ON THE INEQUALITY OF THE DIFFERENCE OF
TWO INTEGRAL MEANS AND APPLICATIONS FOR
PDFs
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
A.I. KECHRINIOTIS AND N.D. ASSIMAKIS
Department of Electronics
Technological Educational Institute of Lamia, Greece
EMail: {kechrin, assimakis}@teilam.gr
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Received:
11 November, 2005
Accepted:
12 April, 2006
Communicated by:
N.S. Barnett
2000 AMS Sub. Class.:
26D15.
Key words:
Ostrowski’s inequality, Probability density function, Difference of integral
means.
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Abstract:
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.
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1
Introduction
3
2
Some Inequalities
5
3
Applications for PDFs
12
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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1.
Introduction
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) −
(1.1)
f (t) dt ≤
+
(b − a) M ,
b−a a
4
(b − a)2
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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:
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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 ,
f
(t)
dt
−
f
(t)
dt
(1.2)
∞
b − a
2
d
−
c
a
c
the constant
1
2
being the best possible.
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
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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.
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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2.
Some Inequalities
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)
f 00
g 00
on (a, b), and that the function
following estimation holds,
(2.2)
f 00 (x)
inf 00
≤
x∈(a,b) g (x)
Complements of Ostrowski’s
Inequality
is bounded on (a, b) . For a < c ≤ d < b the
vol. 8, iss. 1, art. 10, 2007
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
A.I. Kechriniotis and N.D. Assimakis
0
0
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00
(x) = g (x) − g (s) = (x − s) g (σ)
g10
for some σ ∈ (s, x) , which, combined with (2.1), gives (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 )
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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),
0
0
Complements of Ostrowski’s
Inequality
00
f (τ ) − f (s)
f (ξ)
= 00
.
0
0
g (τ ) − g (s)
g (ξ)
(2.5)
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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)
00
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00
(x)
(x)
for all s ∈ (a, c), where m = inf x∈(a,b) fg00 (x)
and M = supx∈(a,b) fg00 (x)
.
By further application of the mean value Theorem and using the assumption (2.1)
we readily get,
(2.7)
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g (b) − g (d) − (b − d) g 0 (s) > 0.
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Multiplying (2.6) by (2.7) ,
(2.8)
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0
0
m (g (b) − g (d) − (b − d) g (s)) ≤ f (b) − f (d) − (b − d) f (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,
(2.9)
(c − a) (g (b) − g (d)) − (b − d) (g (c) − g (a)) > 0
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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))
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
as required.
Remark 1. It is obvious that Theorem 2.1 holds also in the case where g 00 < 0 on
(a, b) .
Corollary 2.2. Let a < c ≤ d < b and F, G be two continuous functions on [a, b]
F0
that are 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
−
F (t) dt
F 0 (x)
F 0 (x)
b−d d
c−a 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
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and
b
c
1
1
1
(b + d − a − c) inf F 0 (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)
Z
(2.12)
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The constant
1
2
in (2.12) is the best possible.
Z
Close
Proof. If we apply Theorem 2.1 for the functions,
Z x
Z x
f (x) :=
F (t) dt,
g (x) :=
G (t) dt,
a
x ∈ [a, b] ,
a
then we immediately obtain (2.11). Choosing G (x) = x in (2.11) we get (2.12) .
Remark 2. 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 c
a
Setting d = c in (2.12) of Corollary 2.2 we get,
Z b
Z c
1
1
b−a
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
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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Z
c
b
Z c
1
F (x) dx −
F (x) dx
c−a a
Z
Z c
1
c−a b
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
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Z
Z b
1
b−a b
=
F (x) dx −
F (x) dx
c−a b−c c
a
Z b
Z b
1
b−a
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)
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
From this we clearly get again inequality (2.13) . Consequently, inequality (2.12)
can be seen as a complement of (1.2).
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Corollary 2.3. Let F, G be two continuous functions on an interval I ⊂ R and
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◦
0
◦
0
differentiable on the interior I of I with the properties G > 0 on I or G < 0 on
◦
0
◦
◦
F
I and G
0 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
(2.15)
inf
≤
≤
sup
,
Rb
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 2.2 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
≤
≤
sup
.
z
w
1
1
t∈(u,z) G0 (t)
G (t) dt − w−u
G (t) dt t∈(u,z) G0 (t)
z−y y
u
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We distinguish two cases:
If x <Ra, then by choosing y = a, zR = b and u = w = x in (2.12) and assuming
w
w
1
1
that w−u
F (t) dt = F (x) and w−u
G (t) dt = G (x) as limiting cases, (2.16)
u
u
reduces to,
Rb
1
F (t) dt − F (x)
F 0 (t)
F 0 (t)
b−a a
inf
≤
sup
≤
.
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.
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 2.4. Let F be a continuous function on an interval I ⊂ R. If F 0 ∈ L∞ I,
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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◦
then for all a, b ∈ I with b > a and all x ∈ I − (a, b) we have:
Z b
|b + a − 2x|
1
(2.17)
F (t) dt ≤
kF 0 k∞, (min{a,x},max{b,x}) .
F (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 12 is the best possible.
(2.17) is now used to obtain an extension of Ostrowski’s inequality (1.1).
Proposition 2.5. Let F be as in Corollary 2.3, then for all a, b ∈ I with b > a and
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for all x ∈ I,
(2.18)
Z b
1
F (x) −
F (t) dt
b−a a
"
2 #
x − a+b
1
2
+
(b − a) kF 0 k∞,(min{a,x},max{b,x}) .
≤
2
4
(b − a)
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
|b + a − 2x|
1
2
(b − a)
≤
+
2
4
(b − a)2
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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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.
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3.
Applications for PDFs
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
Rb
random variable X, that is a f (x) dx = 1, and
Z x
Pr (X ≤ x) =
f (t) dt,
x ∈ [a, b]
a
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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] ,
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(3.1)
1
x−a
(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) .
2
Proof. Apply Theorem 2.1 for f (x) = Pr (X ≤ x), g (x) = x , c = d = x.
Proposition 3.2. Let f be as above, then,
(3.2)
1
(x − a)2 (3b − a − 2x) inf f 0 (x)
x∈(a,b)
12
2
(x − a)
− x Pr (X ≤ x) + Ex (X)
≤
2 (b − a)
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≤
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
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
a
= x Pr (X ≤ x) − Ex (X)
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we easily get the desired result.
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Remark 3. Setting x = b in (3.2) we get,
a+b
1
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 3.3. Let f, Pr (X ≤ x) be as above. If f ∈ L∞ [a, b], then we have,
1
x−a
(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) :=
and identity (3.3).
Rx
a
Pr (X ≤ t) dt, g (x) := x2 , x ∈ [a, b],
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References
[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 Riemann-Stieltjes 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.
Complements of Ostrowski’s
Inequality
A.I. Kechriniotis and N.D. Assimakis
vol. 8, iss. 1, art. 10, 2007
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