EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 3, 2018, 740-750
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

The Product-Normed Linear Space

Benedict Barnes1∗, I. A. Adjei2, C. Sebil3, E. Harris4

1,2,3,4 Department of Mathematics, Kwame Nkrumah University of Science and Technology,
Kumasi, Ghana

Abstract. In this paper, both the product-normed linear space P −NLS (product-Banach space)
and product-semi-normed linear space (product-semi-Banch space) are introduced. These normed
linear spaces are endowed with the first and second product inequalities, which have a lot of
applications in linear algebra and differential equations. In addition, we showed that P − NLS
admits functional properties such as completeness, continuity and the fixed point.

2010 Mathematics Subject Classifications: 4B47, 44B49
Key Words and Phrases: product-normed linear space, product-semi-normed linear space,
completeness, fixed point theorem

1. Introduction

In recent times, the solution space of a mathematical problem has become necessary
condition for its existence. Without the solution space, the qualitative, as well as, quan-
titative property of the mathematical structure cannot be established. In [1], the author
introduced the n−metrics. About 30 decades afterwards, the author in [2] narrowed the
n−metric space to two-dimensional space and also, obtained some topological properties
associated with this space.

Definition 1 (Norm). Let V be a linear space over the real number field 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:

1.‖u‖ ≥ 0, and ‖u‖ = 0, iff u = 0

2.‖αu‖ = |α|‖u‖, ∀ u ∈ V and α ∈ R

3.‖u+ v‖ ≤ ‖u‖+ ‖v‖, ∀ u, v ∈ V

See [3].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3284

Email addresses: ewiekwamina@gmail.com bbarnes.cos@knust.edu.gh (B. Barnes)

http://www.ejpam.com 740 c© 2018 EJPAM All rights reserved.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 741

In [4], the author generalized the normed linear space by replacing the third axiom of
the norm, triangle inequality, with tetrahedral inequality.

The authors in [5] and [6] independently extended the linear normed space to quasi-
normed linear space. They replaced the third axiom of the normed linear space with
quasi-norm

‖u+ v‖ ≤ K
(
‖u‖+ ‖v‖

)
, ∀ u, v ∈ V,

where K ≥ 1 is the quasi-number. In [7], the authors extended the quasi-normed linear
space to the complex version and also, included a parameter p ∈ (0,∞) in this space,
which is used to establish p− uniformly PL− convex. The author in [8], observed that
the quasi-normed linear space becomes the p−normed linear space if the quasi number

K = 2
1
p
−1

. In [9], the author established that the quasi-normed linear space is endowed
with a unique fixed point. In [10], he narrowed the quasi-normed linear space to quasi-2-
normed linear space Q-2-NLS. Addition properties of the Q-2-NLS such as the convergence
of the two Cauchy sequences and a pseudo-quasi-2-norm of the Q-2-NLS were observed
by the authors in [11].

Undoubtedly, the normed linear space is called a p−normed linear if

‖u+ v‖p ≤ ‖u‖p + ‖v‖p, ∀ u, v ∈ V, and 0 < p ≤ 1,

is included to the three axioms of the linear normed space. For example, see the author in
[12]. There are, however, situations where it is more convenient to work with a subspace
than the entire functional space. For example, the imposing of absolute value of a variable
x as an inhomogeneous boundary condition of the Laplace equation does not yield a
solution in the Hilbert space, but in Sobolev spaces. Thus, the continuity of the partial
differential operator is not guaranteed in the Hilbert space. In order for a functional space
to admit such a solution, we impose some continuity conditions on the partial differential
operator which lead to analysis in Sobolev spaces. Again, much attention has been paid
to the extension of the triangle inequality, the third axiom of the normed linear space, to
obtain the different normed linear spaces with their functional properties.

In this paper, we include another axiom to the normed linear space as a fourth axiom,
first product inequality, from which we coined the name product-normed linear space
P − NLS, ‖.‖pn. This functional space narrows the normed linear space which allows
the vectors to be defined on [0, 2]. The P −NLS admits the first product inequality, see
[13], and useful properties which we shall establish them later in this paper. Also, the
product-semi-normed linear space ‖.‖pn is introduced in this paper.

2. Some Preliminary Results

In this section, we give the theorems and definitions regarding to the product-normed
linear space and product-semi-normed linear space.

Definition 2 (Inner Product Space). 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



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 742

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 [14]

Definition 3 (Subspace). Let V be a vector space and Ω a nonempty subset of V . If Ω is
a vector space with respect to the operations in V , then Ω is called a subspace of V . See
[15].

Definition 4 (Continuity at a Point). Let U and V be two normed linear spaces. An
operator A : U → V is continuous at uo ∈ U , if for every ε > 0, there exists a δ = δ(ε, uo)
such that

‖Au−Auo‖ < ε, whenever ‖u− uo‖ < δ.

See [16].

Definition 5 (Convergent Sequence). A sequence {Un}∞n=1 in a normed linear space(
Rn, ‖ · ‖

)
converges to a point u ∈ Rn, if and only if, for every real number ε > 0,

there exists an integer N such that

‖Un − u‖ < ε, n > N.

See [17].

Definition 6 (Closed set). A set u = [0, 2] in a normed linear space
(
Rn, ‖ · ‖

)
is closed

if and only if every sequence of points {Un}∞n=1 ⊂ U convergent in Rn has a limit in U .
For example, see [18]

Definition 7 (Contractive Mapping). Let T : U → U be a mapping from a complete
normed linear space U into itself. The Lipschitz continuity on T is said to be contraction
if

‖T (u)− T (v)‖ ≤ λ‖u− v‖, ∀ u, v ∈ U, and λ < 1.

See [19].

3. Main Result

In this section, we introduce product-normed linear space and product-semi-normed
linear space. The product-normed linear space is presented as follows.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 743

Theorem 1 (Product-normed Linear Space). Let U be a linear space over [0, 2] ⊆ R. A
vector space with a product-norm u→ ‖U‖ satisfying real-valued function ‖ · ‖,

‖ · ‖pn : U → [0,∞),

such that for arbitrary u, v ∈ U, α ∈ [0, 2], the following conditions are satisfied:

P1. ‖u‖ ≥ 0, and ‖u‖ = 0, if and only if u = 0

P2. ‖αu‖ = |α|‖u‖, α ∈ [0, 2], and u ∈ U
P3. ‖u+ v‖pn ≤ ‖u‖pn + ‖v‖pn, ∀u, v ∈ U

We call ‖ · ‖pn a product norm if in addition to P1− P3, the u and v satisfy

P4(a). ‖u‖pn‖y‖pn ≤ ‖u‖pn + ‖v‖pn ∀u, v ∈ [0, 2] (first product inequality)

P4(b).‖u‖+ ‖v‖ ≤ ‖u‖‖v‖, ∀u, v ∈ [2,∞) (second product inequality).

P roof : We start at P1. Suppose that u = 0, then

‖u‖2pn = 〈0, 0〉
⇒ ‖u‖2pn = 0

⇒ ‖u‖pn = 0.

On the other hand, we suppose that ‖u‖pn = 0. But,

‖u‖2pn = 〈u, u〉
⇒ 0 = 〈uu〉.

The above equality holds if u = 0. Hence, ‖u‖ = 0 if and only if u = 0. Without loss of
generality, let the

‖u‖pn = |u|pn,

then |u|pn = 0, which implies that ‖u‖pn = 0. But

‖u‖2pn = 〈u, u〉.

Since the 〈u, u〉 cannot be negative. This implies that ‖u‖pn > 0. Thus, P1 is satisfied.
In order to prove axiom P2, we take the expression on the left hand side yields

‖αu‖2 = 〈αu, αu〉
⇒ ‖αu‖2 = |α|2〈u, u〉
⇒ ‖αu‖2 = |α|2‖u‖2



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 744

⇒ ‖αu‖2 =
(
|α|‖u‖

)2
⇒ ‖αu‖ = |α|‖u‖.

Also, the axiom P3 is proved as follows.

‖u+ v‖2 = 〈u+ v, u+ v〉
⇒ ‖u+ v‖2 = 〈u, u〉+ 〈u, v〉+ 〈v, u〉+ 〈v, v〉
⇒ ‖u+ v‖2 = 〈u, u〉+ 〈u, v〉+ 〈u, v〉+ 〈v, v〉
⇒ ‖u+ v‖2 = 〈u, u〉+ 2|〈u, v〉|+ 〈v, v〉
⇒ ‖u+ v‖2 = ‖u‖2 + 2|〈u, v〉|+ ‖v‖2.

Applying the Cauchy-Schwarz inequality to the middle term on the right hand side, we
obtain

‖u+ v‖2 ≤ ‖u‖2 + 2‖u‖‖v‖+ ‖v‖2

⇒ ‖u+ v‖2 =
(
‖u‖+ ‖v‖

)2
⇒ ‖u+ v‖ ≤ ‖u‖+ ‖v‖.

The axiom P4 has been discussed by the authors in [13]. We present the proof as
follows.

(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].



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 745

The proof of P4(b), see [13].
We show that the product-normed space is valid for 1− norm, 2−norm, p− norms and

∞−norm. We give the proof for p−norm which can be used to generate the other norms.

Theorem 2. Suppose that f(x) and g(x) are measurable functions over the domain [0, 2],
then

‖f‖p‖g‖p ≤ ‖f‖p + ‖g‖p,

where 1
p + 1

q = 1.

Proof : We see that

(‖f‖‖g‖)p = (‖f‖‖g‖)p−1(‖f‖‖g‖)

Applying the first product inequality to the right hand side of the above equation yields

(‖f‖‖g‖)p ≤ (‖f‖+ ‖g‖)p−1(‖f‖+ ‖g‖)
(‖f‖‖g‖)p = (‖f‖+ ‖g‖)p−1‖f‖+ (‖f‖+ ‖g‖)p−1‖g‖

Taking the Lebesgue integral of both sides of the above equation with respect µ, we obtain∫
(‖f‖‖g‖)pdµ =

∫ {
(‖f‖+ ‖g‖)p−1‖f‖+ (‖f‖+ ‖g‖)p−1‖g‖

}
dµ

⇒
∫

(‖f‖‖g‖)pdµ =

∫
(‖f‖+ ‖g‖)p−1‖f‖dµ+

∫
(‖f‖+ ‖g‖)p−1‖g‖dµ

⇒
(∫

(‖f‖‖g‖)p
) 1

p
dµ =

[ ∫
(‖f‖+ ‖g‖)p−1‖f‖dµ+

∫
(‖f‖+ ‖g‖)p−1‖g‖dµ

] 1
p

⇒ ‖f‖p‖g‖p =
[( ∫

((‖f‖+ ‖g‖)p−1‖f‖)pdµ
) 1

p

+
(∫

((‖f‖+ ‖g‖)p−1‖g‖)pdµ
) 1

p
] 1

p

⇒ ‖f‖p‖g‖p ≤
[{(∫

‖f‖pdµ
) 1

p
+
(∫
‖g‖pdµ

) 1
p
}{(∫

(‖f‖+ ‖g‖)pdµ
) 1

p
}p−1] 1

p

⇒ ‖f‖p‖g‖p ≤
[(
‖f‖p + ‖g‖p

)(
‖f‖p + ‖g‖p

)p−1] 1
p

⇒ ‖f‖p‖g‖p =
(
‖f‖p + ‖g‖p

) 1
p
(
‖f‖p + ‖g‖p

) p−1
p

⇒ ‖f‖p‖g‖p ≤ ‖f‖p + ‖g‖p. (1)

Choosing p = 1 and p = 2 in inequality (1), we obtain the following

‖f‖1‖g‖1 ≤ ‖f‖1 + ‖g‖1

and

‖f‖2‖g‖2 ≤ ‖f‖2 + ‖g‖2,



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 746

respectively. But if, p→∞ then the inequality (1) becomes

‖f‖∞‖g‖∞ ≤ ‖f‖∞ + ‖g‖∞.

Theorem 3 (Completeness of the Product-normed linear space). The product-normed
linear space (U, ‖ · ‖pn) is complete.

Proof : Let {Un}∞n=1 be a sequence in (U, ‖ · ‖pn) and u ∈ U . For every ε > 0, there
exists an integer no > N such that ‖Un − u‖ < ε, for all n > no, then {Un}∞n=1 converges
u. Thus, limn→∞ ‖Un − u‖pn = 0. Also, we see that for every ε > 0, there exists integers
m,n > N such that limm,n→∞ ‖Um − Un‖ = 0. In this case, we see that every Cauchy
sequence converges to a point (U, ‖ · ‖pn), then (U, ‖ · ‖pn) is a complete product-normed
linear space. Alternatively, we the following result shows the product-normed linear space
is complete.

Theorem 4 (Completeness of the Product-normed linear space Using Closed Set). Let

U be a linear space over [0, 2] ⊆ R. A subspace
(
U, ‖ · ‖pn

)
of a complete normed linear

space
(
Rn, ‖ · ‖pn

)
is complete if and only if U is a closed set.

Proof : Let {Un}∞n=1 be a sequence of points of U . We see that {Un}∞n=1 has limit u
in U , since U is closed. Thus,

lim
n→∞

‖Un − u‖pn = 0.

Since u ∈ U then
(
U, ‖ · ‖pn

)
is complete.

Conversely, assume that
(
U, ‖ · ‖pn

)
is complete. Let u be an accumulation point of

U . Then each open ball centred at u, B1/n(u) contains a point un ∈ U . That is,

‖Un − u‖pn <
1

n
.

This implies that {Un}∞n=1 converges to u. However,
(
U, ‖·‖pn

)
is complete on the grounds

that u is in U . Hence, U is closed.

Theorem 5 (Continuity). Let (U, ‖ · ‖pn1) and (U, ‖ · ‖pn2) be two product normed linear
spaces, and T : U1 → U2 be a linear operator. Then T is bounded if and only if T is
continuous.

Proof : Let A be a bounded linear operator. Then there exists a constant M such that

‖A(u)‖U2 ≤M‖x‖U1 , ∀ u ∈ U1.

Let {Un}∞n=1 be a sequence in U1 such that

lim
n→∞

‖Un − u‖U1 = 0.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 747

Now,

‖A(Un)−A(u)‖U2 =
∥∥∥A(Un − u)∥∥∥

U2

‖A(Un)−A(u)‖U2 ≤ M
∥∥∥Un − u∥∥∥

U1

lim
n→∞

‖A(Un)−A(u)‖U2 ≤ 0

lim
n→∞

A(Un)− lim
n→∞

A(u) = 0

lim
n→∞

A(Un)−A(u) = 0

A( lim
n→∞

Un) → A(u).

This implies that A is continuous on U1.
Conversely, suppose that A is a continuous linear operator. Then A is continuous at

0. Thus, for every δ > 0, ‖u‖U1 , there exists an ε > 0 such that

‖Au‖ ≤ ε. (2)

Setting uo 6= 0, be an arbitrary element in U1 and u = αuo, where α = δ
‖uo‖U1

. We can

see that

‖Au‖U2 = ‖A(αuo)‖U2

‖Au‖U2 = |α|‖Auo‖U2 , ∀ uo ∈ U1. (3)

Comparing (2) and (3), we obtain

|α|‖Auo‖U2 ≤ ε

⇒ ‖Auo‖U2 ≤ ε

|α|

⇒ ‖Auo‖U2 ≤ ε

|δ|
‖uo‖U1 .

Hence A is bounded. This completes the proof.

Theorem 6 (Unique Fixed Point). Let
(
U, ‖ · ‖pn

)
be a product-normed linear space and

setting T : U → U be a contractive mapping. Then T has a unique fixed point ū ∈ U .

Proof : Letting uo ∈ U , then

Un = T (Un)

⇒ Un = Tn(uo), ∀ n ∈ N.

Setting {U}∞n a Cauchy sequence in
(
U, ‖ · ‖pn

)
. Since

(
U, ‖ · ‖pn

)
is a complete, which

implies that u ∈ U such that

lim
n→∞

Un = u.



Barnes et al. / Eur. J. Pure Appl. Math, 11 (3) (2018), 740-750 748

Again, we see that T is a contractive mapping, then it is continuous. Thus,

lim
n→∞

T (Un) = Tu.

We can see that

Tu = lim
n→∞

T (Un)

⇒ Tu = lim
n→∞

Un+1

⇒ Tu = u.

Hence, u is a fixed point of the mapping T .

In addition, we show that the fixed point is unique in
(
U, ‖ · ‖pn

)
. Setting u1, u2 ∈ U

be the fixed points of T , then

‖T (u1)− T (u2)‖pn = ‖u1 − u2‖pn. (2)

Also, we see that:

‖T (u1)− T (u2)‖pn ≤ λ‖u1 − u2‖pn, ∀ λ < 1 (3)

Substituting equation (2) into inequality (3) yields(
1− λ

)
‖u1 − u2‖pn ≤ 0

⇒ ‖u1 − u2‖pn = 0.

The above equation holds if u1 = u2. Hence, the fixed point x is unique in the product-
normed linear space.

Another normed linear space is introduced as:

Theorem 7 (Product Semi-Normed Linear Space). Let U be a linear space over [2,∞] ⊆
R. A product norm on U is a real-valued function ‖ · ‖,

‖ · ‖ : U → [0,∞),

such that for arbitrary u, v ∈ U, α ∈ [0, 2], the following conditions are satisfied:

1. ‖u‖ > 0, and ‖u‖ = 0, if and only if u = 0

2. ‖αu‖ = |α|‖u‖, α ∈ [0, 2], and u ∈ U
3. ‖u+ v‖ ≤ ‖u‖+ ‖v‖, ∀u, v ∈ U

4. ‖u‖+ ‖v‖ ≤ ‖u‖‖v‖ ∀u, v ∈ V (product inequality).

4. Conclusion

In a nutshell, we have introduced the product-normed linear space and product-semi-
normed linear space (product-semi-Banch space) which are endowed with some functional
space. In addition, P −NLS is endowed with fixed point.



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