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On Geometric Constants for Discrete Morrey Spaces

Adam Adam, Hendra Gunawan∗
Analysis and Geometry Group, Faculty of Mathematics and Natural Sciences,

Bandung Institute of Technology, Bandung 40132, Indonesia
adam_adam@students.itb.ac.id, hgunawan@math.itb.ac.id

∗Correspondence: hgunawan@math.itb.ac.id

Abstract. In this paper we prove that the n-th Von Neumann-Jordan constant and the n-th Jamesconstant for discrete Morrey spaces `pq where 1 ≤ p < q < ∞ are both equal to n. This resulttells us that the discrete Morrey spaces are not uniformly non-`1, and hence they are not uniformly
n-convex.

1. Introduction
Let n ≥ 2 be a non-negative integer and (X, ‖ · ‖) be a Banach space. The n-th Von Neumann-

Jordan constant for X [6] is defined by
C
(n)
NJ (X) := sup

{∑
± ‖u1 ± u2 ± · · · ± un‖2X
2n−1

∑n
i=1 ‖ui‖X

: ui 6= 0, i = 1, 2, . . . , n
}

and the n-th James constant for X [7] is defined by
C
(n)
J (X) := sup{min ‖u1 ± u2 ± · · · ± un‖ : ui ∈ SX , i = 1, 2, . . . , n}.Note that in the definition of C(n)NJ (X), the sum ∑

± is taken over all possible combinations of ±signs. Similarly, in the definition of C(n)J (X), the minimum is taken over all possible combinationsof ± signs, while the supremum is taken over all ui ’s in the unit sphere SX := {u ∈ X : ‖u‖ = 1}.These constants measure some sort of convexity of a Banach space.We say that X is uniformly n-convex [2] if for every ε ∈ (0, n] there exists a δ ∈ (0, 1) such thatfor every u1, u2, . . . , un ∈ SX with ‖u1 ± u2 ± · · · ± un‖ ≥ ε for all combinations of ± signs exceptfor ‖u1 + u2 + · · ·+ un‖, we have
‖u1 + u2 + · · ·+ un‖ ≤ n(1− δ).

Received: 31 Aug 2021.
Key words and phrases. n-th Von Neumann-Jordan constant; n-th James constant; discrete Morrey spaces; uniformlynon-`1 spaces; uniformly n-convex spaces. 1

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https://doi.org/10.28924/ada/ma.2.2
https://orcid.org/0000-0001-7879-8321


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 2Meanwhile, we say that X is uniformly non-`1n [1,5,8] if there exists a δ ∈ (0, 1) such that for every
u1, u2, . . . , un ∈ SX we have

min ‖u1 ± u2 ± · · · ± un‖ ≤ n(1− δ).

Note that for n = 2, uniformly non-`1n spaces are known as uniformly nonsquare spaces, while for
n = 3 they are known as uniformly non-octahedral spaces. One may verify that if X is uniformly
n-convex, then X is uniformly non-`1n [2].Now a few remarks about the two constants, and their associations with the uniformly non-`1nand uniformly n-convex properties.

• 1 ≤ C(n)NJ (X) ≤ n and C(n)NJ (X) = 1 if and only if X is a Hilbert space [6].
• 1 ≤ C(n)J (X) ≤ n. If dim(X) = ∞, then √n ≤ C(n)J (X) ≤ n. Moreover, if X is a Hilbertspace, then C(n)J (X) = √n [7].
• X is uniformly non-`1n if and only if C(n)NJ (X) < n [6].
• X is uniformly non-`1n if and only if C(n)J (X) < n [7].

The last two statements tell us that if C(n)NJ (X) = n or C(n)J (X) = n, then X is not uniformly non-`1nand hence not uniformly n-convex.In this paper, we shall compute the value of the two constants for discrete Morrey spaces. Let
ω := N ∪ {0} and m = (m1, m2, . . . , md) ∈ Zd . Define

Sm,N := {k ∈ Zd : ‖k −m‖∞ ≤ N}

where N ∈ ω and ‖m‖∞ = max{|mi | : 1 ≤ i ≤ d}. Denote by |Sm,N | the cardinality of Sm,N for
m ∈ Zd and N ∈ ω. Then we have |Sm,N | = (2N + 1)d .Now let 1 ≤ p ≤ q < ∞. Define `pq = `pq(Zd) to be the discrete Morrey space as introducedin [3], which consists of all sequences x : Zd → R with

‖x‖`pq := sup
m∈Zd ,N∈ω

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|xk |p
) 1
p

<∞,

where x := (xk) with k ∈ Zd . One may observe that these discrete Morrey spaces are Banachspaces [3]. Note, in particular, that for p = q, we have `pq = `q .From [4] we already know that CNJ(`pq) = CJ(`
p
q) = 2 for 1 ≤ p < q < ∞, which impliesthat `pq are not uniformly nonsquares for those p’s and q’s. In this paper, we shall show that

C
(n)
NJ (`

p
q) = C

(n)
J (`

p
q) = n for 1 ≤ p < q < ∞, which leads us to the conclusion that `pq arenot uniformly non-`1n for those p’s and q’s, which is sharper than the existing result. (If X is notuniformly non-`1n, then X is not uniformly non-`1n−1, provided that n ≥ 3.)

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 32. Main Results
The value of the n-th Von Neumann-Jordan constant and the n-th James constant for discreteMorrey spaces are stated in the following theorems. To understand the idea of the proof, we firstpresent the result for n = 3.

Theorem 2.1. For 1 ≤ p < q <∞, we have C(3)NJ (`
p
q(Zd)) = C(3)J (`

p
q(Zd)) = 3.

Proof. To prove the theorem, it suffices for us to find x (1), x (2), x (3) ∈ `pq such that∑
± ‖x (1) ± x (2) ± x (3)‖2`pq
22
∑3
i=1 ‖x (i)‖`pq

= 3

for the Von Neumann-Jordan constant, and
min ‖x (1) ± x (2) ± x (3)‖`pq = 3

for the James constant.
Case 1: d = 1. Let j ∈ Z be a nonnegative, even integer such that j > 4 q

q−p − 1, or equivalently
(j + 1)

1
q
− 1
p < 4−

1
p .

Construct x (1), x (2), x (3) ∈ `pq(Z) as follows:
• x (1) = (x (1)k )k∈Z is defined by

x
(1)
k =

1, k = 0, j, 2j, 3j,

0, otherwise;
• x (2) = (x (2)k )k∈Z is defined by

x
(2)
k =


1, k = 0, j,

−1, k = 2j, 3j,

0, otherwise;
• x (3) = (x (3)k )k∈Z is defined by

x
(3)
k =


1, k = 0, 2j,

−1, k = j, 3j,

0, otherwise.

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 4The three sequences are in the unit sphere of `pq(Z). Indeed, for the first sequence, we have
‖x (1)‖`pq = sup

m∈Z,N∈ω
|Sm,N |

1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k |
p

) 1
p

= sup
m∈Z∩[0,3j ],N∈Z∩[0,3j/2]

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k |
p

) 1
p

= max{1, (j + 1)
1
q
− 1
p 2

1
p , (2j + 1)

1
q
− 1
p 3

1
p , (3j + 1)

1
q
− 1
p 4

1
p }.

Since (3j + 1) 1q− 1p < (2j + 1) 1q− 1p < (j + 1) 1q− 1p < 4− 1p , we get ‖x (1)‖`pq = 1. Similarly, one mayobserve that ‖x (2)‖`pq = ‖x (3)‖`pq = 1.Next, we observe that
x
(1)
k + x

(2)
k + x

(3)
k =



3, k = 0,

1, k = j, 2j,

−1, k = 3j,

0, otherwise;

x
(1)
k + x

(2)
k − x

(3)
k =



3, k = j,

1, k = 0, 3j,

−1, k = 2j,

0, otherwise;

x
(1)
k − x

(2)
k + x

(3)
k =



3, k = 2j,

1, k = 0, 3j,

−1, k = j,

0, otherwise;

x
(1)
k − x

(2)
k − x

(3)
k =



3, k = 3j,

1, k = j, 2j,

−1, k = 0,

0, otherwise.We first compute that
‖x (1)+ x (2)+ x (3)‖`pq = max{3, (j +1)

1
q
− 1
p (3p+1)

1
p , (2j +1)

1
q
− 1
p (3p+2)

1
p , (3j +1)

1
q
− 1
p (3p+3)

1
p }.

Notice that
• (j + 1)

1
q
− 1
p (3p + 1)

1
p <

(
3p+1p

4

) 1
p
< (3p)

1
p = 3.

• (2j + 1)
1
q
− 1
p (3p + 2)

1
p < (j + 1)

1
q
− 1
p (3p + 2)

1
p <

(
3p+2
4

) 1
p
< 3.

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 5

• (3j + 1)
1
q
− 1
p (3p + 3)

1
p < (j + 1)

1
q
− 1
p (3p + 3)

1
p <

(
3p+3
4

) 1
p
< 3.Hence, we obtain ‖x (1) + x (2) + x (3)‖`pq = 3.Similarly, we have

‖x (1) ± x (2) ± x (3)‖`pq = sup
m∈Z∩[0,3j ],N∈Z∩[0,3j/2]

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k ± x
(2)
k ± x

(3)
k |
p

) 1
p

= 3

for every combination of ± signs.Consequently, ∑± ‖x(1)±x(2)±x(3)‖2`pq
22
∑3
i=1 ‖x(i)‖`pq

= 3 and min ‖x (1) ± x (2) ± x (3)‖`pq = 3, so we come to theconclusion that
C
(3)
NJ (`

p
q(Z)) = C

(3)
J (`

p
q(Z)) = 3.

Case 2: d > 1. Let j ∈ Z be a nonnegative, even integer such that j > 4 q
d(q−p) − 1, which isequivalent to

(j + 1)d(
1
q
− 1
p
) < 4−

1
p .

We then construct x (1), x (2), x (3) ∈ `pq(Zd) as follows:
• x (1) = (x (1)k )k∈Zd is defined by

x
(1)
k =

1, k = (0, 0, . . . , 0), (j, 0, . . . , 0), (2j, 0, . . . , 0), (3j, 0, . . . , 0),

0, otherwise;
• x (2) = (x (2)k )k∈Zd is defined by

x
(2)
k =


1, k = (0, 0, . . . , 0), (j, 0, . . . , 0),

−1, k = (2j, 0, . . . , 0), (3j, 0, . . . , 0),

0, otherwise;
• x (3) = (x (3)k )k∈Zd is defined by

x
(3)
k =


1, k = (0, 0, . . . , 0), (2j, 0, . . . , 0),

−1, k = (j, 0, . . . , 0), (3j, 0, . . . , 0),

0, otherwise.
As in the case where d = 1, one may observe that

‖x (1)‖`pq = sup
m∈Zd ,N∈ω

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k |
p

) 1
p

= max{1, (j + 1)d(
1
q
− 1
p
)2
1
p , (2j + 1)d(

1
q
− 1
p
)3
1
p , (3j + 1)d(

1
q
− 1
p
)4
1
p }

= 1.

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 6We also get ‖x (2)‖`pq = ‖x (3)‖`pq = 1. Moreover, through similar observation as in the 1-dimensionalcase, we have
‖x (1) ± x (2) ± x (3)‖`pq = 3for every possible combinations of ± signs. It thus follows that

C
(3)
J (`

p
q(Zd)) = sup{min ‖x1 ± x2 ± x3‖`pq : x1, x2, x3 ∈ S`pq} = 3

and
C
(3)
NJ (`

p
q(Zd)) = sup

{∑
± ‖x1 ± x2 ± x3‖2`pq
22
∑3
i=1 ‖xi‖`pq

: xi 6= 0, i = 1, 2, 3
}
= 3.

�

We now state the general result for n ≥ 3. (The proof is also valid for n = 2, which amounts tothe work of [3].)
Theorem 2.2. For 1 ≤ p < q <∞, we have C(n)NJ (`

p
q(Zd)) = C(n)J (`

p
q(Zd)) = n.

Proof. As for n = 3, we shall consider the case where d = 1 first, and then the case where d > 1later.
Case 1: d = 1. Let j ∈ Z be a nonnegative, even integer such that j > 2(n−1)( qq−p ) − 1, which isequivalent to

(j + 1)
1
q
− 1
p < 2−

(n−1)
p .

We construct x (i) ∈ `pq ∈ Z for i = 1, 2, . . . , n as follows:
• x (1) = (x (1)k )k∈Z is defined by

x
(1)
k =

1, k ∈ S(1)1 ,

0, otherwise,
where

S
(1)
1 = {0, j, 2j, 3j, . . . , (2

n−1 − 1)j};

• x (i) = (x (i)k )k∈Z for 2 ≤ i ≤ n is defined by
x
(i)
k =


1, k ∈ S(i)1 ,

−1, k ∈ S(i)−1,

0, otherwise,
with the following rules: Write P = {0, j, 2j, . . . , (2n−1 − 1)j} as

P = P
(i)
1 ∪ P

(i)
2 ∪ · · · ∪ P

(i)

2i−1

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 7

where P (i)1 consists of the first 2n−1
2i−1

terms of P , P (i)2 consists of the next 2n−1
2i−1

terms of P ,and so on. Then S(i)1 and S(i)−1 are given by
S
(i)
1 = P

(i)
1 ∪ P

(i)
3 ∪ · · · ∪ P

(i)

2i−1−1,

S
(i)
−1 = P

(i)
2 ∪ P

(i)
4 ∪ · · · ∪ P

(i)

2i−1
.

For example, for i = 2, x (2) = (x (2)k )k∈Z is defined by
x
(2)
k =


1, k ∈ S(2)1 ,

−1, k ∈ S(2)−1 ,

0, otherwise,
where

S
(2)
1 =

{
0, j, 2j, 3j, . . . ,

(2n−1
2
− 1
)
j

}
S
(2)
−1 =

{(2n−1
2

)
j,
(2n−1
2
+ 1
)
j, . . . , (2n−1 − 1)j

}
;

Note that the largest absolute value of the terms of x (i) in the above construction will beequal to 1 for each i = 1, . . . , n. Next, since the number of possible combinations of ± signs in
x (1) ± x (2) ± · · · ± x (n) is 2n−1, the above construction will give us 1 + 1 + · · · + 1 = n as thelargest absolute value of x (1)± x (2)±· · ·± x (n) for every combination of ± signs. This means that,if x (1) ± x (2) ± · · · ± x (n) = (xk)k∈Z, then max

k∈Z
|xk | = n.Let us now compute the norms. For x (1), we have

‖x (1)‖`pq = sup
m∈Z,N∈ω

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k |
p

) 1
p

= sup
m∈Z∩[0,(2n−1−1)j ],N∈Z∩[0,(2n−1−1)j/2]

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k |
p

) 1
p

= max{1, (j + 1)
1
q
− 1
p 2

1
p , (2j + 1)

1
q
− 1
p 3

1
p , . . . , ((2n−1 − 1)j + 1)

1
q
− 1
p 2

n−1
p }.

For each r = 1, 2, . . . , 2n−1 − 1, we have (r j + 1) 1q− 1p ≤ (j + 1) 1q− 1p and (r + 1) 1p ≤ 2 n−1p , so that
(r j + 1)

1
q
− 1
p (r + 1)

1
p ≤ (j + 1)

1
q
− 1
p 2

n−1
p < 2−

n−1
p 2

n−1
p = 1.

Hence we obtain ‖x (1)‖`pq = 1. Similarly, one may verify that
‖x (2)‖`pq = ‖x

(3)‖`pq = · · · = ‖x
(n)‖`pq = 1.

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 8Next, we shall compute the norms of x (1)±x (2)±· · ·±x (n). Write x (1)+x (2)+ · · ·+x (n) = (xk)k∈Zwhere

xk :=



a1, k = 0,

a2, k = j,

a3, k = 2j,...
a2n−1 , k = (2n−1 − 1)j,

0, otherwise,
with a1 = n and |ai | < n for i = 2, 3, . . . , (2n−1)j . Accordingly, we have
‖x (1) + x (2) + · · ·+ x (n)‖`pq = sup

m∈Z,N∈ω
|Sm,N |

1
q
− 1
p

( ∑
k∈Sm,N

|xk |p
) 1
p

= sup
m∈Z∩[0,(2n−1−1)j ],N∈Z∩[0,(2n−1−1)j/2]

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|xk |p
) 1
p

=max
{
n, (j + 1)

1
q
− 1
p (np + ap2)

1
p , (2j + 1)

1
q
− 1
p (np + ap2 + a

p
3)
1
p ,

. . . , ((2n−1 − 1)j + 1)
1
q
− 1
p
(
np +

2n−1∑
i=2

api
) 1
p

}
.

Since (r j + 1) 1q− 1p ≤ (j + 1) 1q− 1p for each r = 1, 2, . . . , 2n−1 − 1, we obtain
(r j + 1)

1
q
− 1
p
(
np +

r+1∑
i=2

api
) 1
p ≤ (j + 1)

1
q
− 1
p
(
np +

r+1∑
i=2

api
) 1
p

< 2−
(n−1)
p
(
np +

r+1∑
i=2

api
) 1
p

< 2−
(n−1)
p (np + np + · · ·+ np︸ ︷︷ ︸

r + 1 times )
1
p

= 2−
(n−1)
p (r + 1)

1
p (np)

1
p

≤ 2−
(n−1)
p 2

(n−1)
p n

= n.

It thus follows that
‖x (1) + x (2) + · · ·+ x (n)‖`pq = n.As we have remarked earlier, the largest absolute value of x (1) ± x (2) ± · · · ± x (n) is equal to

n for every combination of ± signs. Moreover, it is clear that for k /∈ {0, 2j, . . . , (2n−1 − 1)j}, the

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 9

k-th term of x (1) ± x (2) ± · · · ± x (n) is equal to 0. Hence, we obtain
‖x (1) ± x (2) ± · · · ± x (n)‖`pq = sup

m∈Z,N∈ω
|Sm,N |

1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k ± x
(2)
k ± · · · ± x

(n)
k |

p

) 1
p

= sup
m∈Z∩[0,(2n−1−1)j ],N∈Z∩[0,(2n−1−1)j/2]

|Sm,N |
1
q
− 1
p

( ∑
k∈Sm,N

|x (1)k ± x
(2)
k ± · · · ± x

(n)
k |

p

) 1
p

= n.

Consequently, we get ∑
± ‖x (1) ± x (2) ± · · · ± x (n)‖2`pq

2n−1
∑n
i=1 ‖xi‖`pq

=
2n−1n2

2n−1n
= n

and
min ‖x (1) ± x (2) ± · · · ± x (n)‖`pq = n,whence
C
(n)
NJ (`

p
q(Z)) = C

(n)
J (`

p
q(Z)) = n.

Case 2: d > 1. Here we choose j ∈ Z to be a nonnegative, even integer such that j >
2(
n−1
d
)( q
q−p ) − 1 or, equivalently,

(j + 1)d(
1
q
− 1
p
) < 2−

(n−1)
p .

Then, using the sequences
x (i) = (x

(i)
k1
)k1∈Z ∈ `

p
q(Z), i = 1, . . . , n,

in the case where d = 1, we now define x (i) := (x (i)k )k∈Zd ∈ `pq(Zd) for i = 1, . . . , n, where
x
(i)
k =

x (i)k1 , k = (k1, 0, 0, . . . , 0),

0, otherwise.
We shall then obtain

C
(n)
NJ (`

p
q(Zd)) = C

(n)
J (`

p
q(Zd)) = n,as desired. �

Corollary 2.2.1. For 1 ≤ p < q <∞, the space `pq is not uniformly non-`1n.

Corollary 2.2.2. For 1 ≤ p < q <∞, the space `pq is not uniformly n-convex.

Acknowledgement. The work is part of the first author’s thesis. Both authors are supported byP2MI 2021 Program of Bandung Institute of Technology.

https://doi.org/10.28924/ada/ma.2.2


Eur. J. Math. Anal. 10.28924/ada/ma.2.2 10References
[1] B. Beauzamy, Introduction to Banach Spaces and Their Geometry, 2nd Ed., North Holland, Amsterdam- NewYork-Oxford, 1985. https://pascal-francis.inist.fr/vibad/index.php?action=getRecordDetail&idt=

PASCAL82X0319279.[2] H. Gunawan, D.I. Hakim, A.S. Putri, On geometric properties of Morrey spaces, Ufimsk. Mat. Zh. 13 (2021) 131–136.
https://doi.org/10.13108/2021-13-1-131.[3] H. Gunawan, E. Kikianty, C. Schwanke, Discrete Morrey spaces and their inclusion properties, Math. Nachr. 291(2018) 1283–1296. https://doi.org/10.1002/mana.201700054.[4] H. Gunawan, E. Kikianty, Y. Sawano, and C. Schwanke, Three geometric constants for Morrey spaces, Bull. Korean.Math. Soc. 56 (2019) 1569-1575. https://doi.org/10.4134/BKMS.b190010.[5] R.C. James, Uniformly non-square banach spaces, Ann. Math. 80 (1964) 542-550. https://doi.org/10.2307/
1970663.[6] M. Kato, Y. Takahashi, and K. Hashimoto, On n-th Von Neumann-Jordan constants for Banach spaces, Bull. KyushuInst. Tech. 45 (1998), 25-33. https://ci.nii.ac.jp/naid/110000079659.[7] L. Maligranda, L. Nikolova, L.-E. Persson, T. Zachariades, On n-th James and Khintchine constants of Banachspaces, Math. Inequal. Appl. 1 (2007) 1–22. https://doi.org/10.7153/mia-11-01.[8] W.A. Wojczynski, Geometry and martingales in Banach spaces, Part II, in: Probability in Banach Spaces IV, J.Kuelbs, ed., Marcel-Dekker, 1978, 267–517. https://doi.org/10.1201/9780429462153.

https://doi.org/10.28924/ada/ma.2.2
https://pascal-francis.inist.fr/vibad/index.php?action=getRecordDetail&idt=PASCAL82X0319279
https://pascal-francis.inist.fr/vibad/index.php?action=getRecordDetail&idt=PASCAL82X0319279
https://doi.org/10.13108/2021-13-1-131
https://doi.org/10.1002/mana.201700054
https://doi.org/10.4134/BKMS.b190010
https://doi.org/10.2307/1970663
https://doi.org/10.2307/1970663
https://ci.nii.ac.jp/naid/110000079659
https://doi.org/10.7153/mia-11-01
https://doi.org/10.1201/9780429462153

	1. Introduction
	2. Main Results
	References

