Volume 8 (2007), Issue 4, Article 93, 13 pp.
ON AN INTEGRATION-BY-PARTS FORMULA FOR MEASURES
A. ČIVLJAK, LJ. DEDIĆ, AND M. MATIĆ
A MERICAN C OLLEGE OF M ANAGEMENT AND T ECHNOLOGY
ROCHESTER I NSTITUTE OF T ECHNOLOGY
D ON F RANA B ULICA 6, 20000 D UBROVNIK
C ROATIA
acivljak@acmt.hr
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION
U NIVERSITY OF S PLIT
T ESLINA 12, 21000 S PLIT
C ROATIA
ljuban@pmfst.hr
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION
U NIVERSITY OF S PLIT
T ESLINA 12, 21000 S PLIT
C ROATIA
mmatic@pmfst.hr
Received 21 June, 2007; accepted 15 November, 2007
Communicated by W.S. Cheung
A BSTRACT. An integration-by-parts formula, involving finite Borel measures supported by intervals on real line, is proved. Some applications to Ostrowski-type and Grüss-type inequalities
are presented.
Key words and phrases: Integration-by-parts formula, Harmonic sequences, Inequalities.
2000 Mathematics Subject Classification. 26D15, 26D20, 26D99.
1. I NTRODUCTION
In the paper [4], S.S. Dragomir introduced the notion of a w0 -Appell type sequence of functions as a sequence w0 , w1 , . . . , wn , for n ≥ 1, of real absolutely continuous functions defined
on [a, b], such that
wk0 = wk−1 , a.e. on [a, b],
210-07
k = 1, . . . , n.
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A. Č IVLJAK , L J . D EDI Ć , AND M. M ATI Ć
For such a sequence the author proved a generalisation of Mitrinović-Pečarić integration-byparts formula
Z b
(1.1)
w0 (t)g(t)dt = An + Bn ,
a
where
An =
n
X
(−1)k−1 wk (b)g (k−1) (b) − wk (a)g (k−1) (a)
k=1
and
n
b
Z
wn (t)g (n) (t)dt,
Bn = (−1)
a
(n−1)
for every g : [a, b]→R such that g
is absolutely continuous on [a, b] and wn g (n) ∈ L1 [a, b].
Using identity (1.1) the author proved the following inequality
Z b
≤ kwn k kg (n) kq ,
(1.2)
w
(t)g(t)dt
−
A
0
n
p
a
for wn ∈ Lp [a, b], g (n) ∈ Lp [a, b], where p, q ∈ [1, ∞] and 1/p + 1/q = 1, giving explicitly
some interesting special cases. For some similar inequalities, see also [5], [6] and [7]. The aim
of this paper is to give a generalization of the integration-by-parts formula (1.1), by replacing
the w0 -Appell type sequence of functions by a more general sequence of functions, and to
generalize inequality (1.2), as well as to prove some related inequalities.
2. I NTEGRATION - BY- PARTS F ORMULA FOR M EASURES
For a, b ∈ R, a < b, let C[a, b] be the Banach space of all continuous functions f : [a, b]→R
with the max norm, and M [a, b] the Banach space of all real Borel measures on [a, b] with the
total variation norm. For µ ∈ M [a, b] define the function µ̌n : [a, b]→R, n ≥ 1, by
Z
1
µ̌n (t) =
(t − s)n−1 dµ(s).
(n − 1)! [a,t]
Note that
1
µ̌n (t) =
(n − 2)!
Z
t
(t − s)n−2 µ̌1 (s)ds,
n≥2
a
and
(t − a)n−1
kµk , t ∈ [a, b], n ≥ 1.
(n − 1)!
The function µ̌n is differentiable, µ̌0n (t) = µ̌n−1 (t) and µ̌n (a) = 0, for every n ≥ 2, while for
n=1
Z
µ̌1 (t) =
dµ(s) = µ([a, t]),
|µ̌n (t)| ≤
[a,t]
which means that µ̌1 (t) is equal to the distribution function of µ. A sequence of functions
Pn : [a, b] → R, n ≥ 1, is called a µ-harmonic sequence of functions on [a, b] if
Pn0 (t) = Pn−1 (t), n ≥ 2;
P1 (t) = c + µ̌1 (t),
t ∈ [a, b],
for some c ∈ R. The sequence (µ̌n , n ≥ 1) is an example of a µ-harmonic sequence of functions
on [a, b]. The notion of a µ-harmonic sequence of functions has been introduced in [2]. See also
[1].
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
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I NTEGRATION - BY- PARTS FORMULA
3
Remark 2.1. Let w0 : [a, b] → R be an absolutely integrable function and let µ ∈ M [a, b] be
defined by
dµ(t) = w0 (t)dt.
If (Pn , n ≥ 1) is a µ-harmonic sequence of functions on [a, b], then w0 , P1 , . . . , Pn is a w0 Appell type sequence of functions on [a, b].
For µ ∈ M [a, b] let µ = µ+ − µ− be the Jordan-Hahn decomposition of µ, where µ+ and µ−
are orthogonal and positive measures. Then we have |µ| = µ+ + µ− and
kµk = |µ| ([a, b]) = kµ+ k + kµ− k = µ+ ([a, b]) + µ− ([a, b]).
The measure µ ∈ M [a, b] is said to be balanced if µ([a, b]) = 0. This is equivalent to
1
kµ+ k = kµ− k = kµk .
2
Measure µ ∈ M [a, b] is called n-balanced if µ̌n (b) = 0. We see that a 1-balanced measure is
the same as a balanced measure. We also write
Z
mk (µ) =
tk dµ(t), k ≥ 0
[a,b]
for the k-th moment of µ.
Lemma 2.2. For every f ∈ C[a, b] and µ ∈ M [a, b] we have
Z
Z
f (t)dµ̌1 (t) =
f (t)dµ(t) − µ({a})f (a).
[a,b]
[a,b]
Proof. Define I, J : C[a, b] × M [a, b] → R by
Z
I(f, µ) =
f (t)dµ̌1 (t)
[a,b]
and
Z
f (t)dµ(t) − µ({a})f (a).
J(f, µ) =
[a,b]
Then I and J are continuous bilinear functionals, since
|I(f, µ)| ≤ kf k kµk ,
|J(f, µ)| ≤ 2 kf k kµk .
Let us prove that I(f, µ) = J(f, µ) for every f ∈ C[a, b] and every discrete measure µ ∈
M [a, b].
For x ∈ [a, b] let µ = δx be the Dirac measure at x, i.e. the measure defined by
R
f (t)dδx (t) = f (x).
[a,b]
If a < x ≤ b, then
(
µ̌1 (t) = δx ([a, t]) =
0, a ≤ t < x
1, x ≤ t ≤ b
and by a simple calculation we have
Z
R
I(f, δx ) =
f (t)dµ̌1 (t) = f (x) =
f (t)dδx (t) − 0
[a,b]
=
R
[a,b]
f (t)dδx (t) − δx ({a})f (a) = J(f, δx ).
[a,b]
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
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A. Č IVLJAK , L J . D EDI Ć , AND M. M ATI Ć
Similarly, if x = a, then
µ̌1 (t) = δa ([a, t]) = 1,
a≤t≤b
and by a similar calculation we have
Z
I(f, δa ) =
f (t)dµ̌1 (t) = 0 = f (a) − f (a)
[a,b]
R
=
f (t)dδa (t) − δa ({a})f (a) = J(f, δx ).
[a,b]
Therefore, for every f ∈ C[a, b] and every x ∈ [a, b] we have I(f, δx ) = J(f, δx ). Every
discrete measure µ ∈ M [a, b] has the form
X
µ=
ck δxk ,
k≥1
where (ck , k ≥ 1) is a sequence in R such that
X
|ck | < ∞,
k≥1
and {xk ; k ≥ 1} is a subset of [a, b].
By using the continuity of I and J, for every f ∈ C[a, b] and every discrete measure µ ∈
M [a, b] we have
!
X
X
ck I(f, δxk )
ck δxk =
I(f, µ) = I f,
k≥1
k≥1
!
=
X
ck J(f, δxk ) = J
f,
X
ck δxk
k≥1
k≥1
= J(f, µ).
Since the Banach subspace M [a, b]d of all discrete measures is weakly∗ dense in M [a, b]
and the functionals I(f, ·) and J(f, ·) are also weakly∗ continuous we conclude that I(f, µ) =
J(f, µ) for every f ∈ C[a, b] and µ ∈ M [a, b].
Theorem 2.3. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every µ-harmonic sequence (Pn , n ≥ 1) we have
Z
(2.1)
f (t)dµ(t) = µ({a})f (a) + Sn + Rn ,
[a,b]
where
(2.2)
Sn =
n
X
(−1)k−1 Pk (b)f (k−1) (b) − Pk (a)f (k−1) (a)
k=1
and
(2.3)
n
Z
Rn = (−1)
Pn (t)df (n−1) (t).
[a,b]
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
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I NTEGRATION - BY- PARTS FORMULA
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Proof. By partial integration, for n ≥ 2, we have
Z
n
Rn = (−1)
Pn (t)df (n−1) (t)
[a,b]
n
= (−1) Pn (b)f (n−1) (b) − Pn (a)f (n−1) (a)
Z
n
− (−1)
Pn−1 (t)f (n−1) (t)dt
[a,b]
n
= (−1) Pn (b)f (n−1) (b) − Pn (a)f (n−1) (a) + Rn−1 .
By Lemma 2.2 we have
Z
R1 = −
P1 (t)df (t)
[a,b]
Z
= − [P1 (b)f (b) − Pn (a)f (a)] +
f (t)dP1 (t)
[a,b]
Z
= − [P1 (b)f (b) − Pn (a)f (a)] +
f (t)dµ̌1 (t)
[a,b]
Z
= − [P1 (b)f (b) − Pn (a)f (a)] +
f (t)dµ(t) − µ({a})f (a).
[a,b]
Therefore, by iteration, we have
Rn =
n
X
k
(−1) Pk (b)f (k−1) (b) − Pk (a)f (k−1) (a) +
Z
f (t)dµ(t) − µ({a})f (a),
[a,b]
k=1
which proves our assertion.
Remark 2.4. By Remark 2.1 we see that identity (2.1) is a generalization of the integration-byparts formula (1.1).
Corollary 2.5. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every µ ∈ M [a, b] we have
Z
f (t)dµ(t) = Šn + Řn ,
[a,b]
where
Šn =
n
X
(−1)k−1 µ̌k (b)f (k−1) (b)
k=1
and
n
Z
Řn = (−1)
µ̌n (t)df (n−1) (t).
[a,b]
Proof. Apply the theorem above for the µ-harmonic sequence (µ̌n , n ≥ 1) and note that µ̌n (a) =
0, for n ≥ 2.
Corollary 2.6. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every x ∈ [a, b] we have
f (x) =
n
X
(x − b)k−1
k=1
(k − 1)!
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
f (k−1) (b) + Rn (x),
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6
A. Č IVLJAK , L J . D EDI Ć , AND M. M ATI Ć
where
(−1)n
Rn (x) =
(n − 1)!
Z
(t − x)n−1 df (n−1) (t).
[x,b]
Proof. Apply Corollary 2.5 for µ = δx and note that in this case
µ̌k (t) =
(t − x)k−1
,
(k − 1)!
x ≤ t ≤ b,
and
µ̌k (t) = 0,
a ≤ t < x,
for k ≥ 1.
Corollary 2.7. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Further, let (cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then
X
cm f (xm ) =
m≥1
n
XX
m≥1 k=1
cm
X
(xm − b)k−1 (k−1)
f
(b) +
cm Rn (xm ),
(k − 1)!
m≥1
where Rn (xm ) is from Corollary 2.6.
Proof. Apply Corollary 2.5 for the discrete measure µ =
P
m≥1 cm δxm .
3. S OME O STROWSKI - TYPE I NEQUALITIES
In this section we shall use the same notations as above.
Theorem 3.1. Let f : [a, b] → R be such that f (n−1) is L-Lipschitzian for some n ≥ 1. Then
for every µ-harmonic sequence (Pn , n ≥ 1) we have
Z
Z b
f (t)dµ(t) − µ({a})f (a) − Sn ≤ L
(3.1)
|Pn (t)| dt,
[a,b]
a
where Sn is defined by (2.2).
Proof. By Theorem 2.3 we have
Z
|Rn | = Pn (t)df
(n−1)
[a,b]
Z b
(t) ≤ L
|Pn (t)| dt,
a
which proves our assertion.
Corollary 3.2. If f is L-Lipschitzian, then for every c ∈ R and µ ∈ M [a, b] we have
Z
Z b
f (t)dµ(t) − µ([a, b])f (b) − c [f (b) − f (a)] ≤ L
|c + µ̌1 (t)| dt.
[a,b]
a
Proof. Put n = 1 in the theorem above and note that P1 (t) = c + µ̌1 (t), for some c ∈ R.
Corollary 3.3. If f is L-Lipschitzian, then for every c ≥ 0 and µ ≥ 0 we have
Z
f (t)dµ(t) − µ([a, b])f (b) − c [f (b) − f (a)]
[a,b]
≤ L [c(b − a) + µ̌2 (b)]
≤ L(b − a)(c + kµk).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
http://jipam.vu.edu.au/
I NTEGRATION - BY- PARTS FORMULA
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Proof. Apply Corollary 3.2 and note that in this case
Z b
Z b
|c + µ̌1 (t)| dt =
[c + µ̌1 (t)] dt
a
a
= c(b − a) + µ̌2 (b)
≤ c(b − a) + (b − a) kµk
= (b − a)(c + kµk).
Corollary 3.4. Let f be L-Lipschitzian, (cm , m ≥ 1) a sequence in [0, ∞) such that
X
cm < ∞,
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then for every c ≥ 0 we have
"
#
X
X
cm [f (b) − f (xm )] + c [f (b) − f (a)] ≤ L c(b − a) +
cm (b − xm )
m≥1
m≥1
"
#
X
≤ L(b − a) c +
cm .
m≥1
Proof. Apply Corollary 3.3 for the discrete measure µ =
P
m≥1 cm δxm .
Corollary 3.5. If f is L-Lipschitzian and µ ≥ 0, then
Z
f
(t)dµ(t)
−
µ([a,
x])f
(a)
−
µ((x,
b])f
(b)
[a,b]
≤ L [(2x − a − b)µ̌1 (x) − 2µ̌2 (x) + µ̌2 (b)] ,
for every x ∈ [a, b].
Proof. Apply Corollary 3.2 for c = −µ̌1 (x). Then
c + µ̌1 (b) = µ((x, b]),
µ̌1 (x) = µ([a, x])
and
Z
b
Z
|−µ̌1 (x) + µ̌1 (t)| dt =
a
x
Z
(µ̌1 (x) − µ̌1 (t)) dt +
a
b
(µ̌1 (t) − µ̌1 (x)) dt
x
= (2x − a − b)µ̌1 (x) − 2µ̌2 (x) + µ̌2 (b).
Corollary 3.6. Let f : [a, b] → R be such that f (n−1) is L-Lipschitzian for some n ≥ 1. Then
for every µ ∈ M [a, b] we have
Z
Z b
(b − a)n
f
(t)dµ(t)
−
Š
≤
L
|µ̌
(t)|
dt
≤
L kµk ,
n
n
n!
[a,b]
a
where Šn is from Corollary 2.5.
Proof. Apply the theorem above for the µ-harmonic sequence (µ̌n , n ≥ 1).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
http://jipam.vu.edu.au/
8
A. Č IVLJAK , L J . D EDI Ć , AND M. M ATI Ć
Corollary 3.7. Let f : [a, b] → R be such that f (n−1) is L-Lipschitzian for some n ≥ 1. Then
for every x ∈ [a, b] we have
n
X
(x − b)k−1 (k−1) (b − x)n
f
(b) ≤
L.
f (x) −
(k − 1)!
n!
k=1
Proof. Apply Corollary 3.6 for µ = δx and note that in this case
µ̌k (t) =
(t − x)k−1
, x ≤ t ≤ b,
(k − 1)!
µ̌k (t) = 0, a ≤ t < x,
and
for k ≥ 1.
Corollary 3.8. Let f : [a, b] → R be such that f (n−1) is L-Lipschitzian, for some n ≥ 1.
Further, let (cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then
n
X
k−1
X
X
(x
−
b)
m
cm f (xm ) −
cm
f (k−1) (b)
(k − 1)!
m≥1
m≥1 k=1
L X
≤
|cm | (b − xm )n
n! m≥1
X
L
≤ (b − a)n
|cm | .
n!
m≥1
Proof. Apply Corollary 3.6 for the discrete measure µ =
P
m≥1 cm δxm .
Theorem 3.9. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every µ-harmonic sequence (Pn , n ≥ 1) we have
Z
b
_
f (t)dµ(t) − µ({a})f (a) − Sn ≤ max |Pn (t)| (f (n−1) ),
t∈[a,b]
[a,b]
where
Wb
a (f
(n−1)
a
) is the total variation of f (n−1) on [a, b].
Proof. By Theorem 2.3 we have
Z
b
_
(n−1)
|Rn | = Pn (t)df
(t) ≤ max |Pn (t)| (f (n−1) ),
[a,b]
t∈[a,b]
a
which proves our assertion.
Corollary 3.10. If f is a function of bounded variation, then for every c ∈ R and µ ∈ M [a, b]
we have
Z
b
_
f (t)dµ(t) − µ([a, b])f (b) − c [f (b) − f (a)] ≤ max |c + µ̌1 (t)| (f ).
[a,b]
Proof. Put n = 1 in the theorem above.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
t∈[a,b]
a
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I NTEGRATION - BY- PARTS FORMULA
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Corollary 3.11. If f is a function of bounded variation, then for every c ≥ 0 and µ ≥ 0 we
have
Z
b
_
f (t)dµ(t) − µ([a, b])f (b) − c [f (b) − f (a)] ≤ [c + kµk] (f ).
[a,b]
a
Proof. In this case we have
max |c + µ̌1 (t)| = c + µ̌1 (b) = c + kµk .
t∈[a,b]
Corollary 3.12. Let f be a function of bounded variation, (cm , m ≥ 1) a sequence in [0, ∞)
such that
X
cm < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then for every c ≥ 0 we have
"
# b
X
X
_
cm [f (b) − f (xm )] + c [f (b) − f (a)] ≤ c +
cm
(f ).
a
m≥1
m≥1
Proof. Apply Corollary 3.11 for the discrete measure µ =
P
m≥1 cm δxm .
Corollary 3.13. If f is a function of bounded variation and µ ≥ 0, then we have
Z
f
(t)dµ(t)
−
µ([a,
x])f
(a)
−
µ((x,
b])f
(b)
[a,b]
b
≤
_
1
[µ̌1 (b) − µ̌1 (a) + |µ̌1 (a) + µ̌1 (b) − 2µ̌1 (x)|] (f ).
2
a
Proof. Apply Corollary 3.11 for c = −µ̌1 (x). Then
max |c + µ̌1 (t)| = max |µ̌1 (t) − µ̌1 (x)|
t∈[a,b]
t∈[a,b]
= max{µ̌1 (x) − µ̌1 (a), µ̌1 (b) − µ̌1 (x)}
1
= [µ̌1 (b) − µ̌1 (a) + |µ̌1 (a) + µ̌1 (b) − 2µ̌1 (x)|] .
2
Corollary 3.14. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every µ ∈ M [a, b] we have
Z
b
_
f
(t)dµ(t)
−
Š
≤
max
|µ̌
(t)|
(f (n−1) )
n
n
[a,b]
t∈[a,b]
a
b
_
(b − a)n−1
≤
kµk (f (n−1) ),
(n − 1)!
a
where Šn is from Corollary 2.5.
Proof. Apply the theorem above for the µ-harmonic sequence (µ̌n , n ≥ 1).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
http://jipam.vu.edu.au/
10
A. Č IVLJAK , L J . D EDI Ć ,
AND
M. M ATI Ć
Corollary 3.15. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Then for every x ∈ [a, b] we have
n
b
(b − x)n−1 _
k−1
X
(x − b)
(k−1)
f
(b) ≤
(f (n−1) ).
f (x) −
(k − 1)!
(n − 1)! a
k=1
Proof. Apply Corollary 3.14 for µ = δx and note that in this case
max |µ̌n (t)| =
t∈[a,b]
(b − x)n−1
.
(n − 1)!
Corollary 3.16. Let f : [a, b] → R be such that f (n−1) has bounded variation for some n ≥ 1.
Further, let (cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then
n
X
XX
(xm − b)k−1 (k−1) f
(b)
cm f (xm ) −
cm
(k − 1)!
m≥1
m≥1 k=1
b
_
X
1
≤
(f (n−1) )
|cm | (b − xm )n−1
(n − 1)! a
m≥1
≤
b
(b − a)n−1 _ (n−1) X
(f
)
|cm |
(n − 1)! a
m≥1
Proof. Apply Corollary 3.14 for the discrete measure µ =
P
m≥1 cm δxm .
Theorem 3.17. Let f : [a, b] → R be such that f (n) ∈ Lp [a, b] for some n ≥ 1. Then for every
µ-harmonic sequence (Pn , n ≥ 1) we have
Z
≤ kPn k kf (n) kp ,
f
(t)dµ(t)
−
µ({a})f
(a)
−
S
n
q
[a,b]
where p, q ∈ [1, ∞] and 1/p + 1/q = 1.
Proof. By Theorem 2.3 and the Hölder inequality we have
Z
(n−1)
|Rn | = Pn (t)df
(t)
[a,b]
Z
(n)
= Pn (t)f (t)dt
[a,b]
Z
≤
b
1q Z b
p1
(n) p
f (t) dt
|Pn (t)| dt
q
a
a
= kPn kq kf
(n)
kp .
Remark 3.18. We see that the inequality of the theorem above is a generalization of inequality
(1.2).
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
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11
Corollary 3.19. Let f : [a, b] → R be such that f (n) ∈ Lp [a, b] for some n ≥ 1, and µ ∈
M [a, b]. Then
Z
f (t)dµ(t) − Šn ≤ kµ̌n kq kf (n) kp
[a,b]
(b − a)n−1+1/q
≤
(n − 1)! [(n − 1)q + 1]
1/q
kµk kf (n) kp ,
where p, q ∈ [1, ∞] and 1/p + 1/q = 1.
Proof. Apply the theorem above for the µ-harmonic sequence (µ̌n , n ≥ 1).
Corollary 3.20. Let f : [a, b] → R be such that f (n) ∈ Lp [a, b], for some n ≥ 1. Further, let
(cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. Then
n
X
k−1
X
X
(x
−
b)
m
f (k−1) (b)
cm f (xm ) −
cm
(k − 1)!
m≥1
≤
m≥1 k=1
kf
(n)
kp
(n − 1)! [(n − 1)q + 1]
n−1+1/q
≤
X
(b − a)
kf
(n)
1/q
kp
(n − 1)! [(n − 1)q + 1]1/q
|cm | (b − xm )n−1+1/q
m≥1
X
|cm | ,
m≥1
where p, q ∈ [1, ∞] and 1/p + 1/q = 1.
Proof. Apply the theorem above for the discrete measure µ =
P
m≥1 cm δxm .
4. S OME G RÜSS - TYPE I NEQUALITIES
Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b], for some n ≥ 1. Then
mn ≤ f (n) (t) ≤ Mn ,
t ∈ [a, b], a.e.
for some real constants mn and Mn .
Theorem 4.1. Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b], for some n ≥ 1. Further, let
(Pk , k ≥ 1) be a µ-harmonic sequence such that
Pn+1 (a) = Pn+1 (b) ,
for that particular n. Then
Z
Z
Mn − m n b
f (t)dµ(t) − µ({a})f (a) − Sn ≤
|Pn (t)| dt.
2
[a,b]
a
Proof. Apply Theorem 2.3 for the special case when f (n−1) is absolutely continuous and its
derivative f (n) , existing a.e., is bounded a.e. Define the measure νn by
dνn (t) = −Pn (t) dt.
Then
Z
νn ([a, b]) = −
b
Pn (t) dt = Pn+1 (a) − Pn+1 (b) = 0,
a
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
http://jipam.vu.edu.au/
12
A. Č IVLJAK , L J . D EDI Ć ,
AND
M. M ATI Ć
which means that νn is balanced. Further,
Z
kνn k =
b
|Pn (t)| dt
a
and by [1, Theorem 2]
Z b
(n)
|Rn | = Pn (t) f (t)dt
a
Mn − m n
kνn k
2
Z b
Mn − m n
=
|Pn (t)| dt,
2
a
≤
which proves our assertion.
Corollary 4.2. Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b], for some n ≥ 1. Then for every
(n + 1)-balanced measure µ ∈ M [a, b] we have
Z
Z
Mn − m n b
f (t)dµ(t) − Šn ≤
|µ̌n (t)| dt
2
[a,b]
a
≤
Mn − mn (b − a)n
kµk ,
2
n!
where Šn is from Corollary 2.5.
Proof. Apply Theorem 4.1 for the µ-harmonic sequence (µ̌k , k ≥ 1) and note that the condition
Pn+1 (a) = Pn+1 (b) reduces to µ̌n+1 (b) = 0, which means that µ is (n + 1)-balanced.
Corollary 4.3. Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b] for some n ≥ 1. Further, let
(cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b] satisfy the condition
X
cm (b − xm )n = 0.
m≥1
Then
n
X
XX
(xm − b)k−1 (k−1) cm f (xm ) −
cm
f
(b)
(k − 1)!
m≥1
m≥1 k=1
Mn − m n X
≤
|cm | (b − xm )n
2n!
m≥1
X
Mn − m n
≤
(b − a)n
|cm | .
2n!
m≥1
Proof. Apply Corollary 4.2 for the discrete measure µ =
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
P
m≥1 cm δxm .
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13
Corollary 4.4. Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b] for some n ≥ 1. Then for every
µ ∈ M [a, b], such that all k-moments of µ are zero for k = 0, . . . , n, we have
Z
Z
Mn − m n b
|µ̌n (t)| dt
f (t)dµ(t) ≤
2
[a,b]
a
Mn − mn (b − a)n
kµk .
2
n!
Proof. By [1, Theorem 5], the condition mk (µ) = 0, k = 0, . . . , n is equivalent to µ̌k (b) = 0,
k = 1, . . . , n + 1. Apply Corollary 4.2 and note that in this case Šn = 0.
≤
Corollary 4.5. Let f : [a, b] → R be such that f (n) ∈ L∞ [a, b] for some n ≥ 1. Further, let
(cm , m ≥ 1) be a sequence in R such that
X
|cm | < ∞
m≥1
and let {xm ; m ≥ 1} ⊂ [a, b]. If
X
X
X
cm =
cm xm = · · · =
cm xnm = 0,
m≥1
m≥1
m≥1
then
X
M −m X
n
n
cm f (xm ) ≤
|cm | (b − xm )n
2n!
m≥1
m≥1
X
Mn − m n
≤
(b − a)n
|cm | .
2n!
m≥1
P
Proof. Apply Corollary 4.4 for the discrete measure µ = m≥1 cm δxm .
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[1] A. ČIVLJAK, LJ. DEDIĆ AND M. MATIĆ, Euler-Grüss type inequalities involving measures, submitted.
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[3] Lj. DEDIĆ, M. MATIĆ, J. PEČARIĆ AND A. AGLIĆ ALJINOVIĆ, On weighted Euler harmonic
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J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 93, 13 pp.
http://jipam.vu.edu.au/