NEW ZEALAND JOURNAL OF MATHEMATICS
Volume 34 (2005), 31–42
SOME OSTROWSKI TYPE INEQUALITES VIA CAUCHY’S
MEAN VALUE THEOREM
S.S. Dragomir
(Received February 2003)
Abstract. Some Ostrowski type inequalities via Cauchy’s mean value theorem
and applications for certain particular instances of functions are given.
1. Introduction
The following result is known in the literature as Ostrowski’s inequality [1].
Theorem 1.1. Let f : [a, b] → R be a differentiable mapping on (a, b) with the
property that |f 0 (t)| ≤ M for all t ∈ (a, b). Then

!2 
Z b
x − a+b
1
1
2
 (b − a) M,
f (t) dt ≤  +
(1)
f (x) −
b−a a
4
b−a
for all x ∈ [a, b]. The constant
replaced by a smaller constant.
1
4
is best possible in the sense that it cannot be
In [2], the author has proved the following Ostrowski type inequality.
Theorem 1.2. Let f : [a, b] → R be continuous on [a, b] with a > 0 and differentiable on (a, b). Let p ∈ R\ {0} and assume that
Kp (f 0 ) := sup u1−p |f 0 (u)| < ∞.
u∈(a,b)
Then we have the inequality
Z b
1
Kp (f 0 )
f (t)dt ≤
f (x) −
|p| (b − a)
b−a a

2xp (x − A) + (b − x) Lpp (b, x) − (x − a) Lpp (x, a)
if p ∈ (0, ∞) ;





(x − a) Lpp (x, a) − (b − x) Lpp (b, x) − 2xp (x − A)
if p ∈ (−∞, −1) ∪ (−1, 0)
×



2


(x − a) L−1 (x, a) − (b − x) L−1 (b, x) − (x − A) if p = −1,
x
for any x ∈ (a, b), where for a 6= b,
a+b
A = A (a, b) :=
,
2
is the arithmetic mean,
1991 Mathematics Subject Classification Primary 26D15; Secondary 26D10.
Key words and phrases: Ostrowski’s inequality, Cauchy’s mean value theorem.
(2)
32
S.S. DRAGOMIR
bp+1 − ap+1
Lp = Lp (a, b) =
(p + 1) (b − a)
p1
,
is the p–logarithmic mean
p ∈ R\ {−1, 0} ,
and
L = L (a, b) :=
b−a
ln b − ln a
is the logarithmic mean.
Another result of this type obtained in the same paper is:
Theorem 1.3. Let f : [a, b] → R be continuous on [a, b] (with a > 0) and differentiable on (a, b). If
P (f 0 ) :=
sup |uf 0 (x)| < ∞,
u∈(a,b)
then we have the inequality
#
#
" "
Z b
b−x
P (f 0 )
1
[I (x, b)]
ln
f (t) dt ≤
f (x) −
x−a + 2 (x − A) ln x
b−a a
b−a
[I(a, x)]
(3)
for any x ∈ (a, b), where for a 6= b
1
I = I (a, b) :=
e
bb
aa
1
b−a
,
is the identric mean.
If some local information around the point x ∈ (a, b) is available, then we may
state the following result as well [2].
Theorem 1.4. Let f : [a, b] → R be continuous on [a, b] and differentiable on
(a, b). Let p ∈ (0, ∞) and assume, for a given x ∈ (a, b), we have that
Mp (x) :=
sup
n
o
1−p
|x − u|
|f 0 (u)| < ∞.
u∈(a,b)
Then we have the inequality
Z b
1
f (t) dt
f (x) −
b−a a
≤
h
i
1
p+1
(x − a)p+1 + (b − x)
Mp (x) . (4)
p (p + 1) (b − a)
For recent results in connection to Ostrowski’s inequality see the papers [3], [4]
and the monograph [5].
The main aim of this paper is to point out some generalizations of the results
incorporated in Theorems 1.2–1.4 by the use of Cauchy mean value theorem. Applications for other particular instances of functions are given as well.
33
OSTROWSKI TYPE INEQUALITES
2. The Results
We may state the following theorem.
Theorem 2.1. Let f, g : [a, b] → R be continuous on [a, b] and differentiable on
(a, b). If g 0 (t) 6= 0 for each t ∈ (a, b) and
0 0
f := sup f (t) < ∞,
(5)
g 0 (t) g0 t∈(a,b)
∞
then for any x ∈ [a, b] one has the inequality
Z b
1
f (t) dt
f (x) −
b−a a
≤ 2
x − a+b
2
b−a
Rb
!
g (x) +
x
Rx
g (t) dt − a g (t) dt f0 . (6)
·
g 0 ∞
b−a
Proof. Let x, t ∈ [a, b] with t 6= x. Applying Cauchy’s mean value theorem, there
exists a η between t and x such that
(f (x) − f (t)) =
f 0 (η)
(g(x) − g(t))
g 0 (η)
from where we get
0 0
f (η) f |f (x) − f (t)| = 0
|g (x) − g (t)| ≤ g 0 |g(x) − g(t)| ,
g (η)
∞
(7)
for any t, x ∈ [a, b].
Using the properties of the integral, we deduce by (7), that
Z b
Z b
1
1
f (t)dt ≤
|f (x) − f (t)| dt
f (x) −
b−a a
b−a a
≤
0
Z b
f 1
|g(x) − g(t)| dt.
g0 b − a
a
∞
(8)
Since g 0 (t) 6= 0 on (a, b), it follows that either g 0 (t) > 0 or g 0 (t) < 0 for any
t ∈ (a, b).
If g 0 (t) > 0 for all t ∈ (a, b), then g is strictly monotonic increasing on (a, b) and
Z b
Z x
Z b
|g(x) − g(t)| dt =
(g(x) − g(t)) dt +
(g(t) − g(x)) dt
a
a
=
2 x−
x
a+b
2
Z
b
Z
g (t) dt −
g(x) +
x
x
g (t) dt.
a
If g 0 (t) < 0 for all t ∈ (a, b), then
" #
Z b
Z b
Z x
a+b
|g (x) − g (t)| dt = − 2 x −
g (x) +
g (t) dt −
g (t) dt
2
a
x
a
and the inequality (6) is proved.
34
S.S. DRAGOMIR
The following midpoint inequality is a natural consequence of the above result.
Corollary 2.2. With the above assumptions for f and g, one has the inequality
Z
Z a+b
Z b
f0 2
a+b
1 b
1
.
g(t)dt −
f (t) dt ≤
−
g (t) dt f
g 0 ∞
2
b−a a
b − a a+b
a
2
(9)
Remark 2.3.
1. If in the above theorem, we choose g (t) = t, then from (6) we recapture
Ostrowski’s inequality (1).
2. If in Theorem 2.1 we choose g (t) = tp , p ∈ R\ {0}, or g (t) = ln t with
t ∈ (a, b) ⊂ (0, ∞), then we obtain Theorem 1.2 and Theorem 1.3 respectively.
One may obtain many inequalities from Theorem 2.1 on choosing different instances of functions g.
Proposition 2.4. Let f : [a, b] ⊂ R → R be continuous on [a, b] and differentiable
on (a, b). If there exists a constant Γ < ∞ such that
|f 0 (t)| ≤ Γe−t
t ∈ (a, b) ,
for any
(10)
then one has the inequality:
Z b
a+b
1
−
f (t) dt
f
2
b−a a
x − A(a, b) x (b − x)E(x, b) − (x − a)E(a, x)
≤ Γ 2
e +
b−a
b−a
(11)
for any x ∈ (a, b), where A = A (a, b) = a+b
2 and E is the exponential men, i.e.,
 x
y

 e −e
if x 6= y
x−y
E(x, y) :=
,
x, y ∈ R.

 ey
if x = y
In particular, we have
Z b
1
1
f (t) dt ≤ [E(A, b) − E(a, A)] Γ.
f (A) −
b−a a
2
(12)
The proof is obvious by Theorem 2.1 on choosing g (t) = et and we omit the
details.
Another example is considered in the following proposition.
Proposition 2.5. Let f : [a, b] ⊂ 0, π2 → R be continuous on [a, b] and differentiable on (a, b).
(i) If there exists a constant Γ1 < ∞ such that
|f 0 (t)| ≤ Γ1 cos t,
t ∈ (a, b) ,
(13)
OSTROWSKI TYPE INEQUALITES
35
then one has the inequality
Z b
1
f (t) dt
f (x) −
b−a a
x − A (a, b)
(x − a) C (a, x) − (b − x) C (x, b)
≤ Γ1 2
sin x +
b−a
b−a
(14)
for any x ∈ (a, b), where C is the cos–mean value, i.e.,
 cos x − cos y

if x =
6 y,
x−y
C (x, y) :=

− sin y
if x = y.
In particular we have
Z b
1
1
f (t) dt ≤ [C(a, A) − C(A, b)] Γ1 .
f (A) −
b−a a
2
(15)
(ii) If there exists a constant Γ2 < ∞ such that
|f 0 (t)| ≤ Γ1 sin t,
t ∈ (a, b) ,
(16)
then one has the inequality
Z b
1
f (t) dt
f (x) −
b−a a
x − A (a, b)
(b − x) S (x, b) − (x − a) S (a, x)
≤ Γ2 2
cos x +
, (17)
b−a
b−a
for any x ∈ (a, b), where S is the sin–mean value, i.e.,

 sin x − sin y if x =
6 y,
x−y
S (x, y) :=

cos y
if x = y.
In particular, we have
Z b
1
1
f (t) dt ≤ [S(A, b) − S(a, A)] Γ2 .
f (A) −
b−a a
2
(18)
The following result also holds.
Theorem 2.6. Let f, g : [a, b] → R be continuous on [a, b] and differentiable on
(a, b) \ {x}, x ∈ (a, b). If g 0 (t) 6= 0 for t ∈ (a, x)∪(x, b), then we have the inequality
0
Z b
Z x
f 1 1
f (t) dt ≤
g (x) (x − a) −
g (t) dt · f (x) −
g0 b−a a
b−a
a
(a,x),∞
Z b
f0 1 +
g (t) dt · (19)
.
g (x) (b − x) −
g 0 (x,b),∞
b−a
x
36
S.S. DRAGOMIR
Proof. We obviously have:
Z b
1
f (t) dt =
f (x) −
b−a a
1 Z b
(f (x) − f (t)) dt
b − a a
≤
=
b
1
b−a
Z
1
b−a
"Z
|f (x) − f (t)| dt
a
x
Z
#
b
|f (x) − f (t)| dt .
|f (x) − f (t)| dt +
x
a
(20)
Applying Cauchy’s mean value theorem on the interval (a, x), we deduce (see the
proof of Theorem 2.1) that
0
f |f (x) − f (t)| ≤ |g (x) − g (t)|
(21)
g0 (a,x),∞
for any t ∈ (a, x), and similarly
0
f |f (x) − f (t)| ≤ |g (x) − g (t)|
g0 (x,b),∞
for any t ∈ (x, b).
Consequently
Z x
a
and
Z
b
x
(22)
0
Z x
f |f (x) − f (t)| dt ≤ 0 |g (x) − g (t)| dt
g (a,x),∞ a
0
Z b
f |g (x) − g (t)| dt.
|f (x) − f (t)| dt ≤ g0 (x,b),∞ x
Since g 0 has a constant sign in either (a, x) or (x, b), it follows that g is strictly
increasing or strictly decreasing in (a, x) and (x, b).
Thus
Z x



g
(x)
(x
−
a)
−
g (t) dt if g is increasing on [a, x]
Z x

a
|g (x) − g (t)| dt =
Z x

a


g (t) dt − g (x) (x − a) if g is decreasing
a
Z
= g (x) (x − a) −
a
and, in a similar way
Z b
|g (x) − g (t)| dt =
x
x
g (t) dt
Z b
g (t) dt .
g (x) (b − x) −
x
Consequently, by the use of (20), we deduce the desired inequality (19).
The following particular case may be of interest.
37
OSTROWSKI TYPE INEQUALITES
Corollary
2.7.
Let f, g : [a, b] → R be
continuous
on [a, b] and differentiable on
a+b
a+b
0
(a, b) \ a+b
.
If
g
(t)
=
6
0
on
a,
∪
,
b
, then we have the inequality
2
2
2
Z
b
a+b
1
f (t) dt
−
f
2
b−a a
1
≤
2
( Z a+b
f0 2
a+b
2
g (t) dt · −
g
g 0 (a, a+b ),∞
2
b−a a
2
)
Z b
f0 a+b
2
+ g
−
g (t) dt · 0 . (23)
g ( a+b ,b),∞
2
b − a a+b
2
2
The following result also holds.
Proposition 2.8. Let f : [a, b] → R be continuous on [a, b] and differentiable on
(a, b) \ {x}, x ∈ (a, b). Assume that, for p > 0, we have

 M1,p (x) (x − t)1−p for any t ∈ (a, x) ,
0
|f (t)| ≤
(24)
 M (x) (t − x)1−p for any t ∈ (x, b) .
2,p
Then we have the inequality
Z b
1
f (t) dt
f (x) −
b−a a
≤
h
i
1
p+1
p+1
(b − a) M1,p (x) (x − a)
+ M2,p (x) (b − x)
.
p (p + 1)
p
The proof follows by Theorem 2.6 applied for g (x) = |x − t| , p > 0. We omit
the details.
Remark 2.9. If f is as in Proposition 2.8 and

1−p

a+b
a+b
a+b


M
−
t
for
any
t
∈
a,
,

 1
2
2
2
0
|f (t)| ≤
1−p


a+b
a+b
a+b


t−
for any t ∈
,b ,
 M2
2
2
2
(25)
then, by Proposition 2.8 we get
Z b
1
a+b
−
f (t) dt
f
2
b−a a
p+1 a+b
a+b
(b − a)
≤ p+1
M1
+ M2
. (26)
2
p (p + 1)
2
2
Remark 2.10. If f is as in Proposition 2.8 and
1−p
|f 0 (t)| ≤ Mp (x) |x − t|
t ∈ (a, b) ,
38
S.S. DRAGOMIR
then, we get
Z b
h
i
1
1
p+1
p+1
f (t) dt ≤
(x − a)
+ (b − x)
Mp (x) ,
f (x) −
b−a a
p (p + 1) (b − a)
(27)
which is the result obtained in (4).
3. Some Inequalities of Midpoint Type
1. Let 0 < a < b. Consider the function g : [a, b] → R, g (t) = tp , t ∈ R\ {0, −1}.
= Ap (a, b),
Then g 0 (t) = ptp−1 , g a+b
2
Z a+b
2
2
g (t) dt = Lpp (a, A (a, b)) ,
b−a a
2
b−a
Z
b
g (t) dt
a+b
2
= Lpp (A (a, b) , b) ,
and by Corollary 2.7, we may state the following proposition.
Proposition 3.1.
Let f : [a, b] ⊂ (0, ∞) → R be continuous on [a, b] and differentiable on (a, b) \ a+b
. If
2

a+b p
a+b


M
t
,
t
∈
a,
,

 1
2
2
0
(28)
|f (t)| ≤

a+b
a+b p


t , t∈
,b ,
 M2
2
2
then we have the inequality
Z b
a+b
1
−
f (t) dt
f
2
b−a a
≤
1
2p
M1
a+b
2
p
A (a, b) − Lpp (a, A(a, b))
+ M2
a+b
2
p
Lp (A(a, b), b) − Ap (a, b) . (29)
The particular case p = 1 is of interest and so we may state the following corollary.
Corollary 3.2. Let f : [a, b] ⊂ (0, ∞) → R be continuous on [a, b] and differentiable on (a, b) \ a+b
. If
2

a+b
a+b


N
t,
t
∈
a,
,

 1
2
2
0
|f (t)| ≤
(30)


a+b
a+b

 N2
t, t ∈
,b ,
2
2
39
OSTROWSKI TYPE INEQUALITES
then we have the inequality:
Z b
1 a + b
a+b
1
a+b
f (t) dt ≤
−
N1
+ N2
(b − a) . (31)
f
8
2
b−a a
2
2
2. Let 0 < a < b. Consider the function g : [a, b] → R, g (t) =
g 0 (t) = − t12 , g a+b
= A−1 (a, b),
2
2
b−a
a+b
2
Z
g (t) dt =
L−1 (a, A(a, b)) ,
g (t) dt =
L−1 (A(a, b), b) ,
1
t
. Then
a
2
b−a
Z
b
a+b
2
and by Corollary 2.7 we may state the following Proposition.
Proposition 3.3.
Let f : [a, b] ⊂ (0, ∞) → R be continuous on [a, b] and differentiable on (a, b) \ a+b
. If
2

a + b −2
a+b


M
t
,
t
∈
a,
,

 1
2
2
0
(32)
|f (t)| ≤

a
+
b
a
+
b

−2

 M2
,b ,
t , t∈
2
2
then we have the inequality:
Z b
1
a+b
1
a+b
[A (a, b) − L (a, A(a, b))]
−
f (t) dt ≤
M1
·
f
2
2
b−a a
2
L (a, A(a, b)) A (a, b)
+ M2
a+b
2
·
[L (A(a, b), b) − A (a, b)]
.
L (A(a, b), b) A (a, b)
(33)
3. Let 0 < a < b. Consider the function g : [a, b] → R, g (t) = ln t. Then
g 0 (t) = 1t , g a+b
= ln A (a, b),
2
2
b−a
Z
2
b−a
a+b
2
g (t) dt
=
ln I (a, A(a, b)) ,
g (t) dt
=
ln I (A(a, b), b) ,
a
Z
b
a+b
2
and by Corollary 2.7 we may state the following proposition.
40
S.S. DRAGOMIR
Proposition 3.4.
Let f : [a, b] ⊂ (0, ∞) → R be continuous on [a, b] and differentiable on (a, b) \ a+b
. If
2

a+b
a + b −1


t , t ∈ a,
,

 M1
2
2
0
|f (t)| ≤
(34)

a + b −1
a+b


 M2
t , t∈
,b ,
2
2
then we have the inequality:
Z b
a+b
1
f (t) dt
−
f
2
b−a a
(
≤ ln G
A (a, b)
I (a, A(a, b))
M1 ( a+b
M a+b !)
2 )
I (A(a, b), b) 2 ( 2 )
,
. (35)
A (a, b)
4. The Case of Weighed Integrals
We may state the following theorem.
Theorem 4.1. Let f, g : [a, b] → R be continuous on [a, b] and differentiable on
Rb
(a, b) and w : [a, b] → [0, ∞) an integrable function such that a w (s) ds > 0. If
g 0 (t) 6= 0 for each t ∈ (a, b) and
0 0
f := sup f (t) < ∞,
(36)
g 0 (t) g0 t∈(a,b)
∞
then for any x ∈ (a, b) one has the inequality
Z b
1
f (t) w (t) dt
f (x) − R b
w (t) dt a
a
Rb
Rx
Rx
Rb
w (t) g (t) dt − a g (t) w (t) dt w (t) dt − x w (t) dt
f0 x
a
.
+
≤ g (x) ·
·
Rb
Rb
g 0 ∞
w
(t)
dt
w
(t)
dt
a
a
(37)
Proof. Let x, t ∈ [a, b] with t 6= x. Applying Cauchy’s mean value theorem, there
exists a η between t and x such that
f (x) − f (t) =
f 0 (η)
[g(x) − g(t)] ,
g 0 (η)
from where we get
0 0
f (η) f |f (x) − f (t)| = 0
|g (x) − g (t)| ≤ g 0 |g (x) − g (t)|
g (η)
∞
for any t, x ∈ [a, b].
(38)
41
OSTROWSKI TYPE INEQUALITES
Using the properties of the integral, we deduce by (38), that
Z b
1
w (s) f (s) ds
f (x) − R b
a
w (s) ds
a
≤ Rb
a
b
Z
1
w (t) |f (x) − f (t)| dt
w (s) ds
a
0
Z b
f 1
≤ 0 Rb
w (t) |g(x) − g(t)| dt.
g ∞
w (s) ds a
a
(39)
Since g 0 (t) 6= 0 on (a, b), it follows that either g 0 (t) > 0 or g 0 (t) < 0 for any
t ∈ (a, b).
If g 0 (t) > 0 for all t ∈ (a, b), then g is strictly monotonic increasing on (a, b) and
Z b
w (t) |g(x) − g(t)| dt
a
Z
x
Z
b
w (t) (g(x) − g(t)) dt +
=
a
w (t) (g(t) − g(x)) dt
x
x
Z
x
Z
Z
w (t) dt −
= g (x)
a
x
x
#
b
Z
w (t) dt −
= g (x)
w (t) dt
x
Z
w (t) g (t) dt −
x
b
x
b
Z
w (t) dt +
a
Z
w (t) g (t) dt − g (x)
a
"Z
b
w (t) g (t) dt +
w (t) g (t) dt.
x
a
If g 0 (t) < 0 for all t ∈ (a, b), then
Z b
w (t) |g (x) − g (t)| dt
a
"
"Z
= − g (x)
x
Z
b
w (t) dt −
a
#
Z
b
x
w (t) g (t) dt −
w (t) dt +
x
Z
x
w (t) g (t) dt ,
a
and the inequality (6) is proved.
Corollary 4.2. If x0 ∈ [a, b] is a point for which
Z x0
Z b
w (t) dt =
w (t) dt,
a
#
(40)
x0
and f, g, w are as in Theorem 4.1, then we have the inequality
Z b
1
w (t) f (t) dt
f (x0 ) − R b
w (t) dt a
a
R
R x0
b
x0 w (t) g (t) dt − a g (t) w (t) dt
≤
Rb
w (t) dt
a
0
f ·
g 0 . (41)
∞
42
S.S. DRAGOMIR
In a similar manner, we may deduce the following result as well.
Theorem 4.3. Let f, g : [a, b] → R be continuous on [a, b] and differentiable on
Rb
(a, b) \ {x}, x ∈ (a, b). If w : [a, b] → [0, ∞) is integrable and a w (s) ds > 0 and
g 0 (t) 6= 0 for t ∈ (a, x) ∪ (x, b), then we have the inequality
Z b
1
w (t) f (t) dt
f (x) − R b
w (t) dt a
a
Rx
Rx
w (t) dt
w (t) g (t) dt f0 a
a
≤ g (x) · R b
− Rb
·
0
w (t) dt
w (t) dt g (a,x),∞
a
a
Rb
Rb
w (t) dt
w (t) g (t) dt f0 x
x
+ g (x) · R b
− Rb
.
·
g 0 (x,b),∞
w
(t)
dt
w
(t)
dt
a
a
(42)
References
1. A. Ostrowski, Über die Absolutabweichung einer differentienbaren Funktionen
von ihren Integralmittelwert, Comment. Math. Hel. 10 (1938), 226–227.
2. S.S.
Dragomir,
Some
new
inequalities
of
Ostrowski
type,
RGMIA Res. Rep. Coll. 5 (2002), Supplement, Article 11, [ON
LINE:http://rgmia.vu.edu.au/v5(E).html] to apper in Gazette, Austral. Math. Soc.
3. G.A. Anastassiou, Multidimensional Ostrowski inequalities, revisited , Acta
Math. Hungar, 97 no. 4 (2002), 339–353.
4. G.A. Anastassiou, Univariate Ostrowski inequalities, revisited , Monatsh. Math.
135 no. 3 (2002), 175–189.
5. S.S. Dragomir and T.M. Rassias (Eds), Ostrowski Type Inequalities
and Applications in Numerical Integration, Kluwer Academic Publishers,
Dordrecht/Boston/London, 2002.
S.S. Dragomir
School of Computer Science and Mathematics
Victoria University of Technology
PO Box 14428, MCMC 8001
VIC, AUSTRALIA
[email protected]
http://rgmia.vu.edu.au/SSDragomirWeb.html