. Let f : [a, b] → C be an absolutely continuous function on the
interval [a, b] and α ∈ R, α 6= 0. Then for any t, x ∈ [a, b] we have
f (x)
f
(t)
exp (iαx) − exp (iαt) 
kf 0 − iαf k∞ |x − t|
if f 0 − iαf ∈ L∞ [a, b] ,








if f 0 − iαf ∈ Lp [a, b]

p > 1,
kf 0 − iαf kp |x − t|1/q
≤

1
1

+
= 1,

p
q





 kf 0 − iαf k
1
(5.9)
or, equivalently
|exp (iαt) f (x) − f (t) exp (iαx)|

kf 0 − iαf k∞ |x − t|
if f 0 − iαf ∈ L∞ [a, b] ,








if f 0 − iαf ∈ Lp [a, b]

p > 1,
≤
kf 0 − iαf kp |x − t|1/q

1
1

+
= 1,

p
q





 kf 0 − iαf k .
1
(5.10)
28
S.S. DRAGOMIR
5.2. An Inequality Generalizing Ostrowski’s. The following result holds:
Theorem 5.5 (Dragomir, 2013 [8]). Let f, g : [a, b] → C be absolutely continuous
functions on the interval [a, b] . If 0 < m ≤ |g (t)| ≤ M < ∞ for any t ∈ [a, b] ,
then
Z b
Z b
f (x)
f (t) dt
g (t) dt − g (x)
a
a

a+b 2 
t−
2 1
0
0

kf g − f g k∞ (b − a) 4 + b−a2
if f 0 g − f g 0 ∈ L∞ [a, b] ,







2 

M
if f 0 g − f g 0 ∈ Lp [a, b]
h
i
1+1/q
1+1/q
≤
(b−x)
+(x−a)
p > 1,
kf 0 g − f g 0 kp
m 

1+1/q

1

+ 1q = 1,

p






kf 0 g − f g 0 k1 (b − a)
(5.11)
for any x ∈ [a, b] .
Proof. Utilizing (5.2) we have
Z b
Z b
f (x)
f
(t)
dt
g
(t)
dt
−
g
(x)
a
a
Z
b
|g (t) f (x) − f (t) g (x)| dt
≤
a

R
Rb
x 1
0
0

kf
g
−
f
g
k
|g
(x)|
|g
(t)|
ds

t |g(s)|2 dt,
∞

a






R
1/q Rb
x 1
0
0
≤
kf g − f g kp |g (x)| a |g (t)| t |g(s)|2q ds
dt,






o
n

R 
1
 kf 0 g − f g 0 k |g (x)| b |g (t)| sups∈[t,x]([x,t])
dt
2
1
a
|g(s)|
for any x ∈ [a, b] , which is of interest in itself.
Since 0 < m ≤ |g (t)| ≤ M < ∞ for any t ∈ [a, b] , then
Z x
2 Z b
Z b
1
M
|g (x)|
|g (t)| |x − t| dt
2 ds dt ≤
m
a
a
t |g (s)|

!2 
2
a+b
t− 2
M
1 +
,
=
m
4
b−a
(5.12)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
29
1/q !
1
|g (x)|
dt
ds
2q
a
t |g (s)|
2 Z b
2
M
M
(b − x)1+1/q + (x − a)1+1/q
1/q
≤
|x − t| dt =
m
m
1 + 1/q
a
Z
b
Z
|g (t)| x
and
Z
b
|g (t)|
|g (x)|
sup
s∈[t,x]([x,t])
a
1
|g (s)|2
!
dt ≤
M
m
2 Z
b
dt =
a
M
m
2
(b − a)
for any x ∈ [a, b] and by (5.12) we get the desired result (5.11).
Remark 5.6. If we take g (t) = 1, t ∈ [a, b] in the first inequality (5.11) we recapture Ostrowski’s inequality.
Corollary 5.7. With the assumptions in Theorem 5.5 we have the midpoint
inequalities
Z b
Z b
a
+
b
a
+
b
f
g
(t)
dt
−
g
f
(t)
dt
2
2
a
a
≤








2 

M
m









1
4
(b − a)2 kf 0 g − f g 0 k∞
1
21/q (1+1/q)
1+1/q
(b − a)
if f 0 g − f g 0 ∈ L∞ [a, b] ,
0
0
kf g − f g kp
if f 0 g − f g 0 ∈ Lp [a, b]
p > 1,
1
+ 1q = 1.
p
(5.13)
The following result also holds:
Theorem 5.8 (Dragomir, 2013 [8]). Let f, g : [a, b] → C be absolutely continuous
functions on the interval [a, b] , g (x) 6= 0 for x ∈ [a, b] and g −2 ∈ L∞ [a, b] . Then
Z
Z b
f (x) b
g
(t)
dt
−
f
(t)
dt
g (x)
a
a

Rb
kf 0 g − f g 0 k∞ a |g (t)| |x − t| dt, if f 0 g − f g 0 ∈ L∞ [a, b] ,








if f 0 g − f g 0 ∈ Lp [a, b]

−2 R
b
p > 1,
≤ g ∞ ×
kf 0 g − f g 0 kp a |g (t)| |x − t|1/q dt

1
1

+
= 1,

p
q





Rb

kf 0 g − f g 0 k1 a |g (t)| dt
(5.14)
30
S.S. DRAGOMIR
for any x ∈ [a, b] .
Proof. Utilizing (5.2) we have
Z b
Z
f (x) b
g
(t)
dt
−
f
(t)
dt
g (x)
a
a

R
Rb
x 1
0
0

kf g − f g k∞ a |g (t)| t |g(s)|2 ds dt,







R

1/q Rb
x 1
0
0
kf g − f g kp a |g (t)| t |g(s)|2q ds
dt,
≤






n
o

R 
1
 kf 0 g − f g 0 k b |g (t)| sups∈[t,x]([x,t])
dt
1 a
|g(s)|2
(5.15)
for any x ∈ [a, b].
Since
Z x
1/q
Z x
−2 1
1
≤ g −2 |x − t|
|x − t| , 2 ds ≤ g
2q ds
∞
∞
t |g (s)|
t |g (s)|
and
sup
s∈[t,x]([x,t])
1
|g (s)|2
≤ g −2 ∞
for any x, t ∈ [a, b] , then on making use of (5.15) we get the desired result
(5.14).
We have the midpoint inequalities:
Corollary 5.9. With the assumptions of Theorem 5.8 we have
Z b
f a+b Z b
2 g (t) dt −
f (t) dt
a+b
g 2
a
a
≤ g −2 ∞ ×

Rb

kf 0 g − f g 0 k∞ a |g (t)| a+b
− t dt,

2











1/q
Rb


kf 0 g − f g 0 kp a |g (t)| a+b
− t dt


2

if f 0 g − f g 0 ∈ L∞ [a, b] ,
if f 0 g − f g 0 ∈ Lp [a, b]
p > 1,
1
1
+
= 1.
p
q
(5.16)
We have the following exponential version of Ostrowski’s inequality as well:
Theorem 5.10 (Dragomir, 2013 [8]). Let f : [a, b] → C be an absolutely continuous function on the interval [a, b] and α ∈ R, α 6= 0. Then for any x ∈ [a, b] we
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
31
have
Z b
exp (iα (b − x)) − exp (−iα (x − a))
f (t) dt
f (x) −
iα
a

a+b 2 
t−
2 1
0


kf − iαf k∞ (b − a) 4 + b−a2
, if f 0 − iαf ∈ L∞ [a, b] ,









if f 0 − iαf

1+1/q
∈ Lp [a, b]
≤
+(x−a)1+1/q
kf 0 − iαf kp (b−x) 1+1/q
,


p > 1,



1
1

+
= 1,

p
q





 kf 0 − iαf k .
1
(5.17)
Proof. If we write the inequality (5.12) for g (t) = exp (iαt) , t ∈ [a, b] , then we
get
Z b
Z b
f (x)
f
(t)
dt
exp
(iαt)
dt
−
exp
(iαx)
a
a

Rb

kf 0 − iαf k∞ a |x − t| dt, if f 0 − iαf ∈ L∞ [a, b]








if f 0 − iαf


Rb
∈ Lp [a, b]
≤
kf 0 − iαf kp |g (x)| a |x − t|1/q dt,
p > 1,



1

+ 1q = 1

p






kf 0 − iαf k1 ,
which, after simple calculation, is equivalent with (5.17).
The details are omitted.
Corollary 5.11. With the assumptions of Theorem 5.10 we have the midpoint
inequalities
exp iα b−a − exp −iα b−a a + b Z b
2
2
f
−
f (t) dt
iα
2
a
≤















1
4
kf 0 − iαf k∞ (b − a)2 , if f 0 − iαf ∈ L∞ [a, b] ,
(5.18)
1
21/q (1+1/q)
(b − a)1+1/q kf 0 − iαf kp ,
if f 0 − iαf ∈ Lp [a, b]
p > 1, p1 + 1q = 1,
32
S.S. DRAGOMIR
or, equivalently
2 sin α
α
≤















b−a
2
1
4
Z b
a+b
f (t) dt
f
−
2
a
kf 0 − iαf k∞ (b − a)2 , if f 0 − iαf ∈ L∞ [a, b] ,
(5.19)
1
21/q (1+1/q)
(b − a)1+1/q kf 0 − iαf kp ,
if f 0 − iαf ∈ Lp [a, b]
p > 1, p1 + 1q = 1.
5.3. An Application for CBS-Inequality. The following inequality is well
known in the literature as the Cauchy-Bunyakovsky-Schwarz inequality, or the
CBS-inequality, for short:
Z b
2 Z b
Z b
2
|g (t)|2 dt,
|f (t)| dt
f (t) g (t) dt ≤
a
a
(5.20)
a
provided that f, g ∈ L2 [a, b] .
We have the following result concerning some reverses of the CBS-inequality:
Theorem 5.12 (Dragomir, 2013 [8]). Let f, g : [a, b] → C be absolutely continuous functions on the interval [a, b] with g (t) 6= 0 for all t ∈ [a, b] . Then
Z
0≤
b
Z b
2
|f (t)| dt − f (t) g (t) dt
|g (t)| dt
a
a
R
2 R
2
f 0 g − f g 0 ∈ L∞ [a, b] ,
b
b
1
,
if
kf 0 g − f g 0 k2∞ a |g (t)|2 dt
dt
1
2
a |g(t)|
∈ L [a, b]
2
Z
b
2
a






|g|2







f 0 g − f g 0 ∈ Lp [a, b] ,


1
2 R
2/q
1
∈ L [a, b]
b
1
|g|2q
≤ × kf 0 g − f g 0 k2 R b |g (t)|2 dt
, if
2q dt
p
a
a
2 
|g(t)|
p > 1,



1
1

+
= 1,


p
q





n
R
2
o


1
1
 kf 0 g − f g 0 k2 b |g (t)|2 dt ess supt∈[a,b]
, if |g|
∈ L∞ [a, b] .
1
a
|g(t)|4
(5.21)
A SURVEY ON OSTROWSKI TYPE INEQUALITIES
33
Proof. Utilising the inequality (5.2) we have
g
(t)f
(x)
−
f
(t)
g
(x)

R

x 1
0
0

kf
ds
g
−
f
g
k
|g
(t)
g
(x)|

∞

t |g(s)|2










R
1/q
x 1
0
0
≤
kf
g
−
f
g
k
|g
(t)
g
(x)|
ds
t |g(s)|2q p










n
o



1
 kf 0 g − f g 0 k1 |g (t) g (x)| sups∈[t,x]([x,t])
.
|g(s)|2
if f 0 g − f g 0
∈ L∞ [a, b] ,
if f 0 g − f g 0
∈ Lp [a, b]
p > 1,
1
1
+
= 1,
p
q
(5.22)
for any t, x ∈ [a, b] .
Taking the square in (5.22) and integrating over (t, x) ∈ [a, b]2 we have
Z bZ b
2
g (t)f (x) − f (t) g (x) dtdx
a
a
2

RbRb
2 R x
1
0 2
0

kf
|g
(t)
g
(x)|
ds
g
−
f
g
k
dtdx,

∞ a a
t |g(s)|2






2/q
R R
2 R x
1
0 2 b b
0
≤
kf g − f g kp a a |g (t) g (x)| t |g(s)|2q ds dtdx,





n
o


 kf 0 g − f g 0 k2 R b R b |g (t) g (x)|2 sup
1
dtdx.
1
a
s∈[t,x]([x,t])
a
(5.23)
|g(s)|4
Observe that
Z bZ b
2
g (t)f (x) − f (t) g (x) dtdx
a
a
Z bZ b
h
i
2
2
2
2
=
|g (t)| |f (x)| − 2Re g (t)f (x) f (t) g (x) + |g (x)| |f (t)| dtdx
a
a
"Z
Z b
2 #
Z b
b
2
2
=2
|g (t)| dt
|f (t)| dt − f (t) g (t) dt ,
a
Z bZ b"
a
a
a
Z
2
|g (t) g (x)| t
a
x
2 #
Z b
2 Z b
2
1
1
2
ds dtdx ≤
|g (t)| dt
,
2 dt
|g (s)|2 a
a |g (t)|
34
S.S. DRAGOMIR
Z bZ b"
2/q #
1
dtdx
ds
2q
t |g (s)|
a
a
Z b
2 Z b
2/q
1
2
≤
|g (t)| dt
2q dt
a
a |g (t)|
Z
|g (t) g (x)| x
2
and
Z bZ b"
#
1
dtdx
|g (t) g (x)|
sup
|g (s)|4
s∈[t,x]([x,t])
a
a
Z b
2
1
2
,
≤
|g (t)| dt ess sup
|g (t)|4
t∈[a,b]
a
2
then by (5.23) we get the desired result (5.21).
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3, 281–287.
3. S.S. Dragomir, An inequality of Ostrowski type via Pompeiu’s mean value theorem, J. Inequal. Pure Appl. Math. 6 (2005), no. 3, Article 83, 9 pp.
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Dragomir,
Ostrowski
via
a
two
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http://rgmia.org/papers/v16/v16a90.pdf].
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Integralmittelwert, Comment. Math. Hel, 10 (1938), 226–227.
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(2006), no. 4, Article 151 .
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A SURVEY ON OSTROWSKI TYPE INEQUALITIES
35
Mathematics, College of Engineering & Science, Victoria University, P.O.
Box 14428, Melbourne City, MC 8001, Australia.
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