EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 2, 2018, 375-389
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

The Proofs of product inequalities in a generalized
vector space

Benedict Barnes1,∗, E. D. J. Owusu-Ansah1, S. K. Amponsah1, C. Sebil1

1 Department of Mathematics, Kwame Nkrumah University of Science and Technology,
Kumasi, Ghana

Abstract. In this paper, we introduce the proofs of product inequalities:
‖u‖‖v‖ ≤ ‖u‖+ ‖v‖, for all u, v ∈ [0, 2], and ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖, for all u, v ∈ [2,∞). By applying
the first product inequality to the Lp spaces, we observed that if f : Ω → [0, 1], and g : Ω → R,
then ‖f‖p‖g‖p ≤ ‖f‖p + ‖g‖p. Also, if f, g : Ω→ R, then ‖f‖p + ‖g‖p ≤ ‖f‖p‖g‖p.

2010 Mathematics Subject Classifications: 44B45, 44B46

Key Words and Phrases: Product inequality, first product nequality, second product inequality,
generalized vector space, Cauchy-Schwarz inequality

1. Introduction

Inequalities are inevitable tools in the mathematical analysis as they provide bases
for sound arguments. Due to the enormous applications of inequalities, most researchers
are shifting to this line of research by introducing new inequalities. From the historical
standpoint, the triangle inequality was first discovered, see [1]. Since then many researchers
across the globe had obtained different ways of proving triangle inequality. For example,
see a research paper by authors in [2]. After the discovery of triangle inequality the
so-called arithmetic-geometric mean AGM inequality:( n∏

i=1

xi

) 1
n ≤ 1

n

n∑
i=1

xi, ∀ n ∈ N and xi ∈ R+,

with equality occurs when x1 = x2 = . . . = xn, was observed. Specifically, for any two
positive real numbers a and b the AGM inequality becomes

ab ≤
(a+ b

2

)2
,

∗Corresponding author.

Email addresses: bbarnes.cos@knust.edu.gh (B. Barnes),
degraftt@gmail.com (E. D. J. Owusu-Ansah), skamponsah@knust.edu.gh (S. K. Amponsah),
sebicharles@yahoo.com (C. Sebil)

http://www.ejpam.com 375 c© 2018 EJPAM All rights reserved.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 376

see [3]. Several ways of proving the AGM inequality have been observed, see research
papers by authors in [4] and [5].

However, the author in [6] proved that the sum of vector points and one is less than
the product of sum of one and the vector points:

1 +
n∑

i=1

xi < Πn
i=1(1 + xi).

The AGM inequality has been applied to establish the relationships between areas of plane
figures and their perimeters of what is called isoperimetric inequality. For example, see [7] .
Although the inequality of two real numbers has been in history for long time with many
researchers looking at inequalities involving products of real numbers. Newton showed
that the square of a real number between the first and third consecutive real numbers is
greater than their product

Pr−1Pr+1 < P 2
r , ∀ 1 ≤ r < n,

see [8]. In [9], the author obtained another way of proving the Weierstrass inequality by
applying AGM inequality and also, extended the Weierstrass inequalities by the use of
majorization. The author in [10], generalized the Weierstrass inequality in the Euclidean
space. Inequalities involving functions have received much attention in the 21st century.
Jordan as cited in [11] introduced fractional inequality:

2

π
≤ sin(x)

x
< 1, 0 < x ≤ π

2
.

Hilbert constructed double series inequality:

∞∑
m,n=1

ambn
m+ n

≤ π
( ∞∑

m=1

a2
m

) 1
2
( ∞∑

m=1

b2m

) 1
2
,

see [12].
Unlike AGM inequality, the product of two positive real numbers and their sum with-

out any factor multiplier, in Euclidean space, delineate unique inequalities within certain
intervals. In this paper, we provide two inequalities in the interval [0,∞); the first product
inequality ‖u‖‖v‖ ≤ ‖u‖ + ‖v‖ holds for all values of u and v in the interval [0, 2], and
the second product inequality ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖ holds for all u and v in interval [2,∞).
Each of these inequalities is proved by induction. The section 1 contains the introduction.
The section 2 contains the proofs of the first product inequality in generalized linear space
and the second product inequality in the linear space. In section 3, we illustrated the first
product inequality to a real line, norms and trigonometric functions. We extend and also,
apply the second product inequality to the Lp spaces in section 4 of this paper. In section
5, we discuss our main findings and summarize these results in section 6.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 377

2. Main Result

In this section, the two product inequalities are introduced: the first product inequality

‖u‖‖v‖ ≤ ‖u‖+ ‖v‖, ∀ u, v ∈ [0, 2], (1)

and the second product inequality,

‖u‖+ ‖v‖ ≤ ‖u‖‖v‖, ∀ u, v ∈ [0, 2]. (2)

Before we proceed to provide the proofs of the product inequalities some of relevant der-
intions regarding the introduction of product inequalities are given in this paper.

Definition 1 (Inner product). Let V be a linear vector space defined over the real number
field R. A scalar-valued function p : V × V → R that associates with each pair u, v of
vectors in V a scalar, denoted (u, v), is called an inner product on V if and only if

(i) (u, u) > 0 whenever u 6= 0, and (u, u) = 0 if and only if u = 0

(ii) (u, v) = (v, u), ∀ u, v ∈ V

(iii) (αu1 + βu2, v) = α(u1, v) + β(u2, v), ∀ α, β ∈ R, and u1, u2, v ∈ V , see [13].

Definition 2 (Norm). Let V be a linear space over R. A norm on V is a real-valued
function

‖ · ‖ : V → [0,∞)

such that for any u, v ∈ V and α ∈ R the following conditions are met:

‖u‖ ≥ 0, and ‖u‖ = 0, iff u = 0

‖αu‖ = |α|‖u‖, ∀ u ∈ V and α ∈ R

‖u± v‖ ≤ ‖u‖+ ‖v‖, ∀ u, v ∈ V

The norm of a vector u can be generated by the inner product (, )

‖u‖ =
√

(u, u).

See [14].

Definition 3 (Cauchy-Schwarz inequality). Let E be an inner product space. Then

|〈u, v〉|2 ≤ ‖x‖.‖y‖, ∀ x, y ∈ E.

The equality holds if and only if x and y are linearly dependent, see [15].



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 378

Definition 4 (boundedness). Let T : X → Y be a linear map. Then T is said to be a
bounded linear map if there exists K ≥ 0 such that

‖Tx‖ ≤ K‖x‖, ∀ x ∈ X,

and K is the boundedness constant for T . The boundedness of the linear map implies
continuity of T . See [16]

Definition 5 (Integral Operator and Monotonicity of the Integral Operator). Let f(x)
and g(x) be simple functions defined on a set of finite measure E. Then for any α and β,∫

E
(αf(x) + βg(x))dx = α

∫
E
f(x)dx+ β

∫
E
g(x)dx, ∀ f, g ∈ E.

and α, β are scalars. Moreover, if f(x) ≤ g(x), then∫
E
f(x)dx ≤

∫
E
g(x)dx,

see [17].

2.1. The Proof of the First Product Inequality

We provide the proofs of product inequalities, through binomial inequalities, by induc-
tion. Firstly, we consider a positive integer n = 2 as follows.

(u+ v)2 ≥ 0

(u, u) + 2(u, v) + (v, v) ≥ 0

−2(u, v) ≤ {(u, u) + (v, v)}
−2(u, v) ≤ (u, u) + (v, v) + 2(u, v)

−2(u, v)
1

2
(u, v) ≤ (u+ v)2

−(u, v)2 = (u+ v)2

‖ − (u, v)2‖ = ‖(u+ v)2‖
‖(u, v)‖2 ≤ ‖(u+ v)‖2

⇒ ‖u‖‖v‖ ≤ ‖u‖+ ‖v‖.

For n = 4, we observe the following inequalities:

(u+ v)4 ≥ 0

−6(u, u)(v, v) ≤ {(u, u)2 + 4(u, u)(u, v) + 4(u, v)(v, v) + (v, v)2}
−6(u, u)(v, v) ≤ {(u, u)2 + 4(u, u)(u, v) + 6(u, u)(v, v) + 4(u, v)(v, v) + (v, v)2}

−6(u, u)(v, v).
1

6
(u, u)(v, v) ≤ (u+ v)4

−{(u, u)(v, v)}2 = (u+ v)4



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 379

‖ − {(u, u)(v, v)}2‖ = ‖(u+ v)4‖
{‖u‖‖v‖}4 = ‖(u+ v)‖4

⇒ ‖u‖‖v‖ ≤ ‖u‖+ ‖v‖.

In order to generalize the first product inequality, we observed that for any positive integer
n, we obtain:

(u+ v)n ≥ 0

⇒ (u, u)
n
2 +n C1(u, v)(u, u)

n−2
2 +n C2(u, u)

n−2
2 (v, v) +n C3(u, u)

n−4
2 (v, v)(u, v)

+ nC4(u, u)
n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v) +n C6(u, u)

n−6
2 (v, v)3

+ nC7(u, u)
n−8
2 (v, v)3(u, v) + . . .+n Cn

2
(u, u)

n
4 (v, v)

n
4 + . . .+ (u, u)

n
2 ≥ 0

⇒ −nCn
2
(u, u)

n
4 (v, v)

n
4 ≤ {(u, u)

n
2 +n C1(u, v)(u, u)

n−2
2 +n C2(u, u)

n−2
2 (v, v)

+ nC3(u, u)
n−4
2 (v, v)(u, v) +n C4(u, u)

n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v)

+ nC6(u, u)
n−6
2 (v, v)3 +n C7(u, u)

n−8
2 (v, v)3(u, v) + . . .+ (u, u)

n
2 }

⇒ −nCn
2
(u, u)

n
4 (v, v)

n
4 ≤

{
(u, u)

n
2 +n C1(u, v)(u, u)

n−2
2 +n C2(u, u)

n−2
2 (v, v)

+ nC3(u, u)
n−4
2 (v, v)(u, v) +n C4(u, u)

n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v)

+ nC6(u, u)
n−6
2 (v, v)3 +n C7(u, u)

n−8
2 (v, v)3(u, v) + . . .+n Cn

2
(u, u)

n
4 (v, v)

n
4 + . . .+ (u, u)

n
2

}
⇒ −nCn

2
(u, u)

n
4 (v, v)

n
4

1
nCn

2

(u, u)
n
4 (v, v)

n
4 ≤ (u+ v)n

⇒ ‖− (u, u)
n
2 (v, v)

n
2 ‖ = ‖(u+ v)n‖

⇒ {‖u‖‖v‖}n = ‖(u+ v)‖n

⇒ ‖u‖‖v‖ ≤ ‖u‖+ ‖v‖, ∀ u, v ∈ [0, 2]. (3)

2.2. The Proof of the Second Product Inequality

In this subsection, we show that the sum of norms of two vectors less than or equal to
the norms of product of their vectors. Setting n = 2, we obtain:

(u+ v)2 ≥ 0

(u, u) + 2(u, v) + (v, v) ≥ 0

−{(u, u) + (v, v)} ≤ 2(u, v)

−{(u, u) + (v, v) + 2(u, v)} ≤ 2(u, v)

−{(u, u) + (v, v) + 2(u, v)} ≤ 2(u, v).
1

2
(u, v)

−(u+ v)2 = (u, v)2

‖ − (u+ v)2‖ = ‖(u, v)2‖
‖(u+ v)‖2 = ‖(u, v)‖2



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 380

⇒ ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖.

Also, we set n = 4, which gives as the following result:

(u+ v)4 ≥ 0

−{(u, u)2 + 4(u, u)(u, v) + 4(u, v)(v, v) + (v, v)2} ≤ 6(u, u)(v, v)

−{(u, u)2 + 4(u, u)(u, v) + 6(u, u)(v, v) + 4(u, v)(v, v) + (v, v)2} ≤ 6(u, u)(v, v)

−(u+ v)4 ≤ 6(u, u)(v, v).
1

6
(u, u)(v, v)

−(u+ v)4 = {(u, u)(v, v)}2

‖ − (u+ v)4‖ = ‖{(u, u)(v, v)}2‖
‖(u+ v)‖4 = {‖u‖‖v‖}4

⇒ ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖.

Similarly, we generalize the second product inequality by observing the following:

(u+ v)n ≥ 0

⇒ (u, u)
n
2 +n C1(u, v)(u.u)

n−2
2 +n C2(u, u)

n−2
2 (v, v) +n C3(u, u)

n−4
2 (v, v)(u, v)

+ nC4(u, u)
n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v) +n C6(u, u)

n−6
2 (v, v)3

+ nC7(u, u)
n−8
2 (v, v)3(u, v) + . . .+n Cn

2
(u, u)

n
4 (v, v)

n
4 + . . .+ (u, u)

n
2 ≥ 0

⇒ −{(u, u)
n
2 +n C1(u, v)(u, u)

n−2
2 +n C2(u, u)

n−2
2 (v, v)

+ nC3(u, u)
n−4
2 (v, v)(u, v) +n C4(u, u)

n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v)

+ nC6(u, u)
n−6
2 (v, v)3 +n C7(u, u)

n−8
2 (v, v)3(u, v) + . . .+ (u, u)

n
2 }

≤ nCn
2
(u, u)

n
4 (v, v)

n
4

⇒ −
{

(u, u)
n
2 +n C1(u, v)(u, u)

n−2
2 +n C2(u, u)

n−2
2 (v, v)

+ nC3(u, u)
n−4
2 (v, v)(u, v) +n C4(u, u)

n−4
2 (v, v)2 +n C5(u, u)

n−6
2 (v, v)2(u, v)

+ nC6(u, u)
n−6
2 (v, v)3 +n C7(u, u)

n−8
2 (v, v)3(u, v) + . . .+n Cn

2
(u, u)

n
4 (v, v)

n
4 + . . .+ (u, u)

n
2

}
≤ nCn

2
(u, u)

n
4 (v, v)

n
4

⇒ −(u+ v)n ≤n Cn
2
(u, u)

n
4 (v, v)

n
4

1
nCn

2

(u, u)
n
4 (v, v)

n
4

⇒ ‖− (u+ v)n‖ = ‖(u, u)
n
2 (v, v)

n
2 ‖

⇒ ‖(u+ v)‖n = {‖u‖‖v‖}n

⇒ ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖, ∀ u, v ∈ [2,∞). (4)

3. Illustration of the First Product Inequality

In this section of the paper, we illustrate some of the aspects of mathematics where
the first product inequality is feasible. Notwithstanding, the areas of applications of the



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 381

first product inequality are highlighted.

3.1. Illustration of the First Product Inequality to the Real Line

Firstly, we illustrate the first product inequality in (1) to any two real numbers a, b ∈
[0, 2].

Example 1. Let a = 5
6 and b = 13

18 , then∥∥∥5

6

∥∥∥∥∥∥13

18

∥∥∥ ≤
∥∥∥5

6

∥∥∥+
∥∥∥13

18

∥∥∥
65

108
<

28

18
.

Example 2. Let a = −1
2 and b = −3

5 , then∥∥∥−1

2

∥∥∥∥∥∥−3

5

∥∥∥ ≤
∥∥∥−1

2

∥∥∥+
∥∥∥−3

5

∥∥∥
3

10
<

11

10
.

Example 3. Let a = 19
10 and b = 9

5 , then∥∥∥19

10

∥∥∥∥∥∥9

5

∥∥∥ ≤
∥∥∥19

10

∥∥∥+
∥∥∥ 9

10

∥∥∥
171

50
<

37

10
.

3.2. Illustration of the First Product Inequality to the Euclidean Space

Again, the first product inequality is illustrated to the vector points of real numbers as
follows. In this paper, three basic norms: 1− norm, 2− norm and∞-norm are considered.

Example 4. Let a =

(
1
2
2
3

)
and b =

(
4
5
−1
3

)
, then

∥∥∥( 1
2
2
3

)∥∥∥
1

∥∥∥( 4
5
−1
3

)∥∥∥
1
≤

∥∥∥( 1
2
2
3

)∥∥∥
1

+
∥∥∥( 4

5
−1
3

)∥∥∥
1

⇒ 119

90
<

69

30
.

Example 5. Let a =

(
1
2
2
3

)
and b =

(
4
5
−1
3

)
, then

∥∥∥( 1
2
2
3

)∥∥∥
2

∥∥∥( 4
5
−1
3

)∥∥∥
2
≤

∥∥∥( 1
2
2
3

)∥∥∥
2

+
∥∥∥( 4

5
−1
3

)∥∥∥
2

⇒ 13

18
<

28

18
.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 382

Example 6. Let a =

(
1
2
2
3

)
and b =

(
4
5
−1
3

)
, then

∥∥∥( 1
2
2
3

)∥∥∥
∞

∥∥∥( 4
5
−1
3

)∥∥∥
∞
≤

∥∥∥( 1
2
2
3

)∥∥∥
∞

+
∥∥∥( 4

5
−1
3

)∥∥∥
∞

⇒ 8

15
<

22

15
.

3.3. Illustration of the First Product Inequality to Trigonometric Func-
tions

Undoubedly, we construct the inequalities involving the product of sine of an angle
and cosine of an angle. We can see that:

‖ sin(θ)‖ ≤ 1,

and

‖ cos(θ)‖ ≤ 1.

By induction, we start at n = 0, by setting(
sin(θ) cos(θ)

)0

Taking the magnitude of
(

sin(θ) cos(θ)
)0

and applying the Cauchy-Schwarz inequality

to it, we obtain:(
|〈sin(θ), cos(θ)〉|

)0
≤
(
‖ sin(θ)‖‖ cos(θ)‖

)0
. (5)

Applying the first product inequality to the expression on the right hand side of equation
(5) yields (

‖ sin(θ)‖‖ cos(θ)‖
)0
≤

(
‖ sin(θ)‖+ ‖ cos(θ)‖

)0

(
‖ sin(θ)‖‖ cos(θ)‖

)0
= (1 + 1)0(

‖ sin(θ)‖‖ cos(θ)‖
)0

= 20

For n = 1, the following result is obtained:(
sin(θ) cos(θ)

)1

⇒
(
|〈sin(θ), cos(θ)〉|

)1
≤

(
‖ sin(θ)‖‖ cos(θ)‖

)1
. (6)



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 383

Applying the first product inequality to the expression on the right hand side of equation
(6) yields (

‖ sin(θ)‖‖ cos(θ)‖
)1
≤

(
‖ sin(θ)‖+ ‖ cos(θ)‖

)1

(
‖ sin(θ)‖‖ cos(θ)‖

)1
= (1 + 1)1(

‖ sin(θ)‖‖ cos(θ)‖
)1

< 21

Similarly, we observed that when n = 2, the following expression is obtained:(
sin(θ) cos(θ)

)2

⇒
(
|〈sin(θ), cos(θ)〉|

)2
≤

(
‖ sin(θ)‖‖ cos(θ)‖

)2
(7)

Applying the first product inequality to the expression on the right hand side of equation
(7) yields (

‖ sin(θ)‖‖ cos(θ)‖
)2
≤

(
‖ sin(θ)‖+ ‖ cos(θ)‖

)2

(
‖ sin(θ)‖‖ cos(θ)‖

)2
= (1 + 1)2(

‖ sin(θ)‖‖ cos(θ)‖
)2

< 22

Generalizing the product inequality of sin(θ) and cos(θ), we observed that for any
integer n ∈ [0,∞), we obtain:(

sin(θ) cos(θ)
)n

⇒
(
|〈sin(θ), cos(θ)〉|

)n
≤

(
‖ sin(θ)‖‖ cos(θ)‖

)n
. (8)

Applying the first product inequality to the expression on the right hand side of equation
(6), we obtain:(

‖ sin(θ)‖‖ cos(θ)‖
)n

≤
(
‖ sin(θ)‖+ ‖ cos(θ)‖

)n
(
‖ sin(θ)‖‖ cos(θ)‖

)n
= (1 + 1)n(

‖ sin(θ)‖‖ cos(θ)‖
)n

≤ 2n. (9)

Again, the first product inequality is applied to functions which give the following
results.

Theorem 1. Suppose that f be the space of all measurable functions f : [0, 2] → R, and
g : Ω→ [0, 1], then following inequalities hold:

‖ex‖‖ sin(θ)‖ ≤ ‖ex‖+ ‖ sin(θ)‖



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 384

‖ex‖‖ cos(θ)‖ ≤ ‖ex‖+ ‖ cos(θ)‖
‖ tan(θ)‖‖ sin(θ)‖ ≤ ‖ tan(θ)‖+ ‖ sin(θ)‖
‖ tan(θ)‖‖ cos(θ)‖ ≤ ‖ tan(θ)‖+ ‖ cos(θ)‖
‖sec(θ)‖‖ sin(θ)‖ ≤ ‖sec(θ)‖+ ‖ sin(θ)‖

‖cosec(θ)‖‖ cos(θ)‖ ≤ ‖cosec(θ)‖+ ‖ cos(θ)‖
‖cot(θ)‖‖ cos(θ)‖ ≤ ‖cot(θ)‖+ ‖ cos(θ)‖

We observed that the norms of the trigonometric functions: sin(θ) and cos(θ) are seen
as delayed growth norms.

Definition 6. If u and v are any two vectors in the Euclidean space, then

|u+ v| ≤ ‖u‖+ ‖v‖ ≤ 2 max{|u|, |v|}
⇒ |u+ v|p ≤ 2p{|u|p + |v|p},

see [17].

With the above inequalities, the first product inequality is extended to p−normed
spaces as follows:

Theorem 2. If p is any positive integer, then

‖u‖p‖v‖p ≤ 2p{‖u‖p + ‖v‖p}. (10)

Proof : We see from the above inequality that:

|u+ v| ≤ ‖u‖+ ‖v‖

Raising the expression on the both sides of inequality to the power p, we get:

|u+ v|p ≤
∣∣∣‖u‖+ ‖v‖

∣∣∣p
⇒ |u+ v|p ≤ ‖u‖p + ‖v‖p

⇒ |u+ v|p ≤ 2p{‖u‖p + ‖v‖p}
⇒ ‖u‖p + ‖v‖p ≤ 2p{‖u‖p + ‖v‖p}.

Substituting inequality in (3) into the inequality in (10) yields

|‖u‖‖v‖|p ≤ 2p{|u|p + |v|p}
‖u‖p‖v‖p ≤ 2p{‖u‖p + ‖v‖p}, ∀ u, v ∈ [0, 2]

Theorem 3. If u and v are any two vectors in the Euclidean space, then

‖u‖p‖v‖p ≤ ‖u‖p + ‖v‖p, ∀ ‖u‖, ‖v‖ ∈ [0, 2].

P roof : The result in theorem (3) follows from theorem (2).



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 385

Theorem 4. Suppose that at least one of the measurable functions over the domain Ω is
such that

f : Ω→ [0, 1]

and the other measurable function

g : Ω→ R,

then

‖fg‖p ≤ ‖f‖p + ‖g‖p.

P roof : ∣∣∣ ∫
Ω
f(x)g(x)dx

∣∣∣p ≤ ∫
Ω

∣∣∣f(x)g(x)
∣∣∣pdx

Applying the first product inequality, we obtain∣∣∣ ∫
Ω
f(x)g(x)dx

∣∣∣p ≤
∫

Ω

{∣∣∣f(x)|p + |g(x)
∣∣∣p}dx

⇒
(∣∣∣ ∫

Ω
f(x)g(x)dx

∣∣∣p) 1
p

=
(∫

Ω

∣∣∣f(x)|pdx+

∫
Ω
|g(x)

∣∣∣pdx) 1
p

⇒
(∣∣∣ ∫

Ω
f(x)g(x)dx

∣∣∣p) 1
p ≤

(∫
Ω

∣∣∣f(x)|pdx
) 1

p
+
(∫

Ω
|g(x)

∣∣∣pdx) 1
p

⇒ ‖fg‖p ≤ ‖f‖p + ‖g‖p. (11)

Using the Cauchy-Schwarz inequality, we observe that:

‖fg‖p ≤ ‖f‖p‖g‖p. (12)

Substituting inequality (11) into inequality (12) yields

‖f‖p‖g‖p ≤ ‖f‖p + ‖g‖p.

We show how theorem (4) is applied to estimate two smooth functions defined on
domains.

Theorem 5. Suppose that f(x), g(x) are C∞(Ω) such that f : Ω → [0, 1] and g : Ω →
[0, 1], then the following inequality holds:∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ 2(b− a).



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 386

Proof : ∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ ∫ b

a

∣∣∣ cos(nx) sin(nx)
∣∣∣dx∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ ∫ b

a

∣∣∣ cos(nx)
∣∣∣∣∣∣ sin(nx)

∣∣∣dx
Applying the first product inequality, we obtain:∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ ∫ b

a
{
∣∣∣ cos(nx)

∣∣∣+
∣∣∣ sin(nx)

∣∣∣}dx
⇒
∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ =

∫ b

a

∣∣∣ cos(nx)
∣∣∣dx+

∫ b

a

∣∣∣ sin(nx)
∣∣∣dx

⇒
∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ ∫ b

a
1dx+

∫ b

a
1dx

⇒
∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ =
[
x
]b
a

+
[
x
]b
a

⇒
∣∣∣ ∫ b

a
cos(nx) sin(nx)dx

∣∣∣ ≤ 2(b− a).

4. Extension and applications of the Second Product Inequality to the
Lp Spaces

In this section, we apply the second product inequality to Lp spaces.

Theorem 6. If u and v are any two vectors in the Euclidean space, then

‖u‖p + ‖v‖p ≤ ‖u‖p‖v‖p, ∀ u, v ∈ [2,∞)

Proof : We can see that:

‖u‖+ ‖v‖ ≤ ‖u‖‖v‖

⇒
(
‖u‖+ ‖v‖

)p
≤

(
‖u‖‖v‖

)p
. (13)

Also, we observed that:

‖u‖p + ‖v‖p ≤
(
‖u‖+ ‖v‖

)p
. (14)

By the transitivity of the inequalities (13) and (14), we obtain

‖u‖p + ‖v‖p ≤
(
‖u‖‖v‖

)p
‖u‖p + ‖v‖p ≤ ‖u‖p‖v‖p.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (2) (2018), 375-389 387

Theorem 7. Suppose that the measurable functions over the domain Ω are such that

f : Ω→ R

and the other measurable function

g : Ω→ R,

then

‖f‖p + ‖g‖p ≤ ‖f‖p‖g‖p.

P roof : ∣∣∣ ∫
Ω
f(x)g(x)dx

∣∣∣p ≤ ∫
Ω

∣∣∣f(x)g(x)
∣∣∣pdx

Applying the second product inequality, we obtain∫
Ω

(
|f(x)|+ |g(x)|

)
dx ≤

∫
Ω
|f(x)g(x)|dx

⇒
∫

Ω

(
|f(x)|+ |g(x)|

)p
dx =

∫
Ω

(
|f(x)g(x)|

)p
dx. (15)

We see that:∫
Ω
|f(x)|p + |g(x)|pdx ≤

∫
Ω

(
|f(x)|+ |g(x)|

)p
dx. (16)

Applying the transitive law to inequalities (15) and (16), we obtain:∫
Ω
|f(x)|p + |g(x)|pdx ≤

∫
Ω

(
|f(x)g(x)|

)p
dx

⇒
∫

Ω
|f(x)|p + |g(x)|pdx =

∫
Ω
|f(x)|p|g(x)|pdx

⇒
(∫

Ω
|f(x)|p + |g(x)|pdx

) 1
p

=
(∫

Ω
|f(x)|p|g(x)|pdx

) 1
p

⇒
(∫

Ω
|f(x)|pdx

) 1
p

+
(∫
|g(x)|pdx

) 1
p ≤

(∫
Ω
|f(x)|pdx

) 1
p
(∫
|g(x)|pdx

) 1
p

⇒ ‖f‖p + ‖g‖p ≤ ‖f‖p‖g‖p.

5. Discussion

We deduce from (9) that, applying the first product inequality to the trigonometric
functions sin(θ) and cos(θ), another new inequality 1 ≤ 2n, ∀n = 0, 1, 2, . . . is uncovered.
The first and second product inequalities have given birth to new inequalities in the
generalized space. These new inequalities give additional information about embeddings
of one space into another space.



REFERENCES 388

6. Conclusion

In summary, we observed that the first product inequality holds for any two vectors in
a genearalized linear space. On the other hand, the second product inequality holds for
any two vectors only in the Euclidean space. In addition, by applying the first product
inequality to the Lp spaces, we observed that if, f : Ω → [0, 1], and g : Ω → R, then
‖f‖p‖g‖p ≤ ‖f‖p +‖g‖p. But if, f, g : Ω→ R, then ‖f‖p +‖g‖p ≤ ‖f‖p‖g‖p. Last but not
the least, we have shown in this paper that the first product inequality holds for C∞[0, 1].

References

[1] A. M. Fink, An essay on the history of inequalities, Journal of Mathematical Analysis
and Applications, 249 , (2000).pp. 118-134.

[2] B. Barnes, E. D. J. Owusu-Ansah, S. K. Amponsah and I. A. Adjei, The proofs of
triangle inequality using binomial inequalities, European journal of pure and applied
mathematics, 11,1 , (2018). pp. 352-361.

[3] D. S. Mitrinović, Analytic inequalities, Springer-Verlag, New York/Berlin; 1961.

[4] D. Rüthing, Proofs of the arithmetic mean-geometric mean inequality. International
journal of mathematical education in science and technology, 13, no. 1 (1982). pp.
49-54.

[5] G. H. Hardy, J. E. Littlewood and G. Polya, Inequalities. London, New York; Cam-
bridge university press, 1978.

[6] D. S. Mitrinović, J. E. Pec̆arić, and A. M. Fink, Classical and new inequalities in
analysis, Kluwer Academic, Dordrecht (1993).pp. 69-72.

[7] S. Alan, A historical review of the isoperimetric theorem in 2− D, and its place in
Elementary plane geometry, courant institute of mathematical sciences, New York
University; 2013. http : //www.cs.nyu.edu/faculty/siegel/sciam.pdf .

[8] J. V. Grabiner, Was Newton’s calculus a dead end? The continental influence of
Maclaurin’s treatise of fluxions, American mathematics monthly, 104 , (1997).pp.
393-410.

[9] S. Wu, Some results on extending and sharpening the Weierstrass product inequalities,
Journal of Mathematical Analysis and Applications , 308 , (2005).pp. 689-702.

[10] M. S. Klamkin, Extension of the Weierstrass product inequality II, Amer. Math., 82
, (1975).pp. 741-742.

[11] D. S. Mitrinović and P. M. Vasić, Analytic inequalities, Springer-Verlag, New York,
1970



REFERENCES 389

[12] I. Schur, Bemerkungen zur theorie de beschränkten bilinearformen mit unendlich
vielen veränderlichen, Journal of mathematics, 140, (1911).pp. 1-28.

[13] B. Kolman and D. R. Hill, Elementary Linear Algebra, Prentice-Hall, Inc, New Jersey;
2000.

[14] J. T. Scheick, Linear Algebra with Applications, Prentice-Hall, Inc, New York; 1997.

[15] J. T. Oden, Applied Functional Analysis: A first course for students of mechanics
and engineering science, Prentice-Hall, Inc, New Jersey; 1979.

[16] C. E. Chidume, Functional Analysis: An introduction to metric spaces, Longman,
Nigeria; 1989.

[17] H. Royden and P. Fitxpatrick, Real Analysis. Pearson Education, Inc, 4th ed., Upper
saddle River; 2010.