STUDIA UNIV. “BABES¸–BOLYAI”, MATHEMATICA, VolumeLII, Number 1, March 2007
OPTIMAL MULTIVARIATE OSTROWSKI EULER TYPE INEQUALITIES
GEORGE A. ANASTASSIOU
Abstract. In [8] we derived general tight multivariate high order Os- trowski type inequalities for the estimate of the error of a multivariate functionfevaluated at a point from its average. The estimates involve only the single partial derivatives offand are with respect tok · kp, 1≤p≤ ∞.
We give here specific applications of these results to the multivariate trape- zoid and midpoint rules for functionsf differentiable up to order 6. We prove sharpness of these inequalities for differentiation ordersm= 1,2,4 and with respect tok · k∞.
1. Introduction
We mention as inspiration to our work the great Ostrowski inequality, see [12], [3], [5].
Theorem 1. It holds
f(x)− 1
b−a Z b
a
f(t)dt
≤
"
1
4 + x−a+b2 2 (b−a)2
#
(b−a)kf0k∞, (1) wheref ∈C1([a, b]),x∈[a, b], which is a sharp inequality.
Here Bk(x), k ≥0, are the Bernoulli polynomials, Bk =Bk(0), k ≥0, the Bernoulli numbers, andB∗k(x),k≥0, are periodic functions of period one, related to the Bernoulli polynomials as
B∗k(x) =Bk(x), 0≤x <1, (2)
Received by the editors: 24.10.2006.
2000Mathematics Subject Classification.26D10, 26D15.
Key words and phrases. Multivariate Euler identity, multivariate Ostrowski inequality, multivariate trapezoid and midpoint rules, sharp inequality.
and
Bk∗(x+ 1) =Bk∗(x), x∈R. (3) Some basic properties of Bernoulli polynomials follow (see [1, 23.1]). We have
B0(x) = 1, B1(x) =x−1
2, B2(x) =x2−x+1 6, and
Bk0(x) =kBk−1(x), k∈N, (4) Bk(x+ 1)−Bk(x) =kxk−1, k ≥0. (5) Clearly B0∗ = 1, B1∗ is a discontinuous function with a jump of −1 at each integer, andBk∗,k≥2, is a continuous function. Notice that Bk(0) =Bk(1) =Bk,k≥2.
We make
Assumption 1. Let f and the existing ∂∂x`f` j
, all` = 1, . . . , m; j = 1, . . . , n, be continuous real valued functions on
n
Q
i=1
[ai, bi];m, n∈N,ai, bi∈R. A general set of suppositions follow
Assumption 2. Here m∈N,j= 1, . . . , n. We suppose 1) f:
n
Q
i=1
[ai, bi]→Ris continuous.
2) ∂∂x`f` j
are existing real valued functions for allj= 1, . . . , n;`= 1, . . . , m−2.
3) For eachj = 1, . . . , nwe assume that
∂m−1f
∂xm−1j (x1, . . . , xj−1,·, xj+1, . . . , xn) is a continuous real valued function.
4) For eachj = 1, . . . , nwe assume that gj(·) := ∂mf
∂xmj (x1, . . . , xj−1,·, xj+1, . . . , xn)
exists and is real valued with the possibility of being infinite only over an at most countable subset of(aj, bj).
5) Parts #3, #4 are true for all
(x1, . . . , xj−1, xj+1, . . . , xn)∈
n
Y
i=1i6=j
[ai, bi].
6) The functions forj= 2, . . . , n;`= 1, . . . , m−2, qj
j−1
z }| {
·,·,· · · ,·
:= ∂`f
∂x`j
j−1
z }| {
·,·,·,· · ·,·, xj, xj+1, . . . , xn
are continuous on
j−1
Q
i=1
[ai, bi], for each
(xj, xj+1, . . . , xn)∈
n
Y
i=j
[ai, bi].
7) The functions for eachj= 1, . . . , n, ϕj
j
z }| {
·,·,·,· · ·,·,·,·
:=∂mf
∂xmj
j
z }| {
·,·,·,· · ·,·, xj+1, . . . , xn
∈L1
j
Y
i=1
[ai, bi]
! ,
for any (xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi].
Some weaker general suppositions follow.
Assumption 3. Herem∈N,j= 1, . . . , n, and only the Parts #1, #2, #6,
#7 of Assumption 2 remain the same. We further assume that for eachj= 1, . . . , n and over[aj, bj], the function
∂m−1
∂xm−1j f(x1, . . . , xj−1,·, xj+1, . . . , xn) is absolutely continuous, and this is true for all
(x1, . . . , xj−1, xj+1, . . . , xn)∈
n
Y
i=1i6=j
[ai, bi].
In [8] we proved the following multivariate Euler type identity.
Theorem 2. All as in Assumption 1 or 2 or 3 for m, n∈ N, xi ∈ [ai, bi], i= 1,2, . . . , n. Then
Emf(x1, x2, . . . , xn) :=f(x1, x2, . . . , xn) (6)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn−
n
X
j=1
Aj =
n
X
j=1
Bj,
where forj= 1, . . . , n we have
Aj :=Aj(xj, xj+1, . . . , xn) (7)
= 1
j−1
Q
i=1
(bi−ai) (m−1
X
k=1
(bj−aj)k−1
k! Bk
xj−aj
bj−aj
× Z
j−1
Q
i=1
[ai,bi]
∂k−1f
∂xk−1j (s1, s2, . . . , sj−1, bj, xj+1, . . . , xn)
− ∂k−1f
∂xk−1j (s1, s2, . . . , sj−1, aj, xj+1, . . . , xn)
ds1· · ·dsj−1
!) , and
Bj :=Bj(xj, xj+1, . . . , xn) (8)
:= (bj−aj)m−1 m!
j−1
Q
i=1
(bi−ai) (Z
j
Q
i=1
[ai,bi]
Bm
xj−aj
bj−aj
−Bm∗
xj−sj
bj−aj
!
∂mf
∂xmj (s1, s2, . . . , sj, xj+1, . . . , xn)
!
ds1ds2· · ·dsj
) . Whenm= 1 thenAj = 0,j= 1, . . . , n.
Also in [8] we proved the following tight multivariate Ostrowski type inequal- ities.
Theorem 3. Suppose Assumptions 1 or 2 or 3. Let Emf(x1, x2, . . . , xn) as in (6) andAj forj= 1, . . . , n as in (7),m∈N. In particular we assume that
∂mf
∂xmj
j
z}|{· · ·, xj+1, . . . , xn
∈L∞
j
Y
i=1
[ai, bi]
! ,
for any(xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi], all j= 1, . . . , n. Then
|Efm(x1, . . . , xn)| (9)
=
f(x1, . . . , xn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn−
n
X
j=1
Aj
≤ 1 m!
n
X
j=1
"
(bj−aj)m
s(m!)2
(2m)!|B2m|+Bm2
xj−aj
bj−aj
!
×
∂mf
∂xmj
j
z}|{· · ·, xj+1, . . . , xn) ∞,Qj
i=1
[ai,bi]
# .
Theorem 4. Suppose Assumptions 1 or 2 or 3. Let Emf(x1, . . . , xn) as in (6), m∈N. Letpj, qj >1 : p1
j +q1
j = 1;j= 1, . . . , n. In particular we assume that
∂mf
∂xmj (. . . , xj+1, . . . , xn)∈Lqj j
Y
i=1
[ai, bi]
! ,
for any(xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi], for all j= 1, . . . , n. Then
|Efm(x1, . . . , xn)| (10)
≤ 1 m!
n
X
j=1
"
(bj−aj)m−
1 qj
j−1 Y
i=1
(bi−ai)
−qj1 Z 1 0
Bm
xj−aj
bj−aj
−Bm(tj)
pj
dtj 1/pj
∂mf
∂xmj (. . . , xj+1, . . . , xn) q
j,
j
Q
i=1
[ai,bi]
# .
Whenpj =qj = 2, allj = 1, . . . , n, then we get
|Emf(x1, . . . , xn)| ≤ 1 m!
n
X
j=1
"
(bj−aj)m−12 j−1
Y
i=1
(bi−ai) −1/2
(11)
×
s(m!)2
(2m)!|B2m|+Bm2 xj−aj
bj−aj
!
×
∂mf
∂xmj (. . . , xj+1, . . . , xn) 2,Qj
i=1
[ai,bi]
# .
Theorem 5. Suppose Assumptions 1 or 2 or 3. Let Emf(x1, . . . , xn) as in (6), m∈N. In particular we assume forj = 1, . . . , nthat
∂mf
∂xmj (. . . , xj+1, . . . , xn)∈L1
j
Y
i=1
[ai, bi]
! , for any(xj+1, . . . , xn)∈Qn
i=j+1[ai, bi]. Then
|Emf(x1, . . . , xn)| ≤ 1 m!
n
X
j=1
"
(bj−aj)m−1
j−1
Q
i=1
(bi−ai)
(12)
∂mf
∂xmj (. . . , xj+1, . . . , xn) 1,
j
Q
i=1
[ai,bi]
Bm(t)−Bm
xj−aj bj−aj
∞,[0,1]
# . The special cases are calculated and estimated further as follows:
1)Whenm= 2r,r∈N, then
|E2rf (x1, . . . , xn)| (13)
≤ 1 (2r)!
n
X
j=1
((bj−aj)2r−1
j−1
Q
i=1
(bi−ai)
∂2rf
∂x2rj (. . . , xj+1, . . . , xn) 1,
j
Q
i=1
[ai,bi]
!
×
(1−2−2r)|B2r|+
2−2rB2r−B2r
xj−aj
bj−aj
) . 2)Whenm= 2r+ 1,r∈N, then
|E2r+1f (x1, . . . , xn)| (14)
≤ 1
(2r+ 1)!
n
X
j=1
( (bj−aj)2r
j−1
Q
i=1
(bi−ai)
∂2r+1f
∂x2r+1j (. . . , xj+1, . . . , xn) 1,Qj
i=1
[ai,bi]
!
×
2(2r+ 1)!
(2π)2r+1(1−2−2r)+
B2r+1
xj−aj
bj−aj
) . And at last
3)Whenm= 1, then
|E1f(x1, . . . , xn)| (15)
≤
n
X
j=1
( 1
j−1
Q
i=1
(bi−ai)
"
∂f
∂xj(. . . , xj+1, . . . , xn) 1,Qj
i=1
[ai,bi]
#
× 1
2+
xj−
aj+bj 2
#) .
In this article we give lots of specific and important applications of Theo- rems 3, 4, 5. see Theorems 6–28. There are produced many multivariate Ostrowski type inequalities for differentiation ordersm= 1, . . . ,6, mostly related to multivariate trapezoid and midpoint rules. When we impose some basic and natural boundary con- ditions, then inequalities become very simple and elegant, see Theorems 25–28. The surprising fact there is, that only a very small number of sets of boundary conditions
is needed comparely to the higher order of differentiation of the involved functions.
At the end we establish sharpness of our inequalities with respect tok · k∞ and for differentiation ordersm= 1,2,4, see Theorems 30-34.
2. Main Results
Here we apply Theorems 3, 4, 5. We get
Theorem 6. Suppose Assumptions 1 or 2 or 3, casem= 1.
i)Assume
∂f
∂xj
(. . . , xj+1, . . . , xn)∈L∞ j
Y
i=1
([ai, bi])
,
for any(xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi], all j= 1, . . . , n. Then
f(x1, x2, . . . , xn)− 1
n
Q
i=1
(bi−ai) Z
Qn i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(16)
≤
n
X
j=1
"
(bj−aj) s1
12+
xj−aj bj−aj
−1 2
2!
×
×
∂f
∂xj
(. . . , xj+1, . . . , xn) ∞,Qj
i=1
[ai,bi]
# . ii)Let pj, qj>1 : p1
j +q1
j = 1;j= 1, . . . , n. In particular we assume that
∂f
∂xj(. . . , xj+1, . . . , xn)∈Lqj j
Y
i=1
[ai, bi]
! ,
for any(xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi], for all j= 1, . . . , n. Then
f(x1, x2, . . . , xn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(17)
≤
n
X
j=1
"
(bj−aj)1−
1 qj
j−1 Y
i=1
(bi−ai)
−qj1Z 1 0
xj−aj
bj−aj
−tj
pj
dtj 1/pj
×
∂f
∂xj
(. . . , xj+1, . . . , xn) qj,
j
Q
i=1
[ai,bi]
# . Whenpj =qj = 2, allj = 1, . . . , n, then
f(x1, . . . , xn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(18)
≤
n
X
j=1
"
q
(bj−aj)
v u u t
j−1
Y
i=1
(bi−ai)
! v u u t 1
12+ xj−aj
bj−aj
−1 2
2!
×
∂f
∂xj(. . . , xj+1, . . . , xn) 2,Qj
i=1
[ai,bi]
# . iii)Here assume forj= 1, . . . , n that
∂f
∂xj(. . . , xj+1, . . . , xn)∈L1 j
Y
i=1
[ai, bi]
! ,
for any(xj+1, . . . , xn)∈
n
Q
i=j+1
[ai, bi]. Then
f(x1, . . . , xn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(19)
≤
n
X
j=1
( 1
j−1
Q
i=1
(bi−ai)
"
∂f
∂xj(. . . , xj+1, . . . , xn) 1,Qj
i=1
[ai,bi]
#
× 1
2 +
xj−
aj+bj
2
) . Notice 1. We have forj= 1, . . . , n:
λj := xj−aj
bj−aj = 0 iffxj=aj,
λj = 1 iffxj=bj,
λj = 1
2 iffxj= aj+bj 2 .
(20)
We continue with
Theorem 7. Suppose Assumptions 1 or 2 or 3, Case m = 1, all xj =aj, j= 1, . . . , n.
i)Assume
∂f
∂xj
(. . . , aj+1, . . . , an)∈L∞ j
Y
i=1
[ai, bi]
!
, allj= 1, . . . , n.
Then
f(a1, a2, . . . , an)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(21)
≤
√3 3
( n X
j=1
(bj−aj)
∂f
∂xj
(. . . , aj+1, . . . , an) ∞,Qj
i=1
[ai,bi]
) . ii)Let pj, qj>1 : p1
j +q1
j = 1;j= 1, . . . , n. In particular we assume that
∂f
∂xj
(. . . , aj+1, . . . , an)∈Lqj
j
Y
i=1
[ai, bi]
! , for allj= 1, . . . , n. Then
f(a1, . . . , an)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(22)
≤
n
X
j=1
"(bj−aj)1−
1 qj
j−1
Q
i=1
(bi−ai)−1/qj
(pj+ 1)1/pj
×
∂f
∂xj
(. . . , aj+1, . . . , an) q
j,
j
Q
i=1
[ai,bi]
# . Whenpj =qj = 2, allj = 1, . . . , n, then
f(a1, . . . , an)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(23)
≤
√3 3
n
X
j=1
"
q
(bj−aj)
v u u t
j−1
Y
i=1
(bi−ai)
×
∂f
∂xj
(. . . , aj+1, . . . , an) 2,
j
Q
i=1
[ai,bi]
#!
.
iii)Here assume forj= 1, . . . , n, that
∂f
∂xj
(. . . , aj+1, . . . , an)∈L1
j
Y
i=1
[ai, bi]
! .
Then
f(a1, . . . , an)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(24)
≤ 1 2
n
X
j=1
((bj−aj+ 1)
j−1
Q
i=1
(bi−ai)
∂f
∂xj
(. . . , aj+1, . . . , an) 1,Qj
i=1
[ai,bi]
) .
Proof. By Theorem 6.
Theorem 8. Suppose Assumptions 1 or 2 or 3, case m = 1, all xj =bj, j= 1, . . . , n.
i)Assume
∂f
∂xj
(. . . , bj+1, . . . , bn)∈L∞ j
Y
i=1
[ai, bi]
! ,
for allj= 1, . . . , n. Then
f(b1, . . . , bn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(25)
≤
√3 3
n
X
j=1
"
(bj−aj)
∂f
∂xj
(. . . , bj+1, . . . , bn) ∞,Qj
i=1
[ai,bi]
# .
ii)Let pj, qj>1 : p1
j +q1
j = 1;j= 1, . . . , n. In particular we assume that
∂f
∂xj
(. . . , bj+1, . . . , bn)∈Lqj
j
Y
i=1
[ai, bi]
! ,
for allj= 1, . . . , n. Then
f(b1, . . . , bn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(26)
≤
n
X
j=1
"
(pj+ 1)−1/pj(bj−aj)1−
1 qj
j−1 Y
i=1
(bi−ai) −qj1
×
×
∂f
∂xj(. . . , bj+1, . . . , bn) q
j,
j
Q
i=1
[ai,bi]
# .
Whenpj =qj = 2, allj = 1, . . . , n, then
f(b1, . . . , bn)− 1
n
Q
i=1
(bi−ai) Z
Qn i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(27)
≤
√ 3 3
n
X
j=1
"
pbj−aj
v u u t
j−1
Y
i=1
(bi−ai)
∂f
∂xj
(. . . , bj+1, . . . , bn) 2,Qj
i=1
[ai,bi]
# .
iii)Here assume forj= 1, . . . , n that
∂f
∂xj
(. . . , bj+1, . . . , bn)∈L1 j
Y
i=1
[ai, bi]
! . Then
f(b1, . . . , bn)− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(28)
≤ 1 2
n
X
j=1
((1 +bj−aj)
j−1
Q
i=1
(bi−ai)
∂f
∂xj(. . . , bj+1, . . . , bn) 1,Qj
i=1
[ai,bi]
) .
Proof. By Theorem 6.
Next come the multivariate midpoint rule inequalities.
Theorem 9. Suppose Assumptions 1 or 2 or 3, casem= 1, all xj =aj+b2 j, j= 1, . . . , n.
i)Assume
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
∈L∞
j
Y
i=1
[ai, bi]
!
, allj= 1, . . . , n.
Then
f
a1+b1
2 ,a2+b2
2 , . . . ,an+bn 2
(29)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
≤ 1 2√ 3
n
X
j=1
(bj−aj)
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
∞,Qj
i=1
[ai,bi]
. ii)Let pj, qj>1 : p1
j +q1
j = 1;j= 1, . . . , n. In particular we assume that
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
∈Lqj
j
Y
i=1
[ai, bi]
! , for allj= 1, . . . , n. Then
f
a1+b1
2 , . . . ,an+bn
2
− 1
n
Q
i=1
(bi−ai) Z
Qn i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(30)
≤1 2
n
X
j=1
"
(pj+1)−pj1 (bj−aj)1−qj1
j−1
Y
i=1
(bi−ai)
!−
1 qj
×
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
qj,
j
Q
i=1
[ai,bi]
# . Whenpj =qj = 2, allj = 1, . . . , n, then
f
a1+b1
2 , . . . ,an+bn 2
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(31)
≤ 1 2√ 3
n
X
j=1
"
pbj−aj
v u u t
j−1
Y
i=1
(bi−ai)
×
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
2,
j
Q
i=1
[ai,bi]
# .
iii)Here assume forj= 1, . . . , n that
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
∈L1 j
Y
i=1
[ai, bi]
! . Then
f
a1+b1
2 , . . . ,an+bn 2
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
(32)
≤1 2
n
X
j=1
( 1
j−1
Q
i=1
(bi−ai)
∂f
∂xj
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
1,Qj
i=1
[ai,bi]
) .
Proof. By Theorem 6.
Next we treat the case of m= 2 and only for the norms k · k∞, k · k2, and specifically forλj = 0, 1, 12, j = 1, . . . , n. The multivariate trapezoid rule estimates follow immediately.
Theorem 10. Suppose Assumptions 1 or 2 or 3, case m= 2, all xj =aj, j= 1, . . . , n.
i)Assume
∂2f
∂x2j(. . . , aj+1, . . . , an)∈L∞ j
Y
i=1
[ai, bi]
! , allj= 1, . . . , n. Then
K2:=
f(a1, a2, . . . , an) +f(b1, a2, . . . , an) 2
(33)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
+ 1 2
( n X
j=2
1
j−1
Q
i=1
(bi−ai) Z
j−1
Q
i=1
[ai,bi]
f(s1, s2, . . . , sj−1, bj, aj+1, . . . , an)
−f(s1, . . . , sj−1, aj, aj+1, . . . , an)
ds1· · ·dsj−1 )
≤ 1 2√
30 ( n
X
j=1
(bi−aj)2
∂2f
∂x2j(. . . , aj+1, . . . , an) ∞,Qj
i=1
[ai,bi]
) . ii)Assume
∂2f
∂x2j(. . . , aj+1, . . . , an)∈L2 j
Y
i=1
[ai, bi]
! , allj= 1, . . . , n. Then
K2≤ 1
2√ 30
( n X
j=1
(bj−aj)3/2 j−1
Y
i=1
(bi−ai) −1/2
× (34)
×
∂2f
∂x2j(. . . , aj+1, . . . , an) 2,Qj
i=1
[ai,bi]
) . Proof. By Theorem 6.
We continue with trapezoid rule estimates.
Theorem 11. Suppose Assumptions 1 or 2 or 3, case m= 2, all xj =bj, j= 1, . . . , n.
i)Assume
∂2f
∂x2j(. . . , bj+1, . . . , bn)∈L∞
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then Λ2:=
f(b1, . . . , bn) +f(a1, b2, . . . , bn)
2 (35)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
−1 2
( n X
j=2
"
1
j−1
Q
i=1
(bi−ai) (Z
j−1
Q
i=1
[ai,bi]
f(s1, . . . , sj−1, bj, bj+1, . . . , bn)
−f(s1, . . . , sj−1, aj, bj+1, . . . , bn)
ds1· · ·dsj−1
#)
≤ 1 2√
30
n
X
j=1
(bj−aj)2
∂2f
∂x2j(. . . , bj+1, . . . , bn) ∞,Qj
i=1
[ai,bi]
. ii)Assume
∂2f
∂x2j(. . . , bj+1, . . . , bn)∈L2
j
Y
i=1
[ai, bi]
! , allj= 1, . . . , n. Then
Λ2≤ 1 2√
30 ( n
X
j=1
"
(bj−aj)3/2 j−1
Y
i=1
(bi−ai) −1/2
(36)
×
∂2f
∂x2j(. . . , bj+1, . . . , bn
2,Qj
i=1
[ai,bi]
#) .
Proof. By Theorem 6.
The multivariate midpoint rule estimates follow.
Theorem 12. Suppose Assumptions 1 or 2 or 3, casem= 2, allxj =aj+b2 j, j= 1, . . . , n.
i)Assume
∂2f
∂x2j
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
∈L∞
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then
M2:=
f
a1+b1
2 , . . . ,an+bn
2
(37)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
≤ 1 8√ 5
( n X
j=1
"
(bj−aj)2
∂2f
∂x2j
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
∞,
j
Q
i=1
[ai,bi]
#) .
ii)Assume
∂2f
∂x2j
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
∈L2
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then
M2≤ 1
8√ 5
( n X
j=1
"
(bj−aj)3/2 j−1
Y
i=1
(bi−ai) −1/2
× (38)
×
∂2f
∂x2j
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
2,Qj
i=1
[ai,bi]
#) . Proof. By Theorem 6.
We continue with trapezoid and midpoint rules inequalities form= 3.
Theorem 13. Suppose Assumptions 1 or 2 or 3, case m= 3, all xj =aj, j= 1, . . . , n.
i)Assume
∂3f
∂x3j(. . . , aj+1, . . . , an)∈L∞ j
Y
i=1
[ai, bi]
! , allj= 1, . . . , n. Then
K3:=
f(a1, . . . , an) +f(b1, a2, . . . , an) 2
(39)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
−
n
X
j=1 withj=k6=1
( 1
j−1
Q
i=1
(bi−ai) 2
X
k=1
(bj−aj)k−1
k! Bk(0)
× Z
j−1
Q
i=1
[ai,bi]
∂k−1f
∂xk−1j (s1, . . . , sj−1, bj, aj+1, . . . , an)
−∂k−1f
∂xk−1j (s1, . . . , sj−1, aj, aj+1, . . . , an)
ds1· · ·dsj−1
!)
≤ 1
12√ 210
n
X
j=1
(bj−aj)3
∂3f
∂x3j(. . . , an+1, . . . , an) ∞,
j
Q
i=1
[ai,bi]
. ii)Assume
∂3f
∂x3j(. . . , aj+1, . . . , an)∈L2
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then
K3≤ 1
12√ 210
n
X
j=1
"
(bj−aj)5/2 j−1
Y
i=1
(bi−ai) −1/2
× (40)
×
∂3f
∂x3j(. . . , aj+1, . . . , an) 2,Qj
i=1
[ai,bi]
# . Proof. By Theorem 6.
Theorem 14. Suppose Assumptions 1 or 2 or 3, case m= 3, all xj =bj, j= 1, . . . , n.
i)Assume
∂3f
∂x3j(. . . , bj+1, . . . , bn)∈L∞ j
Y
i=1
[ai, bi]
! , allj= 1, . . . , n. Then
Λ3:=
f(b1, . . . , bn) +f(a1, b2, . . . , bn) 2
(41)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
−
n
X
j=1 withj=k6=1
( 1
j−1
Q
i=1
(bi−ai) ( 2
X
k=1
(bj−aj)k−1
k! Bk(1)
× Z
j−1
Q
i=1
[ai,bi]
∂k−1f
∂xk−1j (s1, s2, . . . , sj−1, bj, bj+1, . . . , bn)
−∂k−1f
∂xk−1j (s1, s2, . . . , sj−1, aj, bj+1, . . . , bn)
ds1· · ·dsj−1
!))
≤ 1
12√ 210
n
X
j=1
(bj−aj)3
∂3f
∂x3j(. . . , bj+1, . . . , bn) ∞,Qj
i=1
[ai,bi]
. ii)Assume
∂3f
∂x3j(. . . , bj+1, . . . , bn)∈L2
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then Λ3≤ 1
12√ 210
n
X
j=1
"
(bj−aj)5/2 j−1
Y
i=1
(bi−ai) −1/2
× (42)
×
∂3f
∂x3j(. . . , bj+1, . . . , bn) 2,Qj
i=1
[ai,bi]
# . Proof. By Theorem 6.
Theorem 15. Suppose Assumptions 1 or 2 or 3, casem= 3, allxj =aj+b2 j, j= 1, . . . , n.
i)Assume
∂3f
∂x3j
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
∈L∞
j
Y
i=1
[ai, bi]
! ,
allj= 1, . . . , n. Then
M3:=
f
a1+b1
2 , . . . ,an+bn
2
(43)
− 1
n
Q
i=1
(bi−ai) Z
n
Q
i=1
[ai,bi]
f(s1, . . . , sn)ds1· · ·dsn
+1 24
n
X
j=1
( (bj−aj)
j−1
Q
i=1
(bi−ai)
×
× Z
j−1
Q
i=1
[ai,bi]
∂f
∂xj(s1, s2, . . . , sj−1, bj,aj+1+bj+1
2 , . . . ,an+bn
2
− ∂f
∂xj
s1, . . . , sj−1, aj,aj+1+bj+1
2 , . . . ,an+bn 2
!
ds1· · ·dsj−1 )
≤ 1
12√ 210
n
X
j=1
"
(bj−aj)3
∂3f
∂x3j
. . . ,aj+1+bj+1
2 , . . . ,an+bn
2
∞,
j
Q
i=1
[ai,bi]
# .
ii)Assume
∂3f
∂x3j
. . . ,aj+1+bj+1
2 , . . . ,an+bn 2
∈L2
j
Y
i=1
[ai, bi]
! ,