The geometric theory of Banach spaces is an important branch of functional analysis. It has important applications in many mathematical fields, such as approximation theory, fixed point theory. Since Clarkson [7] put forward the concept of the modulus of convexity in 1936, geometric constants have evolved as a useful way to compare the geometric properties of Banach spaces. Quantifying geometric properties with numbers is an easy intuitive way to understand these properties on a given Banach spaces. Some of the more prominant geometric constants to date are the modulus of smoothness ρ_X(t) proposed by Day [15], the James constant J(X) proposed by Gao and Lau [12], and the von Neumann-Jordan constant CNJ(X) proposed by Jordan and von Neumann [22]. There are, however, many other geometric constants worth noting. For readers interested in this field, we recommend the references mentioned in this paper.

Fitzpatrick and Reznick [26] introduced the skewness s(X) of a Banach space X, which describes the asymmetry of the norm:

s(X)=suplimt→0+‖x+ty‖−‖y+tx‖t : x,y∈SX.

They showed that s(X)=0 if and only if X is a Hilbert space, and calculated the values of s(X) for L_p spaces where 1≤p≤∞. Moreover, they showed that the uniform non-squareness of X can be described in terms of s(X). Further, based on the work in [26], Baronti and Papini [19] established the relationship between s(X) and ρX(1), and Mitani et al. [13] established the relationship between s(X) and J(X). All the results mentioned above illustrate that a geometric constant which describes the asymmetry of norm is worth studying. In this paper, we introduce three constants, E[t,X], CNJ[X] and J[t,X], all of which describe the asymmetry of the norm.

In Section 2, we recall some necessary concepts.

In Section 3, we introduce the constant E[t,X]. Some basic properties of E[t,X] are given and used to calculate the values of E[t,X] for some specific spaces. The relation between E[t,X] and CNJ′(X) is established. Also, E[t,X] is used to study some geometric properties of Banach spaces. Moreover, we discuss the relation of the values of E[t,X] for two isomorphic Banach spaces in terms of Banach-–Mazur distance.

In Section 4, we introduce the constant CNJ[X] and use it to study Hilbert spaces and uniformly non-square spaces. We apply results from Section 3 to establish properties of CNJ[X].

In Section 5, we consider the constant J[t,X], relate it to J[t,X^*], and use it to study the uniformly non-square spaces.

Throughout the paper, let X be a real Banach space with dim X≥2. The unit ball and the unit sphere of X are denoted by B_X and S_X, respectively. We now recall some concepts that we need in this paper.

Definition 2.1. ([23]) A Banach space X is said to be uniformly non-square, if there exists δ∈(0,1) such that if x, y∈SX then

x+y2≤1−δ or  x−y2≤1−δ.

Definition 2.2. ([17]) A Banach space X is said to have (weak) normal structure, if for every (weakly compact) closed bounded convex subset K in X that contains more than one point, there exists a point x₀ ∈ K such that

sup{‖x0−y‖ : y∈K}<d(K)=sup{‖x−y‖ : x,y∈K}.

Remark 2.3. For the reflexive Banach spaces X, the normal structure and weak normal structure coincide.

The above concepts are closely related to the fixed point property, since Kirk [27] proved that every reflexive Banach space with normal structure has the fixed point property in 1965.

Definition 2.4. ([14]) For isomorphic Banach spaces X and Y, the Banach-Mazur distance between X and Y, denoted by d(X, Y ), is defined to be the infimum of ‖T‖‖T−1‖ taken over all isomorphisms T from X onto Y, that is,

d(X,Y)=inf{‖T‖‖T−1‖ : T:X→Y  being an isomorphism}.

Based on the Pythagorean theorem, Gao introduced the following two quadratic parameters called Gao's parameters in [9] and [10], respectively,

E(X)=sup{‖x+y‖2+‖x−y‖2 : x,y∈SX}.
Eϵ(X)=sup{‖x+ϵy‖2+‖x−ϵy‖2 : x,y∈SX},  0≤ϵ≤1.

He showed that both uniformly non-square spaces and normal structures are closely related to them. More results on E(X) and Eϵ(X) can be found in [2].

As mentioned in the Introduction, in this section we consider the following constant,

E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈SX},  t∈ℝ,

which can be regard as the skew version of Eϵ(X).

3.1. Some basic conclusions about E[t,X]

In this section, we will give some basic conclusions about E[t,X], and calculate the values of E[t,X] for some specific spaces. First, we give the bounds of E[t, X].

Proposition 3.1.1. Let X be a Banach space. Then

2(1+t2)≤E[t,X]≤2(1+t)2,  t∈ℝ.

Proof. By taking x,y∈SX such that x=y, we can easily get E[t,X]≥2(1+t2). In addition, by the triangle inequality, it is obvious that E[t,X]≤2(1+t)2.

Next, we give the some equivalent forms of E[t,X], which will be used in our subsequent discussion.

Proposition 3.1.2. Let X be a Banach space. Then

(1) E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈BX},  t∈ℝ.

(2) If X is a reflexive Banach space, then

E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈ext(BX)},  t∈ℝ.

(3) E[t,X]=sup{E[t,X0] : X0⊂X, dim(X0)=2},  t∈ℝ.

Proof. (1) First, for any x,y∈BX, we have

x=1−‖x‖2−x‖x‖+1−1−‖x‖2x‖x‖,
y=1−‖y‖2−y‖y‖+1−1−‖y‖2y‖y‖.

For convenience, we denote 1−‖x‖2,1−1−‖x‖2,1−‖y‖2,1−1−‖y‖2,−x‖x‖, x‖x‖,−y‖y‖,y‖y‖ by a1,a2,b1,b2, x1,x2,y1,y2, respectively. Thus

x=a1x1+a2x2,  y=b1y1+b2y2.

It is obvious that x1,x2,y1,y2∈SX and a1,a2,b1,b2∈[0,1] such that a1+a2=1,b1+b2=1.

Let t∈ℝ. Since f(x)=x2 is a convex function on [0,∞), for any x,y∈BX we have

‖x+ty‖2+‖tx−y‖2=‖∑i=12aixi+t(∑i=12aiy)‖2+‖t(∑i=12aixi)−(∑i=12aiy)‖2=‖∑i=12ai(xi+ty)‖2+‖∑i=12ai(txi−y)‖2≤(∑i=12ai‖xi+ty‖)2+(∑i=12ai‖txi−y‖)2≤∑i=12ai‖xi+ty‖2+∑i=12ai‖txi−y‖2=∑i=12ai‖∑j=12bjxi+t(∑j=12bjyj)‖2+∑i=12ai‖t(∑j=12bjxi)−(∑j=12bjyj)‖2=∑i=12ai‖∑j=12bj(xi+tyj)‖2+∑i=12ai‖∑j=12bj(txi−yj)‖2≤∑i=12ai(∑j=12bj‖xi+tyj‖)2+∑i=12ai(∑j=12bj‖txi−yj‖)2≤∑i=12ai∑j=12bj‖xi+tyj‖2+∑i=12ai∑j=12bj‖txi−yj‖2=∑i=12ai∑j=12bj(‖xi+tyj‖2+‖txi−yj‖2)≤max{‖xi+tyj‖2+‖txi−yj‖2 : i=1,2,  j=1,2}≤sup{‖x+ty‖2+‖tx−y‖2 : x,y∈SX}.

This shows that

E[t,X]≥sup‖x+ty‖2+‖tx−y‖2:x,y∈BX,  t∈ℝ.

In addition, it is obvious that

E[t,X]≤sup‖x+ty‖2+‖tx−y‖2:x,y∈BX,  t∈ℝ.

(2) Since X is a reflexive Banach space, then, according to Krein-Milman theorem, we know that BX=co¯(ext(BX)). Then, by (1), we obtain

E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈co¯(ext(BX))},  t∈ℝ.

Further, according to the continuity of norm, we can easily know

(3.1.1)E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈co(ext(BX))},  t∈ℝ.

Now, for any x,y∈co(ext(BX)), there exist {xi}i=1nx,{yj}j=1ny∈ext(BX) and {ai}i=1nx,{bj}j=1ny∈[0,1] such that

x=∑ i=1 nxaixi,  y=∑ j=1 nybjyj,  ∑ i=1 nxai=1,  ∑ j=1 nybj=1.

Moreover, since f(x)=x² is a convex function on [0,∞), then, for any t∈ℝ and any x,y∈co(ext(BX)), we obtain

‖x+ty‖2+‖tx−y‖2=‖∑i=12aixi+t(∑i=12aiy)‖2+‖t(∑i=12aixi)−(∑i=12aiy)‖2=‖∑i=12ai(xi+ty)‖2+‖∑i=12ai(txi−y)‖2≤(∑i=12ai‖xi+ty‖)2+(∑i=12ai‖txi−y‖)2≤∑i=12ai‖xi+ty‖2+∑i=12ai‖txi−y‖2=∑i=12ai‖∑j=12bjxi+t(∑j=12bjyj)‖2+∑i=12ai‖t(∑j=12bjxi)−(∑j=12bjyj)‖2=∑i=12ai‖∑j=12bj(xi+tyj)‖2+∑i=12ai‖∑j=12bj(txi−yj)‖2≤∑i=12ai(∑j=12bj‖xi+tyj‖)2+∑i=12ai(∑j=12bj‖txi−yj‖)2≤∑i=12ai∑j=12bj‖xi+tyj‖2+∑i=12ai∑j=12bj‖txi−yj‖2=∑i=12ai∑j=12bj(‖xi+tyj‖2+‖txi−yj‖2)≤max{‖xi+tyj‖2+‖txi−yj‖2 : i=1,2,  j=1,2}≤sup{‖x+ty‖2+‖tx−y‖2 : x,y∈SX}.

Thus, by (3.1.1), we have

E[t,X]≤sup{‖x+ty‖2+‖tx−y‖2 : x,y∈ext(BX)},  t∈ℝ.

Further, it is clear that

E[t,X]≥sup{‖x+ty‖2+‖tx−y‖2 : x,y∈ext(BX)},  t∈ℝ.

(3) Let t∈ℝ. First, it is clear that

E[t,X]≥sup{E[t,X0] : X0⊂X,dim(X0)=2}.

Second, for any ε>0, there exist x,y∈SX such that

E[t,X]−ε≤‖x+ty‖2+‖tx−y‖2.

Let X₀ be a two-dimensional subspace of X that contains x and y, we have

E[t,X]−ε≤‖x+ty‖2+‖tx−y‖2≤E[t,X0]≤sup{E[t,X0] : X0⊂X, dim(X0)=2}.

Let ε→0, we can obtain

E[t,X]≤sup{E[t,X0] : X0⊂X, dim(X0)=2}.

This completes the proof.

Now, we will use the above conclusion to calculate the values of E[t,X] for some spaces.

Example 3.1.3. Let X be the space ℝ2 with the norm defined by

x1,x2=maxx1+13x2,x1−13x2,23|x2|.

Then

E[t,X]=(1+|t|)2+1|t|≤1,(1+|t|)2+t2|t|≥1.

Proof. First, notice that the unit ball of the this norm is a regular hexagon (see [14]).

By the above figure and some simple calculations, we can get that

ext(BX)=(1,0),12,32,−12,32,(−1,0),−12,−32,12,−32.

Now, because finite-dimensional Banach spaces must be reflexive spaces, according to Proposition 3.1.2 (2), we can get

E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈ext(BX)},  t∈ℝ.

Thus, by some simple calculations, it is not difficult for us to get

E[t,X]=(1+|t|)2+1|t|≤1,(1+|t|)2+t2|t|≥1.

EXample 3.1.4. Let X be the space ℝ2 with the norm defined by

x1,x2=x1 ,x2 1x1x2≤0,x1 ,x2 ∞x1x2≥0.

Then

E[t,X]=(1+|t|)2+1|t|≤1,(1+|t|)2+t2|t|≥1.

Proof. First, because finite-dimensional Banach spaces must be reflexive spaces, by Proposition 3.1.2 (2) we get

E[t,X]=sup{‖x+ty‖2+‖tx−y‖2 : x,y∈ext(BX)},  t∈ℝ.

Now, since

ext(BX)={(1,0),(1,1),(0,1),(−1,0),(−1,−1),(0,−1)},

by some simple calculations, it is not difficult for us to get

E[t,X]=(1+|t|)2+1|t|≤1,(1+|t|)2+t2|t|≥1.

Example 3.1.5. Let X be the space ℝ2 with the norm defined by

x1,x2=x1 ,x2 1x1x2≤0,x1 ,x2 2x1x2≥0.

Then

E[t,X]=2(1+t2)+2tt≥0,2(1+t2)−2tt≤0.

Proof. First, we will prove that

(3.1.2)‖x+ty‖2+‖tx−y‖2≤2(1+t2)+2t,  t≥0,

holds for any x=(x1,x2),y=(y1,y2)∈ext(BX).

Let t≥0. Then, for any x=(x1,x2),y=(y1,y2)∈ext(BX), we consider the following four cases:

Case 1: x1,x2,y1,y2≥0.

Case 1a: (tx1−y1)(tx2−y2)≥0. Then

‖x+ty‖2+‖tx−y‖2=‖(x1+ty1,x2+ty2)‖22+‖(tx1−y1,tx2−y2)‖22=2(1+t2).

Case 1b: tx1−y1≥0, tx2−y2≤0. Then

‖x+ty‖2+‖tx−y‖2=‖(x1+ty1,x2+ty2)‖22+‖(tx1−y1,tx2−y2)‖12      =2(1+t2)+2tx1y2+2tx2y1−2y1y2−2t2x1x2      ≤2(1+t2)+2tx1y2+2tx2y1      ≤2(1+t2)+2t((x12+x22)12(y12+y22)12)      =2(1+t2)+2t.

Case 1c: tx1−y1≤0, tx2−y2≥0. Similar to the proof of Case 1b, we can get

‖x+ty‖2+‖tx−y‖2≤2(1+t2)+2t.

Thus, we obtain (3.1.2) holds for any x=(x1,x2),y=(y1,y2)∈ext(BX) with x1,x2,y1,y2≥0.

Case 2: x1,x2≥0,y1,y2≤0. Let z=-y. Then, by Case 1, we obtain

‖x+ty‖2+‖tx−y‖2=‖x−tz‖2+‖tx+z‖2=‖z+tx‖2+‖tz−x‖2≤2(1+t2)+2t.

Case 3: x1,x2≤0,y1,y2≥0. Similar to the proof of Case 2, we omit it.

Case 4: x1,x2≤0,y1,y2≤0. Let z=-y and w=-x. Then, by Case 1, we obtain

‖x+ty‖2+‖tx−y‖2=‖−w−tz‖2+‖−tw+z‖2=‖w+tz‖2+‖tw−z‖2≤2(1+t2)+2t.

Consequently, we prove (3.1.2) holds.

Now, according to Proposition 3.1.2 (2) and the fact that finite-dimensional Banach spaces must be reflexive spaces, we can get

E[t,X]≤2(1+t2)+2t,  t≥0.

Further, put x=(1,0) and y=(0,1), we can get

E[t,X]≥‖x+ty‖2+‖tx−y‖2=‖(1,t)‖22+‖(t,−1)‖12=2(1+t2)+2t,  t≥0,

which shows that E[t,X]=2(1+t2)+2t,  t≥0. Since, it is obvious that E[t,X]=E[−t,X],  t∈ℝ, we can also get E[t,X]=2(1+t2)−2t,  t≤0.

3.2. The relation between CNJ′(X) and E[t,X]

Alonso et al. [8] introduced the following constants CNJ'(X) in 2008, according to the characterization of Hilbert spaces called the rhombus law given by Day [16],

CNJ'(X)=sup‖x+y‖2+‖x−y‖24 : x,y∈SX.

This constant is closely related to some geometric properties of Banach spaces and some other constants (see [8]), and also plays an important role in the study of Tingley's problem (see [24]).

It is clear that 4CNJ'(X)=E[1,X]. Thus, it is natural for us to ask whether there is a relationship between CNJ'(X) and E[t,X], t≠1. In order to answer this question, we need to give the following result first.

Proposition 3.2.1. Let X be a Banach space. The following statements hold.

(1) E[t,X] is a convex function of t on ℝ.

(2) E[t,X] is a continuous function of t on ℝ.

(3) E[t,X] is an even function of t on ℝ.

(4) E[t,X] is non-decreasing on [0,+∞) and non-increasing on (−∞,0].

Proof. (1) Let t1,t2∈ℝ,λ∈(0,1). Then, for any x,y∈SX, we can deduce that

x+λt1+(1−λ)t2y2+λt1+(1−λ)t2x−y2⩽λx+t1y+(1−λ)x+t2y2+λt1x−y+(1−λ)t2x−y2⩽λ x+ t 1 y 2+ t 1 x−y 2+(1−λ) x+ t 2 y 2+ t 2 x−y 2⩽λE[t1,X]+(1−λ)E[t2,X],

which implies that

E[λt1+(1−λ)t2,X]⩽λE[t1,X]+(1−λ)E[t2,X].

(2) By (1), we can obtain E[t,X] is continuous on ℝ.

(3) By the definition of E[t,X], it is obvious that E[t,X]=E[-t,X] holds for any t∈ℝ.

(4) Let t2>t1≥0. Then, from (1) and (3), we have

(3.2.1)E[t1,X]=Et2 +t1 2t2 t2+t2 −t1 2t2 −t2 ,X≤E[t2,X].

which deduces that E[t,X] is non-decreasing on [0,+∞). Then, from (3), we can obtain E[t,X] non-increasing on (−∞,0]. This completes the proof.

From the above proposition, we know that we only need to consider E[t,X] on t∈[0,∞], since E[t,X] is an even function. Also, it is obvious that E[0,X]=2 for any Banach spaces. Thus, in the following discussion, we only consider E[t,X] with t∈(0,∞). Now, we use the above proposition to give the relation between CNJ′(X) and E[t,X].

Proposition 3.2.2. Let X be a Banach space. Then

4tCNJ′(X)≤E[t,X]≤4tCNJ′(X)+2max{t,1}|t−1|,  t>0.

Proof. Let t>0. First, notice that, for any x,y∈SX, we have

x+1ty2+1tx−y21+1t2=‖tx+y‖2+‖x−ty‖2t2+1.

From the above equality, we can imply that

(3.2.2)E[t,X]1+t2=E1t ,X1+1t2.

Now, since E[t,X] is convex function and (3.2.2), we obtain

E[1,X]=E11+t⋅t+1−11+t⋅1t,X≤11+tE[t,X]+1−11+tE1t,X=1+t21+tE[t,X]1+t2+1+t2t(1+t)E1t,X1+1t2=1+t21+tE[t,X]1+t2+1+t2t(1+t)E[t,X]1+t2=1tE[t,X],

which implies that 4tCNJ′(X)≤E[t,X].

For the other inequality, we divide the proof into the following two cases.

Case 1 : 0<t<1. Using E[t,X] is a convex function again, we have

E[t,X]=E[(t⋅1+(1−t)⋅0),X]≤tE[1,X]+(1−t)E[0,X]=4tCNJ′(X)+2(1−t),

and hence E[t,X]≤4tCNJ′(X)+2max{t,1}|t−1|.

Case 2. : t≥1. According to (3.2.2) and Case 1, we have

E[t,X]1+t2=E1t ,X1+1t2≤4t1+t2CNJ′(X)+2t(t−1)1+t2,

which means that

E[t,X]≤4tCNJ′(X)+2max{t,1}|t−1|.

This completes the proof.

3.3. The relations between E[t,X] and some geometric properties of Banach spaces

Next, we will discuss the relations between E[t,X] and some geometric properties of Banach spaces. First, we will use the following lemma to characterize Hilbert spaces by E[t,X].

Lemma 3.3.1. ([4]) A normed space X is an inner product space if and only if for all x,y∈SX there exist λ,μ≠0 such that

(3.3.1)‖λx+μy‖2+‖μx−λy‖2≈2(λ2+μ2),

where ≈ means either ≤ or ≥.

Theorem 3.3.2. Let X be a Banach space. Then the following statements are equivalent.

(1) E[t,X]=2(1+t2) for all t∈(0,∞).

(2)E[t,X]=2(1+t2) for some t∈(0,∞).

(3)X is a Hilbert space.

Proof. (1)⇒(2). Obvious.

(2)⇒(3). Since E[t,X]=2(1+t2) holds for some t∈(0,∞), we can deduce that there exists t0∈(0,∞) such that

‖x+t0y‖2+‖t0x−y‖2⩽21+t02,  x,y∈SX.

Then, from Lemma 3.3.1, we know that X is a Hilbert space.

(3)⇒(1). Since X is a Hilbert space, then, for any t∈(0,∞) and x,y∈SX, we obtain

‖x+ty‖2+‖tx−y‖2=‖x‖2+2t〈x,y〉+t2‖y‖+t2‖x‖2−2t〈x,y〉+‖y‖2=2(1+t2).

This shows that E[t,X]=2(1+t2) for all t∈(0,∞).

Remark 3.3.3. The above conclusion also shows that the lower bound of E[t,X] given in Proposition 3.1.1 is sharp.

Second, we will give the relation between E[t,X] and uniformly non-square spaces.

Theorem 3.3.4. Let X be a Banach space. Then the following statements are equivalent.

(1) E[t,X]<2(1+t)2 for all t∈(0,∞).

(2) E[t,X]<2(1+t)2 for some t∈(0,∞).

(3) X is uniformly non-square.

Proof. It is clear that we only need to show that the following statements are equivalent

(i) E[t,X]=2(1+t)2 for all t∈(0,∞).

(ii) E[t,X]=2(1+t)2 for some t∈(0,∞).

(iii) X is not uniformly non-square.

(i) ⇒ (ii). Obvious.

(ii) ⇒ (iii). Since E[t,X]=2(1+t)2 for some t∈(0,∞), we can deduce that there exists t0∈(0,∞) and xn,yn∈SX such that,

‖xn+t0yn‖2+‖t0xn−yn‖2→2(1+t0)2.

Notice that ‖xn+t0yn‖2≤(1+t0)2 and ‖t0xn−yn‖2≤(1+t0)2, thus we obtain

‖xn+t0yn‖2→(1+t0)2,   ‖t0xn−yn‖2→(1+t0)2.

Further, we obtain

‖xn+t0yn‖→1+t0,   ‖t0xn−yn‖→1+t0,

which can deduces that

‖xn+yn‖→2,   ‖xn−yn‖→2.

So, X is not uniformly non-square.

(iii) ⇒ (i). Since X is not uniformly non-square , there exist xn,yn∈SX such that

‖xn+yn‖→2,   ‖xn−yn‖→2.

Thus, for all t∈(0,∞), we obtain

‖xn+tyn‖→1+t,   ‖txn−yn‖→1+t.

Furthermore, from Proposition 3.1.1, then we obtain

2(1+t)2≥E[t,X]≥‖xn+tyn‖2+‖txn−yn‖2,  t∈(0,∞)

Let n→∞, we obtain E[t,X]=2(1+t)2 for all t∈(0,∞).

Remark 3.3.5. The above conclusion also shows that the upper bound of E[t,X] given in Proposition 3.1.1 is sharp.

Finally, we use E[t,X] to give a sufficient condition for normal structure by the following lemma.

Definition 3.3.6. ([11]) Let X be a Banach space, a hexagon H in X is called a normal hexagon if the

length of each side of H is 1 and each pair of opposite sides are parallel.

Lemma 3.3.7. ([11]) Let X be a Banach space without weak normal structure, then for any ϵ, 0<ϵ<1, and x1∈SX, there exists an inscribed normal hexagon with vertices x1,x2,x3,−x1,−x2 and −x3∈SX satisfying

(1) x1=x2−x3.

(2) ‖(x1+x2)/2‖,‖(x3+(−x1))/2‖>1−ϵ.

Theorem 3.3.8. Let X be a Banach space. Then

(1) Let t∈[23,1]. If E[t,X]<5t2−2t+2, then X has normal structure.

(2) Let t∈[1,32]. If E[t,X]<2t2−2t+5, then X has normal structure.

Proof. (1) Let t∈[23,1]. According to Theorem 3.3.4, we know that X is uniformly non-square when E[t,X]<5t2−2t+2. Hence X is reflexive (see [23]), and normal structure and weak normal structure coincide. So, we only need to show that X has weak normal structure.

Now, suppose conversely that X does not have weak normal structure. By Lemma 3.3.7, for any ϵ∈ (0,1), there exist x₁, x₂, x₃∈ S_X such that

x1=x2−x3,  ‖(x1+x2)/2‖>1−ϵ,  ‖(x3+(−x1))/2‖>1−ϵ.

Now, we have

‖x1+tx2‖=‖(x1+x2)−(1−t)x2‖≥‖x1+x2‖−‖(1−t)x2‖≥2−2ϵ−(1−t)=1+t−2ϵ,

and

‖tx1−x2‖=‖tx1+tx2−tx2−x2‖    =‖tx1+tx2−t(x1+x3)−x2‖    =‖t(x1−x3)+(t−1)x2−tx1‖    ≥‖t(x1−x3)‖−‖(t−1)x2−tx1‖    ≥(2−2ϵ)t−(1−t+t)    =(2−2ϵ)t−1.

Thus, we obtain

E[t,X]≥‖x1+tx2‖2+‖tx1−x2‖2≥(1+t−2ϵ)2+((2−2ϵ)t−1)2.

Let ϵ→0, we have

E[t,X]≥(1+t)2+(2t−1)2=5t2−2t+2.

This contradicts E[t,X]<5t2−2t+2.

(2) Let t∈[1,32]. By (3.2.2), we have E1t,X=1t2E[t,X]. Further, since E[t,X]<2t2−2t+5, we get

E1t,X=1t2E[t,X]<1t2(2t2−2t+5)=51t2−21t+2,

which implies that X has normal structure by (1).

Remark 3.3.9. Through simple calculations, we get that 2(1+t2)≤5t2−2t+2 if and only if t∈[23,1]. The reason we take t∈[23,1] in (1) is to ensure that there exists space X satisfying E[t,X]<5t2−2t+2 by Proposition 3.1.1. This is also why we take t∈[1,32] in (2).

3.4. BanachMazur distance and stability

In this section, we will use the Banach-–Mazur distance to discuss the relation of the values of E[t,X] for two isomorphic Banach spaces. This relationship plays a decisive role in our subsequent discussion of the relationship between E[t,X] and E[t,X**].

Theorem 3.4.1. If X and Y are isomorphic Banach spaces, then,

E[t,Y]d(X,Y)2≤E[t,X]≤d(X,Y)2E[t,Y],   t>0.

Proof. Let x,y∈SX and t>0. For each ε>0, there exists an isomorphism T from X onto Y such that ‖T−1‖‖T‖≤(1+ε)d(X,Y). Set

x¯=Tx‖T‖,   y¯=Ty‖T‖,

then x¯,y¯∈BY. Now, according to Proposition {3.1.2} (1), we obtian

‖x+ty‖2+‖tx−y‖2=‖T−1(Tx+tTy)‖2+‖T−1(Ttx+Ty)‖2        =‖T−1(‖T‖x¯+t‖T‖y¯)‖2+‖T−1(‖T‖tx¯+‖T‖y¯)‖2        =‖T−1‖2‖T‖2(‖x¯+ty¯‖2+‖tx¯−y¯‖2)        ≤(1+ε)2d(X,Y)2E[t,Y].

This means that E[t,X]≤(1+ε)2d(X,Y)2E[t,Y]. Let ε→0, we obtain

E[t,X]≤d(X,Y)2E[t,Y].

The other inequality follows by interchanging X and Y.

Corollary 3.4.2. Let X be a Banach space and let X1=(X,‖⋅‖1), where ‖⋅‖1 is an equivalent norm on X satisfying, for a,b>0,

a‖x‖≤‖x‖1≤b‖x‖,   x∈X

Then

a2b2E[t,X]≤E[t,X1]≤b2a2E[t,X],  t>0.

Proof. This follows from Theorem 3.4.1 and the fact that d(X,X1)≤ba.

Finally, we will use Theorem 3.4.1 to show that E[t,X]=E[t,X**]. To do this, we need to recall the definition of finite representability. A Banach space X is finitely representable in a Banach space Y if, for every ε>0 and for every finite-dimensional subspace X₀ of X, there exists a finite-dimensional subspace Y₀ of Y with dim(X0)=dim(Y0) such that d(X0,Y0)≤1+ε.

Theorem 3.4.3. Let X and Y be Banach spaces. The following statements hold.

(1) If X is finitely representable in Y, then E[t,X]≤E[t,Y],  t>0.

(2) E[t,X]=E[t,X**],  t>0.

Proof. (1) Let t>0 and X₀ be a two-dimensional subspace of X. For any ε>0, since X is finitely representable in Y, there exists a two-dimensional subspace Y₀ of Y such that d(X0,Y0)≤1+ε. Applying Theorem 3.4.1 to the pair of X₀ and Y₀, we obtain

E[t,X0]≤(1+ε)E[t,Y0]≤(1+ε)E[t,Y].

Let ε→0, we obtain E[t,X0]≤E[t,Y]. Further, by Proposition 3.1.2 (3), we obtain E[t,X]≤E[t,Y].

(2) For any Banach spaces X, by the principle of local reflexivity, X** is always finitely representable in X. Then, by (1),

E[t,X]≥E[t,X**],  t>0.

On the other hand, X is isometric to a subspace of X**, therefore,

E[t,X]≤E[t,X**],  t>0.

This completes the proof.