Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 26, 2002
SOME INTEGRAL INEQUALITIES INVOLVING TAYLOR’S REMAINDER. I
HILLEL GAUCHMAN
D EPARTMENT OF M ATHEMATICS ,
E ASTERN I LLINOIS U NIVERSITY,
C HARLESTON , IL 61920, USA
cfhvg@ux1.cts.eiu.edu
Received 17 September, 2001; accepted 23 January, 2001.
Communicated by G. Anastassiou
A BSTRACT. In this paper, using Steffensen’s inequality we prove several inequalities involving
Taylor’s remainder. Among the simplest particular cases we obtain Iyengar’s inequality and one
of Hermite-Hadamard’s inequalities for convex functions.
Key words and phrases: Taylor’s remainder, Steffensen’s inequality, Iyengar’s inequality, Hermite-Hadamard’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
AND
S TATEMENT OF M AIN R ESULTS
In this paper, using Steffensen’s inequality we prove several inequalities (Theorems 1.1 and
1.2) involving Taylor’s remainder. In Sections 3 and 4 we give several applications of Theorems
1.1 and 1.2. Among the simplest particular cases we obtain Iyengar’s inequality and one of
Hermite-Hadamard’s inequalities for convex functions. We prove Theorems 1.1 and 1.2 in
Section 2.
In what follows n denotes a non-negative integer, I ⊆ R is a generic interval, and I ◦ is the
interior of I. We will denote by Rn,f (c, x) the nth Taylor’s remainder of function f (x) with
center c, i.e.
n
X
f (k) (c)
Rn,f (c, x) = f (x) −
(x − c)k .
k!
k=0
The following two theorems are the main results of the present paper.
Theorem 1.1. Let f : I → R and g : I → R be two mappings, a, b ∈ I ◦ with a < b, and let
f ∈ C n+1 ([a, b]), g ∈ C ([a, b]). Assume that m ≤ f (n+1) (x) ≤ M , m 6= M , and g(x) ≥ 0 for
all x ∈ [a, b]. Set
(n)
1
λ=
f (b) − f (n) (a) − m(b − a) .
M −m
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068-01
2
H ILLEL G AUCHMAN
Then
(i)
1
(n + 1)!
Z
b
1
(n + 1)!
Z
a+λ
(x − b + λ)n+1 g(x)dx
b−λ
Z b
1
(x − a)n+1
≤
Rn,f (a, x) − m
g(x)dx
M −m a
(n + 1)!
Z b
1
≤
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
(n + 1)! a
Z
(−1)n+1 a+λ
+
(a + λ − x)n+1 g(x)dx;
(n + 1)! a
and
(ii)
(a + λ − x)n+1 g(x)dx
Z (−1)n+1 b
(x − b)n+1
≤
Rn,f (b, x) − m
g(x)dx
M −m a
(n + 1)!
Z b
1
(b − x)n+1 − (b − λ − x)n+1 g(x)dx
≤
(n + 1)! a
Z
(−1)n+1 b
(x − b + λ)n+1 g(x)dx.
+
(n + 1)! b−λ
a
Theorem 1.2. Let f : I → R and g : I → R be two mappings, a, b ∈ I ◦ with a < b,
and let f ∈ C n+1 ([a, b]), g ∈ C ([a, b]). Assume that f n+1 (x) is increasing on [a, b] and
m ≤ g(x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
Z b
1
m
b−a
λ1 =
(x − a)n+1 g(x)dx −
·
,
n+1
(M − m)(b − a)
M −m n+2
a
Z b
1
m
b−a
(b − x)n+1 g(x)dx −
·
.
λ2 =
n+1
(M − m)(b − a)
M −m n+2
a
Then
f
(i)
(n)
(a − λ1 ) − f
(n)
(n + 1)!
(a) ≤
(M − m)(b − a)n+1
Z
b
Rn,f (a, x)(g(x) − m)dx
a
≤ f (n) (b) − f (n) (b − λ1 );
and
(ii)
f
(n)
(a + λ2 ) − f
(n)
n+1
(a) ≤ (−1)
(n + 1)!
(M − m)(b − a)n+1
Z
b
Rn,f (b, x) (g(x) − m) dx
a
≤ f (n) (b) − f (n) (b − λ2 ).
Remark 1.3. It is easy to verify that the inequalities in Theorems 1.1 and 1.2 become equalities
if f (x) is a polynomial of degree ≤ n + 1.
2. P ROOFS OF T HEOREMS 1.1
AND
1.2
The following is well-known Steffensen’s inequality:
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S OME I NTEGRAL I NEQUALITIES I NVOLVING TAYLOR ’ S R EMAINDER . I
3
Theorem 2.1. [4]. Suppose the f and g are integrable functions defined on (a, b), f is decreasRb
ing and for each x ∈ (a, b), 0 ≤ g(x) ≤ 1. Set λ = a g(x)dx. Then
b
Z
b
Z
a+λ
Z
f (x)dx ≤
f (x)g(x)dx ≤
b−λ
a
f (x)dx.
a
Proposition 2.2. Let f : I → R and g : I → R be two maps, a, b ∈ I ◦ with a < b and
let f ∈ C n+1 ([a, b]), g ∈ C[a, b]. Assume that 0 ≤ f (n+1) (x) ≤ 1 for all x ∈ [a, b] and
Rb
(t − x)n g(t)dt is a decreasing function of x on [a, b]. Set λ = f (n) (b) − f (n) (a). Then
x
Z
1
(n + 1)!
(2.1)
b
(x − b + λ)n+1 g(x)dx
b−λ
Z
b
≤
Rn,f (a, x)g(x)dx
Z b
1
≤
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
(n + 1)! a
Z
(−1)n+1 a+λ
(a + λ − x)n+1 g(x)dx.
+
(n + 1)! a
a
Proof. Set
Z
1 b
Fn (x) =
(t − x)n g(t)dt,
n! x
Gn (x) = f n+1 (x),
Z b
λ =
Gn (x)dx = f (n) (b) − f (n) (a).
a
Then Fn (x), Gn (x), and λ satisfy the conditions of Theorem 2.1. Therefore
Z
b
Z
b
Fn (x)dx ≤
(2.2)
b−λ
Z
a+λ
Fn (x)Gn (x)dx ≤
a
Fn (x)dx.
a
It is easy to see that Fn0 (x) = −Fn−1 (x). Hence
Z
b
Z
b
Fn (x)df (n) (x)
a
b Z b
(n)
= f (x)Fn (x) +
f (n) (x)Fn−1 (x)dx
Fn (x)Gn (x)dx =
a
a
=−
f
(n)
J. Inequal. Pure and Appl. Math., 3(2) Art. 26, 2002
(a)
n!
Z
a
b
n
Z
(x − a) g(x)dx +
a
b
Fn−1 (x)Gn−1 (x)dx
a
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H ILLEL G AUCHMAN
f (n) (a)
=−
n!
Z
b
f (n−1) (a)
(x − a) g(x)dx −
(n − 1)!
n
a
b
Z
n−1
(x − a)
Z
b
g(x)dx +
a
Fn−2 (x)Gn−2 (x)dx
a
= ...
Z
f (n−1) (a) b
(x − a) g(x)dx −
(x − a)n−1 g(x)dx
(n
−
1)!
a
a
Z b
Z b
− · · · − f (a)
g(x)dx +
f (x)g(x)dx.
f (n) (a)
=−
n!
Z
b
n
a
a
Thus
Z
b
b
Z
Fn (x)Gn (x)dx =
(2.3)
Rn,f (a, x)g(x)dx.
a
a
In addition
a+λ
Z b
Z
1 a+λ
n
Fn (x)dx =
(t − x) g(t)dt dx.
n! a
a
x
Changing the order of integration, we obtain
Z a+λ
Fn (x)dx
a
Z t
Z a+λ
Z
Z
1 a+λ
1 b
n
n
=
(t − x) g(t)dx dt +
(t − x) g(t)dx dt
n! a
n! a+λ a
a
x=t
x=a+λ
Z a+λ
Z
(t − x)n+1 1 b
(t − x)n+1 1
=−
g(t)
dt −
g(t)
dt
n! a
n + 1 x=a
n! a+λ
n + 1 x=a
Z a+λ
Z b
1
1
n+1
=
(t − a) g(t)dt −
(t − a − λ)n+1 − (t − a)n+1 g(t)dt
(n + 1)! a
(n + 1)! a+λ
Z b
Z b
1
1
n+1
=
(t − a) g(t)dt −
(t − a − λ)n+1 g(t)dt
(n + 1)! a
(n + 1)! a
Z a+λ
1
+
(t − a − λ)n+1 g(t)dt.
(n + 1)! a
Z
Thus,
Z
(2.4)
a
a+λ
1
Fn (x)dx =
(n + 1)!
Z
b
(x − a)n+1 − (x − a − λ)n+1 g(x)dx
a
(−1)n+1
+
(n + 1)!
Z
a+λ
(a + λ − x)n+1 g(x)dx.
a
Similarly we obtain
Z
(2.5)
b
1
Fn (x)dx =
(n + 1)!
b−λ
Z
b
(x − b + λ)n+1 g(x)dx
b−λ
Substituting (2.3), (2.4), and (2.5) into (2.2), we obtain (2.1).
Proposition 2.3. Let f : I → R and g : I → R be two maps, a, b ∈ I ◦ with a < b and
R b let
n+1
(n+1)
f ∈C
([a, b]), g ∈ C ([a, b]). Assume that m ≤ f
(x) ≤ M for all x ∈ [a, b] and x (t −
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S OME I NTEGRAL I NEQUALITIES I NVOLVING TAYLOR ’ S R EMAINDER . I
5
(n)
1
x)n g(t)dt is a decreasing function of x on [a, b]. Set λ = M −m
f (b) − f (n) (a) − m(b − a) .
Then
Z b
1
(2.6)
(x − b + λ)n+1 g(x)dx
(n + 1)! b−λ
Z b
1
(x − a)n+1
≤
Rn,f (a, x) − m
g(x)dx
M −m a
(n + 1)!
Z b
1
≤
(x − a)n+1 − (x − a − λn+1 ) g(x)dx
(n + 1)! a
Z
(−1)n+1 a+λ
+
(a + λ − x)n+1 g(x)dx.
(n + 1)! a
Proof. Set
f˜(x) =
1
(x − a)n+1
f (x) − m
.
M −m
(n + 1)!
Then 0 ≤ f˜(n+1) (x) ≤ 1 and
λ=
(n)
1
f (b) − f (n) (a) − m(b − a) = f˜(n) (b) − f˜(n) (a).
M −m
Hence f˜(x), g(x), and λ satisfy the conditions of Proposition 2.2. Substituting f˜(x) instead of
f (x) into (2.1), we obtain (2.6).
Rb
Proof of Theorem 1.1(i). If g(x) ≥ 0 for all x ∈ [a, b], then x (t − x)n g(t)dt is a decreasing
function of x on [a, b]. Hence Proposition 2.3 implies Theorem 1.1(i).
Proof of Theorems 1.1(ii), 1.2(i), and 1.2(ii). Proofs of Theorems 1.1(ii), 1.2(i), and 1.2(ii) are
similar to the above proof of Theorem 1.1(i). For the proof of Theorem 1.1(ii) we take
Z
1 x
Fn (x) = −
(x − t)n g(t)dt, Gn (x) = f n+1 (x).
n! a
For the proof of Theorem 1.2(i) we take
Fn (x) = −f
(n+1)
1
(x), Gn (x) =
n!
b
Z
(t − x)n g(t)dt.
x
For the proof of Theorem 1.2(ii) we take
Fn (x) = −f
(n+1)
1
(x), Gn (x) =
n!
Z
x
(x − t)n g(t)dt.
a
3. A PPLICATIONS OF T HEOREM 1.1
Theorem 3.1. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C n+1 ([a, b]).
Assume that m ≤ f (n+1) (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
λ=
(n)
1
f (b) − f (n) (a) − m(b − a) .
M −m
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H ILLEL G AUCHMAN
Then
1
m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
≤
Rn,f (a, x)dx
(i)
a
≤
1
M (b − a)n+2 − (M − m)(b − a − λ)n+2 ;
(n + 2)!
and
1
m(b − a)n+2 + (M − m)λn+2
(n + 2)!
Z b
n+1
≤ (−1)
Rn,f (b, x)dx
(ii)
a
1
≤
M (b − a)n+2 − (M − m)(b − a − λ)n+2 .
(n + 2)!
Proof. Take g(x) ≡ 1 on [a, b] in Theorem 1.1.
Two inequalities of the form A ≤ X ≤ B and A ≤ Y ≤ B imply two new inequalities
A ≤ 21 (X + Y ) ≤ B and |X − Y | ≤ B − A. Applying this construction to inequalities (i)
and (ii) of Theorem 3.1, we obtain the following two more symmetric with respect to a and b
inequalities:
Theorem 3.2. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C n+1 ([a, b]).
Assume that m ≤ f n+1 (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
(n)
1
λ=
f (b) − f (n) (a) − m(b − a) .
M −m
Then
1
(i)
[m(b − a)n+2 + (M − m)λn+2 ]
(n + 2)!
Z b
1
Rn,f (a, x) + (−1)n+1 Rn,f (b, x) dx
≤
a 2
1
≤
M (b − a)n+2 − (M − m)(b − a − λ)n+2 ;
(n + 2)!
and
Z b
n
(ii) [Rn,f (a, x) + (−1) Rn,f (b, x)] dx
a
M −m (b − a)n+2 − λn+2 − (b − a − λ)n+2 .
(n + 2)!
We now consider the simplest cases of inequalities (i) and (ii) of Theorem 3.2, namely the
cases when n = 0 or 1.
Case 1. n = 0
Inequality (i) of Theorem 3.2 for n = 0 gives us the following result.
Theorem 3.3. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b and let f ∈ C 1 ([a, b]).
Assume that m ≤ f 0 (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
1
λ=
[f (b) − f (a) − m(b − a)] .
M −m
≤
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S OME I NTEGRAL I NEQUALITIES I NVOLVING TAYLOR ’ S R EMAINDER . I
Then
m+
7
(M − m)λ2
f (b) − f (a)
(M − m)(b − a − λ)2
≤
≤
M
−
.
(b − a)2
b−a
(b − a)2
(a)
Remark 3.4. Theorem 3.3 is an improvement of a trivial inequality m ≤ f (b)−f
≤ M.
b−a
For n = 0, inequality (ii) of Theorem 3.2 gives the following result:
Theorem 3.5. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C 1 ([a, b]).
Assume that m ≤ f 0 (x) ≤ M , m 6= M for all x ∈ [a, b]. Then
Z b
[f (b) − f (a) − m(b − a)] [M (b − a) − f (b) + f (a)]
f
(a)
+
f
(b)
(b − a) ≤
.
f (x)dx −
2
2(M − m)
a
Theorem 3.5 is a modification of Iyengar’s inequality due to Agarwal and Dragomir [1]. If
|f 0 (x)| ≤ M , then taking m = −M in Theorem 3.5, we obtain
Z b
M (b − a)2
1
f
(a)
+
f
(b)
≤
f
(x)dx
−
(b
−
a)
−
[f (b) − f (a)]2 .
2
4
4M
a
This is the original Iyengar’s inequality [2]. Thus, inequality (ii) of Theorem 3.2 can be considered as a generalization of Iyengar’s inequality.
Case 2. n = 1
In the case n = 1, inequality (i) of Theorem 3.2 gives us the following result:
Theorem 3.6. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b and let f ∈ C 2 ([a, b]).
Assume that m ≤ f 00 (x) ≤ M , m 6= M , for all x ∈ [a, b]. Set
1
λ=
[f 0 (b) − f 0 (a) − m(b − a)] .
M −m
Then
Z b
f (a) + f (b)
f 0 (b) − f 0 (a)
1
3
3
m(b − a) + (M − m)λ ≤
f (x)dx −
(b − a) +
(b − a)2
6
2
4
a
1
≤
M (b − a)3 − (M − m)(b − a − λ)3 .
6
In the case n = 1, inequality (ii) of Theorem 3.2 implies that if f ∈ C 2 ([a, b]) and m ≤
f 00 (x) ≤ M , then
f (b) − f (a) f 0 (a) + f 0 (b) [f 0 (b) − f 0 (a) − m(b − a)] [M (b − a) − f 0 (b) + f 0 (a)]
≤
−
.
b−a
2
2(b − a)(M − m)
This result follows readily from Iyengar’s inequality if we take f 0 (x) instead of f (x) in Theorem
3.5.
4. A PPLICATIONS OF T HEOREM 1.2
Take g(x) = M on [a, b] in Theorem 1.2. Then λ1 = λ2 =
b−a
n+2
and Theorem 1.2 implies
Theorem 4.1. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C n+1 ([a, b]).
Assume that f n+1 (x) is increasing on [a, b]. Then
(b − a)n+1 (n)
b−a
(n)
(i)
f
a+
− f (a)
(n + 1)!
n+2
Z b
(b − a)n+1 (n)
b−a
(n)
≤
Rn,f (a, x)dx ≤
f (b) − f
b−
;
(n + 1)!
n+2
a
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H ILLEL G AUCHMAN
and
(b − a)n+1 (n)
b−a
(n)
(ii)
f
a+
− f (a)
(n + 1)!
n+2
Z b
(b − a)n+1 (n)
b−a
n+1
(n)
≤ (−1)
Rn,f (b, x)dx ≤
f (b) − f
b−
.
(n + 1)!
n+2
a
The next theorem follows from Theorem 4.1 in exactly the same way as Theorem 3.2 follows
from Theorem 3.1.
Theorem 4.2. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C n+1 ([a, b]).
Assume that f n+1 (x) is increasing on [a, b]. Then
(b − a)n+1 (n)
b−a
(n)
(i)
f
a+
− f (a)
(n + 1)!
n+2
Z
1 b
≤
Rn,f (a, x) + (−1)n+1 Rn,f (b, x) dx
2 a
(b − a)n+1 (n)
b−a
(n)
≤
f (b) − f
b−
;
(n + 1)!
n+2
Z b
n
(ii) [Rn,f (a, x) + (−1) Rn,f (b, x)] dx
a
(b − a)n+1 (n)
b−a
b−a
(n)
(n)
(n)
≤
f (b) − f
b−
−f
a+
+ f (a) .
(n + 1)!
n+2
n+2
We now consider inequalities (i) and (ii) of Theorem 4.2 in the simplest cases when n = 0 or
1.
Case 1. n = 0.
Inequality (i) of Theorem 4.2 gives a trivial fact: If f 0 (x) increases then f a+b
≤
2
Inequality (ii) of Theorem 4.2 gives the following result: If f 0 (x) is increasing, then
(4.1)
f
a+b
2
1
≤
b−a
Z
b
f (x)dx ≤ f (a) + f (b) − f
a
a+b
2
f (a)+f (b)
.
2
.
The left inequality (2.1) is a half of the famous Hermite-Hadamard’s inequality [3]: If f (x) is
convex, then
Z b
a+b
1
f (a) + f (b)
(4.2)
f
≤
f (x)dx ≤
.
2
b−a a
2
Note that the right inequality (4.1) is weaker than the right inequality (4.2). Thus, inequality (ii)
of Theorem
4.2Rcan be considered as a generalization of the Hermite-Hadamard’s inequality
b
a+b
1
f 2 ≤ b−a
f (x)dx, where f (x) is convex.
a
Case 2. n = 1.
In this case Theorem 4.2 implies the following two results:
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S OME I NTEGRAL I NEQUALITIES I NVOLVING TAYLOR ’ S R EMAINDER . I
9
Theorem 4.3. Let f : I → R be a mapping, a, b ∈ I ◦ with a < b, and let f ∈ C 2 ([a, b]).
Assume that f 00 (x) is increasing on [a, b]. Then
(b − a)2 0
b−a
0
(i)
f a+
+ f (a)
2
3
Z b
f 0 (b) − f 0 (a)
f (a) + f (b)
≤
f (x)dx −
(b − a) +
(b − a)2
2
4
a
b−a
(b − a)2 0
0
≤
f (b) − f
;
2
3
f (b) − f (a) f 0 (a) + f 0 (b) (ii) −
b−a
2
b−a
b−a
1 0
0
0
0
≤
f (a) − f a +
−f b−
+ f (b) .
2
3
3
R EFERENCES
[1] R.P. AGARWAL AND S.S. DRAGOMIR, An application of Hayashi’s inequality for differentiable
functions, Computers Math. Applic., 32(6) (1996), 95–99.
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