EXPONENTIAL POMPEIU’S TYPE INEQUALITIES
WITH APPLICATIONS TO OSTROWSKI’S INEQUALITY
S. S. DRAGOMIR
Abstract. In this paper, some exponential Pompeiu’s type inequalities for complex-valued 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]).
Lemma 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)
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x1 f (x2 ) − x2 f (x1 )
= f (ξ) − ξf 0 (ξ) .
x1 − x2
The following inequality is useful to derive some Ostrowski type inequalities.
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Received December 18, 2013.
2010 Mathematics Subject Classification. Primary 25D10, 25D10.
Key words and phrases. Ostrowski inequality; Pompeiu’s mean inequality; Integral inequalities; Exponetial function;
Integral mean.
Corollary 1 (Pompeiu’s Inequality). With the assumptions of Lemma 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].
In 1938, A. Ostrowski [4] proved the following result by the estimating the integral mean.
Theorem 1 (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
1
 M (b − a) .
(1.3)
f (t)dt ≤  +
f (x) −
b−a a
4
b−a
The constant
1
4
is best possible in the sense that it cannot be replaced by a smaller quantity.
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:
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Theorem 2 (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
·
−
f (t)dt
2
x
b−a a

!2 
(1.4)
x − a+b
b − a 1
2
 kf − `f 0 k ,
≤
+
∞
|x|
4
b−a
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], using a mean value theorem, E. C. Popa obtained a generalization of (1.4) as follows:
Theorem 3 (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
(1.5)
Z
a+b
α−x b
− α f (x) +
f (t)dt
2
b−a a

!2 
x − a+b
1
2
 (b − a) kf − `α f 0 k ,
≤ +
∞
4
b−a
where `α (t) = t − α, t ∈ [a, b].
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In [5], J. Pečarić and S. Ungar proved a general estimate with the p-norm, 1 ≤ p ≤ ∞ which
for p = ∞, gives Dragomir’s result.
Theorem 4 (Pečarić & 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
Close
(1.6)
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Z b
a + b f (x)
1
·
−
f (t)dt ≤ P U (x, p) kf − `f 0 kp
2
x
b−a a
for x ∈ [a, b], where
P U (x, p) := (b − a)
1
p −1
"
1/q
a2−q − x2−q
x2−q − a1+q x1−2q
+
(1 − 2q) (2 − q)
(1 − 2q) (1 + q)
2−q
1/q #
x2−q − b1+q x1−2q
b
− x2−q
+
+
.
(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.
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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 [1] and [2].
In this paper, some exponential Pompeiu’s type inequalities for complex-valued absolutely continuous functions are provided. They are applied to obtain some new Ostrowski type inequalities.
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2. Exponential Inequalities
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We can provide some similar results for complex-valued functions with the exponential instead of
`.
Lemma 2. 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) 
kf 0 − αf k∞
if f 0 −αf ∈ L∞ [a, b],

 |Re(α)|


1
1

 × exp(t Re(α)) − exp(x Re(α)) 




(2.1)

1/q
q 1/q |Re(α)| kf 0 − αf kp
if f 0 −αf ∈ Lp [a, b]
≤
1/q




× exp(tq1Re(α)) − exp(xq1Re(α)) p > 1, p1 + 1q = 1,







 kf 0 − αf k
1
1 min{exp(t Re(α)),exp(x Re(α))} ,
or equivalently,
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(2.2)
|exp (αt) f (x) − f (t) exp (αx)|

|Re(α)| kf 0 − αf k∞




× |exp (x Re(α)) − exp (t Re(α))|







 q 1/q |Re(α)|1/q kf 0 − αf k
p
≤
1/q

×
|exp
(xq
Re(α))
−
exp
(tq
Re(α))|








 kf 0 − αf k1


max {exp (t Re(α)) , exp (x Re (α))} .
if f 0 −αf ∈ L∞ [a, b] ,
if f 0 −αf ∈ Lp [a, b]
p > 1, p1 + 1q = 1,
Proof. If f is absolutely continuous, then f / exp (α·) is absolutely continuous on the interval
[a, b] and
0
Z x
f (s)
f (x)
f (t)
ds =
−
exp
(αs)
exp
(αx)
exp
(αt)
t
for any t, x ∈ [a, b] with x 6= t.
Since
0
Z x 0
Z x
f (s) exp(αs) − αf (s) exp(αs)
f (s)
ds =
ds
exp(αs)
exp (2αs)
t
t
Z x 0
f (s) − αf (s)
=
ds,
exp(αs)
t
then we get the following identity
(2.3)
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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 (2.3) we have
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f (t)
f (x)
−
=
exp (αx) exp (αt)
(2.4)
Z
f (x)
f (t) x f 0 (s) − αf (s) ds
exp (αx) − exp (αt) = exp(αs)
Zt x 0
|f (s) − αf (s)| ≤
ds := I
|exp (αs)|
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
R
1
x |f 0 (s) − αf (s)|p ds1/p x
I≤
,
q ds
t
t |exp(αs)|


 R
n
o


1
 x |f 0 (s) − αf (s)| ds sup
s∈[t,x]([x,t])
|exp(αs)| ,
t
(2.5)
R

x
1
0

ds
kf
−
αf
k
,

∞
|exp(αs)|
t




1/q
R
x
1
≤
kf 0 − αf kp t |exp(αs)|
,
q ds



o
n


1
 kf 0 − αf k sup
s∈[t,x]([x,t]) |exp(αs)| .
1
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Now, since α = Re(α) + i Im (α) and s ∈ [a, b], then
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|exp(αs)| = exp (s Re(α) + i s Im(α)) = |exp (s Re(α)) exp (i s Im(α))|
= |exp (s Re(α))| |exp (i s Im(α))| = exp (s Re(α)) .
We have
Z
t
x
Z x
Z x
1
1
exp (−s Re(α)) ds
ds =
ds =
|exp(αs)|
t exp (s Re(α))
t
x
= − Re(α) exp (−s Re (α))|t
= − Re(α) exp (−x Re(α)) + Re(α) exp (−t Re(α))
= Re(α) [exp (−t Re (α)) − exp (−x Re(α))]
1
1
−
.
= Re(α)
exp (t Re(α)) exp (x Re(α))
By (2.4) and (2.5), we get
f (x)
f (t) 1
1
0
exp (αx) − exp (αt) ≤ kf − αf k∞ |Re(α)| exp (t Re(α)) − exp (x Re(α)) JJ J
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and the first part of (2.1) is proved.
We have
Z x
Z x
1
1
ds
=
ds
q
t exp (sq Re(α))
t |exp(αs)|
1
1
= q Re(α)
−
.
exp (tq Re(α)) exp (xq Re (α))
By (2.4) and (2.5), we get the second part of (2.1).
We have
1
1
sup
=
.
min {exp (t Re(α)) , exp (x Re(α))}
s∈[t,x]([x,t]) |exp(αs)|
By (2.4) and ((2.5), we get the last part of (2.1).
The inequality (2.2) follows from (2.1) by multiplying by |exp (αx) exp (αt)| and performing the
required calculation.
The following particular case is of interest.
Corollary 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) (2.6)

1
1
0

−
−
f
k
kf

∞ exp(t)

exp(x) 


1/q

1
1
q 1/q kf 0 − f kp exp(tq)
− exp(xq)
≤






1
 kf 0 − f k
,
if f 0 − f ∈ L∞ [a, b],
if f 0 − f ∈ Lp [a, b],
p > 1, p1 + 1q = 1,
1 min{exp(t),exp(x)}
or equivalently,
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|exp(t)f (x) − f (t) exp(x)|
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(2.7)

kf 0 − f k∞ |exp (x) − exp(t)|





1/q
≤
q 1/q kf 0 − f kp |exp (xq) − exp (tq)|





kf 0 − f k1 max {exp(t), exp(x)} .
if f 0 − f ∈ L∞ [a, b],
if f 0 − f ∈ Lp [a, b]
p > 1, p1 + 1q = 1,
Remark 1. If Re(α) = 0, then the inequality (2.5) becomes
R

x
1
0

sup
|f
(s)
−
i
Im(α)f
(s)|
ds
,

s∈[t,x]([x,t])
t |exp(i Im(α)s)|



 R
1/q
R
1
x |f 0 (s) − i Im(α)f (s)|p ds1/p x
I≤
,
q ds
|exp(i
Im(α)s)|
t
t



n
o


1
 R x |f 0 (s) − i Im(α)f (s)| ds sup
s∈[t,x]([x,t]) |exp(i Im(α)s)| ,
t
R x 

kf 0 − i Im(α)f k∞ t ds ,
kf 0 − i Im(α)f k∞ |x − t| ,








R x 1/q
1/q
≤
kf 0 − i Im(α)f kp |x − t| ,
kf 0 − i Im(α)f kp t ds , =








kf 0 − i Im(α)f k1 .
kf 0 − i Im(α)f k1 ,
Therefore, we have
(2.8)
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or equivalently,
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
 kf 0 − i Im(α)f k∞ |x − t| ,



f (t)
f (x)
1/q
0
exp (i Im(α)x) − exp (i Im (α) t) ≤  kf − i Im(α)f kp |x − t| ,



kf 0 − i Im(α)f k1 ,
(2.9)

kf 0 − i Im(α)f k∞ |x − t| ,




1/q
| exp(i Im(α)t)f (x) − f (t) exp (i Im(α)x) | ≤
kf 0 − i Im(α)f kp |x − t| ,




kf 0 − i Im(α)f k1
for any t, x ∈ [a, b].
In particular, we have

0

 kf − i f k∞ |x − t| ,


f (x)
f (t) 1/q
0
exp (i x) − exp (i t) ≤  kf − i f kp |x − t| ,



kf 0 − i f k1 ,
(2.10)
or equivalently,
(2.11)
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
kf 0 − i f k∞ |x − t| ,




1/q
|exp (i t) f (x) − f (t) exp (i x)| ≤
kf 0 − i f kp |x − t| ,




kf 0 − i f k1
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for any t, x ∈ [a, b].
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3. Inequalities of Ostrowski Type
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The following result holds.
Theorem 5. 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)
− exp (αx)
f (t)dt
f (x)
α
a

|Re(α)| kf 0 − αf k∞ B1 (a, b, x, α) if f 0 −αf ∈ L∞ [a, b],





 1/q
1/q
q |Re(α)| (b − a)1/p
if f 0 −αf ∈ Lp [a, b]
≤
1/q
0

× kf − αf kp |Bq (a, b, x, α)|
p > 1, p1 + 1q = 1,





kf 0 − αf k1 B∞ (a, b, x, α),
(3.1)
where
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a+b
Bq (a, b, x, α) := 2 exp (xq Re(α)) x −
2
1
exp (bq Re(α)) + exp (aq Re (α))
+
− exp (xq Re(α))
q Re(α)
2
for q ≥ 1 and
B∞ (a, b, x, α) := exp (x Re(α)) (x − a) +
exp (b Re(α)) − exp (x Re(α))
.
Re(α)
Proof. Utilising the first inequality in (2.2), we have
Z b
Z b
exp (αt) dt − exp (αx)
f (t)dt
f (x)
a
a
Z b
(3.2)
≤
|exp (αt) f (x) − f (t) exp (αx)| dt
a
≤ |Re(α)| kf 0 − αf k∞
for any x ∈ [a, b].
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Z
b
|exp (x Re(α)) − exp (t Re (α))| dt
a
Observe that since Re(α) > 0, then
Z b
|exp (x Re(α)) − exp (t Re(α))| dt
a
Z x
=
(exp (x Re(α)) − exp (t Re(α))) dt
a
Z
b
(exp (t Re(α)) − exp (x Re(α))) dt
x
exp (t Re(α)) exp (x Re(α)) (x − a) −
Re(α)
a
b
exp (t Re(α))
(α) − (b − x) exp (x Re(α))
+
Re
x
1
(exp (x Re (α)) − exp (a Re(α)))
exp (x Re(α)) (2x − a − b) −
Re(α)
1
+
(exp (b Re(α)) − exp (x Re(α)))
Re(α)
exp (x Re(α)) (2x − a − b)
1
+
(exp (b Re(α)) + exp (a Re(α)) − 2 exp (x Re(α)))
Re(α)
a+b
2 exp (x Re(α)) x −
2
1
exp (b Re(α)) + exp (a Re (α))
+
− exp (x Re(α))
Re(α)
2
+
x
=
=
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=
for any x ∈ [a, b].
Also
Z b
Z
f (x)
exp (αt) dt − exp (αx)
a
b
f (t)dt = f (x)
a
exp (αb) − exp (αa)
− exp (αx)
α
Z
for any x ∈ [a, b] and by (3.2), we get the first inequality in (3.1).
Using the second inequality in (2.2), we have
Z b
Z b
|f (x)
exp (αt) dt − exp (αx)
f (t)dt
a
a
Z b
(3.3)
≤
|exp (αt) f (x) − f (t) exp (αx)| dt
a
≤q
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1/q
1/q
|Re(α)|
Z
0
kf − αf kp
b
|exp (xq Re(α)) − exp (tq Re(α))|
1/q
dt
a
for any x ∈ [a, b].
By Hölder’s integral inequality, we also have
Z b
1/q
|exp (xq Re(α)) − exp (tq Re(α))| dt
a
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Z
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≤
!1/p "Z
b
dt
a
b
q
1/q
|exp (xq Re(α)) − exp (tq Re(α))|
dt
a
1/p
"Z
= (b − a)
#1/q
b
|exp (xq Re(α)) − exp (tq Re (α))| dt
a
,
#1/q
b
f (t)dt
a
for any x ∈ [a, b].
Observe that as above we have
Z b
a+b
|exp (xq Re(α)) − exp (tq Re(α))| dt = 2 exp (xq Re(α)) x −
2
a
exp (bq Re(α)) + exp (aq Re (α))
1
+
− exp (xq Re(α))
= Bq (a, b, x, α)
q Re(α)
2
for any x ∈ [a, b] and by (3.3), we get the second part of (3.1).
Using the third inequality in (2.2), we have
Z b
Z b
Z b
exp (αt) dt − exp (αx)
f (t)dt ≤
|exp (αt) f (x) − f (t) exp (αx)| dt
f (x)
a
a
a
(3.4)
Z
b
≤ kf 0 − αf k1
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for any x ∈ [a, b].
Observe that
Z b
max {exp (t Re(α)) , exp (x Re(α))} dt
a
Z
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x
Z
Z
=
max {exp (t Re(α)) , exp (x Re(α))} dt
x
x
Z
b
exp (x Re(α)) dt +
a
Close
= exp (x Re(α)) (x − a) +
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b
max {exp (t Re(α)) , exp (x Re(α))} dt +
=
a
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max {exp (t Re(α)) , exp (x Re(α))} dt
a
exp (t Re(α)) dt
x
exp (b Re(α)) − exp (x Re(α))
Re(α)
and by (3.4), we get the third part of (3.1).
Remark 2. If Re(α) < 0, then a similar result may be stated. However, the details are left to
the interested reader.
Corollary 3. 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
(3.5)
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
kf 0 − f k∞ B1 (a, b, x)






 q 1/q (b − a)1/p kf 0 − f k
p
≤
1/q

×
|B
(a,
b,
x)|
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, p1 + 1q = 1,
where
Bq (a, b, x)
a+b
1 exp (bq) + exp (aq)
:= 2 x −
− exp (xq)
exp (xq) +
2
q
2
for q ≥ 1 and
B∞ (a, b, x) := (x − a) exp(x) + exp (b) − exp(x).
Remark 3. The midpoint case is as follows:
Z b
a
+
b
a
+
b
f
[exp
(b)
−
exp
(a)]
−
exp
f
(t)dt
2
2
a
(3.6)
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
kf 0 − f k∞ B1 (a, b)






 q 1/q (b − a)1/p kf 0 − f k
p
≤
1/q

×
|B
(a,
b)|
q






kf 0 − f k1 B∞ (a, b),
if f 0 −f ∈ L∞ [a, b],
if f 0 −f ∈ Lp [a, b],
p > 1, p1 + 1q = 1,
where
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Bq (a, b, x) :=
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2
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 6. 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
exp (i Im (α) b) − exp (i Im(α)a)
| f (x)
− exp (i Im(α)x)
i Im(α)
(3.7)
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
kf0 − i Im(α)f k∞ 

a+b 2


x− 2


(b − a)2
× 14 +

b−a




q
kf 0 − i Im(α)f kp
≤
q+1

q+1
q+1

q+1

q
q
b−x
x−a


×
+
(b − a) q

b−a
b−a





kf 0 − i Im(α)f k1 (b − a).
Z
a
b
f (t)dt
if f 0 − i Im(α)f ∈ L∞ [a, b],
if f 0 − i Im(α)f ∈ Lp [a, b],
p > 1,
1 1
p+q=
1,
Proof. Utilizing the inequality (2.9), we have
Z b
f (x) exp (i Im (α) b) − exp (i Im(α)a) − exp (i Im(α)x)
f (t)dt
i Im(α)
a
b
Z
≤
|exp (i Im(α)t) f (x) − f (t) exp (i Im (α) x)| dt
a
(3.8)

Rb
0

|x − t| dt,
kf
−
i
Im(α)f
k

∞
a



Rb
1/q
≤
kf 0 − i Im(α)f kp a |x − t| dt,




Rb

kf 0 − i Im(α)f k1 a dt.
Since
b
Z
a
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b
|x − t|
a
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x − a+b
2
b−a
1
|x − t| dt =  +
4
!2 
 (b − a)2
and
Z
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
1/q
q
dt =
q+1
"
b−x
b−a
q+1
q
then from (3.8), we get the desired result (3.7).
+
x−a
b−a
#
q+1
q
(b − a)
q+1
q
,
Corollary 4. 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
exp (ib) − exp (i a)
− exp (i x)
f (t)dt
f (x)
i
a
(3.9)
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I II
if f 0 − i f ∈ L∞ [a, b],
if f 0 − i f ∈ Lp [a, b],
p > 1,
1 1
p+q=
1,
Remark 4. The midpoint case is as follows
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(3.10)
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
kf0 − i f k∞


a+b 2 

x− 2
1


(b − a)2
× 4+

b−a




q
kf 0 − i f kp
≤
q+1

q+1
q+1

q+1

q
q
b−x
x−a


× b−a
+ b−a
(b − a) q






kf 0 − if k1 (b − a).
Z b
a + b exp (i b) − exp (i a)
a+b
− exp i
f (t) dt
f
2
i
2
a
 1 0
if f 0 − i f ∈ L∞ [a, b] ,
 4 kf − i f k∞ (b − a)2 ,
≤
q+1
q

kf 0 − i f kp (b − a) q , if f 0 − i f ∈ Lp [a, b].
(q+1)21/q
Close
Similar inequalities may be stated if one uses (2.1) and integrates over t on [a, b]. The details
are left to the interested reader.
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1. Acu A. M. and Sofonea F. D., On an inequality of Ostrowski type, J. Sci. Arts 3(16)(2011), 281–287.
2. Acu A. M., Baboş A. and Sofonea F. D., The mean value theorems and inequalities of Ostrowski type, Sci. Stud.
Res. Ser. Math. Inform. 21(1) (2011), 5–16.
3. Dragomir S. S., An inequality of Ostrowski type via Pompeiu’s mean value theorem, J. Inequal. Pure Appl. Math.
6(3) (2005), Article 83, 9 pp.
4. Ostrowski A., Über die Absolutabweichung einer differentienbaren Funktionen von ihren Integralmittelwert, Comment. Math. Hel. 10 (1938), 226–227.
5. Pečarić J. and Ungar Š., On an inequality of Ostrowski type, J. Ineq. Pure & Appl. Math. 7(4) (2006), Article
151.
6. Pompeiu D., Sur une proposition analogue au théorème des accroissements finis, Mathematica (Cluj), 22 (1946),
143–146.
7. Popa E. C., An inequality of Ostrowski type via a mean value theorem, General Mathematics 15(1) (2007),
93–100.
8. Sahoo P. K. and Riedel T., Mean Value Theorems and Functional Equations, World Scientific, Singapore, New
Jersey, London, Hong Kong, 2000.
S. S. Dragomir, Mathematics, College of Engineering & Science, Victoria University, PO Box 14428 Melbourne City,
MC 8001, Australia and
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School of Computational & Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg
2050, South Africa, e-mail: sever.dragomir@vu.edu.au