©2022 Ada Academica https://adac.eeEur. J. Math. Anal. 2 (2022) 2doi: 10.28924/ada/ma.2.2 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‖ 2 X 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 https://adac.ee 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 2 Meanwhile, 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(`pq) = 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 3 2. 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 ` p q 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 qq−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 4 The 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 p2 1 p ,(2j +1) 1 q −1 p3 1 p ,(3j +1) 1 q −1 p4 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 (`pq(Z))= C (3) J (`pq(Z))=3. Case 2: d > 1. Let j ∈ Z be a nonnegative, even integer such that j > 4 qd(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 6 We 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 (`pq(Z d))= sup{min‖x1 ±x2 ±x3‖`pq : x1,x2,x3 ∈ S`pq}=3 and C (3) NJ (`pq(Z d))= sup {∑ ±‖x1 ±x2 ±x3‖ 2 ` p q 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−12i−1 terms of P , P(i)2 consists of the next 2n−12i−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 p2 1 p ,(2j +1) 1 q −1 p3 1 p , . . . ,((2n−1 −1)j +1) 1 q −1 p2 n−1 p }. For each r =1,2, . . . ,2n−1 −1, we have (rj +1)1q−1p ≤ (j +1)1q−1p and (r +1)1p ≤ 2n−1p , so that (rj +1) 1 q −1 p(r +1) 1 p ≤ (j +1) 1 q −1 p2 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 8 Next, 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 =(2 n−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 +a p 2) 1 p ,(2j +1) 1 q −1 p(np +a p 2 +a p 3) 1 p , . . . ,((2n−1 −1)j +1) 1 q −1 p ( np + 2n−1∑ i=2 a p i )1 p } . Since (rj +1)1q−1p ≤ (j +1)1q−1p for each r =1,2, . . . ,2n−1 −1, we obtain (rj +1) 1 q −1 p ( np + r+1∑ i=2 a p i )1 p ≤ (j +1) 1 q −1 p ( np + r+1∑ i=2 a p i )1 p < 2 −(n−1) p ( np + r+1∑ i=2 a p i )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 ` p q 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 (`pq(Z))= C (n) J (`pq(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 ∈ ` p q(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 (`pq(Z d))= C (n) J (`pq(Z d))= 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. 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Main Results References