Volume 9 (2008), Issue 3, Article 64, 10 pp.
BOUNDS FOR SOME PERTURBED ČEBYŠEV FUNCTIONALS
S.S. DRAGOMIR
R ESEARCH G ROUP IN M ATHEMATICAL I NEQUALITIES AND A PPLICATIONS
S CHOOL OF E NGINEERING AND S CIENCE
V ICTORIA U NIVERSITY
PO B OX 14428, MCMC 8001
VICTORIA AUSTRALIA .
sever.dragomir@vu.edu.au
URL: http://www.staff.vu.edu.au/rgmia/dragomir/
Received 20 May, 2008; accepted 17 August, 2008
Communicated by R.N. Mohapatra
A BSTRACT. Bounds for the perturbed Čebyšev functionals C (f, g) − µC (e, g) and C (f, g) −
µC (e, g) − νC (f, e) when µ, ν ∈ R and e is the identity function on the interval [a, b] , are
given. Applications for some Grüss’ type inequalities are also provided.
Key words and phrases: Čebyšev functional, Grüss type inequality, Integral inequalities, Lebesgue p−norms.
2000 Mathematics Subject Classification. 26D15, 26D10.
1. I NTRODUCTION
For two Lebesgue integrable functions f, g : [a, b] → R, consider the Čebyšev functional:
Z b
Z b
Z b
1
1
(1.1)
C (f, g) :=
f (t)g(t)dt −
f (t)dt
g(t)dt.
b−a a
(b − a)2 a
a
In 1934, Grüss [5] showed that
1
(M − m) (N − n) ,
4
provided that there exists the real numbers m, M, n, N such that
(1.2)
(1.3)
|C (f, g)| ≤
m ≤ f (t) ≤ M
and n ≤ g (t) ≤ N
for a.e. t ∈ [a, b] .
The constant 41 is best possible in (1.1) in the sense that it cannot be replaced by a smaller
quantity.
Another, however less known result, even though it was obtained by Čebyšev in 1882, [3],
states that
1
(1.4)
|C (f, g)| ≤
kf 0 k∞ kg 0 k∞ (b − a)2 ,
12
154-08
2
S.S. D RAGOMIR
provided that f 0 , g 0 exist and are continuous on [a, b] and kf 0 k∞ = supt∈[a,b] |f 0 (t)| . The con1
stant 12
can be improved in the general case.
The Čebyšev inequality (1.4) also holds if f, g : [a, b] → R are assumed to be absolutely
continuous and f 0 , g 0 ∈ L∞ [a, b] while kf 0 k∞ = ess supt∈[a,b] |f 0 (t)| .
A mixture between Grüss’ result (1.2) and Čebyšev’s one (1.4) is the following inequality
obtained by Ostrowski in 1970, [9]:
1
(b − a) (M − m) kg 0 k∞ ,
8
provided that f is Lebesgue integrable and satisfies (1.3) while g is absolutely continuous and
g 0 ∈ L∞ [a, b] . The constant 18 is best possible in (1.5).
The case of euclidean norms of the derivative was considered by A. Lupaş in [7] in which he
proved that
|C (f, g)| ≤
(1.5)
|C (f, g)| ≤
(1.6)
1
kf 0 k2 kg 0 k2 (b − a) ,
π2
provided that f, g are absolutely continuous and f 0 , g 0 ∈ L2 [a, b] . The constant π12 is the best
possible.
Recently, P. Cerone and S.S. Dragomir [1] have proved the following results:
p ! p1
Z b
Z b
1
1
f (t) −
,
(1.7)
|C (f, g)| ≤ inf kg − γkq ·
f (s) ds dt
γ∈R
b−a
b−a a
a
where p > 1 and
(1.8)
1
p
+
1
q
= 1 or p = 1 and q = ∞, and
Z b
1
1
|C (f, g)| ≤ inf kg − γk1 ·
ess sup f (t) −
f (s) ds ,
γ∈R
b−a
b−a a
t∈[a,b]
provided that f ∈ Lp [a, b] and g ∈ Lq [a, b] (p > 1, p1 + 1q = 1; p = 1, q = ∞ or p = ∞, q = 1).
Notice that for q = ∞, p = 1 in (1.7) we obtain
Z b
Z b
1
1
f (t) −
dt
(1.9)
|C (f, g)| ≤ inf kg − γk∞ ·
f
(s)
ds
γ∈R
b−a a b−a a
Z b
Z b
1
f (t) − 1
≤ kgk∞ ·
f (s) ds dt
b−a a
b−a a
and if g satisfies (1.3), then
(1.10)
Z b
Z b
1
1
f (t) −
dt
|C (f, g)| ≤ inf kg − γk∞ ·
f
(s)
ds
γ∈R
b−a a b−a a
Z b
Z b
n+N
1
1
≤ g −
·
f (t) −
f (s) ds dt
2
b−a a
∞ b−a a
Z b
Z b
1
1
1
dt.
≤ (N − n) ·
f
(t)
−
f
(s)
ds
2
b−a a b−a a
The inequality between the first and the last term in (1.10) has been obtained by Cheng and
Sun in [4]. However, the sharpness of the constant 21 , a generalisation for the abstract Lebesgue
integral and the discrete version of it have been obtained in [2].
For other recent results on the Grüss inequality, see [6], [8] and [10] and the references
therein.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 64, 10 pp.
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P ERTURBED Č EBYŠEV F UNCTIONALS
3
The aim of the present paper is to establish Grüss type inequalities for some perturbed Čebyšev functionals. For this purpose, two integral representations of the functionals C (f, g) −
µC (e, g) and C (f, g) − µC (e, g) − νC (f, e) when µ, ν ∈ R and e (t) = t, t ∈ [a, b] are given.
2. R EPRESENTATION R ESULTS
The following representation result can be stated.
Lemma 2.1. If f : [a, b] → R is absolutely continuous on [a, b] and g is Lebesgue integrable
on [a, b] , then
Z bZ b
1
(2.1)
C (f, g) =
Q (t, s) [g (s) − λ] f 0 (t) dsdt
2
(b − a) a a
for any λ ∈ R, where the kernel Q : [a, b]2 → R is given by
(
t − b if a ≤ s ≤ t ≤ b,
(2.2)
Q (t, s) :=
t − a if a ≤ t < s ≤ b.
Proof. We observe that for λ ∈ R we have C (f, λ) = 0 and thus it suffices to prove (2.1) for
λ = 0.
By Fubini’s theorem, we have
Z bZ b
Z b Z b
0
0
(2.3)
Q (t, s) g (s) f (t) dsdt =
Q (t, s) f (t) dt g (s) ds.
a
a
a
a
By the definition of Q (t, s) and integrating by parts, we have successively,
Z b
Z s
Z b
0
0
(2.4)
Q (t, s) f (t) dt =
Q (t, s) f (t) dt +
Q (t, s) f 0 (t) dt
a
a
s
Z s
Z b
=
(t − a) f 0 (t) dt +
(t − b) f 0 (t) dt
a
s
Z s
Z b
= (s − a) f (s) −
f (t) dt + (b − s) f (s) −
f (t) dt
a
s
Z b
= (b − a) f (s) −
f (t) dt,
a
for any s ∈ [a, b] .
Now, integrating (2.4) multiplied with g (s) over s ∈ [a, b], we deduce
Z b Z b
Z b
Z b
0
Q (t, s) f (t) dt g (s) ds =
(b − a) f (s) −
f (t) dt g (s) ds
a
a
a
a
Z
= (b − a)
b
Z
f (s)g(s)ds −
a
b
Z
f (s) ds ·
a
b
g(s)ds
a
= (b − a)2 C (f, g)
and the identity is proved.
Utilising the linearity property of C (·, ·) in each argument, we can state the following equality:
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 64, 10 pp.
http://jipam.vu.edu.au/
4
S.S. D RAGOMIR
Theorem 2.2. If e : [a, b] → R, e (t) = t, then under the assumptions of Lemma 2.1 we have:
Z bZ b
1
(2.5)
C (f, g) = µC (e, g) +
Q (t, s) [g (s) − λ] [f 0 (t) − µ] dtds
(b − a)2 a a
for any λ, µ ∈ R, where
Z b
Z
a+b b
1
(2.6)
C (e, g) =
tg (t) dt −
g (t) dt.
b−a a
2
a
The second representation result is incorporated in
Lemma 2.3. If f, g : [a, b] → R are absolutely continuous on [a, b] , then
Z bZ b
1
(2.7)
C (f, g) =
K (t, s) f 0 (t) g 0 (s) dtds,
2
(b − a) a a
where the kernel K : [a, b] → R is defined by
(
(b − t) (s − a) if a ≤ s ≤ t ≤ b,
(2.8)
K (t, s) :=
(t − a) (b − s) if a ≤ t < s ≤ b.
Proof. By Fubini’s theorem we have
Z bZ b
Z b Z b
0
0
0
(2.9)
K (t, s) f (t) g (s) dtds =
K (t, s) g (s) ds f 0 (t) dt.
a
a
a
a
By the definition of K and integrating by parts, we have successively:
Z b
(2.10)
K (t, s)g 0 (s) ds
a
Z t
Z b
0
=
K (t, s) g (s) ds +
K (t, s) g 0 (s) ds
a
t
Z t
Z b
0
= (b − t)
(s − a) g (s) ds + (t − a)
(b − s) g 0 (s) ds
a
t
Z t
= (b − t) (t − a) g (t) −
g (s) ds
a
Z b
+ (t − a) − (b − t) g (t) +
g (s) ds
t
Z
= (t − a)
b
Z
g (s) ds − (b − t)
t
t
g (s) ds,
a
for any t ∈ [a, b].
Multiplying (2.10) by f 0 (t) and integrating over t ∈ [a, b] , we have:
Z b Z b
0
(2.11)
K (t, s) g (s) ds f 0 (t) dt
a
a
Z b
Z b
Z t
=
(t − a)
g (s) ds − (b − t)
g (s) ds f 0 (t) dt
a
t
a
b
Z b
Z t
= f (t) (t − a)
g (s) ds − (b − t)
g (s) ds t
Z
−
a
a
0
Z b
Z t
b
g (s) ds dt
g (s) ds − (b − t)
f (t) (t − a)
a
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t
a
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P ERTURBED Č EBYŠEV F UNCTIONALS
Z
b
Z
=
b
g (s) ds − (t − a) g (t) +
t
=−
t
Z
f (t)
a
Z
5
b
a
Z
f (t)
a
g (s) ds − (b − t) g (t)
b
g (s) ds − (b − a) g (t) dt
a
Z
= (b − a)
b
Z
b
g(t)f (t)dt −
a
Z
f (t) dt ·
a
b
g(t)dt
a
= (b − a)2 C (f, g) .
By (2.11) and (2.9) we deduce the desired result.
Theorem 2.4. With the assumptions of Lemma 2.3, we have for any ν, µ ∈ R that:
(2.12)
C (f, g) = µC (e, g) + νC (f, e)
Z bZ b
1
+
K (t, s) [f 0 (t) − µ] [g 0 (s) − ν] dtds.
(b − a)2 a a
Proof. Follows by Lemma 2.3 on observing that C (e, e) = 0 and
C (f − µe, g − νe) = C (f, g) − µC (e, g) − νC (f, e)
for any µ, ν ∈ R.
3. B OUNDS IN T ERMS OF L EBESGUE N ORMS OF g
AND
f0
Utilising the representation (2.5) we can state the following result:
Theorem 3.1. Assume that g : [a, b] → R is Lebesgue integrable on [a, b] and f : [a, b] → R is
absolutely continuous on [a, b] , then
(3.1)
|C (f, g) − µC (e, g)|
 1
(b − a) kf 0 − µk∞ inf kg − γk∞
if f 0 , g ∈ L∞ [a, b] ;

3

γ∈R




 21/q (b−a) p−q
pq
kf 0 − µkp inf kg − γkp if f 0 , g ∈ Lp [a, b] ,
[(q+1)(q+2)]1/q
≤
γ∈R


p > 1, p1 + 1q = 1;




 (b − a)−1 kf 0 − µk1 inf kg − γk1
γ∈R
for any µ ∈ R.
Proof. From (2.5), we have
(3.2)
1
|C (f, g) − µC (e, g)| ≤
(b − a)2
Z bZ
a
b
|Q (t, s)| |g (s) − λ| |f 0 (t) − µ| dtds
a
0
≤ kg − λk∞ kf − µk∞
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 64, 10 pp.
1
(b − a)2
Z bZ
b
|Q (t, s)| dtds.
a
a
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6
S.S. D RAGOMIR
However, by the definition of Q we have for α ≥ 1 that
Z bZ b
I (α) :=
|Q (t, s)|α dtds
a
a
Z b Z t
Z b
α
α
=
|t − b| ds +
|t − a| ds dt
a
a
t
b
Z
[(t − a) (b − t)α + (b − t) (t − a)α ] dt.
=
a
Since
(t − a) (b − t)α dt =
(b − a)α+2
(α + 1) (α + 2)
(b − t) (t − a)α dt =
(b − a)α+2
,
(α + 1) (α + 2)
b
Z
a
and
b
Z
a
hence
I (α) =
2 (b − a)α+2
,
(α + 1) (α + 2)
α ≥ 1.
Then we have
Z bZ b
1
b−a
|Q (t, s)| dtds =
,
2
3
(b − a) a a
and taking the infimum over λ ∈ R in (3.2), we deduce the first part of (3.1).
Utilising the Hölder inequality for double integrals we also have
Z bZ b
|Q (t, s)| |g (s) − λ| |f 0 (t) − µ| dtds
a
a
Z b Z
≤
1q Z b Z
b
q
|Q (t, s)| dtds
a
1/q
2
=
a
a
1+ 2q
(b − a)
1/q
b
p
p
|g (s) − λ| |f 0 (t) − µ| dtds
p1
a
kg − λkp kf 0 − µkp ,
[(q + 1) (q + 2)]
which provides, by the first inequality in (3.2), the second part of (3.1).
For the last part, we observe that sup(t,s)∈[a,b]2 |Q (t, s)| = b − a and then
Z bZ b
|Q (t, s)| |g (s) − λ| |f 0 (t) − µ| dtds ≤ (b − a) kg − λk1 kf 0 − µk1 .
a
a
This completes the proof.
Remark 1. The above inequality (3.1) is a source of various inequalities as will be shown in
the following.
≤
(1) For instance, if −∞ < m ≤ g (t) ≤ M < ∞ for a.e. t ∈ [a, b] , then g − m+M
2
∞
1/p
1
m+M 1
(M − m) and g − 2 p ≤ 2 (M − m) (b − a) , p ≥ 1. Then for any µ ∈ R we
2
have
 1
(b − a) (M − m) kf 0 − µk∞
if f 0 ∈ L∞ [a, b] ;

6



 2−1/p (b−a)1/q
(M − m) kf 0 − µkp if f 0 ∈ Lp [a, b] ,
(3.3) |C (f, g) − µC (e, g)| ≤
[(q+1)(q+2)]1/q


p > 1, p1 + 1q = 1;


 1
(M − m) kf 0 − µk1 ,
2
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P ERTURBED Č EBYŠEV F UNCTIONALS
7
which gives for µ = 0 that
 1
(b − a) (M − m) kf 0 k∞
if f 0 ∈ L∞ [a, b] ;

6



 2−1/p (b−a)1/q
(M − m) kf 0 kp if f 0 ∈ Lp [a, b] ,
(3.4)
|C (f, g)| ≤
[(q+1)(q+2)]1/q


p > 1, p1 + 1q = 1;


 1
(M − m) kf 0 k1 .
2
≤ 1 |Γ − γ| and
(2) If −∞ < γ ≤ f 0 (t) ≤ Γ < ∞ for a.e. t ∈ [a, b] , then f 0 − γ+Γ
2
2
∞
0 γ+Γ f −
≤ 1 |Γ − γ| (b − a)1/p , p ≥ 1. Then we have from (3.1) that
2
2
p
(3.5)
γ
+
Γ
C (f, g) −
C
(e,
g)
2
 1
(b − a) (Γ − γ) inf kg − ξk∞
if g ∈ L∞ [a, b] ;

6

ξ∈R



 2−1/p (b−a)1/q
(Γ − γ) inf kg − ξkp if g ∈ Lp [a, b] ,
[(q+1)(q+2)]1/q
≤
ξ∈R


p > 1, p1 + 1q = 1;



 1 (Γ − γ) inf kg − ξk .
1
2
ξ∈R
Moreover, if we also assume that −∞ < m ≤ g (t) ≤ M < ∞ for a.e. t ∈ [a, b] , then
by (3.5) we also deduce:
(3.6)
C (f, g) − γ + Γ C (e, g)
2
 1
(b − a) (Γ − γ) (M − m)

12



21−1/p (b−a)
≤
(Γ − γ) (M − m) p > 1,
[(q+1)(q+2)]1/q



 1
(Γ − γ) (M − m) (b − a) .
4
1
p
+
1
q
= 1;
Observe that the first inequality in (3.6) is better than the others.
4. B OUNDS IN T ERMS OF L EBESGUE N ORMS OF f 0
AND g 0
We have the following result:
Theorem 4.1. Assume that f, g : [a, b] → R are absolutely continuous on [a, b], then
(4.1)
|C (f, g) − µC (e, g) − νC (f, e)|
 1
(b − a)2 kf 0 − µk∞ kg 0 − νk∞
if f 0 , g 0 ∈ L∞ [a, b] ;


12



 h
i1
B(q+1,q+1) q
(b − a)2/q kf 0 − µkp kg 0 − νkp if f 0 , g 0 ∈ Lp [a, b] ,
≤
q+1



p > 1, p1 + 1q = 1;


 1 0
kf − µk1 kg 0 − νk1 ;
4
for any µ, ν ∈ R.
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 64, 10 pp.
http://jipam.vu.edu.au/
8
S.S. D RAGOMIR
Proof. From (2.12), we have
(4.2)
|C (f, g) − µC (e, g) − νC (f, e)|
Z bZ b
1
≤
|K (t, s)| |f 0 (t) − µ| |g 0 (s) − ν| dtds.
(b − a)2 a a
Define
Z bZ
(4.3)
b
|K (t, s)|α dtds
J (α) :=
a
a
t
Z b Z
α
b
Z
α
α
(b − t) (s − a) ds +
=
1
=
α+1
(t − a) (b − s) ds dt
a
a
α
t
Z
b
α
α+1
(b − t) (t − a)
b
Z
α
(t − a) (b − t)
dt +
a
α+1
dt .
a
Since
b
Z
p
q
p+q+1
Z
1
(t − a) (b − t) dt = (b − a)
a
sp (1 − s)q ds
0
p+q+1
= (b − a)
B (p + 1, q + 1) ,
hence, by (4.3),
J (α) =
2 (b − a)2α+2
B (α + 1, α + 2) ,
α+1
α ≥ 1.
As it is well known that
B (p, q + 1) =
q
B (p, q) ,
p+q
then for p = α + 1, q = α + 1 we have B (α + 1, α + 2) = 12 B (α + 1, α + 1) .
Then we have
J (α) =
(b − a)2α+2
B (α + 1, α + 1) ,
α+1
α ≥ 1.
Taking into account that
1
(b − a)2
Z bZ
a
b
|K (t, s)| |f 0 (t) − µ| |g 0 (s) − ν| dtds
a
0
0
≤ kf − µk∞ kg − νk∞
1
(b − a)2
Z bZ
b
|K (t, s)| dtds
a
a
= kf 0 − µk∞ kg 0 − νk∞ (b − a)2 B (2, 3)
1
=
(b − a)2 kf 0 − µk∞ kg 0 − νk∞ ,
12
we deduce from (4.2) the first part of (4.1).
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P ERTURBED Č EBYŠEV F UNCTIONALS
9
By the Hölder integral inequality for double integrals, we have
Z bZ b
(4.4)
|K (t, s)| |f 0 (t) − µ| |g 0 (s) − ν| dtds
a
a
Z b Z
≤
b
q
1q
|K (t, s)| dtds
a
kf 0 − µkp kg 0 − νkp
a
# 1q
(b − a)2q+2
B (q + 1, q + 2) kf 0 − µkp kg 0 − νkp
=
q+1
1q
2+2/q B (q + 1, q + 1)
kf 0 − µkp kg 0 − νkp .
= (b − a)
q+1
"
Utilising (4.2) and (4.4) we deduce the second part of (4.1).
By the definition of K (t, s) we have, for a ≤ s ≤ t ≤ b, that
K (t, s) = (b − t) (s − a) ≤ (b − t) (t − a) ≤
1
(b − a)2
4
and for a ≤ t < s ≤ b, that
K (t, s) = (t − a) (b − s) ≤ (t − a) (b − t) ≤
1
(b − a)2 ,
4
therefore
sup |K (t, s)| =
(t,s)∈[a,b]
1
(b − a)2 .
4
Due to the fact that
Z bZ b
1
|K (t, s)| |f 0 (t) − µ| |g 0 (s) − ν| dtds
(b − a)2 a a
Z bZ b
1
≤ sup |K (t, s)|
|f 0 (t) − µ| |g 0 (s) − ν| dtds
2
(b − a) a a
(t,s)∈[a,b]
1
= kf 0 − µk1 kg 0 − νk1 ,
4
then from (4.2) we obtain the last part of (4.1).
Remark 2. When µ = ν = 0, we obtain from (4.1) the following Grüss type inequalities:
 1
(b − a)2 kf 0 k∞ kg 0 k∞
if f 0 , g 0 ∈ L∞ [a, b] ;


12



 h
i1
B(q+1,q+1) q
(b − a)2/q kf 0 kp kg 0 kp if f 0 , g 0 ∈ Lp [a, b] ,
(4.5)
|C (f, g)| ≤
q+1



p > 1, p1 + 1q = 1;


 1 0
kf k1 kg 0 k1 .
4
1
is the best
Notice that the first inequality in (4.5) is exactly the Čebyšev inequality for which 12
possible constant.
If we assume that there exists γ, Γ, φ, Φ such that −∞ < γ ≤ f 0 (t) ≤ Γ < ∞ and −∞ <
φ ≤ g 0 (t) ≤ Φ < ∞ for a.e. t ∈ [a, b] , then we deduce from (4.1) the following inequality
γ+Γ
φ+Φ
≤ 1 (b − a)2 (Γ − γ) (Φ − φ) .
(4.6)
C
(f,
g)
−
·
C
(e,
g)
−
·
C
(f,
e)
48
2
2
J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 64, 10 pp.
http://jipam.vu.edu.au/
10
S.S. D RAGOMIR
1
We also observe that the constant 48
is best possible in the sense that it cannot be replaced by a
smaller quantity.
The sharpness of the constant follows by the fact that for Γ = −γ, Φ = −φ we deduce from
(4.6) the Čebyšev inequality which is sharp.
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1
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