Journal of Inequalities in Pure and
Applied Mathematics
DIFFERENCE OF GENERAL INTEGRAL MEANS
GEORGE A. ANASTASSIOU
Department of Mathematical Sciences
The University of Memphis
Memphis, TN 38152, U.S.A.
volume 7, issue 5, article 185,
2006.
Received 02 May, 2006;
accepted 02 June, 2006.
Communicated by: S.S. Dragomir
EMail: ganastss@memphis.edu
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
In this paper we present sharp estimates for the difference of general integral
means with respect to even different finite measures. This is achieved by the
use of the Ostrowski and Fink inequalities and the Geometric Moment Theory Method. The produced inequalities are with respect to the supnorm of a
derivative of the involved function.
Difference of General Integral
Means
2000 Mathematics Subject Classification: 26D10, 26D15, 28A25, 60A10, 60E15.
Key words: Inequalities, Averages of functions, General averages or means, Moments.
George A. Anastassiou
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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4
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1.
Introduction
Here our work is motivated by the works of J. Duoandikoetxea [5] and P. Cerone
[4]. We use Ostrowski’s ([8]) and Fink’s ([6]) inequalities along with the Geometric Moment Theory Method, see [7], [1], [3], to prove our results.
We compare general averages of functions with respect to various finite measures over different subintervals of a domain, even disjoint. Our estimates are
sharp and the inequalities are attained. They are with respect to the supnorm of
a derivative of the involved function f .
To the best of our knowledge this type of work is totally new.
Difference of General Integral
Means
George A. Anastassiou
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2.
Results
Part A
As motivation we give the following proposition.
Proposition 2.1. Let µ1 , µ2 be finite Borel measures on [a, b] ⊆ R, [c, d],
[ẽ, g] ⊆ [a, b], f ∈ C 1 ([a, b]). Denote µ1 ([c, d]) = m1 > 0, µ2 ([ẽ, g] = m2 > 0.
Then
Z d
Z g
1
1
≤ kf 0 k∞ (b − a).
(2.1)
f
(x)dµ
−
f
(x)dµ
1
2
m1
m2 ẽ
c
Difference of General Integral
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George A. Anastassiou
Proof. From the mean value theorem we have
|f (x) − f (y)| ≤ kf 0 k∞ (b − a) =: γ,
∀x, y ∈ [a, b],
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that is,
−γ ≤ f (x) − f (y) ≤ γ,
∀x, y ∈ [a, b],
and by fixing y we get
1
−γ ≤
m1
Z
d
f (x)dµ1 − f (y) ≤ γ.
c
The last statement holds ∀y ∈ [ẽ, g]. Hence
Z d
Z g
1
1
−γ ≤
f (x)dµ1 −
f (x)dµ2 ≤ γ,
m1 c
m2 ẽ
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J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
proving the claim.
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As a related result we have
Corollary 2.2. Let f ∈ C 1 ([a, b]), [c, d], [ẽ, g] ⊆ [a, b] ⊆ R. Then we have
Z d
Z g
1
1
(2.2)
f (x)dx −
f (x)dx ≤ kf 0 k∞ · (b − a).
d − c
g − ẽ ẽ
c
We use the following famous Ostrowski inequality, see [8], [2].
Theorem 2.3. Let f ∈ C 1 ([a, b]), x ∈ [a, b]. Then
Z b
0
1
≤ kf k∞ (x − a)2 + (x − b)2 ,
f
(x)dx
(2.3)
f
(x)
−
2(b − a)
b−a a
and inequality (2.3) is sharp, see [2].
We also have
Corollary 2.4. Let f ∈ C 1 ([a, b]), x ∈ [c, d] ⊆ [a, b] ⊆ R. Then
(2.4)
Z b
f (x) − 1
f
(x)dx
b−a a
kf 0 k∞
max ((c − a)2 + (c − b)2 ), ((d − a)2 + (d − b)2 ) .
≤
2(b − a)
Proof. Obvious.
We denote by P([a, b]) the power set of [a, b]. We give the following.
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George A. Anastassiou
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Theorem 2.5. Let f ∈ C 1 ([a, b]), µ be a finite measure on ([c, d], P([c, d])),
where [c, d] ⊆ [a, b] ⊆ R and m := µ([c, d]) > 0. Then
1.
(2.5)
Z
Z b
1
1
f
(x)dµ
−
f
(x)dx
m
b−a a
[c,d]
kf 0 k∞
≤
max ((c − a)2 + (c − b)2 ), ((d − a)2 + (d − b)2 ) .
2(b − a)
2. Inequality (2.5) is attained when d = b.
Difference of General Integral
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George A. Anastassiou
Proof. 1) By (2.4) integrating against µ/m.
2) Here (2.5) collapses to
Z
Z b
1
kf 0 k∞
1
≤
(2.6)
f
(x)dµ
−
f
(x)dx
(b − a).
m
b−a a
2
[c,b]
We prove that (2.6) is attained. Take
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2x − (a + b)
f ∗ (x) =
,
b−a
a ≤ x ≤ b.
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Then f ∗0 (x) =
2
b−a
and kf ∗0 k∞ =
Z
a
2
,
b−a
along with
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b
f ∗ (x)dx = 0.
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Therefore (2.6) becomes
Z
1
∗
f (x)dµ ≤ 1.
m
[c,b]
(2.7)
Finally pick
equality.
µ
m
= δ{b} the Dirac measure supported at {b}, then (2.7) turns to
We further have
Corollary 2.6. Let f ∈ C 1 ([a, b]) and [c, d] ⊆ [a, b] ⊆ R. Let M (c, d) := {µ : µ
a measure on ([c, d], P([c, d])) of finite positive mass}, denoted m := µ([c, d]).
Then
1. The following result holds
Z
Z b
1
1
sup f (x)dµ −
f (x)dx
b−a a
µ∈M (c,d) m [c,d]
kf 0 k∞
(2.8)
≤
max ((c − a)2 + (c − b)2 ), ((d − a)2 + (d − b)2 )
2(b − a)
(
)
(d − a)2 + (d − b)2 , if d + c ≥ a + b
kf 0 k∞
=
×
2(b − a)
(c − a)2 + (c − b)2 ,
if d + c ≤ a + b
0
kf k∞
(2.9)
(b − a).
≤
2
Inequality (2.9) becomes equality if d = b or c = a or both.
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George A. Anastassiou
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2. The following result holds
sup
(2.10)
all c,d
a≤c<d≤b
!
Z
Z b
1
1
f (x)dµ −
sup f (x)dx
b−a a
µ∈M (c,d) m [c,d]
≤
kf 0 k∞
(b − a).
2
Next we restrict ourselves to a subclass of M (c, d) of finite measures µ with
given first moment and by the use of the Geometric Moment Theory Method,
see [7], [1], [3], we produce an inequality sharper than (2.8). For that we need
Difference of General Integral
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George A. Anastassiou
Lemma 2.7. Let ν be a probability measure on ([a, b], P([a, b])) such that
Z
(2.11)
x dν = d1 ∈ [a, b]
[a,b]
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is given. Then
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Z
(2.12)
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U1 :=
sup
ν as in (2.11)
(x − a)2 dν = (b − a)(d1 − a),
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Z
(2.13)
U2 :=
sup
ν as in (2.11)
[a,b]
(x − b)2 dν = (b − a)(b − d1 ).
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Proof. i) We observe the graph
G1 = (x, (x − a)2 ) : a ≤ x ≤ b ,
which is a convex arc above the x-axis. We form the closed convex hull of G1
b1 which has as an upper concave envelope the line segment `1
and we call it G
from (a, 0) to (b, (b − a)2 ). We consider the vertical line x = d1 which cuts `1
at the point Q1 . Then U1 is the distance from (d1 , 0) to Q1 . By using the equal
ratios property of similar triangles related here we get
U1
d1 − a
=
,
b−a
(b − a)2
which proves the claim.
ii) We observe the graph
Difference of General Integral
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George A. Anastassiou
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G2 = (x, (x − b)2 ) : a ≤ x ≤ b ,
which is a convex arc above the x-axis. We form the closed convex hull of G2
b2 which has as an upper concave envelope the line segment `2
and we call it G
from (b, 0) to (a, (b−a)2 ). We consider the vertical line x = d1 which intersects
`2 at the point Q2 .
Then U2 is the distance from (d1 , 0) to Q2 . By using the equal ratios property
of the related similar triangles we obtain
U2
b − d1
=
,
2
(b − a)
b−a
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which proves the claim.
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Furthermore we need
Lemma 2.8. Let [c, d] ⊆ [a, b] ⊆ R and let ν be a probability measure on ([c, d],
P([c, d])) such that
Z
(2.14)
x dν = d1 ∈ [c, d]
[c,d]
is given. Then
Difference of General Integral
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(i)
Z
(2.15)
U1 :=
(x − a)2 dν = d1 (c + d − 2a) − cd + a2 ,
sup
ν as in (2.14)
George A. Anastassiou
[c,d]
and
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(ii)
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Z
(2.16)
U2 :=
(x − b)2 dν = d1 (c + d − 2b) − cd + b2 .
sup
ν as in (2.14)
[c,d]
(iii) The following also holds:
Z
(2.17)
sup
ν as in (2.14)
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(x − a)2 + (x − b)2 dν = U1 + U2 .
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[c,d]
Proof. (i) We see that
Z d
Z d
2
2
(x − a) dν = (c − a) + 2(c − a)(d1 − c) +
(x − c)2 dν.
c
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Using (2.12) which is applied on [c, d], we find
Z d
sup
(x − a)2 dν = (c − a)2 + 2(c − a)(d1 − c)
ν as in (2.14)
c
d
Z
+
(x − c)2 dν
sup
ν as in (2.14)
c
2
= (c − a) + 2(c − a)(d1 − c) + (d − c)(d1 − c)
= d1 (c + d − 2a) − cd + a2 ,
proving the claim.
(ii) We see that
Z d
Z d
2
2
(x − b) dν = (b − d) + 2(b − d)(d − d1 ) +
(x − d)2 dν.
c
c
Using (2.13) which is applied on [c, d], we obtain
Z d
sup
(x − b)2 dν = (b − d)2 + 2(b − d)(d − d1 )
ν as in (2.14)
c
Z
+
sup
ν as in (2.14)
d
(x − d)2 dν
c
2
= (b − d) + 2(b − d)(d − d1 ) + (d − c)(d − d1 )
= d1 (c + d − 2b) − cd + b2 ,
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proving the claim.
(iii) Similar to Lemma 2.7 and above and obvious on noting that (x − a)2 + (x −
b)2 is convex, etc.
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Now we are ready to present
Theorem 2.9. Let [c, d] ⊆ [a, b] ⊆ R, f ∈ C 1 ([a, b]), µ a finite measure on
([c, d], P([c, d])) of mass m := µ([c, d]) > 0. Assume that
Z
1 d
(2.18)
x dµ = d1 , c ≤ d1 ≤ d,
m c
is given.
Then
(2.19)
µ
Z d
Z b
1
1
sup f (x)dµ −
f (x)dx
b−a a
as above m c
0
kf k∞
a2 + b 2
≤
d1 (c + d) − (a + b) − cd +
.
(b − a)
2
George A. Anastassiou
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Proof. Denote
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kf 0 k∞
β(x) :=
(x − a)2 + (x − b)2 ,
2(b − a)
1
−β(x) ≤ f (x) −
b−a
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then by Theorem 2.3 we have
Z
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b
f (x)dx ≤ β(x),
∀x ∈ [c, d].
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Thus
1
−
m
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Z
c
d
1
β(x)dµ ≤
m
Z
c
d
1
f (x)dµ −
b−a
Z
a
b
1
f (x)dx ≤
m
Z
d
β(x)dµ,
c
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and
Z d
Z b
Z
1
1
1 d
f (x)dµ −
f (x)dx ≤
β(x)dµ =: θ.
m
b−a a
m c
c
Here ν :=
µ
m
is a probability measure subject to (2.18) on ([c, d], P([c, d])) and
Z d
Z d
kf 0 k∞
2 dµ
2 dµ
θ=
(x − a)
+
(x − b)
2(b − a)
m
m
c
c
Z d
Z d
0
kf k∞
2
2
=
(x − a) dν +
(x − b) dν .
2(b − a)
c
c
Using (2.14), (2.15), (2.16) and (2.17) we get
kf 0 k∞ (d1 (c + d − 2a) − cd + a2 ) + (d1 (c + d − 2b) − cd + b2 )
2(b − a)
kf 0 k∞
a2 + b 2
=
d1 ((c + d) − (a + b)) − cd +
,
(b − a)
2
θ≤
proving the claim.
George A. Anastassiou
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We make the following remark.
Remark 1 (Remark on Theorem 2.9).
d we obtain
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1. Case of c + d ≥ a + b, using d1 ≤
(d − a)2 + (d − b)2
a2 + b 2
≤
.
(2.20) d1 (c + d) − (a + b) − cd +
2
2
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2. Case of c + d ≤ a + b, using d1 ≥ c we find that
(2.21)
a2 + b 2
(c − a)2 + (c − b)2
d1 (c + d) − (a + b) − cd +
≤
.
2
2
Hence under (2.18) inequality (2.19) is sharper than (2.8).
We also give
Corollary 2.10. Let all the assumptions in Theorem 2.9 hold. Then
(2.22)
Z d
Z b
1
1
f (x)dµ −
f (x)dx
m
b−a a
c
a2 + b 2
kf 0 k∞
d1 (c + d) − (a + b) − cd +
.
≤
(b − a)
2
By Remark 1, inequality (2.22) is sharper than (2.5).
Part B
Here we follow Fink’s work [6]. We require the following theorem.
Theorem 2.11 ([6]). Let f : [a, b] → R, f (n−1) is absolutely continuous on
[a, b], n ≥ 1. Then
n
(2.23) f (x) =
b−a
Z
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b
f (t)dt
a
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+
(k−1)
n−1 X
n−k
f
(b)(x − b)k − f (k−1) (a)(x − a)k
k!
k=1
b−a
Z b
1
(x − t)n−1 k(t, x)f (n) (t)dt,
+
(n − 1)!(b − a) a
where
(
t − a,
a ≤ t ≤ x ≤ b,
t − b,
a ≤ x < t ≤ b.
k(t, x) :=
(2.24)
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For n = 1 the sum in (2.23) is taken as zero.
George A. Anastassiou
We also need Fink’s inequality
Theorem 2.12 ([6]). Let f (n−1) be absolutely continuous on [a, b] and f (n) ∈
L∞ (a, b), n ≥ 1. Then
!
Z b
n−1
1
X
1
(2.25) f (x) +
Fk (x) −
f (x)dx
n
b
−
a
a
k=1
≤
kf (n) k∞
(b − x)n+1 + (x − a)n+1 ,
n(n + 1)!(b − a)
∀x ∈ [a, b],
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where
(2.26)
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Fk (x) :=
n−k
k!
f (k−1) (a)(x − a)k − f (k−1) (b)(x − b)k
b−a
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.
Inequality (2.25) is sharp, in the sense that it is attained by an optimal f for any
x ∈ [a, b].
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We give
Corollary 2.13. Let f (n−1) be absolutely continuous on [a, b] and f (n) ∈ L∞ (a, b),
n ≥ 1. Then ∀x ∈ [c, d] ⊆ [a, b] we have
!
Z b
n−1
1
X
1
f (x) +
Fk (x) −
f (x)dx
n
b−a a
k=1
kf (n) k∞
(b − x)n+1 + (x − a)n+1
n(n + 1)!(b − a)
kf (n) k∞
(b − a)n .
≤
n(n + 1)!
≤
(2.27)
Also we have
George A. Anastassiou
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Proposition 2.14. Let f (n−1) be absolutely continuous on [a, b] and f (n) ∈
L∞ (a, b), n ≥ 1. Let µ be a finite measure of mass m > 0 on
[c, d], P([c, d]) , [c, d] ⊆ [a, b] ⊆ R.
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Then
(2.28)
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1
K := n
!
Z
n−1
X
1
f (x)dµ +
Fk (x)dµ
m [c,d]
[c,d]
k=1
Z b
1
−
f (x)dx
b−a a
1
m
Z
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Z
kf (n) k∞
1
≤
(b − x)n+1 dµ
n(n + 1)!(b − a) m [c,d]
Z
1
n+1
+
(x − a) dµ
m [c,d]
≤
(2.29)
kf (n) k∞
(b − a)n .
n(n + 1)!
Proof. By (2.27).
Similarly, based on Theorem A of [6] we also conclude
Proposition 2.15. Let f (n−1) be absolutely continuous on [a, b] and f (n) ∈
Lp (a, b), where 1 < p < ∞, n ≥ 1. Let µ be a finite measure of mass m > 0 on
([c, d], P([c, d])), [c, d] ⊆ [a, b] ⊆ R.
0
1
p
1
p0
Here p > 1 such that + = 1. Then
!
Z
Z b
n−1
1 1 Z
X
1
1
f
(x)dµ
+
F
(x)dµ
−
f
(x)dx
k
n m [c,d]
m
b
−
a
[c,d]
a
k=1
1/p0 (n) !
B (n − 1)p0 + 1, p0 + 1)
kf kp
≤
n!(b − a)
Z
1
np0 +1
np0 +1 1/p0
·
(x − a)
+ (b − x)
) dµ
m [c,d]
!
1/p0
n−1+ p10
B (n − 1)p0 + 1, p0 + 1)
(b − a)
(2.30)
≤
kf (n) kp .
n!
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We make the following remark.
Remark 2. Clearly we have the following for
(2.31)
g(x) := (b − x)n+1 + (x − a)n+1 ≤ (b − a)n+1 ,
a ≤ x ≤ b,
is the only critical number of g and
where n ≥ 1. Here x = a+b
2
a+b
(b − a)n−1
00
g
= n(n + 1)
> 0,
2
2n−2
(b−a)n+1
giving that g a+b
=
> 0 is the global minimum of g over [a, b]. Also
2
2n
g is convex over [a, b]. Therefore for [c, d] ⊆ [a, b] we have
M := max (x − a)n+1 + (b − x)n+1
c≤x≤d
(2.32)
= max (c − a)n+1 + (b − c)n+1 , (d − a)n+1 + (b − d)n+1 .
We get further that
(
(2.33)
M=
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(d − a)n+1 + (b − d)n+1 , if c + d ≥ a + b
(c − a)n+1 + (b − c)n+1 ,
if c + d ≤ a + b.
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If d = b or c = a or both then
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(2.34)
M = (b − a)n+1 .
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Based on Remark 2 we give
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Theorem 2.16. Let all assumptions, terms and notations be as in Proposition
2.14. Then
1.
K≤
(2.35)
=
kf (n) k∞
max (c − a)n+1 + (b − c)n+1 ,
n(n + 1)!(b − a)
(d − a)n+1 + (b − d)n+1
kf (n) k∞
n(n + 1)!(b − a)


 (d − a)n+1 + (b − d)n+1 , if c + d ≥ a + b, 
×
 (c − a)n+1 + (b − c)n+1 , if c + d ≤ a + b 
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(n)
(2.36)
≤
kf k∞
(b − a)n ,
n(n + 1)!
where K is as in (2.28). If d = b or c = a or both, nthen (2.36) becomes
µ
equality. When d = b, m
= δ{b} and f (x) = (x−a)
, a ≤ x ≤ b, then
n!
inequality (2.35) is attained, i.e. it becomes equality, proving that (2.35) is
a sharp inequality.
2. We also have
(2.37)
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sup K ≤ R.H.S (2.35)
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µ∈M (c,d)
J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
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and
!
sup K
sup
(2.38)
all c,d
a≤c≤d≤b
≤ R.H.S (2.36)
µ∈M (c,d)
Proof. It remains to prove only the sharpness, via attainability of (2.35) when
d = b. In that case (2.35) collapses to
!
Z
Z b
n−1
1 1 Z
X
1
1
(2.39) f (x)dµ +
Fk (x)dµ −
f (x)dx
n m [c,d]
m [c,b]
b−a a
k=1
kf (n) k∞
(b − a)n .
≤
n(n + 1)!
µ
m
The optimal measure here will be = δ{b} and then (2.39) becomes
!
Z b
n−1
1
X
1
kf (n) k∞
(2.40) f (b) +
Fk (b) −
f (x)dx ≤
(b − a)n .
n
b
−
a
n(n
+
1)!
a
k=1
(x − a)n
,
f (x) =
n!
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a ≤ x ≤ b.
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Then we see that
f ∗(k−1) (x) =
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The optimal function here will be
∗
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(x − a)n−k+1
,
(n − k + 1)!
k − 1 = 0, 1, . . . , n − 2,
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and f ∗(k−1) (a) = 0 for k − 1 = 0, 1, . . . , n − 2. Clearly here Fk (b) = 0,
k = 1, . . . , n − 1. Also we have
Z b
1
(b − a)n
f ∗ (x)dx =
and
kf ∗(n) k∞ = 1.
b−a a
(n + 1)!
Putting all these elements in (2.40) we have
n
(b − a)n (b − a)n = (b − a) ,
−
nn!
(n + 1)! n(n + 1)!
Difference of General Integral
Means
proving the claim.
Next, we again restrict ourselves to the subclass of M (c, d) of finite measures
µ with given first moment and by the use of the Geometric Moment Theory
Method, see [7], [1], [3], we produce an inequality sharper than (2.37). For that
we need the follwing result.
Lemma 2.17. Let [c, d] ⊆ [a, b] ⊆ R and ν be a probability measure on ([c, d],
P([c, d])) such that
Z
(2.41)
x dν = d1 ∈ [c, d]
is given, n ≥ 1. Then
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Z
(2.43)
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[c,d]
(2.42)
George A. Anastassiou
sup
(x − a)n+1 dν
ν as in (2.41) [c,d]
!
n
X
=
(d − a)n−k (c − a)k (d1 − d) + (d − a)n+1 .
W1 :=
k=0
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J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
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Proof. We observe the graph
G1 = (x, (x − a)n+1 ) : c ≤ x ≤ d ,
which is a convex arc above the x-axis. We form the closed convex hull of G1
b1 , which has as an upper concave envelope the line segment `1
and we call it G
from (c, (c − a)n+1 ) to (d, (d − a)n+1 ). Call `1 the line through `1 . The line `1
intersects the x-axis at (t, 0), where a ≤ t ≤ c. We need to determine t: the
slope of `1 is
n
(d − a)n+1 − (c − a)n+1 X
m̃ =
=
(d − a)n−k (c − a)k .
d−c
k=0
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George A. Anastassiou
The equation of line `1 is
y = m̃ · x + (d − a)n+1 − m̃d.
Hence m̃t + (d − a)n+1 − m̃d = 0 and
(d − a)n+1
t=d−
.
m̃
Next we consider the moment right triangle with vertices (t, 0), (d, 0) and
(d, (d−a)n+1 ). Clearly (d1 , 0) is between (t, 0) and (d, 0). Consider the vertical
line x = d1 , it intersects `1 at Q. Clearly then W1 = length((d1 , 0), Q), the line
segment of which length we find by the formed two similar right triangles with
vertices {(t, 0), (d1 , 0), Q} and {(t, 0), (d, 0), (d, (d − a)n+1 )}. We have the
equal ratios
W1
d1 − t
=
,
d−t
(d − a)n+1
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i.e.
n+1
W1 = (d − a)
d1 − t
d−t
.
We also need
Lemma 2.18. Let [c, d] ⊆ [a, b] ⊆ R and ν be a probability measure on ([c, d],
P([c, d])) such that
Z
(2.44)
x dν = d1 ∈ [c, d]
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George A. Anastassiou
[c,d]
is given, n ≥ 1. Then
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1.
Z
W2 :=
ν
(2.45)
=
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n+1
sup
(b − x) dν
as in (2.44) [c,d]
!
n
X
(b − c)n−k (b − d)k (c − d1 ) + (b − c)n+1 .
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k=0
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2. The following result holds
Z
(x − a)n+1 + (b − x)n+1 dν = W1 + W2 ,
(2.46)
sup
ν as in (2.44) [c,d]
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J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
where W1 is as in (2.42).
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Proof.
1. We observe the graph
G2 = (x, (b − x)n+1 ) : c ≤ x ≤ d ,
which is a convex arc above the x-axis. We form the closed convex hull
b2 , which has as an upper concave envelope the line
of G2 and we call it G
segment `2 from (c, (b − c)n+1 ) to (d, (b − d)n+1 ). Call `2 the line through
`2 . The line `2 intersects the x-axis at (t∗ , 0), where d ≤ t∗ ≤ b. We need
to determine t∗ : The slope of `2 is
!
n
n+1
n+1
X
(b
−
c)
−
(b
−
d)
m̃∗ =
=−
(b − c)n−k (b − d)k .
c−d
k=0
The equation of line `2 is
y = m̃∗ x + (b − c)n+1 − m̃∗ c.
Hence
m̃∗ t∗ + (b − c)n+1 − m̃∗ c = 0
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and
(b − c)n+1
t∗ = c −
.
m̃∗
Next we consider the moment right triangle with vertices (c, (b − c)n+1 ),
(c, 0), (t∗ , 0). Clearly (d1 , 0) is between (c, 0) and (t∗ , 0). Consider the
vertical line x = d1 , it intersects `2 at Q∗ . Clearly then
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J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
W2 = length((d1 , 0), Q∗ ),
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the line segment of which length we find by the formed two similar right
triangles with vertices {Q∗ , (d1 , 0), (t∗ , 0)} and {(c, (b − c)n+1 ), (c, 0),
(t∗ , 0)}. We have the equal ratios
t∗ − d1
W2
=
,
t∗ − c
(b − c)n+1
i.e.
n+1
W2 = (b − c)
t∗ − d1
t∗ − c
.
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2. Similar to that above and obvious.
George A. Anastassiou
We make the following useful remark.
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Remark 3. By Lemmas 2.17, 2.18 we obtain
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(2.47) λ := W1 + W2
!
n
X
=
(d − a)n−k (c − a)k (d1 − d)
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k=0
+
n
X
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!
n−k
(b − c)
k
(b − d)
n+1
(c − d1 ) + (d − a)
n+1
+ (b − c)
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k=0
> 0,
n ≥ 1.
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J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
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We present the following important result.
Theorem 2.19. Let f (n−1) be absolutely continuous on [a, b] and f (n) ∈ L∞ (a, b),
n ≥ 1. Let µ be a finite measure of mass m > 0 on ([c, d], P([c, d])), [c, d] ⊆
[a, b] ⊆ R. Furthermore we assume that
Z
1
(2.48)
x dµ = d1 ∈ [c, d]
m [c,d]
is given. Then
(2.49)
kf (n) k∞
sup
K≤
λ,
n(n + 1)!(b − a)
µ as above
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George A. Anastassiou
and
(2.50)
K ≤ R.H.S (2.49),
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where K is as in (2.28) and λ is as in (2.47).
Proof. By Proposition 2.14 and Lemmas 2.17 and 2.18.
We make the following remark.
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Remark 4. We compare M as in (2.32) and (2.33) and λ as in (2.47). We easily
obtain that
(2.51)
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As a result we have that (2.50) is sharper than (2.35) and (2.49) is sharper than
(2.37). That is reasonable since we restricted ourselves to a subclass of M (c, d)
of measures µ by assuming the moment condition (2.48).
J. Ineq. Pure and Appl. Math. 7(5) Art. 185, 2006
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We finish with the following comment.
Remark 5.
I) When c = a and d = b then d1 plays no role in the best upper bounds we
found with the Geometric Moment Theory Method. That is, the restriction
on measures µ via the first moment d1 has no effect in producing sharper
estimates as it happens when a < c < d < b. More precisely we notice
that:
Difference of General Integral
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(a)
(2.52)
kf 0 k∞
R.H.S.(2.19) =
(b − a) = R.H.S.(2.9),
2
George A. Anastassiou
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n+1
(b) by (2.47) here λ = (b − a)
(2.53)
R.H.S.(2.49) =
and
kf (n) k∞
(b − a)n = R.H.S.(2.36).
n(n + 1)!
II) Further differences of general means over any [c1 , d1 ] and [c2 , d2 ] subsets of [a, b] (even disjoint) with respect to µ1 and µ2 , respectively, can be
found by straightforward application of the above results and the triangle
inequality.
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References
[1] G.A. ANASTASSIOU, Moments in Probability and Approximation Theory,
Pitman/Longman, #287, UK, 1993.
[2] G.A. ANASTASSIOU, On Ostrowski type inequalities, Proc. AMS, 123(12)
(1995), 3775–3781.
[3] G.A. ANASTASSIOU, General moment optimization problems, in Encyclopedia of Optimation, C. Floudas and P. Pardalos, Eds., Kluwer, pp. 198–
205, Vol. II, 2001.
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[4] P. CERONE, Difference between weighted integral means, in Demonstratio
Mathematica, 35(2) (2002), 251–265.
George A. Anastassiou
[5] J. DUOANDIKOETXEA, A unified approach to several inequalities involving functions and derivatives, Czechoslovak Mathematical Journal, 51
(126) (2001), 363–376.
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[6] A.M. FINK, Bounds on the deviation of a function from its averages,
Czechoslovak Mathematical Journal, 42 (117) (1992), 289–310.
[7] J.H.B. KEMPERMAN, The general moment problem, a geometric approach, The Annals of Mathematical Statistics, 39(1) (1968), 93–122.
[8] A. OSTROWSKI, Über die Absolutabweichung einer differentiebaren
Funktion von ihrem Integralmittelwert, Comment. Math. Helv., 10 (1938),
226–227.
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