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Some Aspects of Geometric Constants in Modular Spaces

Zhijian Yang, Qi Liu, Muhammad Sarfraz, Yongjin Li∗
Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, P. R. China
yangzhj55@mail2.sysu.edu.cn, liuq325@mail2.sysu.edu.cn, sarfraz@mail2.sysu.edu.cn,

stslyj@mail.sysu.edu.cn
∗Correspondence: stslyj@mail.sysu.edu.cn

Abstract. In this paper, we generalize the typical geometric constants of Banach spaces to modularspaces. We study the equivalence between the convexity of modular and normed spaces, and obtainthe relationship between ρ-Neumann-Jordan constant and ρ-James constant. In particular, we extendthe convexity and smoothness modular, and obtain the criterion theorems of the uniform convexity andstrict convexity.

1. Introduction
In the recent years, the geometric theory of Banach spaces has been fully developed, especiallythe geometric constant, which is a powerful tool to characterize the geometric properties of thespace sphere. As early as 1936, Clarkson introduced the convexity modular of space [1]. In 1963,Lindenstrauss introduced the smoothness modular, and obtained the close relationship betweenthe two constants [2]. In 1937, in order to better characterize Jordan and von-Nuemann’s famouswork in inner product spaces, Clarkson defines the von-Nuemann constant [3] which is the minimumconstant C for all x,y ∈ X and (x,y) 6=(0,0) of the following equations:

1

C
≤
‖x +y‖2 +‖x −y‖2

2(‖x‖2 +‖y‖2)
≤ C.

In 1964, James introduced James constant [4] in order to study the normal structure of space.After the appearance of these constants, many scholars paid attention to them and obtained manywonderful properties [5].Modular space problems have been considered by H. Nakano, Musielak and Orlicz [6] underthe additional hypothesis of convexity or subadditivity of the modular ρ : X → [0,+∞). Moreoverthe case of semi-ordered linear spaces and that of B-norms have been chiefly investigated. Under
Received: 13 Sep 2021.
Key words and phrases. Banach spaces; geometric constants; modular spaces.151

https://adac.ee
https://doi.org/10.28924/ada/ma.1.151


Eur. J. Math. Anal. 1 (2021) 152
weaker assumptions, they investigated the structure of the spaces under consideration. Neitherconvexity nor subadditivity of the modular be assumed. In introducing the norm, a certain naturalconnection between the modular and the norm convergence will be required: norm convergenceshould imply modular convergence.Through their researches, they found that although modular spaces are not generally normedspaces, they still have many wonderful properties, such as convergence, completeness, convexity andadditivity. In view of these properties, Poom Kumam extended Jordan Von-Neumann constant andJames constant in Banach spaces to modular spaces, and obtained uniform convexity and uniformnon-squareness of modular spaces [10].In this paper, based on the idea of generalizing geometric constants in Banach spaces to modularspaces, we generalize the properties of von-Neumann constant and James constant in [10]. Bydefining convexity modules and smoothness modular, we derive the relationships between Jamesconstant, convexity modular and the strict convexity of modular spaces.

2. Preliminaries
We first give some basic facts about modular spaces formulated by Musielak and Orlicz [6].

Definition 1.[8] Let X be a vector space over F(R or C). Then a function ρ : X → [0,∞] is calleda modular on X if for arbitrary x,y in X,(i) ρ(x)=0 if and only if x =0,(ii) ρ(αx)= ρ(x) for every scalar α with |α|=1,(iii) ρ(αx +βy)≤ ρ(x)+ρ(y) if α+β =1 and α,β ≥ 0.If (iii) is replaced by (iv): ρ(αx +βy)≤ αρ(x)+βρ(y) if α,β ≥ 0 and α+β =1. We now callthat ρ is a convex modular.A modular ρ can be used to define a corresponding modular space, i.e, the vector space Xρ asgiven by
Xρ = {x ∈ X : ρ(λx)→ 0 as λ → 0},

where Xρ is a linear subspace of X.In general, the modular ρ is not necessarily subadditive and therefore it does not behave as anorm or a distance. But we can associate it to a modular F-norm.The modular space Xρ can be equipped with a F-norm defined by
‖x‖ρ = inf

{
α > 0;ρ(

x

λ
)≤ α

}
,

when ρ is convex. Then norm ‖ · ‖ρ is frequently called the Luxemburg norm. If ρ is convex, thenthe functional ‖x‖ρ = inf{α > 0;ρ(xλ) ≤ 1} is a norm in Xρ which is equivalent to the F-norm
‖ ·‖ρ.



Eur. J. Math. Anal. 1 (2021) 153
Proposition 1. Let Xρ be a modular space. Then ρ is convex if and only if Xρ is a normed spacewith ρ as norm.

Proof. The proof of sufficiency is obvious.Conversely, assume ρ is convex, then we can obtain ρ(x)=0 if and only if x = 0.(i) According to the Definition 1, if α > 0, then
ρ

(
1

α
x

)
= ρ

(
1

α
x +
1−α
α
·0
)
≤
1

α
ρ(x)+

1−α
α

ρ(0)=
1

α
ρ(x)

and
αρ

(
1

α
x

)
= αρ

(
1

α
x

)
+(1−α)ρ(0)≥ ρ(α ·

1

α
+(1−α) ·0)= ρ(x).

This show that ρ(1
α
x)≥ 1

α
ρ(x) and hence ρ(1

α
x

)
=
1

α
ρ(x) for α > 0.Suppose α 6=0, then |α| > 0. According to the Definition 1, we have

ρ

(
|α| ·

1

|α|
αx

)
= |α|ρ

(
1

|α|
αx

)
= |α|ρ(x)

which shows that ρ(αx)= |α|ρ( 1|α|αx) = |α|ρ(x).(ii) Since
ρ(x +y)= ρ

(
2
(x
2
+
y

2

))
=2ρ

(x
2
+
y

2

)
≤ ρ(x)+ρ(y),

then Xρ is a normed space with ρ as norm.
3. The ρ-Neumann–Jordan constant and the ρ-James constant

In 2006, Poom Kumam [10] generalized two typical constants
CNJ(X)= sup{‖x +y‖2 +‖x −y‖2

2‖x‖2 +2‖y‖2
: x,y ∈ X,(x,y) 6=(0,0)

}
and

J(X)= sup{min{‖x +y‖,‖x −y‖} : x,y ∈ X,‖x‖= ‖y‖=1}
and introduced two new geometric constants CNJ(Xρ) and J(Xρ) defined on modular spaces.
Definition 2.[10] The ρ-Neumann-Jordan constant CNJ(Xρ) of a modular space Xρ is defined by

CNJ(Xρ)=2sup

{
ρ2(

x+y
2
)+ρ2(

x−y
2
)

ρ2(x)+ρ2(y)
: x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1

}
.

Definition 3.[10] The ρ-James constant J(Xρ) of a modular space Xρ is defined by
J(Xρ)=2sup

{
min{ρ(

x +y

2
),ρ(

x −y
2
)} : x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1

}
.

In the following section, we extend the Proposition 3.5 in [10] and obtain inequalities of CNJ(Xρ)and J(Xρ).
Theorem 1. Let Xρ be a modular space, then



Eur. J. Math. Anal. 1 (2021) 154
(i) 0 < J (Xρ)≤ 4 and 1≤ CNJ (Xρ)≤ 8, in particular, if ρ is convex, then 1≤ J (Xρ)≤ 2 and

1≤ CNJ (Xρ)≤ 2;(ii)1
2
J2(Xρ) ≤ CNJ(Xρ) ≤ 64J2(Xρ) +4, in particular, if ρ is convex, then 12J2(Xρ) ≤ CNJ(Xρ) ≤

4
J2(Xρ)

+1.
Proof. (i) Let y =0, then

J(Xρ)≥ 2sup{ρ(
x

2
) : x ∈ Xρ,ρ(x)=1}.

Since ρ(x) = 1, then ρ(x
2
) > 0 implies J(Xρ) > 0. Since ρ(x±y2 ) ≤ ρ(x)+ ρ(y) ≤ 2, then

0 < J(Xρ)≤ 4.Let x = y , then
CNJ(Xρ)≥ 2sup

{
ρ2(x+x

2
)+ρ2(x−x

2
)

ρ2(x)+ρ2(x)
: x ∈ Xρ,ρ(x)=1

}
≥ 1.

Since ρ2(x +y
2
)+ρ2(

x −y
2
)≤ 2(1+ρ(y))2, we have

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
ρ2(x)+ρ2(y)

≤ 2
(
1+

2ρ(y)

1+ρ2(y)

)
≤ 4,

thus 1≤ CNJ(Xρ)≤ 8.In particular, if ρ is convex and let x = y , then
J(Xρ)≥ 2sup{ρ(x

2
) : x ∈ Xρ,ρ(x)=1}=1.

Since ρ(x±y
2
)≤ 1

2
ρ(x)+ 1

2
ρ(y)≤ 1, then 1≤ J(Xρ)≤ 2. We also can prove 1≤ CNJ(Xρ)≤ 2by the same way.(ii) Since

ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
≤ 2[1+ρ(y)]2 ≤ 4

(
1+ρ2(y)

)
,

then
ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
ρ2(x)+ρ2(y)

−2≤
2(1+ρ(y))2

1+ρ2(y)
−2=

4ρ(y)

1+ρ2(y)
.

Since 1
4
(ρ2(

x+y
2
)+ρ2(

x−y
2
))≤ 1+ρ2(y), then 4ρ(y)

1+ρ2(y)
≤

16ρ(y)

ρ2(
x+y
2
)+ρ2(

x−y
2
)
, that is

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
ρ2(x)+ρ2(y)

−2

≤
16ρ(y)

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
≤

16

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

).
Finally 1

2
CNJ(Xρ)−2≤ 161

2
J2(Xρ)

implies that CNJ(Xρ)≤ 64J2(Xρ) +4.



Eur. J. Math. Anal. 1 (2021) 155
According to the proof of Proposition 3.5 in [12], we can prove 1

2
J2(Xρ)≤ CNJ(Xρ), thus

1

2
J2(Xρ)≤ CNJ(Xρ)≤

64

J2(Xρ)
+4.

In particular, if ρ is convex, then
ρ2(

x +y

2
)+ρ2(

x −y
2
)≤
1

2
(1+ρ(y))2 ≤ 1+ρ2(y),

thus
ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
ρ2(x)+ρ2(y)

−
1

2
≤

ρ(y)

1+ρ2(y)
≤

1

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

).
Therefore

CNJ (Xρ)≤
4

J2 (Xρ)
+1.

Example 1. (i) Consider X = R2, ρ(x) = { 0,x =0
1
‖x‖1

,x 6=0
, where ‖x‖1 = ‖(x1,x2)‖1 = |x1|+ |x2|.

Obviously, Xρ is a modular space.We choose x0 = (1
2
,
1

2

)
,y0 =

(
1

2
,−
1

2

), then
ρ(x0)= ρ(y0)=1,ρ

(
x0 +y0
2

)
= ρ

(
x0 −y0
2

)
=2,

thus J (Xρ)≥ 4. Since J (Xρ)≤ 4, then J (Xρ)=4. According to (ii) of Theorem 1, we know that
CNJ(Xρ)=8 in this example.(ii) Consider X = R2,ρ(x)= ‖x‖1. Obviously, Xρ is a modular space and ρ is convex. We have

J (Xρ)= sup{min{‖x +y‖1,‖x −y‖1} : x,y ∈ Xρ,‖x‖=1,‖y‖≤ 1}.We choose x0 = (1,0),y0 = (0,1), then ‖x0‖1 = ‖y0‖1 = 1 and ‖x0 + y0‖1 = ‖x0 − y0‖1 = 2,thus J (Xρ)=2. According to (ii) of Theorem 1, we can get that CNJ(Xρ)=2 in this example.
4. The ρ-convex modular and the ρ-smooth modular

In order to study the uniform convexity of Banach spaces, Clarkson introduced the modular ofconvexity
δX(ε)= inf{1− 1

2
‖x +y‖ : ‖x‖= ‖y‖=1,‖x −y‖≥ ε

}
.

Goebel called ε0 = sup{ε ∈ [0,2] : δX(ε)= 0} as the characteristic of convexity. Based on thegeometric intuitionistic meaning of convexity of Banach spaces and its application in fixed pointtheory, this paper gives the ρ-convex modular of modular spaces with reference to the definition of
δX(ε).

Definition 4. The ρ-convex modular δXρ(ε) of a modular space Xρ is defined by
δXρ(ε)= inf

{
1−ρ

(
x+y
2

)
: x,y ∈ Xρ,ρ(x),ρ(y)≤ 1,ρ(x −y)≥ ε

}
,0≤ ε ≤ 2.

In particular, if ρ is convex, the ρ-uniform convexity of Xρ is defined as
ε0 (Xρ)= sup

{
ε ∈ [0,2] : δXρ(ε)=0

}
.



Eur. J. Math. Anal. 1 (2021) 156
Remark 1. We can easily prove that −1≤ δXρ(ε)≤ 1 and δXρ(0)≤ 0.In Banach spaces, the convexity modular δX(ε) and the smoothness modular

ρX(t)= sup{‖x +y‖+‖x −y‖
2

−1 : ‖x‖=1,‖y‖=1,t ≥ 0
}

are conjugate concepts. Therefore, this paper gives the definition of ρ-smooth modular of modularspaces by referring to the definition of smoothness modular ρX(t).
Definition 5. The ρ-smooth modular ρXρ(t) of a modular space Xρ is defined by

ρXρ(t)= sup

{
ρ(
x +y

2
)+ρ(

x −y
2
)−1 : x,y ∈ Xρ,ρ(x)≤ 1,ρ(y)≤ t

}
,t ≥ 0.

Remark 2. It is true that min{0,t −1}≤ ρXρ(t)≤ 1+2t and ρXρ(t) is increasing of t.
Theorem 2. Let X be a modular space, then(i) J(Xρ) < 2� if and only if δXρ(�) > 1− �, in particular, if ρ is convex, then J(Xρ) < � if andonly if δXρ(�) > 1− �2;(ii) J(Xρ)=2sup{� ∈ (0,2) : δXρ(�)≤ 1−�}, in particular, if ρ is convex, then J(Xρ)= sup{� ∈
(0,2) : δXρ(�) < 1−

�
2
}.

Proof. (i) Note α = J(Xρ) < 2�, thus
min

{
ρ(
x +y

2
),ρ(

x −y
2
)

}
≤
α

2
,

shows that 1−ρ(x+y
2
)≥ 1− α

2
> 1− �. Therefore δXρ(�) > 1− �.Note β = δXρ(�) > 1− �, then 1−ρ(x+y2 )≥ β implies ρ(x+y2 )≤ 1−β < �. Thus

min

{
ρ(
x +y

2
),ρ(

x −y
2
)

}
= ρ(

x +y

2
).

Then
J (Xρ)=2sup

{
ρ

(
x +y

2

)
: x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1

}
≤ 2−2β < 2ε.In particular, if ρ is convex and let λ = J(Xρ) < �, then J(Xρ) < � if and only if ∀x,y ∈

Xρ,ρ(x),ρ(y)≤ 1, we have
ρ(x +y)≤ λ or ρ(x −y)≤ λ.

According to the Definition of δXρ(�), we obtain ρ(x+y)≥ � > λ, thus ρ(x−y)≤ λ shows that
δXρ(�)≥ 1−

α

2
> 1−

�

2
.

(ii) Note �0 =sup{� ∈ (0,2) : δXρ(�)≤ 1− �}.Suppose �0 < 2 . ∀� ∈ (�0,2), for any x,y ∈ Xρ and ρ(x),ρ(y)≤ 1, we have
ρ(x −y) > � or ρ(x −y)≤ �.

If ρ(x −y) > �, then δXρ(�)≥ 1− � implies ρ(x+y2 )≤ �. Thus J(Xρ)≤ 2�.



Eur. J. Math. Anal. 1 (2021) 157
Since δXρ(�)≤ 1− �, then J(Xρ)≤ 2�0 shows that

J(Xρ)=2sup{� ∈ (0,2) : δXρ(�)≤ 1− �}.

In particular, if ρ is convex and let α = J(Xρ)∈ [1,2], then ∀x,y ∈ Xρ,ρ(x),ρ(y)≤ 1, we have
ρ(x +y)≤ α or ρ(x −y)≤ α.

What’s more, ∀η > 0, there exist x′,y ′ ∈ Xρ and ρ(x′),ρ(y ′)≤ 1 such that
ρ
(
x′ +y ′

)
> α−η and ρ(x′ −y ′) > α−η.

Fix η > 0, then 1−ρ(x′ +y ′
2
) < 1−

α−η
2

implies δXρ(�) < 1− α−η2 , therefore
sup

{
� ∈ (0,2) : δXρ(�) < 1−

�

2

}
≥ α−η.

∀� ∈ (0,2), if � ≤ α, thus
sup

{
� ∈ (0,2) : δXρ(�) < 1−

�

2

}
≤ α.

If � > α, then ρ(x +y)≤ α shows that δXρ(�)≥ 1− α2 . In (0,2), we know
sup

{
� ∈ (0,2) : δXρ(�) < 1−

�

2

}
≤ α,

thus α−η ≤ sup{� ∈ (0,2) : δXρ(�) < 1− �2}≤ α.Let η → 0, then
sup

{
� ∈ (0,2) : δXρ(�) < 1−

�

2

}
= α.

Theorem 3. Let Xρ be a modular space, then(i) J(Xρ)≤ ρXρ(1)+1;(ii) CNJ(Xρ)≤ 2(√12+(1+ρXρ(1))2 −2)2.
Proof. (i) We can deduce that

J (Xρ)≤ sup
{
ρ

(
x +y

2

)
+ρ

(
x −y
2

)
: x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1

}
= ρXρ(1)+1.

(ii) We know that a2 +b2 ≤ (a+b)2 −4(a+b)+8 for 0 < a,b ≤ 2. Thus
ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
≤
(
ρ

(
x +y

2

)
+ρ

(
x −y
2

))2
−4

(
ρ

(
x +y

2

)
+ρ

(
x −y
2

))
+8.

Since
ρ

(
x +y

2

)
+ρ

(
x −y
2

)
≥

√
ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
,



Eur. J. Math. Anal. 1 (2021) 158
then

ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
+4

√
ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
−8

≤
(
ρ

(
x +y

2

)
+ρ

(
x −y
2

))2
≤
(
1+ρXρ(1)

)2
.Thus

ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
≤
(√
12+

(
1+ρXρ(1)

)2 −2)2
which shows that 1

2
CNJ(Xρ)≤

(√
12+

(
1+ρXρ(1)

)2 −2)2 .
5. Convexity and Non-squareness

Clarkson introduced uniform convexity in 1936, proved that lp(1≤ p < ∞) spaces are uniformlyconvex Banach spaces and uniformly convex Banach spaces have Radon-Nikodym properties. Dueto the geometrical intuitiveness of convexity, Poom Kumam [10] gave the definitions of ρr -uniformlyconvex, ρ-uniformly non-square and ρ-strictly convex of modular spaces in 2006.On the basis of literature [10], this paper studies the relationships between convexity, non-squareness and geometric constants of modular spaces.
Definition 6.[10] For r > 0, a modular space Xρ is said to be ρr -uniformly convex if for each � > 0,there exists δ > 0 such that for any x,y ∈ Xρ, the conditions ρ(x)≤ r , ρ(y)≤ r and ρ(x−y)≥ r�imply that ρ(x+y

2
)≤ (1−δ)r .

Definition 7.[10] The modular space Xρ is said to be ρ-uniformly non-square if there exists δ ∈ (0,1)such that for any x,y ∈ Xρ with ρ(x)=1 and ρ(y)≤ 1, ρ(x+y2 )≤ 1−δ or ρ(x−y2 )≤ 1−δ.
Definition 8.[10] The modular space Xρ is said to be ρ-strictly convex if for any x,y ∈ Xρ, theconditions ρ(x)≤ 1, ρ(y)≤ 1 and x 6= y imply that ρ(x+y

2
) < 1.

Theorem 4. Let Xρ be a modular space, then the following conditions are equivalent.(i) J(Xρ) < 2;(ii) �0(Xρ) < 2 for all 0 < � ≤ 2;(ii) Xρ is ρ-uniformly non-square.
Proof. Suppose J(Xρ) < 2. There exists � > 0, for any x,y ∈ Xρ with ρ(x)= 1 and ρ(y)≤ 1,such that

ρ(
x +y

2
)≤

J(Xρ)

2
− � < 1− � or ρ(x −y

2
)≤

J(Xρ)

2
− � < 1− �,

implies Xρ is ρ-uniformly non-square.Suppose Xρ is ρ-uniformly non-square, then we can prove J(Xρ) < 2 by the same way. Thus(i) and (iii) are equivalent.Next, we know that �0(Xρ) < 2 if and only if δXρ(2) > 0. Let α = δXρ(2), then ∀x,y ∈
Xρ and ρ(x)= 1,ρ(y)≤ 1, we can get ρ(x±y2 )≤ 1−α, thus Xρ is ρ-uniformly non-square. Thus



Eur. J. Math. Anal. 1 (2021) 159
(ii) and (iii) are equivalent.
Remark 3. In fact, this theorem is a generalization of Theorem 3.8 in [11].
Theorem 5. Let X be a modular space, then(i) X is ρ1-uniformly convex if and only if δXρ(�) > 0 for 0 < � ≤ 2;(ii) If δXρ(2)=1, then Xρ is ρ-strictly convex.

Proof. (i) Denote δε = δXρ(�). Then δXρ(�) > 0 if and only if ∀x,y ∈ Xρ,ρ(x),ρ(y) ≤ 1 and
ρ(x −y)≥ �, we have ρ(x+y

2
)≤ 1−δ�. Thus Xρ is ρ1-uniformly convex.(ii) Since δXρ(2)=1, then ∀x,y ∈ Xρ,ρ(x),ρ(y)≤ 1 and ρ(x −y)≥ 2, we have

ρ(
x +y

2
)=0 < 1,

implies Xρ is ρ-strictly convex.
6. Midpoint Convexity

In the following section, we discuss a special type of modular and study its properties in termsof geometric constants.
Definition 9.[6] Let (X,‖ ·‖) be a normed space and Xρ be a modular space. Then ρ is said to bestrongly midpoint convex with non-negtive constant C if

ρ(
x+y
2
)≤ ρ(x)+ρ(y)

2
− C
4
‖x −y‖2.

Theorem 6. Let (X,‖ · ‖) be a normed space and Xρ be a modular space. If there exists C ≥ 0such that
C‖x‖2 ≤

1

2
ρ(x) for all x ∈ BXρand ρ is strongly midpoint convex with constant C, then CNJ (Xρ)≤ 3.

Proof. Since ρ(x+y
2

)
≤ ρ(x)+ρ(y)

2
− C
4
‖x −y‖2 and ρ(x−y

2

)
≤ ρ(x)+ρ(−y)

2
− C
4
‖x +y‖2, then

ρ2
(
x +y

2

)
≤
1

4
(ρ(x)+ρ(y))2 −

C

4
‖x −y‖2(ρ(x)+ρ(y))+

C2

16
‖x −y‖4

and
ρ2
(
x −y
2

)
≤
1

4
(ρ(x)+ρ(y))2 −

C

4
‖x +y‖2(ρ(x)+ρ(y))+

C2

16
‖x +y‖4 .

Therefore, for x ∈ SXρ and y ∈ BXρ, we have
ρ2
(
x +y

2

)
+ρ2

(
x −y
2

)
≤
1

2
(1+ρ(y))2 −

C

4
(1+ρ(y))

(
‖x +y‖2 +‖x −y‖2

)
+

C2

16

(
‖x +y‖4 +‖x −y‖4

) .
Next, we only need to prove
ρ(y)+

C2

16

(
‖x +y‖4 +‖x −y‖4

)
−
C

4

(
‖x +y‖2 +‖x −y‖2

)
(1+ρ(y))−1−ρ2(y)≤ 0.



Eur. J. Math. Anal. 1 (2021) 160
Let t = ‖x + y‖2 + ‖x − y‖2,s = ‖x + y‖‖x − y‖ and I1 = ρ(y)+ C216 (t2 −2s2) − C4t(1+

ρ(y))−1−ρ2(y), then
I1 ≤

t2 −2s2

16
C2 −

t

4
C =

C
(
t2 −2s2

)
16

(
C −

4t

t2 −2s2

)
.

Since C‖x‖2 ≤ 1
2
ρ(x), then C‖x+y

2
‖2 ≤ 1

2
ρ
(
x+y
2

)
≤ 1 and C‖x−y

2
‖2 ≤ 1

2
ρ
(
x−y
2

)
≤ 1.Therefore

4t

t2 −2s2
=

∥∥x+y
2

∥∥2 +‖x−y
2
‖2∥∥x+y

2

∥∥4 +∥∥x−y
2

∥∥4 ≥ 1‖x+y
2
‖2+‖ x−y

2
‖2
≥ C,

then I1 ≤ 0.
Example 3. If ρ is convex, then C =0 which satisfies the condition of Theorem 6, and CNJ (Xρ)≤
2 < 3.
Theorem 7. Let (X,‖·‖) be a normed space and Xρ be a modular space. If there exist C,λ,µ,γ > 0such that

2µγ ≤ 1≤
1

2λ
+

√
6

8µ
and µρ(x)≤ C‖x‖2 ≤ λρ(x) for all x ∈ Xρ.

What’more, ρ is strongly midpoint convex with constant C, then CNJ (Xρ)≤ 2.
Proof. By following the ideas in Theorem 6, we can get

ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
1+ρ2(y)

≤
1

2
+

1

1+ρ2(y)

{
ρ(y)−

C

4
(1+ρ(y))

(
| x +y

∥∥2+∥∥x −y‖2)}
+

1

1+ρ2(y)

{
C2

16

(
‖x +y

∥∥4+∥∥x −y‖4)}.
Let

t = ‖x +y‖2 +‖x −y‖2,s = ‖x +y‖‖x −y‖

and
I2 =

t2 −2s2

16
C2 −

(1+ρ(y))t

4
C −

1

2
(1−ρ(y))2,

thus we only need to prove I2 ≤ 0. Since ρ(y)≤ 1, then
(1+ρ(y))t −

√
(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2 ≤ t

and (1+ρ(y))t +√(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2 ≥ t +√3t2 −4s2.Thus
t

t2 −2s2
=
1

4
·
‖x+y
2

∥∥2 +‖x−y
2
‖2

‖x+y
2
‖4 +

∥∥x−y
2

∥∥4
and

t +
√
3t2 −4s2

t2 −2s2

=
1

4
·
‖x +y‖2 +‖x−y

2
‖2 +

√
3‖x+y

2
‖4 +3‖x−y

2
‖4 +2‖x+y

2
‖2‖x−y

2
‖2

‖x+y
2
‖4 +‖x−y

2
‖4

,



Eur. J. Math. Anal. 1 (2021) 161
then

t+
√
3t2−4s2

t2−2s2 ≥
1

4‖x+y
2
‖2+4‖x−y

2
‖2
+

√
3

4
√
‖x+y
2
‖4+‖x−y

2
‖4

≥ C
λ(ρ(x+y2 )+ρ(

x−y
2 ))
+

√
3C

4µ
√
ρ2(x+y2 )+ρ

2(x−y2 )

≥ C
4λ
+
√
3C

8
√
2µ
≥ C
2
,and

t

t2 −2s2
≤
1

2
·

1∥∥x+y
2

∥∥2 +∥∥x−y
2

∥∥2 ≤ C4µ(ρ(x+y
2

)
+ρ

(
x−y
2

)) ≤ C
4µγ

≤
C

2
.

Therefore
(1+ρ(y))t −

√
(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2

t2 −2s2

≤
C

2
≤

(1+ρ(y))t +
√
(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2

t2 −2s2Thus
I2 =

t2 −2s2

4

(
C

2
−
(1+ρ(y))t +

√
(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2

t2 −2s2

)
(
C

2
−
(1+ρ(y))t −

√
(1+ρ(y))2t2 +2(t2 −2s2)(1−ρ(y))2

t2 −2s2

)
≤ 0.

Example 4. Consider ρ(x) = 4C‖x‖2 and let λ = µ = 1
4
, then µ2ρ(x) ≤ C‖x‖2 ≤ λρ(x) and

µ2 = λ
4
. What’more,

CNJ (Xρ)=2C sup

{
‖x +y‖4 +‖x −y‖4

‖x +y‖2 +‖x −y‖2
: x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1

}
≤ 2C sup

{
‖x +y

∥∥2+∥∥x −y‖2 : x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1}
≤ 4C sup{‖x‖2 +‖y‖2 : x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1}

=sup{ρ(x)+ρ(y) : x,y ∈ Xρ,ρ(x)=1,ρ(y)≤ 1}=2.

Theorem 8. Let (X,‖ · ‖) be a normed space, Xρ be a modular space and α0 ∈ (0,2√2]. If thereexists C > 0 such that
C ≥

4α0√
‖x0 +y0‖4 +‖x0 −y0‖4

for some x0 ∈ SXρ,y0 ∈ BXρ
and ρ is strongly midpoint convex with positive constant C, then CNJ (Xρ) ≥ α20. In particular, if
α0 =2

√
2, then CNJ (Xρ)=8.

Proof. Since 2ρ(x)≥ C‖x‖2, then ρ2(x±y
2

)
≥ C
2

∥∥x±y
2

∥∥2. Therefore
ρ2
(
x+y
2

)
+ρ2

(
x−y
2

)
1+ρ2(y)

≥
C2

16

(
‖x +y‖4 +‖x −y‖4

)
1+ρ2(y)

,



Eur. J. Math. Anal. 1 (2021) 162
shows that ρ2(x+y2 )+ρ2(x−y2 )

1+ρ2(y)
≥ α

2
0

1+ρ2(y)
, then CNJ (Xρ) ≥ α20. If α0 = 2√2, then CN (Xρ) ≥ 8implies CNJ (Xρ)=8.
7. Data Availability

No data were used to support this study.
8. Conflicts of Interest

The author(s) declare(s) that there is no conflict of interest regarding the publication of this paper.
9. Funding Statement

This work was supported by the National Natural Science Foundation of P. R. China (Nos.11971493 and 12071491).
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https://doi.org/10.1307/mmj/1028998906
https://doi.org/10.1090/S0002-9947-1936-1501880-4
https://doi.org/10.1090/S0002-9947-1936-1501880-4
https://doi.org/10.2307/1968512
https://doi.org/10.2307/1970663
https://doi.org/10.2307/1970663
https://www.uv.es/llorens/Documento.pdf
https://doi.org/10.4064/sm-18-1-49-65
https://doi.org/10.4064/sm-18-1-49-65
https://doi.org/10.1007/978-3-319-14051-3
https://doi.org/10.1016/j.jmaa.2009.04.002
https://doi.org/10.1016/j.jmaa.2012.07.063


Eur. J. Math. Anal. 1 (2021) 163
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https://doi.org/10.1007/BF02591317
https://doi.org/10.1007/BF02591317
https://doi.org/10.15352/bjma/1313362982

	1. Introduction
	2. Preliminaries
	3. The -Neumann–Jordan constant and the -James constant 
	4. The -convex modular and the -smooth modular 
	5. Convexity and Non-squareness
	6. Midpoint Convexity
	7. Data Availability
	8. Conflicts of Interest
	9. Funding Statement
	References

