volume 3, issue 2, article 29, 2002.
Received 21 May, 2001;
accepted 01 February, 2002.
Communicated by:N.S. Barnett
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Journal of Inequalities in Pure and Applied Mathematics
NOTE ON THE PERTURBED TRAPEZOID INEQUALITY
XIAO-LIANG CHENG AND JIE SUN
Department of Mathematics Zhejiang University,
Xixi Campus, Zhejiang 310028, The People’s Republic of China.
EMail:xlcheng@mail.hz.zj.cn
c
2000Victoria University ISSN (electronic): 1443-5756 046-01
Littlewood’s Inequality for p−Bimeasures Xiao-liang Cheng and Jie Sun
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Abstract
In this paper, we utilize a variant of the Grüss inequality to obtain some new perturbed trapezoid inequalities. We improve the error bound of the trapezoid rule in numerical integration in some recent known results. Also we give a new Iyengar’s type inequality involving a second order bounded derivative for the perturbed trapezoid inequality.
In the literature [2], [4] – [8], [11], [12] on numerical integration, the follow- ing estimation is well known as the trapezoid inequality:
(1)
Z b a
f(x)dx− 1
2(b−a)(f(a) +f(b))
≤ 1
12M2(b−a)3,
where the mapping f : [a, b] → R is supposed to be twice differentiable on the interval (a, b), with the second derivative bounded on (a, b) by M2 = supx∈(a,b)|f00(x)| < +∞. In [5], the authors derived the error bounds for the trapezoid inequality (1) by different norm of mapping f. In [2, 7,11], the au- thors obtained the trapezoid inequality by the difference of sup and inf bound of the first derivative, that is,
Z b a
f(x)dx− 1
2(b−a)(f(a) +f(b))
≤ 1
8(Γ1−γ1)(b−a)2, whereΓ1 = supx∈(a,b)f0(x)<+∞andγ1 = infx∈(a,b)f0(x)>−∞.
Littlewood’s Inequality for p−Bimeasures Xiao-liang Cheng and Jie Sun
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For the perturbed trapezoid inequality, S. Dragomir et al. [5] obtained the following inequality by an application of the Grüss inequality:
(2)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
≤ 1
32(Γ2−γ2)(b−a)3, where f is supposed to be twice differentiable on the interval (a, b), with the second derivative bounded on(a, b)byΓ2 = supx∈(a,b)f00(x)<+∞andγ2 = infx∈(a,b)f00(x) > −∞. The constant 321 is smaller than 6√15 given in [11] and
1 18√
3 given in [2].
In this note we first improve the constant 321 in the inequality (2) to the best possible one of 1
36√
3. Then we give two new perturbed trapezoid inequalities for high-order differentiable mappings. We need the following variant of the Grüss inequality:
Theorem 1. Let h, g : [a, b] → R be two integrable functions such that φ ≤ g(x)≤Φfor some constantsφ, Φfor allx∈[a, b], then
(3)
1 b−a
Z b a
h(x)g(x)dx− 1 (b−a)2
Z b a
h(x)dx Z b
a
g(x)dx
≤ 1 2
Z b a
h(x)− 1 b−a
Z b a
h(y)dy
dx
(Φ−φ).
Littlewood’s Inequality for p−Bimeasures Xiao-liang Cheng and Jie Sun
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Proof. We write the left hand of inequality (3) as
Z b a
h(x)g(x)dx− 1 b−a
Z b a
h(x)dx Z b
a
g(x)dx
= Z b
a
(h(x)− 1 b−a
Z b a
h(y)dy)g(x)dx.
Denote
I+ = Z b
a
max(h(x)− 1 b−a
Z b a
h(y)dy,0)dx and
I− = Z b
a
min(h(x)− 1 b−a
Z b a
h(y)dy,0)dx.
ObviouslyI++I− = 0. Forφ≤g(x)≤Φ, then Z b
a
(h(x)− 1 b−a
Z b a
h(y)dy)g(x)dx≤I+Φ +I−φ and
− Z b
a
(h(x)− 1 b−a
Z b a
h(y)dy)g(x)dx ≤ −I+φ−I−Φ and hence the obtained result (3) follows.
Theorem 2. Letf : [a, b]→Rbe a twice differentibale mapping on(a, b)with Γ2 = supx∈(a,b)f00(x) < +∞andγ2 = infx∈(a,b)f00(x) > −∞, then we have
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the estimation
(4)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
≤ 1 36√
3(Γ2−γ2)(b−a)3, where the constant 361√3 is the best one in the sense that it cannot be replaced by a smaller one.
Proof. We choose in (3),h(x) =−12(x−a)(b−x)andg(x) = f00(x), we get 1
2 Z b
a
h(x)− 1 b−a
Z b a
h(y)dy
dx =
Z x2
x1
h(x) + 1
12(b−a)2
dx
= 1
36√
3(b−a)3, where x1 = a+ 3−
√3
6 (b−a)and x2 = a+ 3+
√3
6 (b−a). Thus from (3), we derive
Z b a
f(x)dx− 1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
=
Z b a
−1
2(x−a)(b−x)f00(x)dx
− 1 b−a
Z b a
−1
2(x−a)(b−x)dx Z b
a
f00(x)dx
≤ 1 36√
3(Γ2−γ2)(b−a)3.
Littlewood’s Inequality for p−Bimeasures Xiao-liang Cheng and Jie Sun
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To explain the best constant 1
36√
3 in the inequality (4), we can construct the functionf(x) =Rx
a
Ry
a j(z)dz
dyto attain the inequality in (4),
j(x) =
γ2, a≤x < x1 =a+3−√ 3
6 (b−a), Γ2, x1 ≤x < x2 =a+3 +√
3
6 (b−a), γ2, x2 ≤x≤b.
The proof is complete.
Theorem 3. Letf : [a, b]→Rbe a third-order differentibale mapping on(a, b) withΓ3 = supx∈(a,b)f000(x) <+∞andγ3 = infx∈(a,b)f000(x)> −∞, then we have the estimation
(5)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
≤ 1
384(Γ3−γ3)(b−a)4, where the constant 3841 is the best one in the sense that it cannot be replaced by a smaller one.
Proof. We choose in (3),h(x) = 121 (x−a)(2x−a−b)(b−x),g(x) = f000(x), to get
1 2
Z b a
|h(x)− 1 b−a
Z b a
h(y)dy|dx= Z a+b2
a
h(x)dx= 1
384(b−a)4,
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Thus from (3) in Theorem 1, we can derive the inequality (5) immediately.
Finally, we construct the function f(x) = Rx a
Ry a
Rz
j(s)ds dz
dy, where j(x) = Γ3 fora ≤ x < a+b2 andj(x) = γ3 for a+b2 ≤ x ≤ b, then the equality holds in (5).
Theorem 4. Let f : [a, b] → R be a fourth-order differentibale mapping on (a, b)withM4 = supx∈(a,b)|f(4)(x)|<+∞, then
(6)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
≤ 1
720M4(b−a)5, where 7201 is the best possible constant.
Proof. We may write the remainder of the perturbed trapezoid inequality in the kernel form
(7) Z b
a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
= Z b
a
f(4)(x)k4(x)dx, wherek4(x) = 241(x−a)2(b−x)2. Then we get
(8)
Z b a
|k4(x)|dx= 1 24
Z 1 0
x2(1−x)2dx= 1 720.
Littlewood’s Inequality for p−Bimeasures Xiao-liang Cheng and Jie Sun
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Then (7) – (8) imply (6). The equality holds for f(x) = x4, a ≤ x ≤ b in inequality (6).
Remark 0.1. We also can prove Theorem2and3in the kernel form
(9) Z b
a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
12(b−a)2(f0(b)−f0(a))
= Z b
a
f(n)(x)kn(x)dx, wherek2(x) = −12(x−a)(b−x) +121 andk3(x) = 121(x−a)(2x−a−b)(b−x).
By the formula (7) and (9), and derive the perturbed trapezoid inequality for different norms as shown in [5].
Now we present the composite perturbed trapezoid quadrature for an equidis- tant partitioning of interval[a, b]intonsubintervals. Applying Theorems2–4, we obtain
Z b a
f(x)dx=Tn(f) +Rn(f), where
Tn(f) = b−a 2n
n−1
X
i=0
f
a+ib−a n
+f
a+ (i+ 1)b−a n
−(b−a)2
12n2 (f0(b)−f0(a)),
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and the remainderRn(f)satisfies the error estimate
(10) |Rn(f)| ≤
(b−a)3 36√
3n2(Γ2−γ2), ifγ2 ≤f00(x)≤Γ2, ∀x∈(a, b), (b−a)4
384n3 (Γ3−γ3), ifγ3 ≤f000(x)≤Γ3, ∀x∈(a, b) (b−a)5
720n4 M4, if|f(4)(x)| ≤M4, ∀x∈(a, b).
Then we can use (10) to get different error estimates of the composite perturbed trapezoid quadrature.
As in [5], we may also apply the Theorems 2, 3 and 4 to special means.
In this case we may improve some of the bounds related to inequalities about special means as given in [5, p. 492-494].
Furthermore, we discuss the Iyengar’s type inequality for the perturbed trape- zoidal quadrature rule for functions whose first and second order derivatives are bounded. In [1, 3, 9, 10] they proved the following interesting inequality in- volving bounded derivatives.
Iff is a differentiable function on(a, b)and|f0(x)| ≤M1, then
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b))
≤ M1(b−a)2
4 − (f(b)−f(a))2 4M1 .
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If|f00(x)| ≤M2, x∈[a, b]for positive constantM2 ∈R, then
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
8(b−a)2(f0(b)−f0(a))
≤ M2
24 (b−a)3− |∆|
M2 3!
,
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
8(b−a)2(f0(b)−f0(a))
≤ M2
24(b−a)3− ∆21(b−a) 16M2
, where
∆ =f0(a)−2f0
a+b 2
+f0(b), (11)
∆1 =f0(a)−2(f(b)−f(a))
b−a +f0(b).
We will prove the following inequality.
Theorem 5. Let f : I → R, where I ⊆ R is an interval. Suppose that f is twice differentiable in the interior
◦
I of I, and let a, b ∈
◦
I with a < b. If
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|f00(x)| ≤M2, x∈[a, b]for positive constantM2 ∈R. Then (12)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
8(b−a)2(f0(b)−f0(a))
≤ 1
24M2(b−a)3− s
|∆1|3(b−a)3 72M2 , where∆1is defined as (11).
Proof. Denote
Jf = Z b
a
f(x)dx− 1
2(b−a)(f(a) +f(b)) + 1
8(b−a)2(f0(b)−f0(a)).
It is easy to see that
Jf = Z b
a
1 2
x− a+b 2
2
f00(x)dx, and
∆1 =f0(a)−2(f(b)−f(a))
b−a +f0(b) (13)
= 1
b−a Z b
a
2
x− a+b 2
f00(x)dx.
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For any|ε| ≤ 18, we get for|f00(x)| ≤M2, x∈[a, b], Jf +ε(b−a)2∆1
= Z b
a
1 2
x− a+b 2
2
+ 2ε(b−a)
x−a+b 2
!
f00(x)dx
≤F(ε)M2(b−a)3, where
F(ε) = 1 (b−a)3
Z b a
1 2
x− a+b 2
2
+ 2ε(b−a)
x− a+b 2
dx
= Z 1
0
1 2
x− 1
2 2
+ 2ε
x−1 2
dx.
For the case0≤ε≤ 18, we have F(ε) =
Z 1 0
1 2
x−1
2 2
+ 2ε
x− 1 2
dx
=
(Z 12−4ε
0
1 2
x− 1
2 2
+ 2ε
x−1 2
! dx
− Z 12
1 2−4ε
1 2
x− 1
2 2
+ 2ε
x− 1 2
! dx
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+ Z 1
1 2
1 2
x− 1
2 2
+ 2ε
x−1 2
! dx
)
= 1 24+ 32
3 ε3.
For the case−18 ≤ε≤0, we have similarly F(ε) = 1
24− 32 3 ε3.
We can prove|∆1| ≤ 12(b−a)M2easily from (13). Thus we choose the param- eter
ε∗ =sign(∆1) s
|∆1|
32(b−a)M2, |ε∗| ≤ 1 8. By the above inequalities, we obtain
Jf ≤F(ε∗)M2 −ε∗(b−a)2∆1 ≤ 1
24(b−a)3M2− s
(b−a)3|∆1|3 72M2 . Replacingf with−f, we have
J−f =−Jf ≤ 1
24(b−a)3M2− s
(b−a)3|∆1|3 72M2 . Thus we obtain bounds for|Jf|and prove the inequality (12).
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Remark 0.2. As|∆1| ≤ 12M2(b−a), we have s
|∆1| M2 ≥
r 2 b−a
|∆1| M2 ,
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b)) + 1
8(b−a)2(f0(b)−f0(a))
≤ M2
24(b−a)3− ∆21(b−a) 6M2 . For the casef0(a) = f0(b) = 0, we have
(14)
Z b a
f(x)dx−1
2(b−a)(f(a) +f(b))
≤ M2
24(b−a)3−2 3
|f(b)−f(a)|2 M2(b−a) . The inequality (14) is sharper than that stated in [9, p. 69].
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