EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 1, 2023, 144-155
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Inner Products on Discrete Morrey Spaces

Muhammad Jakfar1,∗, Manuharawati1, Agung Lukito1, Shofan Fiangga1

1 Department of Mathematics, Faculty of Mathematics and Science, Universitas Negeri
Surabaya, Surabaya, East Java, Indonesia

Abstract. The discrete Morrey space mu,p is a generalization of the p-summable sequence space
ℓp. It is known that the discrete Morrey space is a normed space. Furthermore, for p ̸= 2, the space
mu,p equipped with the usual norm is not an inner product space. In this paper, we shall show
that this space is actually contained in an inner product space. That means this space equipped
with the inner product is an inner product space. The relationship between a standard norm on
mu,p and the inner product is studied.

2020 Mathematics Subject Classifications: 46A45

Key Words and Phrases: Discrete Morrey Space, inner product, normed space, p-summable
space

1. Introduction

The Discrete Morrey Space (mu,p) is defined as follows. Let m ∈ Z, N ∈ ω := N
⋃
{0},

and we write Sm,N ; = {m−N, . . . ,m, . . . ,m+N}. Hence, |Sm,N | = 2N + 1. Now, let K
be R or C and 1 ≤ p ≤ u ≤ ∞. The discrete Morrey space, denoted by mu,p, is the set of
sequences λ = (λk)k∈Z taking values in K such that

∥λ∥mu,p
:= sup

m∈Z,N∈ω
|Sm,N |

1
u
− 1

p

( ∑
k∈Sm,N

|λk|p
) 1

p
< ∞.

Clearly mu,p is a vector space. We remark that when u = p, we have mu,p = ℓp, the space
of p-summable sequences with integer indices.
We also have some notes among mu,p as follows.

Theorem 1. [5] For 1 ≤ p ≤ u < ∞, the space (mu,p, ∥ · ∥mu,p) is a normed space.
Moreover, the space is a Bannach space.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4612

Email addresses: muhammadjakfar@unesa.ac.id (M. Jakfar), manuharawati@unesa.ac.id
(Manuharawati), agunglukito@unesa.ac.id (A. Lukito), shofanfiangga@unesa.ac.id (S. Fiangga)

https://www.ejpam.com 144 © 2023 EJPAM All rights reserved.



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 145

Theorem 2. [5] For 1 ≤ p ≤ u < ∞, we have mu,p ⊂ ℓp and ∥λ∥mu,p ≤ ∥λ∥ℓp for every
λ ∈ mu,q.

Many mathematicians had discussed about the discrete Morrey space (see [1], [2], [4],
[5], [6], [7], [8], [9], [10], [11], [12], [13], [14], [20], [21], and [22]). We have known that the
space (mu,p, ∥ · ∥mu,p) is a normed space. However, for p ̸= 2, the discrete Morrey space is
not an inner product space, because if we take u = p then the norm ∥ · ∥mu,p is equal to
the norm ∥ · ∥ℓp which do not satisfy the parallelogram’s law (see [15] ).
In this paper, we will show that we can define an inner product on the discrete Morrey
space. So, this is the first time to say that the discrete Morrey space is an inner product
space. We also discuss the relationship between a standard norm and the inner product
on the space.
To define an inner product on mu,p, the first, we will show for p = 2, the space mu,p or
mu,2 is an inner product space. Then we begin try to define an inner product on mu,p for
p > 2. The next, we construct an inner product on mu,p for p < 2.
Throughout the paper, we assume that X is a real vector space, as in [16], the norm on X
is a mapping ∥ · ∥ : X → R+ such that for all vector x, y ∈ X and a scalar a ∈ R we have:

(i) ∥x∥ ≥ 0 and ∥x∥ = 0 if and only if x = 0

(ii) ∥ax∥ = |a|∥x∥

(iii) ∥x+ y∥ ≤ ∥x∥+ ∥y∥

The inner product onX is a mapping ⟨·, ·⟩ : X×X → R such that for all vectors x, y, z ∈ X
and a scalar a ∈ R we have:

(i) ⟨x, x⟩ ≥ 0 and ⟨x, x⟩ = 0 if and only if x = 0

(ii) ⟨x, y⟩ = ⟨y, x⟩

(iii) ⟨ax, y⟩ = a⟨x, y⟩

(iv) ⟨x+ y, z⟩ ≤ ⟨x, z⟩+ ⟨y, z⟩

Note: The space X which is equipped by a norm ∥ · ∥, (X, ∥ · ∥), is called a normed space.
The space X which is equipped by an inner product ⟨·, ·⟩, (X, ⟨·, ·⟩), is called an inner
product space.

2. Result

2.1. Inner Product on mu,p for p = 2

In mu,2, the norm of λ = (λk)k∈Z ∈ mu,2 is defined as

∥λ∥mu,2
:= sup

m∈Z,N∈ω
|Sm,N |

1
u
− 1

2

( ∑
k∈Sm,N

|λk|2
) 1

2
.



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 146

We can observe that the norm satisfies the parallelogram’s law

∥λ+ µ∥2mu,2
+ ∥λ− µ∥2mu,2

= ∥λ∥2mu,2
+ ∥µ∥2mu,2

for every λ = (λk)k∈Z, µ = (µk)k∈Z ∈ mu,2. That means, the norm is formed from an
inner product. So, we know that mu,2 is an inner product space, equipped with the inner
product

⟨λ, µ⟩mu,2
:= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk

)
.

then ∥λ∥2mu,2
= ⟨λ, λ⟩mu,2 .

Proposition 1. In the space mu,2, the mapping ⟨·, ·⟩mu,2 defines an inner product on mu,2.

Proof. We will show that the mapping ⟨·, ·⟩mu,2 satisfy all conditions of an inner prod-
uct on mu,2.
(i) Clear, for every λ = (λk)k∈Z ∈ mu,2, we get ⟨λ, λ⟩mu,2 = ∥λ∥2mu,2

≥ 0.

If λ = 0 then ⟨λ, λ⟩mu,2 = 0. And, if ⟨λ, λ⟩mu,2 = 0 then ∥λ∥2mu,2
= 0, So, it must be λ = 0.

(ii) Given λ = (λk)k∈Z, µ = (µk)k∈Z ∈ mu,2. Then we have

⟨λ, µ⟩mu,2
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk

)
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

µkλk

)
= ⟨µ, λ⟩mu,2

(iii) Given a ∈ R and λ = (λk)k∈Z, µ = (µk)k∈Z ∈ mu,2. We get

a⟨λ, µ⟩mu,2
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

aλkµk

)
= a sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk

)
= a⟨λ, µ⟩mu,2

(iv) Give λ = (λk)k∈Z, γ = (γk)k∈Z, µ = (µk)k∈Z ∈ mu,2. Thus

⟨λ+ γ, µ⟩mu,2
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

(λk + γk)µk

)
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk +
∑

k∈Sm,N

γkµk

)
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk

)



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 147

+ sup
m∈Z,N∈ω

|Sm,N |2
(

1
u
− 1

2

)( ∑
k∈Sm,N

γkµk

)
= ⟨λ, µ⟩mu,2

+ ⟨γ, µ⟩mu,2

■

Remark 1. The space (mu,2, ∥ · ∥mu,2) is an inner product space. Moreover, by Theorem
1, the space is a complete. Accordingly,

(
mu,2, ⟨·, ·⟩mu,2

)
is a Hilbert space.

For p ̸= 2, the space mu,p is not an inner product space. Because the norm is not
satisfy the parallelogram’s law. So, in the next section, we will try to define an inner
product on mu,p for p ̸= 2.

2.2. Inner Product on mu,p for 2 < p < ∞

In this section, we let 2 < p ≤ u < ∞, unless otherwise stated. We will define an inner
product on the space mu,p. First, before we define an inner product on the space mu,p,
we observe that mu,p ⊂ mu,2 (as set). Indeed, if λ = (λk)k∈Z is a sequence in mu,p, then
by Holder’s inequality, we have∑

k∈Sm,N

|λk|2 ≤
( ∑
k∈Sm,N

|λk|p
) 2

p
( ∑
k∈Sm,N

1
)1− 2

p

= |Sm,N |1−
2
p

( ∑
k∈Sm,N

|λk|p
) 2

p

=
∣∣Sm,N

∣∣( 1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|p
) 2

p
.

Thus
1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|2 ≤
( 1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|p
) 2

p
.

So, taking the square roots of both sides, we get( 1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|2
) 1

2 ≤
( 1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|p
) 1

p

or ∣∣Sm,N

∣∣− 1
2

( ∑
k∈Sm,N

|λk|2
) 1

2 ≤
∣∣Sm,N

∣∣− 1
p

( ∑
k∈Sm,N

|λk|p
) 1

p
.

Then

sup
m∈Z,N∈ω

∣∣Sm,N

∣∣ 1
u
− 1

2

( ∑
k∈Sm,N

|λk|2
) 1

2 ≤ sup
m∈Z,N∈ω

∣∣Sm,N

∣∣ 1
u
− 1

p

( ∑
k∈Sm,N

|λk|p
) 1

p
.



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 148

So, we have
∥λ∥mu,2 ≤ ∥λ∥mu,p

which means that λ is in mu,2.
Thus, we realize that mu,p can actually be considered as a subspace of mu,2, equipped
with the inner product

⟨λ, µ⟩mu,2
:= sup

m∈Z,N∈ω
|Sm,N |2

(
1
u
− 1

2

)( ∑
k∈Sm,N

λkµk

)
and the norm

∥λ∥mu,2
:= sup

m∈Z,N∈ω
|Sm,N |

1
u
− 1

2

( ∑
k∈Sm,N

|λk|2
) 1

2

for every λ, µ ∈ mu,p.
A more general result is formulated in the following proposition.

Proposition 2. For 1 ≤ q ≤ p ≤ u < ∞ then mu,p ⊂ mu,q and ∥λ∥mu,q ≤ ∥λ∥mu,p for
every λ ∈ mu,p.

Proof. Let 1 ≤ q ≤ p ≤ u < ∞. For every λ = (λk)k∈Z ∈ mu,p, we have

∑
k∈Sm,N

|λk|q ≤
( ∑
k∈Sm,N

|λk|p
) q

p
( ∑
k∈Sm,N

1
)1− q

p

= |Sm,N |1−
q
p

( ∑
k∈Sm,N

|λk|p
) q

p

=
∣∣Sm,N

∣∣( 1∣∣Sm,N

∣∣ ∑
k∈Sm,N

|λk|p
) q

p
.

Taking the q-roots of both sides, multiplying by |Sm,N |
1
u of both sides, and taking supre-

mum of both sides we get ∥λ∥mu,q ≤ ∥λ∥mu,p , which tell us mu,p ⊂ mu,q.
■

Corollary 1. For 2 ≤ p ≤ u < ∞ then mu,p ⊂ mu,2 and ∥λ∥mu,2 ≤ ∥λ∥mu,p for every
λ ∈ mu,p.

Proof. Applying Proposition 2, take q = 2, then the corollary is proved. ■

Remark 2. For 2 < p ≤ u < ∞, the space
(
mu,p, ∥ · ∥mu,2

)
is an inner product space.

For 2 < p < ∞, we finish to define an inner product on mu,p, in the next section, we
will try to define an inner product on mu,p for 1 ≤ p < 2.



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 149

2.3. Inner Product on mu,p for 1 ≤ p < 2

In this section, we let 1 ≤ p < 2. First for all, we begin to discuss the following
proposition.

Proposition 3. For 1 ≤ p ≤ u < 2 and v = 2up
2p−2u+up , then mu,p ⊂ mv,2 and ∥λ∥mv,2 ≤

∥λ∥mu,p for every λ ∈ mu,p.

Proof. Let λ = (λk)k∈Z ∈ mu,p. Then

∥λ∥2mv,2
= sup

m∈Z,N∈ω
|Sm,N |2

(
1
v
− 1

2

)( ∑
k∈Sm,N

|λk|2
)

= sup
m∈Z,N∈ω

|Sm,N |2
(

1
v
− 1

2

)( ∑
k∈Sm,N

|λk|2−p|λk|p
)

≤ sup
m∈Z,N∈ω

|Sm,N |2
(

1
v
− 1

2

)(
sup

k∈Sm,N

|λk|2−p
∑

k∈Sm,N

|λk|p
)

≤ sup
m∈Z,N∈ω

|Sm,N |2
(

1
v
− 1

2

)( ∑
k∈Sm,N

|λk|p
) 2−p

p
(

∑
k∈Sm,N

|λk|p
)

= sup
m∈Z,N∈ω

|Sm,N |2
(

1
v
− 1

2

)( ∑
k∈Sm,N

|λk|p
) 2

p

Take the square roots of both sides, we get

∥λ∥mv,2 ≤ sup
m∈Z,N∈ω

|Sm,N |
1
v
− 1

2

( ∑
k∈Sm,N

|λk|p
) 1

p

Because v = 2up
2p−2u+up then we find

1

v
− 1

2
=

1(
2up

2p−2u+up

) − 1

2

=
2p− 2u+ up

2up
− 1

2

=
2p− 2u+ up

2up
− up

2up

=
2p− 2u

2up

=
p− u

up

=
1

u
− 1

p



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 150

Because 1 ≤ p ≤ u < 2, then 2 ≤ v and 0 < 1
v ≤ 1

2 . And also, we get

∥λ∥mv,2 ≤ sup
m∈Z,N∈ω

|Sm,N |
1
u
− 1

p

( ∑
k∈Sm,N

|λk|p
) 1

p
= ∥λ∥mu,p

Consequently, we have
∥λ∥mv,2 ≤ ∥λ∥mu,p

Since λ ∈ mu,p is arbitrary, then from the inequality, every element of mu,p is also in mv,2.
It means mu,p ⊂ mv,2.

■

Remark 3. For 1 ≤ p ≤ u < 2 and v = 2up
2p−2u+up , the space

(
mu,p, ∥ · ∥mv,2

)
is an inner

product space.

A more general result of Proposition 3 is formulated in the following theorem. But,
before discussing the theorem, we need to introduce the following proposition.

Proposition 4. If 1 ≤ p ≤ u ≤ q < ∞ such that 1
p −

1
u < 1

q , then there is v > 0 such that

q ≤ v and 1
v − 1

q = 1
u − 1

p . That is, v = upq
pq−uq+up .

Proof. Let 1 ≤ p ≤ u ≤ q < ∞ and
∣∣ 1
u − 1

p

∣∣ < 1
q . Take v = upq

pq−uq+up . Then

1

v
− 1

q
=

1(
upq

pq−uq+up

) − 1

q
=

pq − uq + up

upq
− 1

q
=

pq − uq

upq
=

1

u
− 1

p
.

Because p ≤ u then 1
u ≤ 1

p . Hence,
1
v ≤ 1

q and q ≤ v. Since 1
p −

1
u < 1

q then 1
q −

1
v < 1

q . So,
1
v > 0 and v ≤ 0.

■

Theorem 3. For 1 ≤ p ≤ u ≤ q < ∞ and v = upq
pq−uq+up , then mu,p ⊂ mv,q and

∥λ∥mv,q ≤ ∥λ∥mu,p for every λ ∈ mu,p.

Proof. Let λ = (λk)k∈Z ∈ mu,p. Then

∥λ∥qmv,q
= sup

m∈Z,N∈ω
|Sm,N |q

(
1
v
− 1

q

)( ∑
k∈Sm,N

|λk|q
)

= sup
m∈Z,N∈ω

|Sm,N |q
(

1
v
− 1

q

)( ∑
k∈Sm,N

|λk|q−p|λk|p
)

≤ sup
m∈Z,N∈ω

|Sm,N |q
(

1
v
− 1

q

)(
sup

k∈Sm,N

|λk|q−p
∑

k∈Sm,N

|λk|p
)

≤ sup
m∈Z,N∈ω

|Sm,N |q
(

1
v
− 1

q

)( ∑
k∈Sm,N

|λk|p
) q−p

p
(

∑
k∈Sm,N

|λk|p
)



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 151

= sup
m∈Z,N∈ω

|Sm,N |q
(

1
v
− 1

q

)( ∑
k∈Sm,N

|λk|p
) q

p

Take the q-roots of both sides and apply Proposition 5, we get

∥λ∥mv,q ≤ sup
m∈Z,N∈ω

|Sm,N |
1
v
− 1

q

( ∑
k∈Sm,N

|λk|p
) 1

p
= ∥λ∥mu,p

Since λ ∈ mu,p is arbitrary, then from the inequality, we get mu,p ⊂ mv,q.
■

3. Furthermore Results

Now, we can define two norms in mu,p, the usual norm ∥ · ∥mu,p and another new

norm (∥ · ∥mu,2 if 2 < p ≤ u < ∞, or ∥ · ∥mv,2 if 1 ≤ p ≤ u < 2 with v = 2up
2p−2u+up).

One might ask whether ∥ · ∥mv,2 is equivalent to ∥ · ∥mu,p . The answer is negative. We
already have ∥λ∥mv,2 ≤ ∥λ∥mu,p for every λ ∈ mu,p. The following proposition say that we
cannot control ∥λ∥mv,2 and ∥λ∥mu,p for every λ ∈ mu,p such that the norm ∥ · ∥mv,2 is not
equivalent to the norm ∥ · ∥mu,p .

Proposition 5. Let 1 ≤ p ≤ u < 2 and v = 2up
2p−2u+up . There is no constant C > 0 such

that ∥λ∥mv,2 ≥ C∥λ∥mu,p for every λ ∈ mu,p.

Proof. For each n ∈ N, take u = p and λ(n) =
(

1

k
1
p+ 1

n

)
with

λk(n) =

{
1

k
1
p+ 1

n
, if k ∈ N

0, if k /∈ N
.

Then v = 2 and we have

∥λ(n)∥2mv,2
= ∥λ(n)∥2ℓ2 =

∑
k∈N

1

k
2
p
+ 2

n

≤
∑
k∈N

1

k
1
p

< ∞

while

∥λ(n)∥pmu,p
= ∥λ(n)∥

p
ℓp =

∑
k∈N

1

k1+
p
n

< ∞

We can see that ∥λ(n)∥mv,2 is bounded by a fix number independent of n, while ∥λ(n)∥mu,p

is dependent on n and tends to ∞ as n → ∞. Hence

∥λ(n)∥mv,2

∥λ(n)∥mu,p

→ 0

as n → ∞. So, there is no constant C > 0 such that ∥λ∥mv,2 ≥ C∥λ∥mu,p for every
λ ∈ mu,p.

■



M. Jakfar et al. / Eur. J. Pure Appl. Math, 16 (1) (2023), 144-155 152

Proposition 6. Let 2 < p ≤ u < ∞. There is no constant C > 0 such that ∥λ∥mu,p ≤
C∥λ∥mu,2 for every λ ∈ mu,p.

Proof. Let 2 < p ≤ u < ∞. Take u = p. Suppose that a constant exists. Then, for
λ(n) = (. . . , 0, 0, 1, 0, 0, . . . , 0, 1, 0, . . .), where the first 1 is at k = 1 and the second 1 is the

(n+ 1)th-term, we have

2
1
p ≤ C

1

(2n+ 1)

(
1
2
− 1

p

)
√
2.

But this cannot be true, since 1
(2n+1)

(
1
2
− 1

p

)
→ 0 as n → ∞.

■

Remark 4. Proposition 5 and 6 say that in the discrete Morrey space, the usual norm
and the second norm are not equivalent.

Let 1 ≤ p ≤ u < 2 and v = 2up
2p−2u+up . As we have seen in the previous section, every

sequence λ ∈ mu,p has ∥λ∥v,2 < ∞. This suggests that mu,p ⊂ mv,2. We shall now discuss
some properties of this space. First, we have the following proposition, which describes
the relationship between mu,p and mv,2.

Proposition 7. As a set, we have mu,p ⊂ mv,2 and the inclusion is strict.

Proof. Let 1 ≤ p ≤ u < 2, v = 2up
2p−2u+up , and λ ∈ mv,2. It follows from Corollary 3

that
∥λ∥mv,2 ≤ ∥λ∥mu,p

which means that λ ∈ mv,2.
To show that the inclusion is strict, we need to find λ = (λk)k∈Z such that ∥λ∥mv,2 < ∞
but ∥λ∥mu,p = ∞. Choose u = p and λ = (λk)k∈Z with

λk =

{ (
1
k

) 1
p , if k ̸= 0

0, if k = 0
.

Hence

∥λ∥mv,2 = ∥λ∥m2,2 =
( ∑
k∈Sm,N

∣∣(1
k

) 1
p ∣∣2) 1

2
=

(
2
∑
k∈N

(1
k

) 2
p
) 1

2
< ∞

while

∥λ∥mu,p = ∥λ∥mp,p =
( ∑
k∈Sm,N

∣∣(1
k

) 1
p ∣∣p) 1

p
=

(
2
∑
k∈N

(1
k

)) 1
p
= ∞.

This means that λ is in mv,2 but not in mu,p.
■

Proposition 8. The space
(
mv,2, ∥ · ∥mv,2

)
is complete. Accordingly,

(
mv,2, ⟨·, ·⟩mv,2

)
is a

Hilbert space.



REFERENCES 153

Proof. Based on Proposition 1 and Theorem 1, the space mv,2 is a complete. So,(
mv,2, ⟨·, ·⟩mv,2

)
is a Hilbert space.

■

4. Concluding Remarks

We have shown the space mu,p can be equipped with an inner product and its induced
norm. So, we can define two norms in mu,p, the standard norm ∥ · ∥mu,p and another new

norm (∥ · ∥mu,2 if 2 < p ≤ u < ∞, or ∥ · ∥mv,2 if 1 ≤ p ≤ u < 2 with v = 2up
2p−2u+up).

But, we have to know that the two norms are not equivalent. Using the inner product,
one may define angles on mu,p, define orthogonality on mu,p, carry out the Gram-Schmidt
process to get an orthogonal set (see [3]), define the volume of an n-dimensional paral-
lelepiped on mu,p, define an n-inner product on mu,p, discuss the concept of convergence
for n-dimensional subspace of mu,p (see [19], [18], and [17]), and so on.

Conflict of Interest

The authors confirm that there is no conflict of interest to declare for this publication

Acknowledgements

This research received a grant from Faculty of Mathematics and Science, Universitas
Negeri Surabaya. The authors would like to thank the editor and anonymous reviewers
for their comments that help improve the quality of this work

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