Volume 8 (2007), Issue 1, Article 10, 6 pp.
ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS
AND APPLICATIONS FOR PDFS
A.I. KECHRINIOTIS AND N.D. ASSIMAKIS
D EPARTMENT OF E LECTRONICS
T ECHNOLOGICAL E DUCATIONAL I NSTITUTE OF L AMIA
G REECE
kechrin@teilam.gr
assimakis@teilam.gr
Received 11 November, 2005; accepted 12 April, 2006
Communicated by N.S. Barnett
A BSTRACT. A new inequality is presented, which is used to obtain a complement of recently
obtained inequality concerning the difference of two integral means. Some applications for pdfs
are also given.
Key words and phrases: Ostrowski’s inequality, Probability density function, Difference of integral means.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In 1938, Ostrowski proved the following inequality [5].
Theorem 1.1. Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with
|f 0 (x)| ≤ M for all x ∈ (a, b) , then,
#
"
Z b
a+b 2
x
−
1
1
2
f (x) −
(b − a) M ,
(1.1)
f (t) dt ≤
+
b−a a
4
(b − a)2
for all x ∈ [a, b]. The constant
1
4
is the best possible.
In [3] N.S. Barnett, P. Cerone, S.S. Dragomir and A.M. Fink obtained the following inequality
for the difference of two integral means:
Theorem 1.2. Let f : [a, b] → R be an absolutely continuous mapping with the property that
f 0 ∈ L∞ [a, b] , then for a ≤ c < d ≤ b,
Z b
Z d
1
1
1
≤ (b + c − a − d) kf 0 k ,
(1.2)
f
(t)
dt
−
f
(t)
dt
∞
b − a
2
d−c c
a
the constant
335-05
1
2
being the best possible.
2
A.I. K ECHRINIOTIS AND N.D. A SSIMAKIS
For c = d = x this can be seen as a generalization of (1.1) .
In recent papers [1], [2], [4], [6] some generalizations of inequality (1.2) are given. Note that
estimations of the difference of two integral means are obtained also in the case where a ≤ c <
b ≤ d (see [1], [2]), while in the case where (a, b) ∩ (c, d) = ∅, there is no corresponding result.
In this paper we present a new inequality which is used to obtain some estimations for the
difference of two integral means in the case where (a, b) ∩ (c, d) = ∅, which in limiting cases
reduces to a complement of Ostrowski’s inequality (1.1). Inequalities for pdfs (Probability
density functions) related to some results in [3, p. 245-246] are also given.
2. S OME I NEQUALITIES
The key result of the present paper is the following inequality:
Theorem 2.1. Let f, g be two continuously differentiable functions on [a, b] and twice differentiable on (a, b) with the properties that,
g 00 > 0
(2.1)
on (a, b), and that the function
estimation holds,
(2.2)
f 00
g 00
is bounded on (a, b) . For a < c ≤ d < b the following
f 00 (x)
inf 00
≤
x∈(a,b) g (x)
f (b)−f (d)
b−d
g(b)−g(d)
b−d
−
−
f (c)−f (a)
c−a
g(c)−g(a)
c−a
f 00 (x)
.
00
x∈(a,b) g (x)
≤ sup
Proof. Let s be any number such that a < s < c ≤ d < b. Consider the mappings f1 , g1 :
[d, b] → R defined as:
(2.3)
f1 (x) = f (x) − f (s) − (x − s) f 0 (s) ,
g1 (x) = g (x) − g (s) − (x − s) g 0 (s) .
Clearly f1 , g1 are continuous on [d, b] and differentiable on (d, b) . Further, for any x ∈ [d, b] ,
by applying the mean value Theorem,
g10 (x) = g 0 (x) − g 0 (s) = (x − s) g 00 (σ)
for some σ ∈ (s, x) , which, combined with (2.1), gives g10 (x) 6= 0, for all x ∈ (d, b) . Hence,
we can apply Cauchy’s mean value theorem to f1 , g1 on the interval [d, b] to obtain,
f1 (b) − f1 (d)
f 0 (τ )
= 10
g1 (b) − g1 (d)
g1 (τ )
for some τ ∈ (d, b) which can further be written as,
(2.4)
f (b) − f (d) − (b − d) f 0 (s)
f 0 (τ ) − f 0 (s)
=
.
g (b) − g (d) − (b − d) g 0 (s)
g 0 (τ ) − g 0 (s)
Applying Cauchy’s mean value theorem to f 0 , g 0 on the interval [s, τ ] , we have that for some
ξ ∈ (s, τ ) ⊆ (a, b),
f 0 (τ ) − f 0 (s)
f 00 (ξ)
= 00
.
g 0 (τ ) − g 0 (s)
g (ξ)
(2.5)
Combining (2.4) and (2.5) we have,
(2.6)
m≤
f (b) − f (d) − (b − d) f 0 (s)
≤M
g (b) − g (d) − (b − d) g 0 (s)
for all s ∈ (a, c), where m = inf x∈(a,b)
f 00 (x)
g 00 (x)
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 10, 6 pp.
and M = supx∈(a,b)
f 00 (x)
.
g 00 (x)
http://jipam.vu.edu.au/
C OMPLEMENTS OF O STROWSKI ’ S INEQUALITY
3
By further application of the mean value Theorem and using the assumption (2.1) we readily
get,
g (b) − g (d) − (b − d) g 0 (s) > 0.
(2.7)
Multiplying (2.6) by (2.7) ,
(2.8)
m (g (b) − g (d) − (b − d) g 0 (s)) ≤ f (b) − f (d) − (b − d) f 0 (s)
≤ M (g (b) − g (d) − (b − d) g 0 (s)) .
Integrating the inequalities (2.7) and (2.8) with respect to s from a to c we obtain respectively,
(c − a) (g (b) − g (d)) − (b − d) (g (c) − g (a)) > 0
(2.9)
and
(2.10)
m ((c − a) (g (b) − g (d)) − (b − d) (g (c) − g (a)))
≤ (c − a) (f (b) − f (d)) − (b − d) (f (c) − f (a))
≤ M ((c − a) (g (b) − g (d)) − (b − d) (g (c) − g (a))) .
Finally, dividing (2.10) by (2.9) ,
m≤
(c − a) (f (b) − f (d)) − (b − d) (f (c) − f (a))
≤M
(c − a) (g (b) − g (d)) − (b − d) (g (c) − g (a))
as required.
Remark 2.2. It is obvious that Theorem 2.1 holds also in the case where g 00 < 0 on (a, b) .
Corollary 2.3. Let a < c ≤ d < b and F, G be two continuous functions on [a, b] that are
F0
differentiable on (a, b) . If G0 > 0 on (a, b) or G0 < 0 on (a, b) and G
0 is bounded (a, b) , then,
Rb
Rc
1
1
F (t) dt − c−a
F (t) dt
F 0 (x)
F 0 (x)
b−d d
a
(2.11)
inf
≤
≤
sup
R
R
b
c
1
x∈(a,b) G0 (x)
G (t) dt − 1
G (t) dt x∈(a,b) G0 (x)
b−d
d
c−a
a
and
(2.12)
The constant
Z b
Z c
1
1
1
0
(b + d − a − c) inf F (x) ≤
F (t) dt −
F (t) dt
x∈(a,b)
2
b−d d
c−a a
1
≤ (b + d − a − c) sup F 0 (x) .
2
x∈(a,b)
1
2
in (2.12) is the best possible.
Proof. If we apply Theorem 2.1 for the functions,
Z x
Z
f (x) :=
F (t) dt,
g (x) :=
a
x
G (t) dt,
x ∈ [a, b] ,
a
then we immediately obtain (2.11). Choosing G (x) = x in (2.11) we get (2.12) .
Remark 2.4. Substituting d = b in (1.2) of Theorem 1.2 we get,
Z b
Z b
1
1
1
≤ (c − a) kF 0 k .
(2.13)
F
(x)
dx
−
F
(x)
dx
∞
b − a
2
b
−
c
a
c
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 10, 6 pp.
http://jipam.vu.edu.au/
4
A.I. K ECHRINIOTIS AND N.D. A SSIMAKIS
Setting d = c in (2.12) of Corollary 2.3 we get,
Z b
Z c
b−a
1
1
0
(2.14)
inf F (x) ≤
F (x) dx −
F (x) dx
2 x∈(a,b)
b−c c
c−a a
b−a
≤
sup F 0 (x) .
2 x∈(a,b)
Now,
1
b−c
Z
c
b
Z c
1
F (x) dx −
F (x) dx
c−a a
Z
Z c
c−a b
1
=
F (x) dx −
F (x) dx
c−a b−c c
a
Z
Z b
Z b
1
c−a b
=
F (x) dx −
F (x) dx +
F (x) dx
c−a b−c c
a
c
Z
Z b
1
b−a b
=
F (x) dx −
F (x) dx
c−a b−c c
a
Z b
Z b
b−a
1
1
=
F (x) dx −
F (x) dx .
c−a b−c c
b−a a
Using this in (2.14) we derive the inequality,
Z b
Z b
c−a
1
1
c−a
0
inf F (x) ≤
F (x) dx −
F (x) dx ≤
sup F 0 (x) .
2 x∈(a,b)
b−c c
b−a a
2 x∈(a,b)
From this we clearly get again inequality (2.13) . Consequently, inequality (2.12) can be seen
as a complement of (1.2).
Corollary 2.5. Let F, G be two continuous functions on an interval I ⊂ R and differentiable
◦
◦
◦
on the interior I of I with the properties G0 > 0 on I or G0 < 0 on I and
◦
F0
G0
◦
bounded on I.
Let a, b be any numbers in I such that a < b, then for all x ∈ I − (a, b) , that is, x ∈ I but
x∈
/ (a, b), we have the estimation:
Rb
1
F (t) dt − F (x)
F 0 (t)
F 0 (t)
b−a a
≤
sup
,
≤
(2.15)
inf
R
b
1
t∈({a,b,x}) G0 (t)
G (t) dt − G (x) t∈({a,b,x}) G0 (t)
b−a
a
where ({a, b, x}) := (min {a, x, } , max {x, b}) .
Proof. Let u, w, y, z be any numbers in I such that u < w ≤ y < z. According to Corollary 2.3
we then have the inequality,
Rz
Rw
1
1
F (t) dt − w−u
F (t) dt
F 0 (t)
F 0 (t)
z−y y
u
R
R
(2.16)
inf
≤ 1 z
≤ sup
.
w
1
t∈(u,z) G0 (t)
G (t) dt − w−u
G (t) dt t∈(u,z) G0 (t)
z−y y
u
We distinguish two cases:
If Rx < a, then by choosing y R= a, z = b and u = w = x in (2.12) and assuming that
w
w
1
1
F (t) dt = F (x) and w−u
G (t) dt = G (x) as limiting cases, (2.16) reduces to,
w−u u
u
Rb
1
F (t) dt − F (x)
F 0 (t)
F 0 (t)
b−a a
≤
sup
.
inf
≤
R
b
1
t∈(x,b) G0 (t)
G (t) dt − G (x) t∈(x,b) G0 (t)
b−a
a
Hence (2.15) holds for all x < a.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 10, 6 pp.
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C OMPLEMENTS OF O STROWSKI ’ S INEQUALITY
5
If x > b, then by choosing u = a, w = b and y = z = x, in (2.16) , similarly to the above,
we can prove that for all x > b the inequality (2.15) holds.
◦
Corollary 2.6. Let F be a continuous function on an interval I ⊂ R. If F 0 ∈ L∞ I, then for all
◦
a, b ∈ I with b > a and all x ∈ I − (a, b) we have:
Z b
|b + a − 2x|
1
≤
F (x) −
(2.17)
F
(t)
dt
kF 0 k∞, (min{a,x},max{b,x}) .
b−a
2
a
The inequality (2.17) is sharp.
Proof. Applying (2.15) for G (x) = x we readily get (2.17) . Choosing F (x) = x in (2.17) we
see that the equality holds, so the constant 21 is the best possible.
(2.17) is now used to obtain an extension of Ostrowski’s inequality (1.1).
Proposition 2.7. Let F be as in Corollary 2.5, then for all a, b ∈ I with b > a and for all x ∈ I,
Z b
1
F (x) −
(2.18)
F
(t)
dt
b−a a
"
2 #
x − a+b
1
2
≤
+
(b − a) kF 0 k∞,(min{a,x},max{b,x}) .
4
(b − a)2
Proof. Clearly, the restriction of inequality (2.18) on [a, b] is Ostrowski’s inequality (1.1) .
Moreover, a simple calculation yields
"
2 #
x − a+b
1
|b + a − 2x|
2
≤
+
(b − a)
2
4
(b − a)2
for all x ∈ R.
Combining this latter inequality with (2.17) we conclude that (2.18) holds also for x ∈
I − (a, b) and so (2.18) is valid for all x ∈ I.
3. A PPLICATIONS FOR PDF S
We now use inequality (2.2) in Theorem 2.1 to obtain improvements of some results in [3, p.
245-246].
Assume that f : [a, b] → R+ is a probability density function (pdf) of a certain random
Rb
variable X, that is a f (x) dx = 1, and
Z x
Pr (X ≤ x) =
f (t) dt,
x ∈ [a, b]
a
is its cumulative distribution function. Working similarly to [3, p. 245-246] we can state the
following:
Proposition 3.1. With the previous assumptions for f , we have that for all x ∈ [a, b] ,
1
x−a
(3.1)
(b − x) (x − a) inf f 0 (x) ≤
− Pr (X ≤ x)
x∈(a,b)
2
b−a
1
≤ (b − x) (x − a) sup f 0 (x) ,
2
x∈(a,b)
provided that f ∈ C [a, b] and f is differentiable and bounded on (a, b) .
Proof. Apply Theorem 2.1 for f (x) = Pr (X ≤ x), g (x) = x2 , c = d = x.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 10, 6 pp.
http://jipam.vu.edu.au/
6
A.I. K ECHRINIOTIS AND N.D. A SSIMAKIS
Proposition 3.2. Let f be as above, then,
(3.2)
1
(x − a)2
2
0
(x − a) (3b − a − 2x) inf f (x) ≤
− x Pr (X ≤ x) + Ex (X)
x∈(a,b)
12
2 (b − a)
1
≤
(x − a)2 (3b − a − 2x) sup f 0 (x) ,
12
x∈(a,b)
for all x ∈ [a, b] , where
Z
x
t Pr (X ≤ t) dt,
Ex (X) :=
x ∈ [a, b] .
a
Proof. Integrating (3.1) from a to x and using, in the resulting estimation, the following identity,
Z x
Z x
(3.3)
Pr (X ≤ x) dx = x Pr (X ≤ x) −
x (Pr (X ≤ x))0 dx
a
a
= x Pr (X ≤ x) − Ex (X)
we easily get the desired result.
Remark 3.3. Setting x = b in (3.2) we get,
1
a+b
1
(b − a)3 inf f 0 (x) ≤ E (X) −
≤
(b − a)3 sup f 0 (x) .
x∈(a,b)
12
2
12
x∈(a,b)
Proposition 3.4. Let f, Pr (X ≤ x) be as above. If f ∈ L∞ [a, b], then we have,
x−a
1
(b − x) (x − a) inf f (x) ≤
(b − E (X)) − x Pr (X ≤ x) + Ex (X)
x∈[a,b]
2
b−a
1
≤ (b − x) (x − a) sup f (x)
2
x∈[a,b]
for all x ∈ [a, b] .
Proof. Apply Theorem 2.1 for f (x) :=
(3.3).
Rx
a
Pr (X ≤ t) dt, g (x) := x2 , x ∈ [a, b], and identity
R EFERENCES
[1] A. AGLIC ALJINOVIĆ, J. PEČARIĆ AND I. PERIĆ, Estimates of the difference between two
weighted integral means via weighted Montgomery identity, Math. Inequal. Appl., 7(3) (2004), 315–
336.
[2] A. AGLIC ALJINOVIĆ, J. PEČARIĆ AND A. VUKELIĆ, The extension of Montgomery identity
via Fink identity with applications, J. Inequal. Appl., 2005(1), 67–79.
[3] N.S. BARNETT, P. CERONE, S.S. DRAGOMIR AND A. M. FINK, Comparing two integral means
for absolutely continuous mappings whose derivatives are in L∞ [a, b] and applications, Comput.
Math. Appl., 44(1-2) (2002), 241–251.
[4] P. CERONE AND S.S. DRAGOMIR, Differences between means with bounds from a RiemannStieltjes integral, Comp. and Math. Appl., 46 (2003), 445–453.
[5] A. OSTROWSKI, Uber die Absolutabweichung einer differenzierbaren funktion von ihren integralmittelwert, Comment. Math. Helv., 10 (1938), 226–227 (German).
[6] J. PEČARIĆ, I. PERIĆ AND A. VUKELIĆ, Estimations of the difference between two integral
means via Euler-type identities, Math. Inequal. Appl., 7(3) (2004), 365–378.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 10, 6 pp.
http://jipam.vu.edu.au/