# On Generalized Moduli of Quasi-Banach Space

Authors: Young Chel Kwun; Hussain Minhaj Uddin Ahmad Qadri; Waqas Nazeer; Absar Ul Haq; Shin Min Kang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We shall discuss three generalized moduli such as generalized modulus of convexity, modulus of smoothness, and modulus of Zou-Cui of quasi-Banach spaces and give some important properties of these moduli. Furthermore, we establish relationships of these generalized moduli with each other.


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## Body

## 1. Introduction

The study on Banach space geometry provides many fundamental notions and interesting aspects and sometimes has surprising results. The basic geometric properties such as convexity, smoothness, and nonsquareness have made great contributions to various fields of Banach space theory. Strict convexity of Banach spaces was first introduced in 1936 by Clarkson [1] (and independently by Akhiezer and Krein) as the property that the unit sphere contains no nontrivial line segments; that is, 1-2-1(x+y)>0 whenever x=y=1. Clarkson [1] made use of these values to define the “uniform” version of convexity to look at how “convex” the unit ball is in a space. And the modulus of convexity provides a quantification of the geometric structure of the space from the viewpoint of convexity. A situation similar to this also occurs in smoothness and other properties. A Banach space X is said to be smooth if each unit vector has a unique norm one support functional. In fact, this is equivalent to the statement that the norm is Gateaux differentiable. This allows us to quantify the geometric structure of the space from the viewpoint of smoothness, namely, the modulus of smoothness of a Banach space X. An advantage of these quantifications is that the complete duality between uniform convexity and uniform smoothness can be easily deduced by the well-known Lindenstrauss formulas; that is, a Banach space X is uniformly convex if and only if its dual space X∗ is uniformly smooth. The same statement still holds if X is replaced with X∗. Thus quantifying geometric structures might lead to better results. Note that the same duality does not hold between strict convexity and smoothness in general, though one of those two properties of X∗ implies the other of X. There are some other ideas to quantify geometric structures of Banach spaces.In [2], the authors claim that modulus of convexity and generalized convexity mold have dual relationship, and generalized convexity mold has many excellent properties.In [3], the authors study a generalized modulus of convexity where certain related geometrical properties of this modulus are analyzed in Banach spaces.In [4, 5], the modulus of Yang-Wang was introduced in Banach spaces.In [6], the modulus of Zuo-Cui was introduced in Banach spaces. The author proved many results with this special type of modulus.The most recent research work at this topic can be consulted from [7, 8].


## 2. Preliminaries

There are lots of quantitative descriptions of geometrical properties of quasi-Banach spaces. The most common way for creating these descriptions is to define a real function (a modulus) and a suitable coefficient or constant closely related to this function, depending on the space structure under consideration. Some of the moduli and their related coefficients (or characteristics) for quasi-Banach spaces have also been investigated so far. These moduli are the attempts in order to get a better understanding of the two facts about the space:(i)
The shape of the unit ball of the concerned space.(ii)
The conditions and relations for convergence of sequences.The most recent research work with these moduli is investigated by [7, 8].Definition 1.
For a quasi-Banach spaceB, themodulus of convexity is a function δB:(0,2]→[0,1] defined as(1)δBϵ=inf⁡1-x1+x22C:ϵ∈0,2,C≥1,x1,x2∈SB,x1-x2C≥ϵ.
A characteristic or related coefficient of this modulus is(2)δ0B=sup⁡ϵ∈0,2:δBϵ=0.Definition 2.
Letα∈(0,1) and ϵ∈[0,2]. For a quasi-Banach space B, thegeneralized modulus of convexity is a function δB(α):(0,2]→[0,1] defined as(3)δBαϵ=inf⁡1-αx1+1-αx2C:C≥1,x1,x2∈SB,x1-x2C≥ϵ.
A characteristic or related coefficient of this modulus is(4)δθB=sup⁡ϵ∈0,2:δBαϵ=0.Definition 3.
For a quasi-Banach spaceB, themodulus of smoothness is a function ρB:[0,∞)→[0,∞) defined as(5)ρBt=sup⁡1-x1+x22C:x1,x2∈SB,C≥1,x1-x2C≤t,t≥0.
A characteristic or related coefficient of this modulus is(6)ρ0B=limt→0+ρBtt=limt→0+ρ1Btt=limt→0+ρ2Btt.Definition 4.
Letα∈(0,1) and t∈[0,2]. For a quasi-Banach space B, thegeneralized modulus of smoothness is a function ρB(α):[0,∞)→[0,∞) defined as(7)ρBαt=sup⁡1-αx1+1-αx2C:C≥1,x1,x2∈SB,x1-x2C≤t,ρBαt=sup⁡αx1-t/2x2+1-αx1-t/2x2C-1C, where x1,x2∈SB, x1≤1, x2≤1, and C≥1.
A characteristic or related coefficient of this modulus is(8)ρθB=limt→0+ρBαtt=limt→0+ρ1Bαtt.Definition 5.
Lett∈[0,∞) and p∈[1,∞). For a quasi-Banach space B, themodulus of Zuo-Cui is a function ξB(p):(0,∞)→(0,∞) defined as(9)ξBpt=supx1+tx2p+x1-tx2p2Cp1/p,x1,x2∈SB,C≥1.
A characteristic or related coefficient of this modulus is(10)ξθB=limt→0ξBpt-1/Ct.


## 3. Relations Concerning Generalized Modulus of Convexity

Lemma 6 (see [9]).
Every convex functionf with convex domain in R is continuous.Proposition 7 (see [10]).
LetB be a uniformly convex space. Then for every d>0, ε>0, and for arbitrary vectors, x1,x2∈B with x1≤d, x2≤d, and x1-x2/C≥ε, where C≥1, there exists δ>0 such that(11)x1+x22C≤1-δεdd.Proposition 8 (see [10]).
A nontrivial quasi-Banach spaceB is uniformly nonsquare if and only if δB(ϵ)>0.Proposition 9.
LetδB(ϵ) be the modulus of convexity of a quasi-Banach space B; then(12)2αδBϵ≤δBαϵ≤21-αδBϵ.Proof.
Letx,y∈B, y=C, and x-y/C≥ϵ. Then we have(13)1-αx+1-αyC=1-αx+y+1-2αyC=1-αx+yC-1-2αyC=1-2αx+y2C-1-2α≤2αsup⁡1-x+y2C=2αδBϵ,and we have obtained the first inequality; now to prove the second one, for any η>0, there exist x,y∈SB with x=1, y=C, and x-y/C≥ϵ such that x+y/2C≤δ(ϵ)+η; then we have(14)1-αx+1-αyC=1-αx+y+1-2αyC≤1-1-αx+y+1-2αy-CC≤1-21-αx+y2C+1-2αy-CC≤1-21-αx+y2C+21-α-1≤21-α-21-αx+y2C≤21-α1-x+y2C≤21-αδBϵ.This completes the proof.Corollary 10.
Letϵθ be the characteristic of generalized modulus of convexity δB(α)(ϵ) of a quasi-Banach space B. Then (15)ϵθB=sup⁡ϵ∈0,2:δBαϵ=0.Theorem 11.
A quasi-Banach spaceB is uniformly nonsquare if and only if (16)δBαϵ>0.Proof.
By Proposition8, B is uniformly nonsquare if and only if δB(ϵ)>0, where ϵ∈(0,2), and by Proposition 9, δB(ϵ)>0 if and only if δB(α)>0(ϵ)>0.
Combining both of these results we get the proof.Lemma 12.
LetB be a quasi-Banach space and ϵ∈(0,2]. Then the following statements hold:(1)
δB(α)(ϵ) is convex and continuous function.(2)
δB(α)(ϵ) is a nondecreasing function.(3)
δB(α)(ϵ)/ϵ is a nondecreasing function.Proof.
( 1 ) Let(17)ϵ2=2ϵ2-ϵ12-ϵ1+ϵ11-ϵ2-ϵ12-ϵ1.Then(18)δBαϵ2=δBα2ϵ2-ϵ12-ϵ1+ϵ11-ϵ2-ϵ12-ϵ1≤δBα2ϵ2-ϵ12-ϵ1+δBαϵ11-ϵ2-ϵ12-ϵ1≤δBα2ϵ2-ϵ12-ϵ1+δBαϵ1-δBαϵ1ϵ2-ϵ12-ϵ1=ϵ2-ϵ12-ϵ1δBα2-δBαϵ1+δBαϵ1.Now(19)δBαϵ2-δBαϵ1≤ϵ2-ϵ12-ϵ11-δBαϵ1≤ϵ2-ϵ12-ϵ121-α,by using Proposition 9. Hence we get (20)δBαϵ2-δBαϵ1≤21-αϵ2-ϵ12-ϵ1.Since δB(α)(ϵ) is convex, so it is continuous by Lemma 6.
( 2 ) Let 0<ϵ1≤ϵ2<0 and x,y∈SB with x-y≥ϵ2. Let us consider (21)z=αx+1-αyC,t≤ϵ1ϵ2,u=tx+1+tz,v=ty+1-tz. Then u,v∈SB with u-v≥ϵ1; we have(22)1-αu+1-αvC≤t1-αx+1-αyC=ϵ1ϵ21-αx+1-αyC;thus (23)sup⁡1-αu+1-αvC≤ϵ1ϵ2sup⁡1-αx+1-αyC,which implies that (24)δBαϵ1≤ϵ1ϵ2δBαϵ2δBαϵ1ϵ1≤δBαϵ2ϵ2.
( 3 ) Let 0≤t1≤t2≤2; we have (25)t1δBαt2≤t2δBαt1≤t1δBαt1. Since δB(α)(t1)>0, so we have (26)δBαt2<δBαt1. This completes the proof.Theorem 13.
LetB be a uniformly convex space and α∈(0,1). Then for every d>0, ε>0 and for arbitrary vectors, x1,x2∈B with x1≤d, x2≤d, and x1-x2/C≥ε, there exists δ=ε/d>0 such that(27)αx1+1-αx2C≤1-21-C+Cδϵd·min⁡α,1-αdC.Proof.
Without loss of generality, assume thatα∈(0,1/2] so(28)αx1+1-αx2C=αx1+x2+1-2αx2C=2αx1+x22C+1-2αx2C.Now by Proposition 7, there exist δ>0 such that, for y≤d, (29)x1+x22C≤1-δϵdd.Therefore, we get(30)αx1+1-αx2C≤2α1-δϵdd+1-2αdC=2α1-δϵd+1-2αCd=2α-2αδϵd+1-2αCd=1-2α+2αCC-2αδϵdd=1C-2αC+2αCC-2αδϵdd=1C-2C1-C+Cδϵd·αd=1-21-C+Cδϵd·min⁡α,1-αdC.This completes the proof.


## 4. Relations Concerning Generalized Modulus of Smoothness

Theorem 14.
LetB be a quasi-Banach space. Then for every x∈B, x=1, and x∗∈X∗, x∗=1(31)ρBαt=suptϵ2-δB∗αϵ:t>0,ϵ∈0,2,ρB∗αt=suptϵ2-δBαϵ:t>0,ϵ∈0,2.Proof.
Throughout the proof of the first part, we take(32)x,y∈SB,x≤1,y≤1,α∈0,1,t≥0,C≥1,f∗≤1,g∗≤1,f,g∈SB∗,x∗≤1,y∗≤1,x∗,y∗∈SB∗.Take(33)ρBαt=sup⁡αCx+t2Cy+1-αCx-t2Cy-1=supfαCx+t2Cy+g1-αCx-t2Cy-1=1Csup⁡fαx+t2y+g1-αx-t2y-C=1Csup⁡αf+t2gx+t2f-gy-C=1Csup⁡αf+1-αg+t2f-g-C=sup⁡αf+1-αgC+t2Cf-g-1=sup⁡t2Cf-g-1-αf+1-αgC=sup⁡tϵC2C-δB∗αϵ=sup⁡tϵ2-δB∗αϵ.Hence, we get (34)ρBαt=sup⁡tϵ2-δB∗αϵ:t>0,ϵ∈0,2. This completes the first part of the proof.
To prove the second part, let us start as(35)ρB∗αt=sup⁡αCf+t2Cg+1-αCg-t2Cg-1=sup⁡αfC+tg2Cx+1-αfC-tg2Cy-1=1Csup⁡fαx+1-αy+gt2x-y-C=1Csup⁡αf+t2gx+t2gx-y-C=1Csup⁡αx+1-αy+t2x-y-C=sup⁡αx+1-αyC+t2Cx-y-1=sup⁡t2Cx-y-1-αx+1-αyC=sup⁡tϵC2C-δBαϵ=sup⁡tϵ2-δBαϵ.Lemma 15.
LetB be a quasi-Banach space and τ∈(0,2]. Then the following statements hold:(1)
ρB(α)(τ) is convex and continuous function.(2)
ρB(α)(τ) is a nondecreasing function.(3)
ρB(α)(τ)/τ is a nondecreasing function.Proof.
( 1 ) Since ρB(α)(τ) is convex, so it is continuous by Lemma 6.
( 2 ) Let x1,x2∈SB, 0<τ1<τ2, and α∈(0,1).
Now let(36)x1=ατ2-τ1τ2+ατ1-τ21-αx+τ12Cy,x2=1-αx+τ12Cy,x3=α1-ατ2-τ1τ2+ατ1-τ2x+τ2τ2-τ11-α2τ2+ατ1-τ2y,x4=1-αx+τ22Cy,x5=αx+τ12Cy,x6=ατ1τ2+ατ1-τ2x+τ1τ22τ2+ατ1-τ2y,x7=αx+τ22Cy.Now (37)x3+x4=x1+x2,(38)x3+x4≥x1+x2,(39)x5-x1=x6,(40)x5-x1≥x6.From (38) and (40), we get (41)x3+x4+x6≤x5+x2.Similarly, (42)x3-x6=x7,x3-x6≥x7.Therefore, we have(43)x5+x2≤x4-x7,αx+τ12Cy+1-αx+τ12Cy≤1-αx+τ22Cy+αx+τ22Cy,which implies that(44)αx+τ12Cy+1-αx-τ12Cy≤αx+τ22Cy+1-αx-τ22Cy;thus we have(45)ρBατ1≤ρBατ2.This shows that ρB(α)(τ) is a nondecreasing function.
( 3 ) This part is an immediate consequence of Theorem 14.Theorem 16.
For a nontrivial quasi-Banach spaceB with t∈[0,∞) and α∈(0,1), one has (46)max⁡0,t+2α-1-1≤ρBα<t.Proof.
Letx,y∈SB and α∈(0,1). Then, we have(47)ρBαt=sup⁡αx+t2Cy+1-αx-t2Cy-1C≤αx+t2Cy+1-αx-t2Cy-1C≤x+t≤t.This proves the first inequality; now to prove the second inequality, we proceed as(48)αx+t2Cy-1-αx+t2Cy-1C=2α-1+t-1C.Also(49)αx+t2Cy-1-αx+t2Cy-1C≤αx+t2Cy+1-αx-t2Cy-1C.Therefore,(50)2α-1+t-1C≤αx+t2Cy+1-αx-t2Cy-1C≤ρBαt.This completes the second inequality. Now, combining both of the above inequalities and taking the max of the latter, we get (51)max⁡0,t+2α-1-1≤ρBα<t. This completes the proof.Theorem 17.
A quasi-Banach spaceB is uniformly smooth if and only if(52)ρθB=limt→0+ρBαtt=0.Proof.
One has(53)αx+21-αt2y=12x+t2y+αx+21-αt2y-12x+t2yαx+21-αt2y+1-αx+21-αt2y≤12x+t2y+αx+21-αt2y-12x+t2y+1-αx+21-αt2y≤12x+t2y+α-12x+1-2αt2y+1-αx+21-αt2y≤12x+t2y+α-12x+1-2αt2y1-αx+21-αt2y≤12x+t2y+12x-t2y.Hence we get(54)αx+21-αt2y+1-αx+21-αt2y≤12x+t2y+12x-t2y.Dividing by “C” and taking supremum on both sides, we get(55)sup⁡αx+21-αt/2y+1-αx+21-αt/2y-1C≤sup⁡1/2x+t/2y+1/2x-t/2y-1C;thus(56)sup⁡αx+21-αt/2y+1-αx+21-αt/2yC-1C≤sup⁡x+ty+x-ty2C-1C,which implies that(57)ρBα21-αt≤ρBt.Now(58)12x-12αt2y=t2y-1-αx+1-αx-t2y12x+12αt2y+12x-12αt2y=12x+12αt2y+t2y-1-αx+1-αx-t2y.This can be expressed as(59)x+1/2αty+x-1/2αty2=12x+12αt2y+t2y-1-αx+1-αx-t2y≤12x+12αt2y+t2y-1-αx+1-αx-t2y=αx+t2y+1-αx+t2y.Dividing by “C” and taking supremum on both sides(60)supx+1/2αty+x-1/2αty-12C=αx+t/2y+1-αx+t/2y-1C,thus(61)ρB12αt≤ρBαt.Therefore, from (57) and (61) we get(62)limt→0+ρBtt=0=limt→0+ρBαtt. This completes the proof.


### 4.1. Relations Concerning Modulus of Zuo-Cui

Theorem 18.
LetB be a nontrivial quasi-Banach space and p∈[1,∞). Then (63)ξBpt≥ρBt+1C.Proof.
By using convexity of the functionf(u)=up on (0,∞), one can easily obtain (64)x1+tx2+x1-tx22p≤x1+tx2p+x1-tx2p2,so,(65)x1+tx2+x1-tx22Cp≤x1+tx2p+x1-tx2p2Cp,which implies that (66)1C+ρBt≤ξBpt.
Forp=1, we have ξB1(t)=1/C+ρB(t).
Forp=2, we have 2CξB(2)(t)2=E(t,B), where (67)Et,B=supx1+tx22+x1-tx22:x1,x2∈SB.Proposition 19.
For a Banach spaceB,(68)ξBt=sup⁡x1+tx2p+x1-tx2p2C1/p:x1,x2∈SB,t∈0,1,C≥1.Proof.
Consider a convex function defined by(69)ft=x1+tx2p+x1-tx2p2C.Let 0<t1≤t2<1 and x1,x2∈SB. Then(70)ξBt1=x1+t1x2p+x1-t1x2p=2Cft1=2Cft2+t12t2t2+t2-t12t2-t2≤x1+t2x2p+x1-t2x2p=2Cft2=ξBt2.This shows that ξB(t1)≤ξB(t2). Therefore, we have (71)1psupx1∈SBsupx2∈BBx1+tx2p+x1-tx2p2C≤ξBtx2≤ξBt.Since the opposite inequality holds obviously, to get the first inequality let (72)gλ=λx1+tx2p+λx1-tx2p2C.Since g(λ) is a convex and an even function, therefore g(λ)≥g(1). Now for x1,x2∈BB we have(73)x1x1+tx2p+x1x1-tx2p≥x1+tx2p+x1-tx2p≥supx1∈SBsupx2∈BBx1+tx2p+x1-tx2p≥supx1,x2∈BBx1+tx2p+x1-tx2p≥supx1,x2∈BBx1+tx2p+x1-tx2p2C,which obtains the second inequality. This completes the proof.Lemma 20.
For any quasi-Banach spaceB and any 1≤p<∞, then the following statements hold:(1)
ξB(p)(t) is a nondecreasing function.(2)
ξB(p)(t) is convex and continuous function.(3)
(ξBp(t)-1)/t is a nondecreasing function.Proof.
( 1 ) Let g(t)=x1+tx2p+x1-tx2p be a convex and even function. Let 0<t1≤t2 and x1,x2∈SB. Then we have(74)x1+t1x2p+x1-t1x2p=gt1=gt2+t12t2t2+t2-t12t2-t2≤t2+t12t2gt2+t2-t12t2gt2=gt2=x1+t2x2p+x1-t2x2p≤2CpξBpt2,which implies that (75)x1+t1x2p+x1-t1x2p2Cp≤ξBpt2.Hence ξB(p)(t1)≤ξB(p)(t2). Since ξB(p)(t) is convex, so it is continuous by Lemma 6.
( 2 ) Let x1,x2∈SB, t1,t2>0, and λ∈(0,1) and rs=sgn[sin⁡(2πs)]. Then we have(76)ξBpλt1+1-λt2=12Cp∫01x+rsλt1+1-λt2ypdt1/p≤12Cp∫01λx+rst1y+1-λx+rst2ypdt1/p≤λ2Cp∫01x+rst1ypdt11/p+1-λ2Cp∫01x+rst2ypdt21/p≤λξBpt1+1-λξBpt2.This shows that ξB(p)(t) is convex.
( 3 ) Let 0<t1≤t2<0 and λ∈(0,1) with t1=λt2. Then(77)ξBpt1-1t1≤ξBpλt2-1λt2≤ξBpt2-1t2. This shows that (ξBp(t)-1)/t is nondecreasing. This completes the proof.Proposition 21.
LetB be a quasi-Banach space, p∈[1,∞), and t>0. Then the following conditions are equivalent:(1)
ξB(p)(t)<1+t.(2)
ξ(t,x1)<1+t.Proof.
( 1 ) ⇒ ( 2 ) Suppose on the contrary that ξ(t,x1)≥1+t; it is enough to take (t,x1)=1+t. Since (78)JB=sup1Cmin⁡x1+x2,x1-x2:x1,x2∈SB,by using the definition of sup, for any ϵ>0, there exist x1,x2∈SB such that(79)x1+tx2+x1-tx22≥min⁡x1+tx2,x1-tx2≥c1+t-ϵ.Applying convexity of the function f(u)=up, we get (80)x1+tx2+x1-tx22p1/p≤x1+tx2p+x1-tx22p1/p;therefore from (79) (81)x1+tx2p+x1-tx22p1/p≥min⁡x1+tx2,x1-tx2≥c1+t-ϵ.Since ϵ is any arbitrary so (82)ξBpt≥1+t,which leads a contradiction.
( 2 ) ⇒ ( 1 ) Suppose on the contrary that ξB(p)(t)≥1+t; it is enough to take ξB(p)(t)=1+t. Again using the definition of sup, for all ϵ>0, there exist x1,x2∈SB such that (83)x1+tx2p+x1-tx2p≥2Cp1+t-ϵp,also using (84)x1+tx2p+x1-tx2p≤2Cp1+tp,thus (85)2Cp1+tp≥x1+tx2p+x1-tx2p≥2Cp1+t-ϵp;since ϵ is arbitrary, so (86)x1+tx2p+x1-tx2p=2Cp1+tp,which implies that (87)x1+tx2=x1-tx2=C1+t.So using the definition of ξ(t,x1) we get ξ(t,x1)≥1+t, which leads to a contradiction.Corollary 22.
LetB be a quasi-Banach space, p∈[1,∞), and t>0. Then the following conditions are equivalent:(1)
B is uniformly nonsquare.(2)
ξB(p)(t)<1+t.(3)
ξ(t,x1)<1+t.Theorem 23.
A quasi-Banach apaceB is uniformly smooth if (88)limt→0ξBpt-1/Ct=0.Proof.
Suppose that(89)limt→0ξBpt-1/Ct=0.From Theorem 18, we know that (90)ξBpt≥ρBt+1C,which implies that (91)ξBpt-1C≥ρBt;dividing both sides by t and applying the limt→0(92)limt→0ρBtt≤limt→0ξBpt-1/Ct=0.So by definition B is uniformly smooth.Theorem 24.
LetB be a quasi-Banach space. Then for τ∈(0,1)(93)ξBp2<12ξBp1+21/2.Proof.
Ifa,b∈[0,2], then (a+b)/22≥a2+b2-4 holds true.
Letx,y∈SB and τ∈[0,1]. Then we have(94)x+τyC≤2,x-τyC≤2 and we have(95)x+τy+x-τy2C2≥x+τy2+x-τy2-4x+τy2+x-τy22C≤12x+τy+x-τy2C2+2.Therefore, we have(96)ξBp2<12ξBp1+21/2.Theorem 25.
LetB be a quasi-Banach space. Then (97)ξBp1<ξp+2p2C1/p.Proof.
Since(98)x+yp+x-yp2C1/p≤minx+yp,x-yp+2t2C1/p,therefore, we have(99)ξBp1<ξp+2p2C1/p.


## 4.1. Relations Concerning Modulus of Zuo-Cui

Theorem 18.
LetB be a nontrivial quasi-Banach space and p∈[1,∞). Then (63)ξBpt≥ρBt+1C.Proof.
By using convexity of the functionf(u)=up on (0,∞), one can easily obtain (64)x1+tx2+x1-tx22p≤x1+tx2p+x1-tx2p2,so,(65)x1+tx2+x1-tx22Cp≤x1+tx2p+x1-tx2p2Cp,which implies that (66)1C+ρBt≤ξBpt.
Forp=1, we have ξB1(t)=1/C+ρB(t).
Forp=2, we have 2CξB(2)(t)2=E(t,B), where (67)Et,B=supx1+tx22+x1-tx22:x1,x2∈SB.Proposition 19.
For a Banach spaceB,(68)ξBt=sup⁡x1+tx2p+x1-tx2p2C1/p:x1,x2∈SB,t∈0,1,C≥1.Proof.
Consider a convex function defined by(69)ft=x1+tx2p+x1-tx2p2C.Let 0<t1≤t2<1 and x1,x2∈SB. Then(70)ξBt1=x1+t1x2p+x1-t1x2p=2Cft1=2Cft2+t12t2t2+t2-t12t2-t2≤x1+t2x2p+x1-t2x2p=2Cft2=ξBt2.This shows that ξB(t1)≤ξB(t2). Therefore, we have (71)1psupx1∈SBsupx2∈BBx1+tx2p+x1-tx2p2C≤ξBtx2≤ξBt.Since the opposite inequality holds obviously, to get the first inequality let (72)gλ=λx1+tx2p+λx1-tx2p2C.Since g(λ) is a convex and an even function, therefore g(λ)≥g(1). Now for x1,x2∈BB we have(73)x1x1+tx2p+x1x1-tx2p≥x1+tx2p+x1-tx2p≥supx1∈SBsupx2∈BBx1+tx2p+x1-tx2p≥supx1,x2∈BBx1+tx2p+x1-tx2p≥supx1,x2∈BBx1+tx2p+x1-tx2p2C,which obtains the second inequality. This completes the proof.Lemma 20.
For any quasi-Banach spaceB and any 1≤p<∞, then the following statements hold:(1)
ξB(p)(t) is a nondecreasing function.(2)
ξB(p)(t) is convex and continuous function.(3)
(ξBp(t)-1)/t is a nondecreasing function.Proof.
( 1 ) Let g(t)=x1+tx2p+x1-tx2p be a convex and even function. Let 0<t1≤t2 and x1,x2∈SB. Then we have(74)x1+t1x2p+x1-t1x2p=gt1=gt2+t12t2t2+t2-t12t2-t2≤t2+t12t2gt2+t2-t12t2gt2=gt2=x1+t2x2p+x1-t2x2p≤2CpξBpt2,which implies that (75)x1+t1x2p+x1-t1x2p2Cp≤ξBpt2.Hence ξB(p)(t1)≤ξB(p)(t2). Since ξB(p)(t) is convex, so it is continuous by Lemma 6.
( 2 ) Let x1,x2∈SB, t1,t2>0, and λ∈(0,1) and rs=sgn[sin⁡(2πs)]. Then we have(76)ξBpλt1+1-λt2=12Cp∫01x+rsλt1+1-λt2ypdt1/p≤12Cp∫01λx+rst1y+1-λx+rst2ypdt1/p≤λ2Cp∫01x+rst1ypdt11/p+1-λ2Cp∫01x+rst2ypdt21/p≤λξBpt1+1-λξBpt2.This shows that ξB(p)(t) is convex.
( 3 ) Let 0<t1≤t2<0 and λ∈(0,1) with t1=λt2. Then(77)ξBpt1-1t1≤ξBpλt2-1λt2≤ξBpt2-1t2. This shows that (ξBp(t)-1)/t is nondecreasing. This completes the proof.Proposition 21.
LetB be a quasi-Banach space, p∈[1,∞), and t>0. Then the following conditions are equivalent:(1)
ξB(p)(t)<1+t.(2)
ξ(t,x1)<1+t.Proof.
( 1 ) ⇒ ( 2 ) Suppose on the contrary that ξ(t,x1)≥1+t; it is enough to take (t,x1)=1+t. Since (78)JB=sup1Cmin⁡x1+x2,x1-x2:x1,x2∈SB,by using the definition of sup, for any ϵ>0, there exist x1,x2∈SB such that(79)x1+tx2+x1-tx22≥min⁡x1+tx2,x1-tx2≥c1+t-ϵ.Applying convexity of the function f(u)=up, we get (80)x1+tx2+x1-tx22p1/p≤x1+tx2p+x1-tx22p1/p;therefore from (79) (81)x1+tx2p+x1-tx22p1/p≥min⁡x1+tx2,x1-tx2≥c1+t-ϵ.Since ϵ is any arbitrary so (82)ξBpt≥1+t,which leads a contradiction.
( 2 ) ⇒ ( 1 ) Suppose on the contrary that ξB(p)(t)≥1+t; it is enough to take ξB(p)(t)=1+t. Again using the definition of sup, for all ϵ>0, there exist x1,x2∈SB such that (83)x1+tx2p+x1-tx2p≥2Cp1+t-ϵp,also using (84)x1+tx2p+x1-tx2p≤2Cp1+tp,thus (85)2Cp1+tp≥x1+tx2p+x1-tx2p≥2Cp1+t-ϵp;since ϵ is arbitrary, so (86)x1+tx2p+x1-tx2p=2Cp1+tp,which implies that (87)x1+tx2=x1-tx2=C1+t.So using the definition of ξ(t,x1) we get ξ(t,x1)≥1+t, which leads to a contradiction.Corollary 22.
LetB be a quasi-Banach space, p∈[1,∞), and t>0. Then the following conditions are equivalent:(1)
B is uniformly nonsquare.(2)
ξB(p)(t)<1+t.(3)
ξ(t,x1)<1+t.Theorem 23.
A quasi-Banach apaceB is uniformly smooth if (88)limt→0ξBpt-1/Ct=0.Proof.
Suppose that(89)limt→0ξBpt-1/Ct=0.From Theorem 18, we know that (90)ξBpt≥ρBt+1C,which implies that (91)ξBpt-1C≥ρBt;dividing both sides by t and applying the limt→0(92)limt→0ρBtt≤limt→0ξBpt-1/Ct=0.So by definition B is uniformly smooth.Theorem 24.
LetB be a quasi-Banach space. Then for τ∈(0,1)(93)ξBp2<12ξBp1+21/2.Proof.
Ifa,b∈[0,2], then (a+b)/22≥a2+b2-4 holds true.
Letx,y∈SB and τ∈[0,1]. Then we have(94)x+τyC≤2,x-τyC≤2 and we have(95)x+τy+x-τy2C2≥x+τy2+x-τy2-4x+τy2+x-τy22C≤12x+τy+x-τy2C2+2.Therefore, we have(96)ξBp2<12ξBp1+21/2.Theorem 25.
LetB be a quasi-Banach space. Then (97)ξBp1<ξp+2p2C1/p.Proof.
Since(98)x+yp+x-yp2C1/p≤minx+yp,x-yp+2t2C1/p,therefore, we have(99)ξBp1<ξp+2p2C1/p.



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