Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 1, Article 12, 2002
ON WEIGHTED OSTROWSKI TYPE INEQUALITIES FOR OPERATORS AND
VECTOR-VALUED FUNCTIONS
1
N.S. BARNETT, 2 C. BUŞE, 1 P. CERONE, AND 1 S.S. DRAGOMIR
1
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428
M ELBOURNE C ITY MC 8001,
V ICTORIA , AUSTRALIA .
2
D EPARTMENT OF M ATHEMATICS
W EST U NIVERSITY OF T IMI ŞOARA
B D . V. PÂRVAN 4
1900 T IMI ŞOARA , ROMÂNIA .
neil@matilda.vu.edu.au
URL: http://sci.vu.edu.au/staff/neilb.html
buse@hilbert.math.uvt.ro
URL: http://rgmia.vu.edu.au/BuseCVhtml/
pc@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/cerone
sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 22 June, 2001; accepted 25 October, 2001.
Communicated by Th. M. Rassias
A BSTRACT. Some weighted Ostrowski type integral inequalities for operators and vector-valued
functions in Banach spaces are given. Applications for linear operators in Banach spaces and
differential equations are also provided.
Key words and phrases: Ostrowski’s Inequality, Vector-Valued Functions, Bochner Integral.
2000 Mathematics Subject Classification. Primary 26D15; Secondary 46B99.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
050-01
2
N.S. BARNETT , C. B U ŞE , P. C ERONE ,
AND
S.S. D RAGOMIR
1. I NTRODUCTION
In [12], Pečarić and Savić obtained the following Ostrowski type inequality for weighted
integrals (see also [7, Theorem 3]):
Theorem 1.1. Let w : [a, b] → [0, ∞) be a weight function on [a, b] . Suppose that f : [a, b] → R
satisfies
(1.1)
|f (t) − f (s)| ≤ N |t − s|α , for all t, s ∈ [a, b] ,
where N > 0 and 0 < α ≤ 1 are some constants. Then for any x ∈ [a, b]
Rb
Rb
|t − x|α w (t) dt
w
(t)
f
(t)
dt
a
(1.2)
.
f (x) − R b
≤ N · a Rb
w
(t)
dt
w
(t)
dt
a
a
Further, if for some constants c and λ
0 < c ≤ w (t) ≤ λc, for all t ∈ [a, b] ,
then for any x ∈ [a, b], we have
Rb
w
(t)
f
(t)
dt
λL (x) J (x)
a
,
(1.3)
f (x) − R b
≤N·
L (x) − J (x) + λJ (x)
w (t) dt a
where
α
1
a
+
b
L (x) :=
(b − a) + x −
2
2 and
J (x) :=
(x − a)1+α + (b − x)1+α
.
(1 + α) (b − a)
The inequality (1.2) was rediscovered in [4] where further applications for different weights
and in Numerical Analysis were given.
For other results in connection to weighted Ostrowski inequalities, see [3], [8] and [10].
In the present paper we extend the weighted Ostrowski’s inequality for vector-valued functions and Bochner integrals and apply the obtained results to operatorial inequalities and linear
differential equations in Banach spaces. Some numerical experiments are also conducted.
2. W EIGHTED I NEQUALITIES
Let X be a Banach space and −∞ < a < b < ∞. We denote by L (X) the Banach algebra
of all bounded linear operators acting on X. The norms of vectors or operators acting on X will
be denoted by k·k .
A function f : [a, b] → X is called measurable if there exists a sequence of simple functions
fn : [a, b] → X which converges punctually almost everywhere on [a, b] at f. We recall also
that a measurable function f : [a, b] → X is Bochner integrable if and only if its norm function
(i.e. the function t 7→ kf (t)k : [a, b] → R+ ) is Lebesgue integrable on [a, b].
The following theorem holds.
Theorem 2.1. Assume that B : [a, b] → L (X) is Hölder continuous on [a, b], i.e.,
(2.1)
kB (t) − B (s)k ≤ H |t − s|α for all t, s ∈ [a, b] ,
where H > 0 and α ∈ (0, 1].
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
3
If f : [a, b] → X is Bochner integrable on [a, b], then we have the inequality:
Z b
Z b
(2.2) B (t)
f (s) ds −
B (s) f (s) ds
a
a
Z
b
|t − s|α kf (s)k ds
a

(b − t)α+1 + (t − a)α+1


k|f |k[a,b],∞
if
f ∈ L∞ ([a, b] ; X) ;



α+1



"
#1


 (b − t)qα+1 + (t − a)qα+1 q
k|f |k[a,b],p if
p > 1, p1 + 1q = 1
≤H×
qα + 1





and f ∈ Lp ([a, b] ; X) ;

α



1
a
+
b

k|f |k

(b − a) + t −
[a,b],1
2
2 ≤H
for any t ∈ [a, b], where
|kf k|[a,b],∞ := ess sup kf (t)k
t∈[a,b]
and
Z
|kf k|[a,b],p :=
b
p1
kf (t)k dt ,
p
p ≥ 1.
a
Proof. Firstly, we prove that the X−valued function s 7→ B (s) f (s) is Bochner integrable on
[a, b]. Indeed, let (fn ) be a sequence of X−valued, simple functions which converge almost
everywhere on [a, b] at the function f . The maps s 7→ B (s) fn (s) are measurable (because
they are continuous with the exception of a finite number of points s in [a, b]). Then
kB (s) fn (s) − B (s) f (s)k ≤ kB (s)k kfn (s) − f (s)k → 0 a.e. on [a, b]
when n → ∞ so that the function s 7→ B (s) f (s) : [a, b] → X is measurable. Now, using the
estimate
kB (s) f (s)k ≤ sup kB (ξ)k · kf (s)k , for all s ∈ [a, b] ,
ξ∈[a,b]
it is easy to see that the function s 7→ B (s) f (s) is Bochner integrable on [a, b].
We have successively
Z b
Z b
Z b
B (t)
f (s) ds −
B (s) f (s) ds
(B (t) − B (s)) f (s) ds
= a
a
Z
a
b
≤
k(B (t) − B (s)) f (s)k ds
a
Z
b
≤
k(B (t) − B (s))k kf (s)k ds
Z b
≤ H
|t − s|α kf (s)k ds
a
a
=: M (t)
for any t ∈ [a, b], proving the first inequality in (2.2).
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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4
N.S. BARNETT , C. B U ŞE , P. C ERONE ,
S.S. D RAGOMIR
AND
Now, observe that
b
Z
|t − s|α ds
M (t) ≤ H k|f |k[a,b],∞
a
(b − t)α+1 + (t − a)α+1
α+1
= H k|f |k[a,b],∞ ·
and the first part of the second inequality is proved.
Using Hölder’s integral inequality, we may state that
Z
M (t) ≤ H
1q Z
b
qα
|t − s|
ds
= H
p1
kf (s)k ds
p
a
a
"
b
qα+1
(b − t)
qα+1
+ (t − a)
qα + 1
# 1q
k|f |k[a,b],p ,
proving the second part of the second inequality.
Finally, we observe that
α
Z
M (t) ≤ H sup |t − s|
s∈[a,b]
b
kf (s)k ds
a
= H max {(b − t)α , (t − a)α } k|f |k[a,b],1
α
1
a + b = H
(b − a) + t −
k|f |k[a,b],1
2
2 and the theorem is proved.
The following corollary holds.
Corollary 2.2. Assume that B : [a, b] → L (X) is Lipschitzian with the constant L > 0. Then
we have the inequality
Z b
Z b
(2.3) B (t)
f (s) ds −
B (s) f (s) ds
a
Z
a
b
≤L
|t − s| kf (s)k ds
 "
2 #
1
a
+
b

2


(b − a) + t −
k|f |k[a,b],∞ if
f ∈ L∞ ([a, b] ; X) ;


4
2



"
#1


 (b − t)q+1 + (t − a)q+1 q
k|f |k[a,b],p
if
p > 1, p1 + 1q = 1
≤L×
q
+
1





and f ∈ Lp ([a, b] ; X) ;




1
a + b


(b − a) + t −
k|f |k[a,b],1
2
2 a
for any t ∈ [a, b].
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
in (2.2) and (2.3), then we
Remark 2.3. If we choose t = a+b
2
inequalities:
Z b
Z b
a
+
b
B
(2.4)
f (s) ds −
B (s) f (s) ds
2
a
a
α
Z b
a + b ≤H
s − 2 kf (s)k ds
a

α+1
1
k|f |k[a,b],∞
if

α (α+1) (b − a)

2




α+ 1q
1
k|f |k[a,b],p if
1 (b − a)
≤H×
2α (qα+1) q



and


 1 (b − a)α k|f |k
[a,b],1
2α
5
get the following midpoint
f ∈ L∞ ([a, b] ; X) ;
p > 1,
1
p
+
1
q
=1
f ∈ Lp ([a, b] ; X) ;
and
(2.5)
Z b
Z b
a
+
b
B
f
(s)
ds
−
B
(s)
f
(s)
ds
2
a
a
Z b
s − a + b kf (s)k ds
≤L
2 a

1

(b − a)2 k|f |k[a,b],∞
if
f ∈ L∞ ([a, b] ; X) ;

4




1+ 1q
1
k|f |k[a,b],p if
p > 1, p1 + 1q = 1
1 (b − a)
≤L×
q
2(q+1)



and f ∈ Lp ([a, b] ; X) ;


 1 (b − a) k|f |k
[a,b],1
2
respectively.
Rb
Remark 2.4. Consider the function Ψα : [a, b] → R, Ψα (t) := a |t − s|α kf (s)k ds, α ∈
(0, 1). If f is continuous on [a, b], then Ψα is differentiable and
Z t
Z b
dΨα (t)
d
α
α
=
(t − s) kf (s)k ds +
(s − t) kf (s)k ds
dt
dt a
t
Z t
Z b
kf (s)k
kf (s)k
= α
1−α ds −
1−α ds .
t (s − t)
a (t − s)
If t0 ∈ (a, b) is such that
Z
a
t0
kf (s)k
ds =
(t0 − s)1−α
Z
b
t0
kf (s)k
ds
(s − t0 )1−α
and Ψ0 (·) is negative on (a, t0 ) and positive on (t0 , b) , then the best inequality we can get in the
first part of (2.2) is the following one
Z b
Z b
Z b
B (t0 )
(2.6)
f (s) ds −
B (s) f (s) ds
|t0 − s|α kf (s)k ds.
≤H
a
a
a
If α = 1, then, for
Z
b
|t − s| kf (s)k ds,
Ψ (t) :=
a
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N.S. BARNETT , C. B U ŞE , P. C ERONE ,
AND
S.S. D RAGOMIR
we have
Z t
Z b
dΨ (t)
=
kf (s)k ds −
kf (s)k ds, t ∈ (a, b) ,
dt
a
t
d2 Ψ (t)
= 2 kf (t)k ≥ 0, t ∈ (a, b) ,
dt2
which shows that Ψ is convex on (a, b).
If tm ∈ (a, b) is such that
Z tm
Z
kf (s)k ds =
a
b
kf (s)k ds,
tm
then the best inequality we can get from the first part of (2.3) is
Z b
Z b
Z b
(2.7)
f (s) ds −
B (s) f (s) ds ≤ L
sgn (s − tm ) s kf (s)k ds.
B (tm )
a
a
a
Indeed, as
Z b
inf
|t − s| kf (s)k ds
t∈[a,b]
a
Z
b
|tm − s| kf (s)k ds
=
a
Z
tm
Z
b
(tm − s) kf (s)k ds +
=
a
Z
tm
Z
b
kf (s)k ds −
= tm
a
Z
(s − tm ) kf (s)k ds
tm
Z
Z
b
kf (s)k ds +
tm
b
Z
tm
s kf (s)k ds
s kf (s)k ds −
a
tm
tm
s kf (s)k ds
s kf (s)k ds −
=
a
tm
Z b
sgn (s − tm ) s kf (s)k ds,
=
a
then the best inequality we can get from the first part of (2.3) is obtained for t = tm ∈ (a, b).
We recall that a function F : [a, b] → L (X) is said to be strongly continuous if for all
x ∈ X, the maps s 7→ F (s) x : [a, b] → X are continuous on [a, b]. In this case the function
s 7→ kB (s)k : [a, b] → R+ is (Lebesgue) measurable and bounded ([6]). The linear operator
Rb
Rb
L = a F (s) ds (defined by Lx := a F (s) xds for all x ∈ X) is bounded, because
Z b
kLxk ≤
kF (s)k ds · kxk for all x ∈ X.
a
In a similar manner to Theorem 2.1, we may prove the following result as well.
Theorem 2.5. Assume that f : [a, b] → X is Hölder continuous, i.e.,
(2.8)
kf (t) − f (s)k ≤ K |t − s|β for all t, s ∈ [a, b] ,
where K > 0 and β ∈ (0, 1].
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
7
If B : [a, b] → L (X) is strongly continuous on [a, b] , then we have the inequality:
Z b
Z b
(2.9) B
(s)
ds
f
(t)
−
B
(s)
f
(s)
ds
a
a
Z
b
|t − s|β kB (s)k ds
a

(b−t)β+1 +(t−a)β+1

k|B|k[a,b],∞

β+1




 h
i1
(b−t)qβ+1 +(t−a)qβ+1 q
≤K×
k|B|k[a,b],p
qβ+1






 1 (b − a) + t − a+b β k|B|k
≤K
2
2
if
kB (·)k ∈ L∞ ([a, b] ; R+ ) ;
if
p > 1, p1 + 1q = 1
and kB (·)k ∈ Lp ([a, b] ; R+ ) ;
[a,b],1
for any t ∈ [a, b].
The following corollary holds.
Rb
Corollary 2.6. Assume that f and B are as in Theorem 2.5. If, in addition, a B (s) ds is
invertible in L (X) , then we have the inequality:
Z b
−1 Z b
(2.10) f (t) −
B (s) ds
B (s) f (s) ds
a
a
Z
−1 Z b
b
≤K
B (s) ds
|t − s|β kB (s)k ds
a
a
for any t ∈ [a, b].
Remark 2.7. It is obvious that the inequality (2.10) contains as a particular case what is the so
called Ostrowski’s inequality for weighted integrals (see (1.2)).
3. I NEQUALITIES FOR L INEAR O PERATORS
Let 0 ≤ a < b < ∞ and A ∈ L (X). We recall that the operatorial norm of A is given by
kAk = sup {kAxk : kxk ≤ 1} .
The resolvent set of A (denoted by ρ (A)) is the set of all complex scalars λ for which λI − A
is an invertible operator. Here I is the identity operator in L (X). The complementary set of
ρ (A) in the complex plane, denoted
is the spectrum of A. It is known that σ (A) is
P by σ (A),
(tA)n
converges absolutely and locally uniformly for
a compact set in C. The series
n≥0 n!
t ∈ R. If we denote by etA its sum, then
tA e ≤ e|t|kAk , t ∈ R.
Proposition 3.1. Let X be a real or complex Banach space, A ∈ L (X) and β be a non-null
real number such that −β ∈ ρ (A). Then for all 0 ≤ a < b < ∞ and each s ∈ [a, b], we have
βb
e − eβa sA
b(βI+A)
−1
a(βI+A) (3.1) ·
e
−
(βI
+
A)
e
−
e
β
"
2 #
1
a
+
b
≤ kAk ebkAk ·
(b − a)2 + s −
· max eβb , eβa .
4
2
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8
N.S. BARNETT , C. B U ŞE , P. C ERONE ,
AND
S.S. D RAGOMIR
Proof. We apply the second inequality from Corollary 2.2 in the following particular case.
B (τ ) := eτ A , f (τ ) = eβτ x, τ ∈ [a, b] , x ∈ X.
For all ξ, η ∈ [a, b] there exists an α between ξ and η such that
∞
X
(ξ n − η n ) n kB (ξ) − B (η)k = A n!
n=1
∞
X
(αA)n = (ξ − η) A
n! n=0
≤ kAk eαA · |ξ − η|
≤ kAk ebkAk · |ξ − η| .
The function τ 7→ eτ A is thus Lipschitzian on [a, b] with the constant L := kAk eb·kAk . On the
other hand we have
Z b
Z b
τA
βτ
e
e x dτ =
eτ A eβτ Ix dτ
a
a
Z b
=
eτ (A+βI) xdτ
a
= (A + βI)−1 eb(A+βI) − ea(A+βI) x,
and
|kf k|[a,b],∞ = sup eτ β x = max eβb , eβa · kxk .
τ ∈[a,b]
Placing all the above results in the second inequality from (2.3) and taking the supremum for
all x ∈ X, we will obtain the desired inequality (3.1).
Remark 3.2. Let A ∈ L (X) such that 0 ∈ ρ (A). Taking the limit as β → 0 in (3.1), we get
the inequality
"
2 #
1
a
+
b
(b − a) esA − A−1 ebA − eaA ≤ kAk ebkAk ·
(b − a)2 + s −
,
4
2
where a, b and s are as in Proposition 3.1.
Proposition 3.3. Let A ∈ L (X) be an invertible operator, t ≥ 0 and 0 ≤ s ≤ t. Then the
following inequality holds:
2
t
2s3 + 2t3 − 3st2
−2
≤
(3.2)
sin
(sA)
−
A
[sin
(tA)
−
tA
cos
(tA)]
kAk .
2
6
In particular, if X = R, A = 1 and s = 0 it follows the scalar inequality
t3
|sin t − t cos t| ≤ , for all t ≥ 0.
3
Proof. We apply the inequality from (2.3) in the following particular case:
B (τ ) = sin (τ A) :=
∞
X
n=0
(−1)n
(τ A)2n+1
, τ ≥ 0,
(2n + 1)!
and
(3.3)
f (τ ) = τ · x, for fixed x ∈ X.
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
9
For each ξ, η ∈ [0, t] , we have
!
∞
2n
X
n (ξ − η) α
2n kB (ξ) − B (η)k = A
(−1)
A
(2n)!
n=0
≤ kAk |ξ − η| · kcos (αA)k
≤ kAk |ξ − η| ,
where α is a real number between ξ and η, i.e., the function τ 7−→ B (τ ) : R+ → L (X) is
kAk −Lipschitzian.
Moreover, it is easy to see that
Z
t
B (τ ) f (τ ) dτ = A−2 [sin (tA) − tA cos (tA)] x
0
and
Z
t
|s − τ | |f (τ )| dτ =
(3.4)
0
2s3 + 2t3 − 3st2
kxk .
6
Applying the first inequality from (2.3) and taking the supremum for x ∈ X with kxk ≤ 1, we
get (3.2).
4. Q UADRATURE F ORMULAE
Consider the division of the interval [a, b] given by
In : a = t0 < t1 < · · · < tn−1 < tn = b
(4.1)
and hi := ti+1 − ti , ν (h) := max hi . For the intermediate points ξ := (ξ0 , . . . , ξn−1 ) with
i=0,n−1
ξi ∈ [ti , ti+1 ] , i = 0, n − 1, define the sum
Sn(1)
(4.2)
(B, f ; In , ξ) :=
n−1
X
Z
ti+1
B (ξi )
i=0
f (s) ds.
ti
Then we may state the following result in approximating the integral
Z
b
B (s) f (s) ds,
a
based on Theorem 2.1.
Theorem 4.1. Assume that B : [a, b] → L (X) is Hölder continuous on [a, b], i.e., it satisfies
the condition (2.1) and f : [a, b] → X is Bochner integrable on [a, b]. Then we have the
representation
Z
(4.3)
b
B (s) f (s) ds = Sn(1) (B, f ; In , ξ) + Rn(1) (B, f ; In , ξ) ,
a
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10
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
(1)
(1)
where Sn (B, f ; In , ξ) is as given by (4.2) and the remainder Rn (B, f ; In , ξ) satisfies the
estimate
(1)
Rn (B, f ; In , ξ)

n−1
P
1
α+1
α+1 

k|f
|k
(t
−
ξ
)
+
(ξ
−
t
)
i+1
i
i
i

[a,b],∞

α+1

i=0



n−1
1


q
P
1

qα+1
qα+1

,
(ti+1 − ξi )
+ (ξi − ti )
1 k|f |k[a,b],p
≤H×
i=0
(qα + 1) q


p > 1, p1 + 1q = 1





α

 1
t
+
t

i+1
i

k|f |k

ν (h) + max ξi −
[a,b],1
2
2
i=0,n−1

n−1
P α+1
1


k|f
|k
hi

[a,b],∞

α
+
1

i=0


n−1
1

P qα+1 q
1
≤H×
hi
,
p > 1,
1 k|f |k[a,b],p

 (qα + 1) q
i=0

α



1
t
+
t
i+1
i

k|f |k

ν (h) + max ξi −
[a,b],1
2
2
i=0,n−1
1
p
+
1
q
=1

1


k|f |k[a,b],∞ [ν (h)]α


α+1



1
(b − a) q
≤H×
α

1 k|f |k[a,b],p [ν (h)]


q

(qα + 1)


 k|f |k
[ν (h)]α .
[a,b],1
Proof. Applying Theorem 4.1 on [xi , xi+1 ] i = 0, n − 1 , we may write that
Z
ti+1
ti
ti+1
B (s) f (s) ds − B (ξi )
f (s) ds
ti
#
 "
α+1
α+1

(t
−
ξ
)
+
(ξ
−
t
)
i+1
i
i
i


k|f |k[ti ,ti+1 ],∞


α+1







#1
 "
qα+1
qα+1 q
(t
−
ξ
)
+
(ξ
−
t
)
i+1
i
i
i
≤H×
k|f |k[ti ,ti+1 ],p


qα
+
1






α



1
ti+1 + ti 

(ti+1 − ti ) + ξi −
k|f |k[ti ,ti+1 ],1 .
2
2
Z
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
11
Summing over i from 0 to n − 1 and using the generalised triangle inequality we get
(1)
Rn (B, f ; In , ξ)
Z ti+1
n−1 Z ti+1
X
≤
B
(s)
f
(s)
ds
−
B
(ξ
)
f
(s)
ds
i
ti
i=0
ti

P
1 n−1


(ti+1 − ξi )α+1 + (ξi − ti )α+1 k|f |k[ti ,ti+1 ],∞



α + 1 i=0



n−1
1q

P
1
qα+1
qα+1
≤H×
k|f |k[ti ,ti+1 ],p
(ti+1 − ξi )
+ (ξi − ti )
1

q

i=0
(qα
+
1)


α

 n−1
P 1
ti+1 + ti 

hi + ξi −

k|f |k[ti ,ti+1 ],1 .
2
2
i=0
Now, observe that
n−1
X
(ti+1 − ξi )α+1 + (ξi − ti )α+1 k|f |k[ti ,ti+1 ],∞
i=0
n−1
X
≤ k|f |k[a,b],∞
(ti+1 − ξi )α+1 + (ξi − ti )α+1
i=0
≤ k|f |k[a,b],∞
n−1
X
hα+1
i
i=0
≤ k|f |k[a,b],∞ (b − a) [ν (h)]α .
Using the discrete Hölder inequality, we may write that
" n−1
# 1q
X
(ti+1 − ξi )qα+1 + (ξi − ti )qα+1 k|f |k[ti ,ti+1 ],p
i=0
" n−1
# 1q " n−1
# p1
X X
1 q
≤
(ti+1 − ξi )qα+1 + (ξi − ti )qα+1 q
×
k|f |kp[ti ,ti+1 ],p
i=0
i=0
( n−1
) 1q Z
p1
b
X
p
qα+1
qα+1
=
(ti+1 − ξi )
+ (ξi − ti )
kf (t)k ds
a
i=0
≤
n−1
X
! 1q
hqα+1
i
k|f |kp[a,b],p
i=0
1
≤ (b − a) q k|f |k[a,b],p [ν (h)]α .
Finally, we have
n−1 X
1
i=0
α
t
+
t
i+1
i
k|f |k
hi + ξi −
[ti ,ti+1 ],1
2
2
α
1
ti+1 + ti ≤
max hi + max ξi −
k|f |k[a,b],1
2 i=0,n−1
2
i=0,n−1
≤ [ν (h)]α k|f |k[a,b],1
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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12
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
and the theorem is proved.
The following corollary holds.
Corollary 4.2. If B is Lipschitzian with the constant L, then we have the representation (4.3)
(1)
and the remainder Rn (B, f ; In , ξ) satisfies the estimates:
(4.4)
(1)
Rn (B, f ; In , ξ)
"

2 #
n−1
n−1

P
P
1
t
+
t
i+1
i


k|f |k[a,b],∞
h2i +
ξi −


4
2

i=0
i=0







n−1
1q
P
1
q+1
q+1 ≤L×
(ti+1 − ξi )
+ (ξi − ti )
,
1 k|f |k[a,b],p


q
i=0
(q
+
1)




p > 1, p1 + 1q = 1




1
ti+1 + ti 

ν (h) + max ξi −

k|f |k[a,b],1
2
2
i=0,n−1

n−1
P 2

 1 k|f |k
hi

[a,b],∞


2
i=0






n−1
1

P q+1 q
1
≤L×
hi
1 k|f |k[a,b],p

q
i=0

(q
+
1)







1
t
+
t

i+1
i
k|f |k

ν (h) + max ξi −

[a,b],1
2
2
i=0,n−1

1


k|f |k[a,b],∞ (b − a) ν (h)


2





1
(b − a) q
≤L×
1 k|f |k[a,b],p ν (h)


q

(q
+
1)






k|f |k[a,b],1 ν (h) .
The second possibility we have for approximating the integral
in the following theorem based on Theorem 2.5.
Rb
a
B (s) f (s) ds is embodied
Theorem 4.3. Assume that f : [a, b] → X is Hölder continuous, i.e., the condition (2.8) holds.
If B : [a, b] → L (X) is strongly continuous on [a, b], then we have the representation:
Z
(4.5)
b
B (s) f (s) ds = Sn(2) (B, f ; In , ξ) + Rn(2) (B, f ; In , ξ) ,
a
where
(4.6)
Sn(2)
(B, f ; In , ξ) :=
n−1 Z
X
i=0
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
ti+1
B (s) ds f (ξi )
ti
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
13
(2)
and the remainder Rn (B, f ; In , ξ) satisfies the estimate:
(4.7)
(2)
Rn (B, f ; In , ξ)

i
n−1
Ph
1
β+1
β+1


k|B|k
(t
−
ξ
)
+
(ξ
−
t
)
i+1
i
i
i

[a,b],∞

β+1

i=0





n−1 h

i 1q


P
1
qβ+1
qβ+1

,
(ti+1 − ξi )
+ (ξi − ti )
1 k|B|k[a,b],p
≤K×
q
i=0
(qβ
+
1)



p > 1, p1 + 1q = 1






β



1
ti+1 + ti 

ν (h) + max ξi −
k|B|k[a,b],1
2
2
i=0,n−1

n−1
P β+1
1


k|B|k
hi

[a,b],∞

β+1

i=0





n−1
1


P qβ+1 q
1
hi
,
p > 1, p1 +
≤K×
1 k|B|k[a,b],p

q
i=0
(qβ + 1)






β


1
t
+
t

i+1
i

k|B|k

ν (h) + max ξi −
[a,b],1
2
2
i=0,n−1

1


k|B|k[a,b],∞ (b − a) [ν (h)]β


β+1






1
(b − a) q
β
≤K×
p > 1, p1 + 1q = 1
1 k|B|k[a,b],p [ν (h)] ,



(qβ + 1) q






β
 k|B|k
[a,b],1 [ν (h)] .
1
q
= 1;
If we consider the quadrature
(4.8)
Mn(1)
(B, f ; In ) :=
n−1
X
i=0
B
ti + ti+1
2
Z
ti+1
f (s) ds,
ti
then we have the representation
Z
(4.9)
b
B (s) f (s) ds = Mn(1) (B, f ; In ) + Rn(1) (B, f ; In ) ,
a
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14
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
(1)
and the remainder Rn (B, f ; In ) satisfies the estimate:
(1)
Rn (B, f ; In )
(4.10)

n−1
P α+1

1


k|f
|k
hi
α

[a,b],∞
2 (α+1)


i=0




n−1
1
≤H×
P qα+1 q
1

, p > 1,
hi
1 k|f |k[a,b],p


2α (qα+1) q

i=0





 1 [ν (h)]α k|f |k
2α
≤H×















1
q
1
p
+
1
q
= 1;
= 1;
[a,b],1
1
2α (α+1)
(b − a) k|f |k[a,b],∞ [ν (h)]α
1
(b−a) q
1
2α (qα+1) q
1
2α
k|f |k[a,b],p [ν (h)]α ,
p > 1,
1
p
+
k|f |k[a,b],1 [ν (h)]α ,
provided that B and f are as in Theorem 4.1.
Now, if we consider the quadrature
n−1 Z
X
(2)
(4.11)
Mn (B, f ; In ) :=
i=0
ti+1
ti
ti + ti+1
,
B (s) ds f
2
then we also have
Z
(4.12)
b
B (s) f (s) ds = Mn(2) (B, f ; In ) + Rn(2) (B, f ; In ) ,
a
and in this case the remainder satisfies the bound
(2)
Rn (B, f ; In )
(4.13)

n−1
P β+1
1


k|B|k
hi

[a,b],∞
β (β + 1)

2

i=0



n−1
1
P qβ+1 q
1
≤K×
hi
, p > 1, p1 + 1q = 1;
1 k|B|k[a,b],p


β
q
i=0

2 (qβ + 1)




1

[ν (h)]β k|B|k[a,b],1
β
2

1


(b − a) k|B|k[a,b],∞ [ν (h)]β

β (β + 1)

2




1
(b − a) q
β
≤K×
p > 1, p1 + 1q = 1;
1 k|B|k[a,b],p [ν (h)] ,


β
q

2 (qβ + 1)




β
 1 k|B|k
[a,b],1 [ν (h)] ,
2β
provided B and f satisfy the hypothesis of Theorem 4.3.
Now, if we consider the equidistant partitioning of [a, b],
b−a
En : ti := a +
· i, i = 0, n,
n
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
15
(1)
then Mn (B, f ; En ) becomes
(4.14)
Mn(1)
(B, f ) :=
n−1
X
i=0
1
B a+ i+
2
b−a
·
n
Z
a+ b−a
·(i+1)
n
f (s) ds
·i
a+ b−a
n
and then
Z
(4.15)
b
B (s) f (s) ds = Mn(1) (B, f ) + Rn(1) (B, f ) ,
a
where the remainder satisfies the bound

(b − a)α+1



k|f |k[a,b],∞


2α (α + 1) nα






1
(1)
(b − a)α+ q
(4.16)
Rn (B, f ) ≤ H ×
k|f |k[a,b],p , p > 1,


2α (α + 1) nα





α



 (b − a) k|f |k
[a,b],1 .
2α nα
Also, we have
Z b
(4.17)
B (s) f (s) ds = Mn(2) (B, f ) + Rn(2) (B, f ) ,
1
p
+
1
q
= 1;
a
where
Mn(2) (B, f ) :=
Z
n−1
X
i=0
a+ b−a
·(i+1)
n
a+ b−a
·i
n
! b−a
1
B (s) ds f a + i +
·
,
2
n
(2)
and the remainder Rn (B, f ) satisfies the estimate

(b − a)β+1


kBk[a,b],∞



2β (β + 1) nβ






β+ 1q
(2)
(b
−
a)
(4.18)
Rn (B, f ) ≤ K ×
k|B|k[a,b],p ,

β
β

 2 (β + 1) n





β


 (b − a) k|B|k
[a,b],1 .
2β nβ
p > 1,
1
p
+
1
q
= 1;
5. A PPLICATION FOR D IFFERENTIAL E QUATIONS IN BANACH S PACES
We recall that a family of operators U = {U (t, s) : t ≥ s} ⊂ L (X) with t, s ∈ R or t, s ∈
R+ is called an evolution family if:
(i) U (t, t) = I and U (t, s) U (s, τ ) = U (t, τ ) for all t ≥ s ≥ τ ; and
(ii) for each x ∈ X, the function (t, s) 7−→ U (t, s) x is continuous for t ≥ s.
Here I is the identity operator in L (X) .
An evolution family {U (t, s) : t ≥ s} is said to be exponentially bounded if, in addition,
(iii) there exist the constants M ≥ 1 and ω > 0 such that
(5.1)
kU (t, s)k ≤ M eω(t−s) , t ≥ s.
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
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16
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
Evolution families appear as solutions for abstract Cauchy problems of the form
(5.2)
u̇ (t) = A (t) u (t) , u (s) = xs , xs ∈ D (A (s)) , t ≥ s, t, s ∈ R (or R+ ),
where the domain D (A (s)) of the linear operator A (s) is assumed to be dense in X. An evolution family is said to solve the abstract Cauchy problem (5.2) if for each s ∈ R there exists a
dense subset Ys ⊆ D (A (s)) such that for each xs ∈ Ys the function
t 7−→ u (t) := U (t, s) xs : [s, ∞) → X,
is differentiable, u (t) ∈ D (A (t)) for all t ≥ s and
d
u (t) = A (t) u (t) , t ≥ s.
dt
This later definition can be found in [15]. In this definition the operators A (t) can be unbounded. The Cauchy problem (5.2) is called well-posed if there exists an evolution family
{U (t, s) : t ≥ s} which solves it.
It is known that the well-posedness of (5.2) can be destroyed by a bounded and continuous
perturbation [13]. Let f : R →X be a locally integrable function. Consider the inhomogeneous
Cauchy problem:
(5.3)
u̇ (t) = A (t) u (t) + f (t) , u (s) = xs ∈ X, t ≥ s, t, s ∈ R (or R+ ).
A continuous function t 7−→ u (t) : [s, ∞) → X is said to a mild solution of the Cauchy
problem (5.3) if u (s) = xs and there exists an evolution family {U (t, τ ) : t ≥ τ } such that
Z t
(5.4)
u (t) = U (t, s) xs +
U (t, τ ) f (τ ) dτ, t ≥ s, xs ∈ X, t, s ∈ R (or R+ ).
s
The following theorem holds.
Theorem 5.1. Let U = {U (ν, η) : ν ≥ η} ⊂ L (X) be an evolution family and f : R →X be
a locally Bochner integrable and locally bounded function. We assume that for all ν ∈ R (or
R+ ) the function η 7−→ U (ν, η) : [ν, ∞) → L (X) is locally Hölder continuous (i.e. for all
a, b ≥ ν, a < b, there exist α ∈ (0, 1] and H > 0 such that
kU (ν, t) − U (ν, s)k ≤ H |t − s|α , for all t, s ∈ [a, b] ).
We use the notations in Section 4 for a = 0 and b = t > 0. The map u (·) from (5.4) can be
represented as
Z ti+1
n−1
X
f (s) ds + Rn(1) (U, f, In , ξ)
(5.5)
u (t) = U (t, 0) x0 +
U (t, ξi )
i=0
ti
(1)
where the remainder Rn (U, f, In , ξ) satisfies the estimate
n−1
(1)
Rn (U, f, In , ξ) ≤
X
H
|kf k|[0,t],∞
(ti+1 − ξi )α+1 + (ξi − ti )α+1 .
α+1
i=0
Proof. It follows by representation (4.3) and the first estimate after it.
Moreover, if n is a natural number, i ∈ {0, . . . , n}, ti := t·i
and ξi := (2i+1)t
, then
n
2n
Z t·(i+1)
n−1
X
n
(2i + 1) t
(5.6)
u (t) = U (t, 0) x0 +
U t,
f (s) ds + Rn(1)
t·i
2n
i=0
n
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
17
(1)
and the remainder Rn satisfies the estimate
α+1
(1) Rn ≤ H · t
(5.7)
|kf k|[0,t],∞ .
α + 1 2α · nα
The following theorem also holds.
Theorem 5.2. Let U = {U (ν, η) : ν ≥ η} ⊂ L (X) be an exponentially bounded evolution
family of bounded linear operators acting on the Banach space X and f : R →X be a locally
Hölder continuous function, i.e., for all a, b ∈ R, a < b there exist β ∈ (0, 1] and K > 0 such
that (2.8) holds. We use the notations of Section 4 for a = 0 and b = t > 0. The map u (·) from
(5.4) can be represented as
n−1 Z ti+1
X
(5.8)
u (t) = U (t, 0) x0 +
U (t, τ ) dτ f (ξi ) + Rn(2) (U, f, In , ξ)
i=0
ti
(2)
where the remainder Rn (U, f, In , ξ) satisfies the estimate
n−1 h
i
X
(2)
β+1
β+1
Rn (U, f, In , ξ) ≤ KM eωt
(ti+1 − ξi )
+ (ξi − ti )
.
β+1
i=0
Proof. It follows from the first estimate in (4.7) for B (s) := U (t, s), using the fact that
|kB (·)k|[0,t],∞ = sup kU (t, τ )k ≤ sup M eω(t−τ ) ≤ M eωt .
τ ∈[0,t]
τ ∈[0,t]
Moreover, if n is a natural number, i ∈ {0, . . . , n} , ti := t·i
and ξi := (2i+1)t
then
n
2n
!
Z t·(i+1)
n−1
X
n
(2i + 1) t
+ Rn(2)
(5.9)
u (t) = U (t, 0) x0 +
U (t, τ ) dτ f
t·i
2n
i=0
n
(2)
and the remainder Rn satisfies the estimate
β+1
(2) Rn ≤ KM eωt · t
.
β+1
2β · nβ
(5.10)
6. S OME N UMERICAL E XAMPLES
p
1. Let X = R2 , x = (ξ, η) ∈ R2 , kxk2 = ξ r + η 2 . We consider the linear 2-dimensional
system


u̇1 (t) = −1 − sin2 t u1 (t) + (−1 + sin t cos t) u2 (t) + e−t ;




(6.1)
u̇2 (t) = (1 + sin t cos t) u1 (t) + (−1 − cos2 t) u2 (t) + e−2t ;




 u (0) = u (0) = 0.
1
2
If we denote

−1 − sin2 t
A (t) := 
1 + sin t cos t
J. Inequal. Pure and Appl. Math., 3(1) Art. 12, 2002
−1 + sin t cos t
−1 − cos2 t

,
f (t) = e−t , e−2t , x = (0, 0)
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18
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
ξ
and we identify (ξ, η) with
η
family associated with A (t) is
, then the above system is a Cauchy problem. The evolution
U (t, s) = P (t) P −1 (s) , t ≥ s, t, s ∈ R,
where

(6.2)
P (t) = 
e−t cos t
−t
−e
sin t
e−2t sin t
e
−2t

 , t ∈ R.
cos t
The exact solution of the system (6.1) is u = (u1 , u2 ) , where
u1 (t) = e−t cos t E1 (t) + e−2t sin t E2 (t)
u2 (t) = − e−t sin t E1 (t) + e−2t cos t E2 (t) , t ∈ R,
and
1
1
E1 (t) = sin t + e−t (cos t + sin t) − ,
2
2
1 t
1
E2 (t) = sin t + e (sin t − cos t) + ,
2
2
see [2, Section 4] for details. The function t 7→ A (t) is bounded on R and therefore there exist
M ≥ 1 and ω > 0
kU (t, s)k ≤ M eω|t−s| , for all t, s ∈ R.
Let ξ ≥ 0 be fixed and t, s ≥ ξ. Then there exists a real number µ between t and s such that
kU (ξ, t) − U (ξ, s)k = |t − s| kU (ξ, µ) A (µ)k ≤ M eωµ |kA (·)k|∞ · |t − s| ,
that is, the function η 7→ U (ξ, η) is locally Lipschitz continuous on [ξ, ∞).
Using (6.2), it follows


a11 (t, s)
a12 (t, s)
,
U (t, s) = 
a21 (t, s)
a22 (t, s)
where
a11 (t, s) = e(s−t) cos t cos s + e2(s−t) sin t sin s;
1
a12 (t, s) = −e(s−t) cos t sin s + e2(s−t) sin t cos s;
2
(s−t)
a21 (t, s) = −e
sin t cos s + e2(s−t) cos t sin s;
1
a22 (t, s) = e(s−t) sin t sin s + e2(s−t) cos t cos s.
2
Then from (5.6) we obtain the following approximating formula for u (·) :
n−1 X
(2i + 1) t − t(i+1)
− ti
n
n
u1 (t) = −
a11 t,
e
−e
2n
i=0
1
(2i + 1) t − 2t(i+1)
(1)
− 2ti
n
n
+ a12 t,
e
−e
+ R1,n
2
2n
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W EIGHTED O STROWSKI T YPE I NEQUALITIES
19
(1) (1) Figure 6.1: The behaviour of the error εn (t) := R1,n , R2,n for n = 200.
2
and
n−1 X
ti
(2i + 1) t − t(i+1)
u2 (t) = −
a21 t,
e n − e− n
2n
i=0
2ti
(2i + 1) t − 2t(i+1)
1
(1)
−
+ a22 t,
+ R2,n ,
e n −e n
2
2n
(1)
(1)
(1)
where the remainder Rn = R1,n , R2,n satisfies the estimate (5.7) with α = 1, H =
M eωt |kA (·)k|∞ and |kf k|[0,t],∞ ≤ 2.
(1) (1) Figure 6.1 contains the behaviour of the error εn (t) := R1,n , R2,n for n = 200.
2
t+1
2. Let X = R and U (t, s) := s+1
, t ≥ s ≥ 0. It is clear that the family {U (t, s) : t ≥ s ≥ 0} ⊂
L (R) is an exponentially bounded evolution family which solves the Cauchy problem
u̇ (t) =
1
u (t) , u (s) = xs ∈ R, t ≥ s ≥ 0.
t+1
Consider the inhomogeneous Cauchy problem

1
u (t) + cos [ln (t + 1)] , t ≥ 0
 u̇ (t) = t+1
(6.3)

u (0) = 0.
The solution of (6.3) is given by
Z t
t+1
u (t) =
cos (ln (τ + 1)) dτ = (t + 1) sin [ln (t + 1)] , t ≥ 0.
0 τ +1
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20
N.S. BARNETT , C. B U ŞE , P. C ERONE , AND S.S. D RAGOMIR
Figure 6.2: The behaviour of the error εn (t) := |Rn | for n = 400.
From (5.9) we obtain the approximating formula for u (·) as,
n−1
X
n + ti + t
(2i + 1) t
u (t) = (t + 1)
ln
cos ln 1 +
+ Rn ,
n + ti
2n
i=0
where Rn satisfies the estimate (5.10) with K = M = ω = β = 1. Indeed,
t+1
≤ et , for all t ≥ s ≥ 0
s+1
and
1
|cos [ln (t + 1)] − cos [ln (s + 1)]| = |t − s| sin [ln (c + 1)] ≤ |t − s|
c+1
for all t ≥ s ≥ 0, where c is some real number between s and t.
The following Figure 6.2 contains the behaviour of the error εn (t) := |Rn | for n = 400.
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