Let X be a Banach space and ψ a continuous convex function on ∆K+1satisfying certain conditions

Bull. Korean Math. Soc. 50 (2013), No. 2, pp. 417–430 http://dx.doi.org/10.4134/BKMS.2013.50.2.417

CHARACTERIZATIONS OF GEOMETRICAL PROPERTIES OF BANACH SPACES USING ψ-DIRECT SUMS

Zhihua Zhang, Lan Shu, Jun Zheng, and Yuling Yang

Abstract. Let X be a Banach space and ψ a continuous convex function on ∆K+1satisfying certain conditions. Let (XLXL

· · ·L

X)ψbe the ψ-direct sum of X. In this paper, we characterize the K strict convexity, K uniform convexity and uniform non-l^N₁-ness of Banach spaces using ψ-direct sums.

1. Introduction A norm k · k on Cⁿ is said to be absolute if

k(x1, x2, . . . , xn)k = k(|x1|, |x2|, . . . , |xn|)k for any (x1, x2, . . . , xn) ∈ Cⁿ and normalized if

k(1, 0, . . . , 0)k = k(0, 1, 0, . . . , 0)k = · · · = k(0, . . . , 0, 1)k.

The lp-norms are such examples:

k(x1, x2, . . . x_n)kp=

((|x1|^p+ |x2|^p+ · · · + |xn|^p)^1p, 1 ≤ p < ∞ max{|x1|, |x2|, . . . , |xn|}, p = ∞.

Let ANn be the family of all absolute normalized norms on Cⁿ. When n = 2 Bonsall and Duncan [2] showed the following characterization of absolute normalized norms on C². Namely, the set AN2 of all absolute normalized norms on C²is in one-to-one correspondence with the set Ψ2of all continuous convex functions on [0, 1] satisfying ψ(0) = ψ(1) = 1 and max{1−t, t} ≤ ψ(t) ≤ 1, 0 ≤ t ≤ 1. The correspondence is given by

(1) ψ(t) = k(1 − t, t)k, 0 ≤ t ≤ 1.

Received September 8, 2011; Revised December 8, 2011.

2010 Mathematics Subject Classification. 46B25, 46b20, 46B99.

Key words and phrases. absolute norm, K strict convexity, K uniform convexity, uniform non-l^N₁ -ness.

The first and second authors are supported by Nation Natural Science Foundation of China(No. 11071178).

c

2013 The Korean Mathematical Society 417

Indeed, for any ψ ∈ Ψ2, define

k(z, w)kψ=

((|z| + |w|)ψ(_|z|+|w|^|w| ), (z, w) 6= (0, 0)

0, (z, w) = (0, 0).

By calculation we have k · kψ∈ AN2 and k · kψ satisfies (1). From this result, there are plenty of concrete absolute normalized norms of C² which are not lp-type.

In [13] K.-S. Saito, M. Kato and Y. Takahashi generalized the result to Cⁿ. Before stating it, we give some notations. For each n ∈ N with n ≥ 2, we put

∆n =







(t1, t2, t3, . . . , tn−1) ∈ Rⁿ⁻¹: tj ≥ 0,

n−1

X

j=1

tj ≤ 1







and define the set Ψn of all continuous convex functions on ∆n satisfying the following conditions:

(A0) ψ(0, 0, . . . , 0) = ψ(1, 0, . . . , 0) = · · · = ψ(0, . . . , 0, 1), (A1)

ψ(t1, t2, . . . , t_n−1) ≥ (t1+ t2+ · · · + t_n−1)ψ







 t1 n−1

P

i=1

ti

, . . . , tn−1 n−1

P

i=1

ti







 , if

n−1

X

i=1

t_i6= 0,

(A2) ψ(t1, t2, . . . , t_n−1) ≥ (1 − t1)ψ

 0, t2

1 − t1

, . . . , tn−1

1 − t1



, if t16= 1,

(A3) ψ(t1, t2, . . . , tn−1) ≥ (1 − t2)ψ

 t1

1 − t2

,0, . . . , tn−1

1 − t2



, if t26= 1,

... (An)

ψ(t1, t2, . . . , t_n−1) ≥ (1 − t_n−1)ψ

 t1

1 − tn−1

, . . . , t_n−2 1 − tn−1

,0



, if t_n−16= 1.

K.-S. Saito, M. Kato and Y. Takahashi in [13] showed that, for each n ∈ N with n ≥ 2, ANn and Ψn are in one-to-one correspondence under the following equation:

(2) ψ(t1, . . . , t_n−1) =

(1 −

n−1

X

j=1

t_j, t1, . . . , t_n−1)

,(t1, . . . , t_n−1) ∈ ∆n.

Indeed, for any ψ ∈ Ψn, the norm k · kψ on Cⁿ is defined as

k(x0, x1, . . . , xn−1)kψ =















_n−1 P

i=0

|xi|

 ψ



 ^|x

1|

n−1P

i=0|xi|

, . . . ,_n−1^|x_P^n−1|

i=0|xi|



, (x0, x1, . . . , xn−1) 6= (0, . . . , 0) 0, (x0, x1, . . . , xn−1) = (0, . . . , 0).

Moreover, M. Kato, K.-S. Saito and Tamura in [6] introduced the ψ-direct sums (X1L X2L · · · L Xn)ψas follows. Let X1, X2, . . . , X_nbe Banach spaces and let ψ ∈ Ψn. Then the product space X1× X2× · · · × Xn with the norm

k(x1, x2, . . . , xn)kψ= k(kx1k, kx2k, . . . , kxnk)kψ, xi∈ Xi, 1 ≤ i ≤ n, is a Banach space which is denoted by (X1L X2L · · · L Xn)ψ. They showed that (X1L X₂L · · · L Xn)ψ is strictly convex (uniformly convex) if and only if X1, X2, . . . , Xn is strictly convex (uniformly convex) and ψ ∈ Ψn is strictly convex. In [7] the authors presented that XL

ψY is uniformly non-square if and only if X and Y are uniformly non-square and ψ 6= ψ1, ψ_∞. Since the introduction of ψ-direct sums of Banach spaces, it has attracted plenty of attention and been treated by several authors (cf. [3, 4, 5, 12, 16]).

In particular, K.-I. Mitani and K.-S. Saito in [11] characterized the strict convexity, uniform convexity and uniform non-squareness of Banach spaces using ψ-direct sums XL

ψX. They showed that, if t0 is a unique minimal point, a Banach space X is strictly convex if and only if, for each x, y ∈ X with x6= y, then

k(1 − t0)x + t0yk < 1

ψ(t0)k((1 − t0)x, t0y)kψ, ψ∈ Ψ2.

As for the cases of uniform convexity and uniform non-squareness, they gained some similar results.

Our main purpose of this paper is to give the characterization of K strict convexity, K uniform convexity and uniform non-l₁^N-ness using ψ-direct sums (XL X L · · · L X)ψ, we first characterize the K strict convexity using ψ- direct sums. We show that, if ψ has a minimal point s0= (t1, t2, . . . , t_K), and 0 < ti <1, i = 1, 2, . . . , K and 0 <PK

i=1ti<1, then a Banach space X is K strictly convex if and only if for any x0, x1, . . . , xK ∈ X, with x0, x1, . . . , xK

linearly independent, we have

kt0x0+ t1x1+ · · · + tKxKk < 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ, where PK

i=0ti = 1. As a result, we can give different characterization by choosing different ψ. In contrast with the result of K.-I. Mitani and K.-S. Saito [11], the uniqueness of t0 is not required, but the linear independence of x and y is necessary. Moreover when K = 1, we get the characterization of strict convexity. In Section 3, we also characterize the K uniform convexity and

make Theorem 8 in [11] as our Corollary 3.5. In Section 4, the characterization of uniform non-l₁^N-ness is gained by adding the uniqueness of minimal point.

2. K strict convexity

A Banach space X is said to be K strictly convex (cf. [14]) if and only if for any K + 1 elements x0, x1, . . . , x_K in X, whenever kPK

i=0x_ik =PK i=0kxik, then x0, x1, . . . , xK are linearly dependent.

The closed unit ball of a Banach space X is {x ∈ X : kxk ≤ 1} and is denoted by BX, the unit sphere of X is {x ∈ X : kxk = 1} and is denoted by SX. It is obvious that when K = 1, X is strictly convex.

Proposition 2.1(cf. [8]). Let X be a Banach space. For all non-zero elements x1, x2, . . . , xn ∈ X, the following inequality holds:

k

n

X

j=1

x_jk +



n− k

n

X

j=1

x_j kxjkk



 min

1≤j≤nkxjk ≤

n

X

j=1

kxjk

≤ k

n

X

j=1

xjk +



n− k

n

X

j=1

xj

kxjkk



 max

1≤j≤nkxjk.

Lemma 2.2. Let X be a Banach space. Then the following assertions are equivalent.

(1) X is K strictly convex.

(2) For any x0, x1, . . . , xK ∈ SX, whenever kPK

i=0xik = K + 1, then x0, x1, . . . , xK are linearly dependent.

(3) If x0, x1, . . . , xK ∈ SX and x0, x1, . . . , xK are linearly independent, then for any {ti}^K_i=0 satisfying0 < ti<1,PK

i=0ti= 1, there holds kPK

i=0tixik < 1.

(3^′) If x0, x1, . . . , xK∈ SX and x0, x1, . . . , xK are linearly independent, then there exists {ti}^K_i=0 with 0 < ti<1,PK

i=0ti= 1, such that kPK

i=0tixik < 1.

Proof. (1) ⇒ (2) is obvious.

(2) ⇒ (1) Let any x0, x1, . . . , xK ∈ X\{0}, and kPK

i=0xik = PK i=0kxik.

By Proposition 2.1 we have kPK i=0 xi

kxikk = K + 1. Hence _kx^x0

0k,_kx^x1

1k, . . . ,_kx^xK

Kk

are linearly dependent, so do x0, x1, . . . , xK.

(2) ⇒ (3) Assume that the conclusion falls to hold. Then there exists {ti}^K_i=0 satisfying 0 < ti<1,PK

i=0ti= 1, but kPK

i=0tixik = 1. Using Proposition 2.1 we have

k

K

X

i=0

tixik + K+ 1 − k

K

X

i=0

xi

kxikk

!

0≤i≤Kmin ktixik ≤

K

X

i=0

tikxik

≤ k

K

X

i=0

tixik + K+ 1 − k

K

X

i=0

x_i kxikk

!

0≤i≤Kmax ktixik.

Hence kPK i=0 xi

kxikk = K + 1. So x0, x1, . . . , xK are linearly dependent. Con- tradiction.

(3) ⇒ (2) Clearly.

(2) ⇒ (3^′) We just need to let ti= _K+1¹ , i = 0, 1, . . . , K.

(3^′) ⇒ (2) If there are x0, x1, . . . , x_K ∈ SX and satisfying kPK

i=0x_ik = K+ 1, but x0, x1, . . . , xK are linearly independent. Then there exists {ti}^K_i=0, with 0 < ti <1,PK

i=0ti = 1, and kPK

i=0tixik < 1. Considering Proposition 2.1 we have

1 =

K

X

i=0

tikxik ≤ k

K

X

i=0

tixik + (K + 1 − k

K

X

i=0

xik) max

0≤i≤Kktixik, that is kPK

i=0t_ix_ik ≥ 1, contradiction. 

Theorem 2.3. Let ψ ∈ ΨK+1. Assume that ψ has a minimal point s0 = (t1, t2, . . . , tK), and 0 < ti <1, i = 1, 2, . . . , K and 0 <PK

i=1ti <1. Then a Banach space X is K strictly convex if and only if for any x0, x1, . . . , xK∈ X, with x0, x1, . . . , xK linearly independent, we have

kt0x0+ t1x1+ · · · + tKxKk < 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ, where PK

i=0t_i= 1.

Proof. Assume that X is K strictly convex. Since ψ(s) ≥ ψ(s0) for all s ∈

∆K+1, and t0x0, t1x1, . . . , t_Kx_K are linearly independent, then we have kt0x₀+ t1x₁+ · · · + tKx_Kk

<kt0x0k + kt1x1k + · · · + ktKxKk

= k(t0x0, t1x1, . . . , t_Kx_K)k1

≤ max

s∈∆K+1

ψ1(s)

ψ(s)k(t0x0, t1x1, . . . , t_Kx_K)kψ

= 1

s∈∆minK+1

ψ(s)k(t0x0, t1x1, . . . , t_Kx_K)kψ

= 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ.

Conversely for any xi ∈ SX, i = 0, 1, . . . , K with x0, x1, . . . , xK linearly inde- pendent. We have

kt0x0+ t1x1+ · · · + tKx_Kk

< 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ

= 1

ψ(s0)k(t0, t1, . . . , tK)kψ

= 1

ψ(s0)k(1 −

K

X

i=1

t_i, t1, . . . , t_K)kψ= 1.

 Corollary 2.4. Let ψ∈ Ψ2. Assume that ψ has a minimal point t0. Then a Banach space X is strictly convex if and only if, for each x, y ∈ X with x, y linearly independent we have

k(1 − t0)x + t0yk < 1

ψ(t0)k((1 − t0)x, t0y)kψ.

Corollary 2.5. If ψ= ψp∈ ΨK+1, when1 < p < ∞, ψp(t1, t2, . . . , t_K) = ((1−

PK

i=1t_i)^p+ t^p₁+ · · · + t^p_K)^1p. Note that for any s6= (_K+1¹ , . . . ,_K+1¹ ), s ∈ ∆K+1, ψp(s) > ψp(_K+1¹ , . . . ,_K+1¹ ) = (K + 1)^p1−1. Then a Banach space X is K strictly convex if and only if for any x0, x1, . . . , x_K ∈ X with x0, x1, . . . , x_K linearly independent, we have

kx0+ x1+ · · · + xK

K+ 1 k^p<kx0k^p+ · · · + kxKk^p

K+ 1 .

Theorem 2.3 does not require that ψ is strictly convex. This should be contrasted with the result of [6], i.e., (X1L X2L · · · L Xn)ψis strictly convex if and only if X1, X2, . . . , X_n are strictly convex respectively and ψ is a strictly convex function on ∆n. Thus, let k·k = max{k·k2, λk·k1} (^√_K+1¹ < λ <1). Let ψbe the corresponding convex function of k·k. Then for any s = (s1, . . . , s_K) ∈

∆K+1,

ψ(s) = k(1 −

K

X

i=1

si, s1, . . . , sK)k

= max (

k(1 −

K

X

i=1

s_i, s1, . . . , s_K)k2, λk(1 −

K

X

i=1

s_i, s1, . . . , s_K)k1

)

= max{ψ2(s), λ}.

Since mins∈∆K+1ψ2(s) = √¹

K+1. Then ψ(s) =

 λ, √¹

K+1 ≤ ψ2(s) ≤ λ ψ2(s), λ < ψ2(s) ≤ 1.

For ψ2(s) is continuous on ∆K+1, we have mins∈∆K+1ψ(s) = λ and ψ is not strictly convex on ∆K+1. Applying Theorem 2.3, we can give the following characterization using ψ above.

Corollary 2.6. Let √¹

K+1 < λ ≤ 1. Then a Banach space X is K strictly convex if and only if for any x0, x1, . . . , x_K ∈ X, with x0, x1, . . . , x_K linearly independent, we have

kx0+ x1+ · · · + xK

K+ 1 k

< 1 λmax

((kx0k²+ · · · + kxKk²)¹²

K+ 1 , λkx0k + · · · + kxKk K+ 1

)

= max

((kx0k²+ · · · + kxKk²)¹²

λ(K + 1) ,kx0k + · · · + kxKk K+ 1

) .

3. K uniform convexity

We say that a Banach space X is K uniformly convex (or K uniformly rotund see [15]) if for any ε > 0, there exists some δ = δ(ε) > 0, such that whenever x0, x1, . . . , x_K ∈ SX and kx0+ x1+ · · · + xKk > (K + 1) − δ, we have

A(x0, x1, . . . , xK)

≡ sup









        

1 1 · · · 1

f1(x0) f1(x1) · · · f1(xK)

· · · · fK(x0) fK(x1) · · · fK(xK)

,{fi}^K_i=1⊂ BX^∗











< ε.

In the case of K = 1, X is uniformly convex.

Proposition 3.1(cf. [17]). Let X be a Banach space. Then X is K uniformly convex if and only if for any K+ 1 sequences {xⁿ₀}, {xⁿ₁}, . . . , {xⁿ_K} in X, if kxⁿ_ik → a, n → ∞, i = 0, 1, 2, . . . , K and kxⁿ₀ + xⁿ₁ + · · · + xⁿ_Kk → (K + 1)a, then

nlim→∞A(xⁿ₀, xⁿ₁, . . . , xⁿ_K) = 0.

Proposition 3.2 (cf. [9]). Let {x^k₁}k,{x^k₂}k, . . . ,{x^k_n}k be n sequences in a Banach space X for which the sequences of their norms are convergent. Then the following are equivalent.

(1) lim

k→∞kPn

j=1x^k_jk = lim

k→∞

Pn

j=1kx^k_jk.

(2) lim

k→∞kαx^k₁+Pn

j=2x^k_jk = lim

k→∞(αkx^k₁k +Pn

j=2kx^k_jk) for all α > 0.

(3) lim

k→∞kαx^k₁+Pn

j=2x^k_jk = lim

k→∞(αkx^k₁k +Pn

j=2kx^k_jk) for some α > 0.

Proposition 3.3 (cf. [13]). Let ψ ∈ Ψn and let x = (x1, x2, . . . , x_n), y = (y1, y2, . . . , y_n) ∈ Cⁿ. Then

(1) If |x| ≤ |y|, then kxkψ≤ kykψ.

(2) If ψ is strictly convex and |x| < |y|, then kxkψ<kykψ.

For x= (x1, x2, . . . , xn) ∈ Cⁿ, denote|x| by |x| = (|x1|, |x2|, . . . , |xn|). We say that |x| ≤ |y| if |xj| ≤ |yj| for 1 ≤ j ≤ n. Further, we say that |x| < |y| if

|x| ≤ |y| and |xj| < |yj| for some j.

Theorem 3.4. Let ψ ∈ ΨK+1. Assume that ψ has a unique minimal point s0 = (t1, t2, . . . , tK), with 0 < ti<1, PK

i=1ti <1. Then a Banach space X is K uniformly convex if and only if for any ε >0, there exists some δ > 0, such

that for any x0, x1, . . . , xK ∈ BX, satisfying kt0x0+ t1x1+ · · · + tKxKk > (1 − δ) 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ, where PK

i=0ti= 1, then we have A(x0, x1, . . . , xK) < ε.

Proof. Let X be a K uniformly convex Banach space. Assume that there exists ε0 >0, for any n ∈ N, there are sequences {xⁿ₀}, {xⁿ₁}, . . . , {xⁿ_K} in BX

satisfying

(3) kt0xⁿ₀+ t1xⁿ₁+ · · · + tKxⁿ_Kk > (1 −1 n) 1

ψ(s0)k(t0xⁿ₀, t1xⁿ₁, . . . , t_Kxⁿ_K)kψ. But A(xⁿ₀, xⁿ₁, . . . , xⁿ_K) ≥ ε0.

Since {kxⁿ_ik}^∞_n=1, i = 0, 1, . . . , K and {kPK

i=0tixⁿ_ik}^∞_n=1 are bounded se- quences, without loss of generality we can let kxⁿ_ik → ai (n → ∞), i = 0, 1, . . . , K and kt0xⁿ₀ + t1xⁿ₁ + · · · + tKxⁿ_Kk → b (n → ∞). Moreover, we can choose {kxⁿ_ik}^∞_n=1 such that max{kxⁿ_ik, 0 ≤ i ≤ K} = 1. Thus max{ai,0 ≤ i≤ K} = 1. From this,PK

i=0tiai >0. It is clear that 0 ≤ ai ≤ 1, 0 ≤ b ≤ 1.

Considering the equality (2), we have (1 − 1

n) 1

ψ(s0)k(t0xⁿ₀, t1xⁿ₁, . . . , t_Kxⁿ_K)kψ

= (1 − 1 n) 1

ψ(s0)k(t0kxⁿ₀k, t1kxⁿ₁k, . . . , tKkxⁿ_Kk)kψ

<kt0xⁿ₀+ t1xⁿ₁+ · · · + tKxⁿ_Kk

≤ t0kxⁿ₀k + t1kxⁿ₁k + · · · + tKkxⁿ_Kk.

Let n → ∞. Then _ψ(s¹

0)k(t0a0, t1a1, . . . , t_Ka_K)kψ ≤ t0a0+ t1a1+ · · · + tKa_K holds. Hence

ψ t1a1

PK

i=0t_ia_i, . . . , t_Ka_K PK

i=0t_ia_i

!

≤ ψ(s0) = ψ(t1, t2, . . . , t_K).

From the uniqueness of s0, we get a0= a1= · · · = aK. Let us denote them as a. Moreover,

nlim→∞k

K

X

i=0

tixⁿ_ik = lim

n→∞

K

X

i=0

ktixⁿ_ik.

Using Proposition 3.2 we get

n→∞lim k1 t0

t0xⁿ₀ +

K

X

i=1

tixⁿ_ik = lim

n→∞(kxⁿ₀k +

K

X

i=1

ktixⁿ_ik).

Repeat the similar process above for K + 1 times, we have

n→∞lim k

K

X

i=0

xⁿ_ik = lim

n→∞

K

X

i=0

kxⁿ_ik = (K + 1)a.

Hence there is limn→∞A(xⁿ₀, xⁿ₁, . . . , xⁿ_K) = 0. By Proposition 3.1, it is a con- tradiction.

Conversely, for any ε > 0 there exists some δ > 0, such that for any x0, x1, . . . , xK in SX with A(x0, x1, . . . , xK) ≥ ε, we have

kt0x0+ t1x1+ · · · + tKx_Kk

≤ (1 − δ) 1

ψ(s0)k(t0x0, t1x1, . . . , tKxK)kψ

≤ (1 − δ) 1

ψ(s0)k(t0, t1, . . . , tK)kψ= 1 − δ.

By Proposition 2.1 we have 1 =

K

X

i=0

t_ikxik ≤ (K + 1 − k

K

X

i=0

x_ik) max

0≤i≤Kt_ikxik + k

K

X

i=0

t_ix_ik

≤ (K + 1 − k

K

X

i=0

xik) + 1 − δ.

Hence kPK

i=0x_ik ≤ (K + 1) − δ. 

Corollary 3.5 (cf. [11]). Let ψ ∈ Ψ2. Assume that ψ has a unique minimal point t0. Then a Banach space X is uniformly convex if and only if, for every ε >0, there exists some δ > 0 such that kx − yk ≥ ε, x, y ∈ BX implies

k(1 − t0)x + t0yk ≤ (1 − δ) 1

ψ(t0)k((1 − t0)x, t0y)kψ. Corollary 3.6. Let ψ(s) = ψp(s) = [(1−PK

i=1si)^p+s^p₁+· · ·+s^p_K]^1p,1 < p < ∞.

Then ψ_p(s) has a unique minimal point s0= (_K+1¹ ,_K+1¹ , . . . ,_K+1¹ ). A Banach space X is K uniformly convex if and only if for every ε >0, there exists some δ >0 such that for any x0, x1, . . . , xK in BX satisfying

kx0+ x1+ · · · + xK

K+ 1 k^p>(1 − δ)kx0k^p+ · · · + kxKk^p K+ 1 implies A(x0, x1, . . . , x_K) < ε.

4. Uniform non-l^N

1 -ness

A Banach space X is said to be uniformly non-l^N₁ (cf. [1, 10]) provided there exists δ(0 < δ < 1) such that for any x0, x1, . . . , xN−1 in SX, there exists an N-tuple of signs θ = (θj) for which

k

N−1

X

j=0

θ_jx_jk ≤ N (1 − δ).

In the case of N = 2, X is called uniform non-squareness. As is well known, we may take x0, x1, . . . , xN−1 from BX in the definition (see [8]).

Lemma 4.1. A Banach space X is uniformly non-l^N₁ if and only if there exist some s= (s0, s1, . . . , sN−1), withPN−1

i=0 si= 1, 0 < si<1, i = 0, 1, . . . , N − 1, and some δ(0 < δ < 1), such that for any x0, x1, . . . , xN−1 in BX, there exists an N -tuple of signs θ= (θj), for which kPN−1

j=0 θjsjxjk ≤ 1 − δ.

Proof. Assume that X is uniformly non-l^N₁. Let si = _N¹, i= 0, 1, . . . , N − 1.

For any x0, x1, . . . , x_N−1 in SX, there exists an N -tuple of signs θ = (θj) and s= (s0, . . . , sN−1) withPN−1

i=0 si= 1, kPN−1

j=0 θjsjxjk ≤ 1−δ. Use Proposition 2.1 we have

1 =

N−1

X

j=0

kθjs_jx_jk

≤ k

N−1

X

j=0

θjsjxjk + (N − k

N−1

X

j=0

θjxjk) max

0≤i≤N−1kθjsjxjk

≤ 1 − δ + N − k

N−1

X

j=0

θ_jx_jk.

Let δ^′ = _N^δ. Then for any x0, x₁, . . . , x_N−1 in SX, there exists an N -tuple of signs θ = (θj), for which kPN−1

j=0 θjxjk ≤ N (1 − δ^′). 

Lemma 4.2. Let X be a Banach space. Then X is uniformly non-l^N₁ if and only if for any N sequences{xⁿ₀}, . . . , {xⁿ_N−1} in X and kxⁿ_jk → a(a > 0), n →

∞, j = 0, 1, . . . , N − 1, kPN−1

j=0 θ_jxⁿ_jk → Aθ for any θ= (θj), then there exists an N -tuple of signs θ= (θj) for which

n→∞lim

N−1

X

j=0

θjxⁿ_j

< N a.

Proof. It is equivalent to prove that: X is not uniformly non-l₁^N if and only if there exist N sequences {xⁿ₀}, . . . , {xⁿ_N−1} in X and kxⁿ_jk → a(a > 0), n →

∞, j = 0, 1, . . . , N − 1, for any N -tuple of signs θ = (θj) there holds

nlim→∞

N−1

X

j=0

θ_jxⁿ_j

= N a.

Without loss of generality, let a = 1. On one hand, since kxⁿ_jk → 1, j = 0, 1, . . . , N − 1, we can assume that kxⁿ_jk > 0, thenn _xn

j

kxⁿjk

o

⊆ SX. In addition, we have

      k

N−1

X

j=0

θ_j xⁿ_j kxⁿ_jkk − k

N−1

X

j=0

θ_jxⁿ_jk      

≤

N−1

X

j=0

θj( xⁿ_j

kxⁿ_jk − xⁿ_j)

≤

N−1

X

j=0

xⁿ_j kxⁿ_jk − xⁿ_j

=

N−1

X

j=0

1 kxⁿ_jk− 1

· xⁿ_j

→ 0 (n → ∞).

Hence

n→∞lim

N−1

X

j=0

θj

xⁿ_j kxⁿ_jk

= lim

n→∞

N−1

X

j=0

θjxⁿ_j

= N.

By definition X is not uniformly non-l₁^N.

The converse is obvious from the definition of uniform non-l^N₁-ness.  Theorem 4.3. Let ψ ∈ ΨN. Assume that ψ has a unique minimal point s= (s1, s2, . . . , sN−1) with PN−1

i=1 si<1, 0 < si <1, i = 1, 2, . . . , N − 1. Then a Banach space X is uniformly non-l^N₁ if and only if there exists δ(0 < δ < 1) such that for any x0, x1, . . . , xN−1 in BX, there exists an N -tuple of signs θ= (θj), for which

k

N−1

X

j=0

s_jθ_jx_jk ≤ (1 − δ) 1

ψ(s)k(s0x0, s1x1, . . . , s_N−1x_N−1)k_ψ, where s0= 1 −PN−1

i=1 sj.

Proof. Let X be a uniformly non-l^N₁ Banach space. Assume that the conclusion fails to hold. Then for δn = _n¹, n ∈ N, there exist sequences {xⁿ_j} in BX, j = 0, 1, . . . , N − 1, for any N -tuple of signs θ = (θj), we have

N−1

X

j=0

s_jθ_jxⁿ_j

>(1 − 1 n) 1

ψ(s)

(s0xⁿ₀, s1xⁿ₁, . . . , s_N−1xⁿ_N−1) ψ

= (1 −1 n) 1

ψ(s)

(s0kxⁿ₀k, s1kxⁿ₁k, . . . , sN−1kxⁿ_N−1k) ψ. (4)

Because {kxⁿ_jk}^∞_n=1, j = 0, 1, . . . , N − 1 are bounded sequences, we just let kxⁿ_jk → aj(n → ∞), j = 0, 1, . . . , N − 1. Without loss of generality, we can choose {kxⁿ_jk}^∞_n=1such that max{kxⁿ_jk, 0 ≤ j ≤ N − 1} = 1. As aj is the limit of {kxⁿ_jk}^∞_n=1, we get max{aj,0 ≤ j ≤ N − 1} = 1. ThusPN−1

j=0 sjaj >0. In (4) let n → ∞, then there is

1

ψ(s)k(s0a0, s1a1, . . . , s_N−1a_N−1)kψ≤

N−1

X

j=0

s_ja_j.

From this we get ψ s1a1

PN−1

j=0 s_ja_j, . . . , s_N−1a_N−1 PN−1

j=0 s_ja_j

!

≤ ψ(s1, . . . , s_N−1).

By the uniqueness of s = (s1, s2, . . . , s_N−1), we get a0 = a1 = · · · = aN−1, denote them as a. In addition, from (4) we get limn→∞

P_N−1

j=0 sjθjxⁿ_j = 1 = limn→∞PN−1

j=0

sjθjxⁿ_j

. Using Proposition 3.2 there holds

nlim→∞

N−1

X

j=0

θ_jxⁿ_j

= lim

n→∞

N−1

X

j=0

kθjxⁿ_jk = N a.

It’s a contradiction by Lemma 4.2.

On the other hand, for any x0, x1, . . . , xN−1 in BX

N−1

X

j=0

s_jθ_jx_j

≤ (1 − δ) 1

ψ(s)k(s0x0, s1x1, . . . , s_N−1x_N−1)k_ψ

≤ (1 − δ) 1

ψ(s)k(s0, s1, . . . , sN−1)k_ψ

= 1 − δ.

We claim that X is uniformly non-l^N₁ by Lemma 4.1.  Corollary 4.4. Let ψ ∈ Ψ2. Assume that ψ has the unique minimum at t = t0(0 < t0 < 1). Then a Banach space X is uniformly non-square if and only if there exists some δ(0 < δ < 1) such that for any x, y ∈ BX implies

min {k(1 − t0)x + t0yk, k(1 − t0)x − t0yk} ≤ (1 − δ) 1

ψ(t0)k((1 − t0)x, t0y)kψ. Corollary 4.5. A Banach space X is uniformly non-l₁^N if and only if there exists some δ(0 < δ < 1) such that for any x0, x1, . . . , xN−1 in BX, there exists an N -tuple of signs θ= (θj) for which

PN−1 j=0 θjxj

N

p

≤ (1 − δ)kx0k^p+ · · · + kxN−1k^p

N ,

where 1 < p < ∞.

Proof. We only need to let ψ(t) = ψp(t) =h

(1 −PN−1

i=1 ti)^p+ t^p₁+ · · · + t^p_N−1i¹_p

in Theorem 4.3. 

Acknowledgment. The authors thank the anonymous referees for the con- structive comments and suggestions. Especially, thanks for bringing forward Keni-Ichi Mitani and Kichi-Suke Saito’s research results.

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Nature Science Edition. 1 (1981), 1–8.

Zhihua Zhang

School of Mathematical Sciences

University of Electronic Science and Technology of China Chengdu 611731, Sichuan Province, P. R. China

E-mail address: zhihuamath@yahoo.cn

Lan Shu

School of Mathematical Sciences

University of Electronic Science and Technology of China Chengdu 611731, Sichuan Province, P. R. China

E-mail address: shul@uestc.edu.cn

Jun Zheng

School of Mathematics and Statistics Lanzhou University

Lanzhou 730000, Gansu Province, P. R. China E-mail address: zhengj 2010@lzu.edu.cn

Yuling Yang

School of Mathematical Sciences

University of Electronic Science and Technology of China Chengdu 611731, Sichuan Province, P. R. China

E-mail address: yulingkathy@sina.com