Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 172, 2006
A UNIFIED TREATMENT OF SOME SHARP INEQUALITIES
ZHENG LIU
I NSTITUTE OF A PPLIED M ATHEMATICS , FACULTY OF S CIENCE
A NSHAN U NIVERSITY OF S CIENCE AND T ECHNOLOGY
A NSHAN 114044, L IAONING , C HINA
lewzheng@163.net
Received 24 July, 2006; accepted 11 December, 2006
Communicated by S.S. Dragomir
A BSTRACT. A generalization of some recent sharp inequalities by N. Ujević is established.
Applications in numerical integration are also considered.
Key words and phrases: Quadrature formula, Sharp error bounds, Generalization, Numerical integration.
2000 Mathematics Subject Classification. Primary 26D15.
1. I NTRODUCTION
In [1] we can find a generalization of the pre-Grüss inequality as:
Lemma 1.1. Let f, g, Ψ ∈ L2 (a, b). Then we have
SΨ (f, g)2 ≤ SΨ (f, f )SΨ (g, g),
(1.1)
where
Z
(1.2) SΨ (f, g) =
a
b
1
f (t)g(t) dt −
b−a
Z
b
Z
b
f (t) dt
g(t) dt
Z b
Z b
1
f (t)Ψ(t) dt
g(t)Ψ(t) dt
−
kΨk22 a
a
a
a
and Ψ satisfies
Z
b
Ψ(t) dt = 0,
(1.3)
a
while as usual, k · k2 is the norm in L2 (a, b). i.e.,
kΨk22
Z
=
b
Ψ2 (t) dt.
a
Using the above inequality, Ujević in [1] obtained the following interesting results:
ISSN (electronic): 1443-5756
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195-06
2
Z HENG L IU
Theorem 1.2. Let f : [a, b] → R be an absolutely continuous function whose derivative f 0 ∈
L2 (a, b). Then
Z b
(b − a) 23
a
+
b
√ C1
(1.4)
(b − a) −
f (t) dt ≤
f
2
2 3
a
where
12
[f (b) − f (a)]2
2
0 2
(1.5)
C1 = kf k2 −
− [Q(f ; a, b)]
b−a
and
Z b
a+b
2
3
(1.6)
Q(f ; a, b) = √
f (a) + f
+ f (b) −
f (t) dt .
2
b−a a
b−a
Theorem 1.3. Let the assumptions of Theorem 1.2 hold. Then
Z b
(b − a) 32
f (a) + f (b)
√ C2 ,
(1.7)
(b − a) −
f (t) dt ≤
2
2 3
a
where
C2 =
(1.8)
kf 0 k22
[f (b) − f (a)]2
−
− [P (f ; a, b)]2
b−a
12
and
Z b
6
1
a+b
f (a) + 4f
+ f (b) −
f (t) dt .
(1.9)
P (f ; a, b) = √
2
b−a a
b−a
Theorem 1.4. Let the assumptions of Theorem 1.2 hold. Then
Z b
f (a) + 2f a+b + f (b)
(b − a) 32
2
√ C3 ,
(1.10)
(b − a) −
f (t) dt ≤
4
4 3
a
where
(
(1.11)
C3 =
kf 0 k22 −
2
[f (b) − f (a)]
1
−
b−a
b−a
f (a) − 2f
a+b
2
2 ) 12
+ f (b)
2
2 ) 12
2
a+b
2
a+b
kf 0 k22 −
f
− f (a) −
f (b) − f
.
b−a
2
b−a
2
(
=
In [2], Ujević further proved that the above all inequalities are sharp.
In this paper, we will derive a new sharp inequality with a parameter for absolutely continuous
functions with derivatives belonging to L2 (a, b), which not only provides a unified treatment of
all the above sharp inequalities, but also gives some other interesting results as special cases.
Applications in numerical integration are also considered.
2. M AIN R ESULTS
Theorem 2.1. Let the assumptions of Theorem 1.2 hold. Then for any θ ∈ [0, 1] we have
Z b
a+b
f (a) + f (b)
(2.1) (b − a) (1 − θ)f
+θ
−
f (t) dt
2
2
a
3
(b − a) 2 √
√
≤
1 − 3θ + 3θ2 C(θ),
2 3
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
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S HARP I NEQUALITIES
3
where
21
[f (b) − f (a)]2
0 2
2
C(θ) = kf k2 −
− [N (f ; a, b; θ)]
b−a
(2.2)
and
2
(2.3) N (f ; a, b; θ) = p
(1 − 3θ + 3θ2 )(b − a)
Z b
a
+
b
f
(a)
+
f
(b)
3
−
6θ
× (1 − 3θ)f
+ (2 − 3θ)
−
f (t) dt .
2
2
b−a a
The inequality (2.1) with (2.2) and (2.3) is sharp in the sense that the constant
replaced by a smaller one.
1
√
2 3
cannot be
Proof. Let us define the functions
t − a, t ∈ a, a+b
,
2
t − b, t ∈ a+b
,b ,
2
(
p(t) =
and
t − a + θ b−a
, t ∈ a, a+b
,
2
2
t − b − θ b−a
, t ∈ a+b
,b ,
2
2
(
Ψ(t) =
where θ ∈ [0, 1].
It is not difficult to verify that
b
Z
Z
p(t) dt =
(2.4)
b
Ψ(t) dt = 0.
a
a
i.e., Ψ satisfies the condition (1.3).
We also have
kpk22
(2.5)
Z
=
b
p2 (t) dt =
a
(b − a)3
12
and
kΨk22
(2.6)
Z
=
b
Ψ2 (t) dt =
a
(b − a)3
(1 − 3θ + 3θ2 ).
12
We now calculate
Z b
(2.7)
p(t)Ψ(t) dt
a
a+b
2
b−a
=
(t − a) t − a − θ
2
a
1
θ
−
(b − a)3 .
=
12 8
Z
b
b−a
dt +
(t − b) t − b + θ
a+b
2
2
Z
dt
Integrating by parts, we have
Z b
Z b
a+b
0
(2.8)
f (t)p(t) dt = f
(b − a) −
f (t) dt
2
a
a
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
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4
Z HENG L IU
and
Z
b
f 0 (t)Ψ(t) dt
(2.9)
a
a+b
2
Z b b−a
b−a
0
t−b+θ
f 0 (t) dt
t−a−θ
f (t) dt +
=
a+b
2
2
a
2
Z b
a+b
f (a) + f (b)
= (b − a) (1 − θ)f
+θ
−
f (t) dt.
2
2
a
Z
From (2.4), (2.6) – (2.9) and (1.2) we get
(2.10)
SΨ (f 0 , p)
Z b
Z b
Z b
1
0
0
=
f (t)p(t) dt −
f (t) dt
p(t) dt
b−a a
a
a
Z b
Z b
1
0
−
f (t)Ψ(t) dt
p(t)Ψ(t) dt
kΨk22 a
a
Z b
2 − 3θ
a+b
=f
(b − a) −
f (t) dt −
2
2(1 − 3θ + 3θ2 )
a
Z b
a+b
f (a) + f (b)
× (b − a) (1 − θ)f
+θ
−
f (t) dt
2
2
a
θ
a+b
=
(b − a) (1 − 3θ)f
2(1 − 3θ + 3θ2 )
2
Z b
f (a) + f (b)
+ (2 − 3θ)
− (3 − 6θ)
f (t) dt .
2
a
From (2.4) – (2.7) and (1.2) we also have
(2.11)
Z b
2
Z b
2
1
1
SΨ (p, p) =
−
p(t) dt −
p(t)Ψ(t) dt
b−a
kΨk22
a
a
θ2 (b − a)3
=
16(1 − 3θ + 3θ2 )
kpk22
and
(2.12)
Z b
2
Z b
2
1
1
0
0
SΨ (f , f ) =
−
f (t) dt −
f (t)Ψ(t) dt
b−a
kΨk22
a
a
[f (b) − f (a)]2
12
= kf 0 k22 −
−
b−a
(1 − 3θ + 3θ2 )(b − a)
2
Z b
a+b
f (a) + f (b)
1
× (1 − θ)f
+θ
−
f (t) dt
2
2
b−a a
0
0
kf 0 k22
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
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S HARP I NEQUALITIES
5
Thus from (2.10) – (2.12) and (1.1) we can easily get
2
Z b
a
+
b
f
(a)
+
f
(b)
(2.13) (b − a) (1 − 3θ)f
+ (2 − 3θ)
− (3 − 6θ)
f (t) dt
2
2
a
2
3
2
(1 − 3θ + 3θ )(b − a)
[f (b) − f (a)]
12
≤
kf 0 k22 −
−
4
b−a
(1 − 3θ + 3θ2 )(b − a)
2 )
Z b
a+b
f (a) + f (b)
1
× (1 − θ)f
+θ
−
f (t) dt
.
2
2
b−a a
It is equivalent to
2
Z b
a+b
f (a) + f (b)
1
(2.14) 3(b − a) (1 − θ)f
+θ
−
f (t) dt
2
2
b−a a
[f (b) − f (a)]2
4
1 − 3θ + 3θ2
3
(b − a) kf 0 k22 −
−
≤
4
b−a
(1 − 3θ + 3θ2 )(b − a)
2 )
Z b
a
+
b
f
(a)
+
f
(b)
3
−
6θ
× (1 − 3θ)f
+ (2 − 3θ)
−
f (t) dt .
2
2
b−a a
2
Consequently, inequality (2.1) with (2.2) and (2.3) follow from (2.14).
In order to prove that the inequality (2.1) with (2.2) and (2.3) is sharp for any θ ∈ [0, 1], we
define the function

 12 t2 − 2θ t,
t ∈ 0, 12 ,
(2.15)
f (t) =
 1 t2 − 1 − θ t + 1−θ , t ∈ 1 , 1
2
2
2
2
The function given in (2.15) is absolutely continuous since it is a continuous piecewise polynomial function.
We now suppose that (2.1) holds with a constant C > 0 as
Z b
a+b
f (a) + f (b)
(2.16)
(b
−
a)
(1
−
θ)f
+
θ
−
f
(t)
dt
2
2
a
√
3
≤ C(b − a) 2 1 − 3θ + 3θ2 C(θ),
where C(θ) is as defined in (2.2) and (2.3).
Choosing a = 0, b = 1, and f defined in (2.15), we get
Z 1
1
θ
f (t) dt =
− ,
24 8
0
1 θ
1
f (0) = f (1) = 0,
f
= − ,
2
8 4
Z 1
1 − 3θ + 3θ2
(f 0 (t))2 dt =
12
0
and
N (f ; a, b; θ) = 0
such that the left-hand side becomes
(2.17)
L.H.S.(2.16) =
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
1 − 3θ + 3θ2
.
12
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6
Z HENG L IU
We also find that the right-hand side is
C(1 − 3θ + 3θ2 )
√
.
2 3
From (2.16) – (2.18), we find that C ≥ 2√1 3 , proving that the constant
in (2.1).
(2.18)
R.H.S.(2.16) =
1
√
2 3
is the best possible
Remark 2.2. If we take θ = 0, θ = 1 and θ = 21 in (2.1) with (2.2) and (2.3), we recapture
the sharp midpoint type inequality (1.4) with (1.5) and (1.6), the sharp trapezoid type inequality
(1.7) with (1.8) and (1.9) and the sharp averaged midpoint-trapezoid type inequality (1.10) with
(1.11), respectively. Thus Theorem 2.1 may be regarded as a generalization of Theorem 1.2,
Theorem 1.3 and Theorem 1.4.
Remark 2.3. If we take θ = 31 , we get a sharp Simpson type inequality as
Z b
b − a
(b − a) 32
a
+
b
(2.19)
f (a) + 4f
+ f (b) −
f (t) dt ≤
C4 ,
6
2
6
a
where
12
[f (b) − f (a)]2
0 2
2
− [R(f ; a, b)]
(2.20)
C4 = kf k2 −
b−a
and
√ Z b
1
2 3 f (a) + f (b)
1
(2.21)
R(f ; a, b) = N f ; a, b;
=√
−
f (t) dt .
2
b−a a
3
b−a
3. A PPLICATIONS IN N UMERICAL I NTEGRATION
We restrict further considerations to the averaged midpoint-trapezoid quadrature rule. We
also emphasize that similar considerations may be made for all the quadrature rules considered
in the previous section.
Theorem 3.1. Let π = {x0 = a < x1 < · · · < xn = b} be a given subdivision of the interval
[a, b] such that hi = xi+1 − xi = h = b−a
and let the assumptions of Theorem 1.4 hold. Then
n
we have
Z
n−1 b
X
h
xi + xi+1
(3.1) f (t) dt −
f (xi ) + 2f
+ f (xi+1 ) a
4 i=0
2
3
3
(b − a) 2
(b − a) 2
λn (f ).
≤ √
δn (f ) ≤ √
4 3n
4 3n
where
[f (b) − f (a)]2
0 2
(3.2) δn (f ) = kf k2 −
b−a
1
"
# 2  2
n−1
n−1 X
X
1
xi + xi+1
−
f (x0 ) + f (xn ) + 2
f (xi ) − 2
f

b−a
2
i=1
i=0
and
(3.3)
λn (f ) =
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
[f (b) − f (a)]2
kf 0 k22 −
b−a
12
.
http://jipam.vu.edu.au/
S HARP I NEQUALITIES
7
Proof. From (1.10) and (1.11) in Theorem 1.4 we obtain
Z xi+1
h
x
+
x
i
i+1
(3.4) f (xi ) + 2f
+ f (xi+1 ) −
f (t) dt
4
2
xi
Z xi+i
3
h2
1
≤ √
(f 0 (t))2 dt − [f (xi+1 ) − f (xi )]2
h
4 3
xi
2 ) 12
1
xi + xi+1
.
f (xi ) − 2f
+ f (xi+1 )
−
h
2
By summing (3.4) over i from 0 to n − 1 and using the generalized triangle inequality, we get
Z
n−1 b
X
h
x
+
x
i
i+1
(3.5) f (xi ) + 2f
+ f (xi+1 ) f (t) dt −
a
4 i=0
2
3
n−1 Z xi+i
h2 X
1
≤ √
(f 0 (t))2 dt − [f (xi+1 ) − f (xi )]2
h
4 3 i=0
xi
2 ) 12
xi + xi+1
1
f (xi ) − 2f
+ f (xi+1 )
−
.
h
2
By using the Cauchy inequality twice, we can easily obtain
n−1 Z xi+1
X
1
(3.6)
(f 0 (t))2 dt − [f (xi+1 ) − f (xi )]2
h
xi
i=0
2 ) 12
1
xi + xi+1
− f (xi ) − 2f
+ f (xi+1 )
h
2
≤
√
(
n
n−1
kf 0 k22
n X
[f (xi+1 ) − f (xi )]2
−
b − a i=0
2 ) 12
n−1 X
n
xi + xi+1
−
f (xi ) − 2f
+ f (xi+1 )
b − a i=0
2
≤
√
[f (b) − f (a)]2
n kf 0 k22 −
b−a
1
"
# 2  2
n−1
n−1 X
X
1
xi + xi+1
−
f (x0 ) + f (xn ) + 2
f (xi ) − 2
f
.

b−a
2
i=1
i=0
Consequently, the inequality (3.1) with (3.2) and (3.3) follow from (3.5) and (3.6).
Remark 3.2. It should be noticed that Theorem 3.1 seems to be a revision and an improvement
of the corresponding result in [2, Theorem 6.1].
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
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8
Z HENG L IU
R EFERENCES
[1] N. UJEVIĆ, A generalization of the pre-Grüss inequality and applications to some qradrature formulae, J. Inequal. Pure Appl. Math., 3(1) (2002), Art. 13. [ONLINE: http://jipam.vu.edu.
au/article.php?sid=165].
[2] N. UJEVIĆ, Sharp error bounds for some quadrature formulae and applications, J. Inequal. Pure
Appl. Math., 7(1) (2006), Art. 8. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=622].
J. Inequal. Pure and Appl. Math., 7(5) Art. 172, 2006
http://jipam.vu.edu.au/