Some Moduli of Angles in Banach Spaces

Abstract

In this paper, we mainly discuss the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$ and the angle modulus of smoothness ${\rho }_X^a\left(ϵ\right)$ in a real normed linear space X, which are closely related to the classical modulus of convexity ${\delta }_X\left(ϵ\right)$ and the modulus of smoothness ${\rho }_X\left(ϵ\right)$. Some geometric properties of the two moduli were investigated. In particular, we obtained a characterization of uniform non-squareness in terms of ${\rho }_X^a\left(1\right)$. Meanwhile, we studied the relationships between ${\delta }_X^a\left(ϵ\right)$, ${\rho }_X^a\left(ϵ\right)$ and other geometric constants of real normed linear spaces through some equalities and inequalities. Moreover, these two coefficients were computed for some concrete spaces.

1. Introduction and Preliminaries

Throughout this paper, we will adopt such notations. Suppose that the real normed linear space is represented by X with $\dim X\ge 2$ and the dual space by $X^{*}$ of X, then the unit ball, as well as the unit sphere of X are respectively symbolized by $B_X$ and $S_X$.

Recall that space X is uniformly convex [1], if, for any $ϵ>0$, there exist $\delta >0$, such that for any $x,y\in S_X$ with $\parallel x-y\parallel >ϵ$, then $∥\frac{x+y}{2}∥<1-\delta$.

For $ϵ\in [0,2]$, the Clarkson modulus of convexity [1] of X is defined in the following way:

$$\begin{array}{cc}\hfill {\delta }_X\left(ϵ\right)& =inf\left\{1-\frac{\parallel x+y\parallel }{2}:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}\hfill \\ & =inf\left\{1-\frac{\parallel x+y\parallel }{2}:x,y\in B_X,\parallel x-y\parallel \ge ϵ\right\}.\hfill \end{array}$$

The characteristic of convexity, ${ϵ}_0\left(X\right)=sup\{ϵ\in [0,2]:{\delta }_X\left(ϵ\right)=0\}$, is defined in [2]. Space X is said to be uniformly convex if ${\delta }_X\left(ϵ\right)>0$ for all $ϵ\in (0,2]$, or equivalently ${ϵ}_0\left(X\right)=0$. The meaning of the modulus of convexity is as follows: if we take $x,y\in S_X$ far apart, then this modulus measures “how far” the middle point of the segment joining them must be from $S_X$. In [3,4], some scholars further studied a generalized modulus of convexity and obtained some interesting geometrical properties related to this modulus.

Recall that space X is uniformly smooth [5], if, for any $ϵ>0$, there exist $\delta >0$, such that if $x\in S_X$ and $\parallel y\parallel \le \delta$, then

$$\parallel x+y\parallel +\parallel x-y\parallel <2+ϵ\parallel y\parallel .$$

If we want to measure “how far” the same point can be from $S_X$, we are led to consider (for any $ϵ\in [0,2]$) the following parameter:

$$\begin{array}{cc}\hfill {\rho }_X\left(ϵ\right)& =sup\left\{1-\frac{\parallel x+y\parallel }{2}:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}\hfill \\ & =sup\left\{1-\frac{\parallel x+y\parallel }{2}:\parallel x\parallel \ge 1,\parallel y\parallel \ge 1,\parallel x-y\parallel \le ϵ\right\}.\hfill \end{array}$$

This modulus was considered first in [6]. It turns out to be a “modulus of smoothness” in the sense that the following holds. Space X is uniformly smooth if and only if

$$\underset{ϵ\to 0^{+}}{lim}\frac{{\rho }_X\left(ϵ\right)}{ϵ}=0.$$

Similarly, Baronti and Luigi-Papini also considered a generalized modulus of smoothness and obtained several interesting properties of this modulus in [7]. We refer the readers to the papers ([8,9,10,11,12,13,14,15,16,17,18,19]) for more information about the modulus of convexity and modulus of smoothness.

The concept and question on how to measure angles are old and interesting mathematical topics. Without them, the pyramids (and other monumental buildings of the ancient world) could not have been built. An angle function defined in a normed space preserves a certain orthogonality type when it attains the “Euclidean value”, $\frac{\pi }{2}$, only for orthogonal pairs of vectors. Now, let us recall the notion of P-orthogonality (see [20]): x is P-orthogonal to y, denoted by $x{\perp }_{P}y$, if and only if

$${\parallel x-y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2.$$

Motivated by P-orthogonality, the authors [21] defined the P-angle between two non-zero vectors $x,y\in X$ to be

$${\mathrm{ang}}_{\mathrm{p}}(x,y):=arccos\frac{{\parallel x\parallel }^2+{\parallel y\parallel }^2-{\parallel x-y\parallel }^2}{2\parallel x\parallel ·\parallel y\parallel }.$$

For more details on the P-angle, readers can consult [21,22].

Inspired by the excellent works mentioned above, in this paper, we define the angle modulus of convexity of X, ${\delta }_X^a\left(ϵ\right)$, and the angle modulus of smoothness of X, ${\rho }_X^a\left(ϵ\right)$ for $ϵ\in [0,2]$ by:

$$\begin{array}{cc}\hfill {\delta }_X^a\left(ϵ\right)& =inf\{cos{\mathrm{ang}}_{\mathrm{p}}(x,-y):x,y\in S_X,\parallel x-y\parallel =ϵ\}\hfill \\ & =inf\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\rho }_X^a\left(ϵ\right)& =sup\{cos{\mathrm{ang}}_{\mathrm{p}}(x,-y):x,y\in S_X,\parallel x-y\parallel =ϵ\}\hfill \\ & =sup\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}.\hfill \end{array}$$

The meaning of the modulus of angle is as follows: if we take $x,y\in S_X$ far apart, then this modulus measures “how large” the angle of x and $-y$ must be.

In Section 2, we study the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$. First, we present several equivalent forms of ${\delta }_X^a\left(ϵ\right)$ and obtain an important property of ${\delta }_X^a\left(ϵ\right)$ as a function via these equivalent forms. Meanwhile, we establish the relationship between ${\delta }_X^a\left(ϵ\right)$ and ${\delta }_X\left(ϵ\right)$. Similar to the classical modulus of convexity, we have that ${\delta }_X^a\left(ϵ\right)$ is a nondecreasing function on $[0,2]$. Further, some estimates of ${\delta }_X^a\left(ϵ\right)$ in terms of other constants will also be discussed. In particular, the relationships between ${\delta }_X^a\left(ϵ\right)$ and some geometric properties of X will be studied, including non-squareness, uniform non-squareness, uniform convexity, strict convexity, and uniform normal structure.

In Section 3, we study the angle modulus of smoothness ${\rho }_X^a\left(ϵ\right)$. First, we study the relationship between ${\rho }_X^a\left(ϵ\right)$ and $\rho \left(ϵ\right)$. Further, we obtain that ${\rho }_X^a\left(ϵ\right)$ is continuous on $(0,2)$. Meanwhile, some estimates of ${\rho }_X^a\left(ϵ\right)$ in terms of other constants will also be discussed. In particular, we obtain a characterization of uniform non-squareness in terms of ${\rho }_X^a\left(1\right)$.

2. The Angle Modulus of Convexity ${\mathbf{\delta }}_{\mathbf{X}}^{\mathit{a}}\left(\mathbf{ϵ}\right)$

In this section, we will study the new geometric constant ${\delta }_X^a\left(ϵ\right)$:

$$\begin{array}{cc}\hfill {\delta }_X^a\left(ϵ\right)& =inf\{cos{\mathrm{ang}}_{\mathrm{p}}(x,-y):x,y\in S_X,\parallel x-y\parallel =ϵ\}\hfill \\ & =inf\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}.\hfill \end{array}$$

To obtain the equivalent forms of the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$, we need the following lemma:

Lemma 1.

([3]). Let X be a real normed linear space, $x\in S_X$, $y\in B_X$. Then there exist $u,v\in X$, such that $u,v\in S_X$, $u-v=x-y$ and $\frac{\parallel u+v\parallel }{2}\ge \frac{\parallel x+y\parallel }{2}$.

Proposition 1.

Let X be a real normed linear space,

$${\delta }_X^{a^1}\left(ϵ\right)=inf\left\{1-\frac{{\parallel x+y\parallel }^2}{2}:x,y\in S_X,\parallel x-y\parallel \ge ϵ\right\}$$

and

$${\delta }_X^{a^2\left(ϵ\right)}=inf\left\{1-\frac{{\parallel x+y\parallel }^2}{2}:x,y\in B_X,\parallel x-y\parallel \ge ϵ\right\}.$$

Then ${\delta }_X^a\left(ϵ\right)={\delta }_X^{a^1}\left(ϵ\right)={\delta }_X^{a^2}\left(ϵ\right)$.

Proof. 

It is easy for us to see that ${\delta }_X^{a^2}\left(ϵ\right)\le {\delta }_X^{a^1}\left(ϵ\right)\le {\delta }_X^a\left(ϵ\right)$.

On the other hand, let $x,y\in S_X$ and $\parallel x-y\parallel \ge ϵ$, then assume that $\parallel x-y\parallel =bϵ(b\ge 1)$. By Hahn–Banach theorem, there exist $x^{*}\in S_{X^{*}}$, such that $x^{*}\left(\frac{x+y}{2}\right)=\frac{\parallel x+y\parallel }{2}$. Without loss of generality, assume that $x^{*}\left(x\right)\ge x^{*}\left(y\right)$, and let $x_1=x$ and $y_1=x+\frac{1}{b}(y-x)$, then $x_1\in S_X$, $y_1\in B_X$ and $\parallel x_1-y_1\parallel =ϵ$. From Lemma 1, there exist $u,v\in S_X$ and $u-v=x_1-y_1$, then we can obtain $\parallel u-v\parallel =ϵ$. Moreover, by Lemma 1, we can also have

$$\begin{array}{cc}\hfill \frac{{\parallel u+v\parallel }^2}{2}& =2·{\left(\frac{\parallel u+v\parallel }{2}\right)}^2\ge 2·{\left(\frac{\parallel x_1+y_1\parallel }{2}\right)}^2\ge 2·{\left(x^{*}\left(\frac{x_1+y_1}{2}\right)\right)}^2\hfill \\ & =2{\left(x^{*}\left(x\right)+\frac{1}{2b}{x}^{*}(y-x)\right)}^2=2{\left(\frac{1}{2b}{x}^{*}(2bx+y-x)\right)}^2\hfill \\ & =2{\left(\frac{1}{2b}{x}^{*}\left(bx\right)+\frac{1}{2b}{x}^{*}(bx+y-x)\right)}^2\hfill \\ & =2{\left(\frac{1}{2}{x}^{*}\left(x\right)+\frac{1}{2b}{x}^{*}((b-1)x+y)\right)}^2\hfill \\ & =2{\left(\frac{1}{2}{x}^{*}\left(x\right)+\frac{b-1}{2b}{x}^{*}\left(x\right)+\frac{1}{2b}{x}^{*}\left(y\right)\right)}^2\hfill \\ & \ge 2{\left(\frac{1}{2}{x}^{*}\left(x\right)+\frac{b-1}{2b}{x}^{*}\left(y\right)+\frac{1}{2b}{x}^{*}\left(y\right)\right)}^2\hfill \\ & =2{\left(\frac{1}{2}{x}^{*}\left(x\right)+\frac{1}{2}{x}^{*}\left(y\right)\right)}^2=\frac{{\parallel x+y\parallel }^2}{2},\hfill \end{array}$$

which implies that ${\delta }_X^a\left(ϵ\right)\le {\delta }_X^{a^1}\left(ϵ\right)$.

Next, let $x,y\in B_X$, and $\parallel x-y\parallel \ge ϵ$. Without loss of generality, assume that $\parallel y\parallel \le \parallel x\parallel \le 1$. Let $x_1=\frac{x}{\parallel x\parallel }$ and $y_1=\frac{y}{\parallel x\parallel }$, by Lemma 1, there exist $u,v\in S_X$, such that $u-v=x_1-y_1$, then we can obtain $\parallel u-v\parallel \ge ϵ$ and

$$\frac{{\parallel u+v\parallel }^2}{2}=2{\left(\frac{\parallel u+v\parallel }{2}\right)}^2\ge 2{\left(\frac{\parallel x_1+y_1\parallel }{2}\right)}^2\ge 2{\left(\frac{\parallel x+y\parallel }{2}\right)}^2=\frac{{\parallel x+y\parallel }^2}{2},$$

which implies that ${\delta }_X^{a^1}\left(ϵ\right)\le {\delta }_X^{a^2}\left(ϵ\right)$. This completes the proof. □

In the following, we will discuss an important property of the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$ as a function.

Proposition 2.

The function $\frac{{\delta }_X^a\left(ϵ\right)}{{ϵ}^2}$ is nondecreasing on $(0,1]$.

Proof. 

Let $0<{ϵ}_1\le {ϵ}_2\le 1$. Take $x,y\in S_X$ and $\parallel x-y\parallel ={ϵ}_2$, then we can obtain

$$2=\parallel 2x\parallel =\parallel (x+y)+(x-y)\parallel \le \parallel x+y\parallel +\parallel x-y\parallel =\parallel x+y\parallel +{ϵ}_2,$$

so we have $\parallel x+y\parallel \ge 2-{ϵ}_2$.

Now, assume that $t=\frac{{ϵ}_1}{{ϵ}_2}$, so we can obtain $0<t\le 1$. Let $z=\frac{x+y}{\parallel x+y\parallel }$, $u=tx+(1-t)z$, $v=ty+(1-t)z$, then it is obvious that $u,v\in B_X$ and $\parallel u-v\parallel ={ϵ}_1$. Thus, we can obtain

$$\begin{array}{cc}\hfill 1-\frac{{\parallel u+v\parallel }^2}{2}& =1-\frac{1}{2}{∥t(x+y)+2(1-t)\frac{x+y}{\parallel x+y\parallel }∥}^2\hfill \\ & =1-\frac{1}{2}{[t\parallel x+y\parallel +2(1-t)]}^2\hfill \\ & =1-\frac{1}{2}[t^2{\parallel x+y\parallel }^2{+4t(1-t)\parallel x+y\parallel +4(1-t)}^2]\hfill \\ & \le 1-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2-2t(1-t)(2-{ϵ}_2)-2{(1-t)}^2\hfill \\ & =1-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2-[-2t^2-2t{ϵ}_2+2t^2{ϵ}_2+2]\hfill \\ & =2t^2+2t{ϵ}_2(1-t)-1-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2\hfill \\ & \le 2t^2+2t(1-t)-1-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2=2t-1-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2\hfill \\ & \le t^2-\frac{1}{2}{t}^2{\parallel x+y\parallel }^2=t^2\left(1-\frac{1}{2}{\parallel x+y\parallel }^2\right),\hfill \end{array}$$

then from Proposition 1, we can obtain

$$\frac{{\delta }_X^a\left({ϵ}_1\right)}{{ϵ}_1^2}\le \frac{1-\frac{{\parallel u+v\parallel }^2}{2}}{{ϵ}_1^2}\le \frac{1-\frac{{\parallel x+y\parallel }^2}{2}}{{ϵ}_2^2}.$$

Since x, y are arbitrary, we have $\frac{{\delta }_X^a\left({ϵ}_1\right)}{{ϵ}_1^2}\le \frac{{\delta }_X^a\left({ϵ}_2\right)}{{ϵ}_2^2}$, and we complete the proof. □

In the following, we shall state the relationship between the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$ and the classical modulus ${\delta }_X\left(ϵ\right)$ and discuss the geometric properties of ${\delta }_X^a\left(ϵ\right)$.

Theorem 1.

Let ${\delta }_X\left(ϵ\right)$ be the modulus of convexity of X. Then,

$${\delta }_X^a\left(ϵ\right)=-2{(1-{\delta }_X\left(ϵ\right))}^2+1.$$

Proof. 

It is easy for us to obtain

$$\begin{array}{cc}\hfill {\delta }_X^a\left(ϵ\right)& =inf\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}\hfill \\ & =1-2sup{\left\{\frac{1}{2}\parallel x+y\parallel :x,y\in S_X,\parallel x-y\parallel =ϵ\right\}}^2\hfill \\ & =-2{(1-{\delta }_X\left(ϵ\right))}^2+1.\hfill \end{array}$$

This completes the proof. □

Below, we give the bounds of ${\delta }_X^a\left(ϵ\right)$.

Corollary 1.

Let X be a real normed linear space. Then $-1\le {\delta }_X^a\left(ϵ\right)\le -1+\frac{1}{2}{ϵ}^2$, and the second inequality attains equality if X is an inner product space.

Proof. 

Since $0\le {\delta }_X\left(ϵ\right)\le 1-{\left(1-\frac{{ϵ}^2}{4}\right)}^{\frac{1}{2}}$ and the second inequality attains equality if X is an inner product space for every $0\le ϵ\le 2$ (see [18]), then we can obtain $-1\le {\delta }_X^a\left(ϵ\right)\le -1+\frac{1}{2}{ϵ}^2$ and ${\delta }_X^a\left(ϵ\right)=-1+\frac{1}{2}{ϵ}^2$ if X is an inner product space from Theorem 1 by a simple calculation. □

Recall that space X is called non-square [23], if for every $x,y\in S_X$, $min\{\parallel x+y\parallel ,\parallel x-y\parallel \}<2$. Now, we discuss the relationship between ${\delta }_X^a\left(ϵ\right)$ and non-squareness.

Corollary 2.

Let X be a real normed linear space. If X is not non-square, then ${\delta }_X^a\left(2\right)=-1$.

Proof. 

Since X is not non-square, there exist $x,y\in S_X$, such that $\parallel x+y\parallel =\parallel x-y\parallel =2$. Thus, by Corollary 1, we can obtain

$$-1\le {\delta }_X^a\left(2\right)\le 1-\frac{1}{2}{\parallel x+y\parallel }^2=-1,$$

which implies the desired conclusion. □

From Theorem 1, ${\delta }_X^a\left(ϵ\right)>-1$ if and only if ${\delta }_X\left(ϵ\right)>0$. In [2], the space X is called uniformly convex if and only if ${\delta }_X\left(ϵ\right)>0$ for all $ϵ\in (0,2]$. Now, we give the following corollary:

Corollary 3.

X is uniformly convex if and only if ${\delta }_X^a\left(ϵ\right)>-1$ for any $ϵ\in (0,2]$.

Recall that space X is called uniformly non-square [24], if there exist a $\delta \in (0,1)$, such that for any $x,y\in S_X$, either $\frac{\parallel x+y\parallel }{2}\le 1-\delta$ or $\frac{\parallel x-y\parallel }{2}\le 1-\delta$. From Proposition 1 in [25], X is uniformly non-square if and only if ${\delta }_X\left(ϵ\right)>0$ for some $ϵ\in (0,2)$. By Theorem 1, ${\delta }_X^a\left(ϵ\right)>-1$ if and only if ${\delta }_X\left(ϵ\right)>0$, we can obtain the following corollary:

Corollary 4.

X is uniformly non-square if and only if there exist ϵ, $0<ϵ<2$, such that ${\delta }_X\left(ϵ\right)>-1$.

Recall that space X is called strictly convex, if for any $x,y\in S_X$ and $x\ne y$, then $\parallel x+y\parallel <2$. From Theorem 1, we can have ${\delta }_X^a\left(2\right)=1$ if and only if ${\delta }_X\left(2\right)=1$. Since X is strictly convex if and only if ${\delta }_X\left(2\right)=1$ (see Lemma 4 in [2]), then we can obtain the following corollary:

Corollary 5.

X is strictly convex if and only if ${\delta }_X^a\left(2\right)=1$.

Corollary 6.

${\delta }_X^a\left(ϵ\right)$ is nondecreasing on the interval $[0,2]$.

Proof. 

From Theorem 1, we can obtain

$${\delta }_X^a\left(ϵ\right)=-2{(1-{\delta }_X\left(ϵ\right))}^2+1,$$

then by Corollary 5 in [15], ${\delta }_X\left(ϵ\right)$ is nondecreasing on the interval $[0,2]$. Hence, we can obtain that ${\delta }_X^a\left(ϵ\right)$ is nondecreasing on the interval $[0,2]$. □

Proposition 3.

Let X be a real normed linear space, if X is not strictly convex, then there exist $\parallel x\parallel =\parallel y\parallel =1$ with $\parallel x-y\parallel ={ϵ}_0(0<{ϵ}_0\le 2)$, such that ${\delta }_X^a\left({ϵ}_0\right)=-1$.

Proof. 

Suppose that X is not strictly convex, there exist $x,y\in S_X$ with $\parallel x-y\parallel ={ϵ}_0$ for some ${ϵ}_0>0$, then we can obtain $\parallel x+y\parallel =2$, which implies ${\delta }_X^a\left({ϵ}_0\right)\le -1$. Hence, we can obtain ${\delta }_X^a\left({ϵ}_0\right)=-1$ by Corollary 1. □

Proposition 4.

Let X be a finite-dimensional real normed linear space. If ${\delta }_X^a\left(ϵ\right)=-1$ for some $0<ϵ\le 2$, then X is not strictly convex.

Proof. 

Assume that ${\delta }_X^a\left(ϵ\right)=-1$, there exist $x_n,y_n\in S_X$ such that $\parallel x_n-y_n\parallel =ϵ$ and

$$\underset{n\to \infty }{lim}\left(1-\frac{1}{2}\parallel x_n+y_n{\parallel }^2\right)=-1.$$

Since the unit sphere of finite-dimensional real normed linear space is compact, there exist $x_0,y_0\in S_X$, such that

$$1-\frac{1}{2}{\parallel x_0+y_0\parallel }^2=-1.$$

Thus, $\parallel x_0+y_0\parallel =2$ and $\parallel x_0-y_0\parallel =ϵ$, which implies that X is not strictly convex. □

The normal structure and the uniform normal structure play important roles in fixed point theory. Many articles have been devoted to investigating the relationship between the modulus of the Banach space X and uniform normal structure. Inspired by the excellent works, we studied the relationship between the angle modulus of convexity and uniform normal structure. Now, we present the following basic definition:

Definition 1

([26]). Let K be a non-singleton subset of a Banach space X, if K is closed, bounded as well as convex, then X holds the normal structure, whenever $r\left(K\right)<\mathrm{diam}\left(K\right)$ for every K, and consequently defined mathematically as is

$$\mathrm{diam}\left(K\right):=sup\{\parallel x-y\parallel :x,y\in K\}$$

and

$$r\left(K\right):=inf\{sup\{\parallel x-y\parallel :y\in K\}:X\in K\},$$

where $\mathrm{diam}\left(K\right)$ and $r\left(K\right)$ are, respectively, symbolized for the diameter as well as for the Chebyshev radius. A Banach space X is said to have a uniform normal structure if

$$inf\left\{\frac{\mathrm{diam}\left(K\right)}{r\left(K\right)}\right\}>1,$$

with $\mathrm{diam}\left(K\right)>0$.

Theorem 2.

Let X be a Banach space. If ${\delta }_X^a(1+ϵ)>-\frac{{ϵ}^2}{2}+2ϵ-1$ for any ϵ, $0\le ϵ\le 1$, then X has a uniform normal structure.

Proof. 

It is easy for us to obtain

$$-\frac{{ϵ}^2}{2}+2ϵ-1=-2{\left(1-\frac{ϵ}{2}\right)}^2+1,$$

and by Theorem 1, we can obtain

$${\delta }_X^a(1+ϵ)=-2{(1-{\delta }_X(1+ϵ))}^2+1.$$

Thus, assume that ${\delta }_X^a(1+ϵ)>-\frac{{ϵ}^2}{2}+2ϵ-1$, we can obtain ${\delta }_X(1+ϵ)>\frac{ϵ}{2}$. Therefore, we can have that X has a uniform normal structure by Theorem 8 in [8]. □

Next, we will give a certain example.

Example 1.

Let $X={\ell }_p$, and $ϵ\in [0,2]$. Then,

(1) If $p\ge 2$, we have ${\delta }_X^a\left(ϵ\right)=1-2{[1-{\left(\frac{ϵ}{2}\right)}^p]}^{\frac{2}{p}}$;

(2) If $1<p\le 2$, we have ${\delta }_X^a\left(ϵ\right)\ge 1-2{[1-{\left(\frac{ϵ}{2}\right)}^q]}^{\frac{2}{q}}$, where $q>0,\frac{1}{p}+\frac{1}{q}=1$.

Proof. 

(1) If $p\ge 2$, then we can obtain ${\delta }_X\left(ϵ\right)=1-{[1-{\left(\frac{ϵ}{2}\right)}^p]}^{\frac{1}{p}}$ from Theorem 2 in [19]. Thus, by Theorem 1, we have

$${\delta }_X^a\left(ϵ\right)=1-2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}}.$$

(2) For $1<p\le 2$, the following formula holds (see Theorem 2 in [1]):

$${2(\parallel x\parallel }^p+{\parallel y\parallel }^p{)}^{q-1}\ge {\parallel x+y\parallel }^q+{\parallel x-y\parallel }^q,$$

thus for any $\parallel x\parallel =\parallel y\parallel =1$ with $\parallel x-y\parallel =ϵ$, we can obtain

$$\frac{{\parallel x+y\parallel }^2}{2}\le 2{\left[1-{\left(\frac{ϵ}{2}\right)}^q\right]}^{\frac{2}{q}}.$$

Hence, we have ${\delta }_X^a\left(ϵ\right)\ge 1-2{\left[1-{\left(\frac{ϵ}{2}\right)}^q\right]}^{\frac{2}{q}}$. □

To study the relationship between the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$ and the modified von Neumann–Jordan constant $C_{NJ}^{{}^{\prime }}\left(X\right)$, we recall the definition of the constant $C_{NJ}^{{}^{\prime }}\left(X\right)$ as follows (see [27]):

$$C_{NJ}^{{}^{\prime }}\left(X\right)=sup\left\{\frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}:x,y\in S_X\right\}.$$

The authors also obtained that X is an inner product space if and only if $C_{NJ}^{{}^{\prime }}\left(X\right)=1$ in [27].

Proposition 5.

Let X be a real normed linear space. Then,

$$C_{NJ}^{\prime }\left(X\right)=sup\left\{\frac{1}{2}(1-{\delta }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\}.$$

Proof. 

For any $\parallel x\parallel =\parallel y\parallel =1$, we can obtain

$${\delta }_X^a(\parallel x-y\parallel )\le 1-\frac{1}{2}{\parallel x+y\parallel }^2.$$

Hence, we can obtain

$$\begin{array}{cc}\hfill \frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}& \le \frac{1}{2}(1-{\delta }_X^a(\parallel x-y\parallel \left)\right)+\frac{{\parallel x-y\parallel }^2}{4}\hfill \\ & \le sup\left\{\frac{1}{2}(1-{\delta }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\},\hfill \end{array}$$

from which it follows that $C_{NJ}^{\prime }\left(X\right)\le sup\left\{\frac{1}{2}(1-{\delta }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\}$.

On the other hand, let $0\le ϵ\le 2$. Then, for any $\mu >0$, there exist $x,y\in S_X$, such that $\parallel x-y\parallel =ϵ$ and

$$1-\frac{1}{2}{\parallel x+y\parallel }^2\le {\delta }_X^a\left(ϵ\right)+\mu .$$

Hence, we can obtain

$$C_{NJ}^{\prime }\left(X\right)\ge \frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}\ge \frac{1}{2}(1-{\delta }_X^a\left(ϵ\right)-\mu )+\frac{{ϵ}^2}{4}.$$

Since $\mu$ is arbitrary, then for any $0\le ϵ\le 2$, we can obtain

$$C_{NJ}^{\prime }\left(X\right)\ge \frac{1}{2}(1-{\delta }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}.$$

This completes the proof. □

Corollary 7.

Let X be a real normed linear space. Then ${\delta }_X^a\left(ϵ\right)\ge (1+\frac{{ϵ}^2}{2})-2C_{NJ}^{{}^{\prime }}\left(X\right)$ for any $0\le ϵ\le 2$, and the inequality attains equality if X is an inner product space.

From Theorem 1 and Proposition 5, we can obtain Corollary 8, which coincides with Proposition 4 in [27].

Corollary 8.

Let X be a real normed linear space. Then,

$$C_{NJ}^{\prime }\left(X\right)=sup\left\{\frac{{ϵ}^2}{4}+{(1-{\delta }_X\left(ϵ\right))}^2:0\le ϵ\le 2\right\}.$$

Alonso and Llorens-Fuster introduced the geometric constant $T\left(X\right)$ as follows [28]:

$$T\left(X\right)=sup\left\{\sqrt{\parallel x+y\parallel \parallel x-y\parallel }:x,y\in S_X\right\}.$$

Now, we will discuss the relationship between ${\delta }_X^a\left(ϵ\right)$ and $T\left(X\right)$.

Proposition 6.

Let X be a real normed linear space. Then,

$$T{\left(X\right)}^2=sup\left\{ϵ\sqrt{2(1-{\delta }_X^a\left(ϵ\right))}:0\le ϵ\le 2\right\}.$$

Proof. 

Let $s=sup\left\{ϵ\sqrt{2(1-{\delta }_X^a\left(ϵ\right))}:0\le ϵ\le 2\right\}$. For each $x,y\in S_X$, ${\delta }_X^a(\parallel x-y\parallel )\le 1-\frac{{\parallel x+y\parallel }^2}{2}$, which implies that

$$\parallel x-y\parallel \parallel x+y\parallel \le \parallel x-y\parallel \sqrt{2(1-{\delta }_X^a(\parallel x-y\parallel )}\le s,$$

so we can obtain $T{\left(X\right)}^2\le s$.

On the other hand, for $\eta >0$ arbitrarily small, there exist $0\le \overline{ϵ}\le 2$, such that

$$s-\eta <\overline{ϵ}\sqrt{2(1-{\delta }_X^a\left(\overline{ϵ}\right))}.$$

Since $0<T{\left(X\right)}^2<s$, then we can obtain $\overline{ϵ}>0$ according to the above inequality. Therefore, ${\delta }_X^a\left(\overline{ϵ}\right)<1-\frac{{(s-\eta )}^2}{2{\overline{ϵ}}^2}$, which implies that there exist $x_0,y_0\in S_X$, such that $\parallel x_0-y_0\parallel =\overline{ϵ}$ and $1-\frac{\parallel x_0+y_0{\parallel }^2}{2}<1-\frac{{(s-\eta )}^2}{2{\overline{ϵ}}^2}$; that is, $\parallel x_0+y_0\parallel >\frac{s-\eta }{\overline{ϵ}}$. Hence, we can obtain

$$T{\left(X\right)}^2\ge \parallel x_0-y_0\parallel \parallel x_0+y_0\parallel =\overline{ϵ}\parallel x_0+y_0\parallel >s-\eta .$$

Let $\eta \to 0^{+}$, we can obtain $T{\left(X\right)}^2\ge s$. This completes the proof. □

From Theorem 1 and Proposition 6, we can obtain Corollary 9, which coincides with Theorem 11 in [28].

Corollary 9.

Let X be a real normed linear space. Then $T^2\left(X\right)=2sup\{ϵ(1-{\delta }_X\left(ϵ\right)):0\le ϵ\le 2\}$.

3. The Angle Modulus of Smoothness ${\mathbf{\rho }}_{\mathbf{X}}^{\mathit{a}}\left(\mathbf{ϵ}\right)$

In this section, we will study the new geometric constant:

$$\begin{array}{cc}\hfill {\rho }_X^a\left(ϵ\right)& =sup\{cos{\mathrm{ang}}_{\mathrm{p}}(x,-y):x,y\in S_X,\parallel x-y\parallel =ϵ\}\hfill \\ & =sup\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}.\hfill \end{array}$$

First, we will provide an example.

Example 2.

Let $X={\ell }_p$ and $ϵ\in [0,2]$. If $1<p<2$, then ${\rho }_X^a\left(ϵ\right)=1-2{[1-{\left(\frac{ϵ}{2}\right)}^p]}^{\frac{2}{p}}$.

Proof. 

For any $x,y\in S_X$, by Clarkson’s inequality (see Theorem 1 in [19]), we have

$${\parallel x+y\parallel }^p+{\parallel x-y\parallel }^p\ge 2^p.$$

Let $ϵ\in [0,2]$, if $\parallel x-y\parallel =ϵ$, we can obtain $\frac{{\parallel x+y\parallel }^2}{2}\ge 2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}}$, which implies

$$1-\frac{{\parallel x+y\parallel }^2}{2}\le 1-2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}}.$$

Hence, we can obtain

$${\rho }_X^a\left(ϵ\right)\le 1-2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}}.$$

Now take $x=(\frac{ϵ}{2},\alpha )$ and $y=(-\frac{ϵ}{2},\alpha )$ with $\alpha \ge 0$, ${\alpha }^p=1-\frac{{ϵ}^p}{2^p}$, so we have $x,y\in S_X$ and $\parallel x-y\parallel =ϵ$, and we can obtain

$$\frac{{\parallel x+y\parallel }^2}{2}=2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}},$$

which implies that ${\rho }_X^a\left(ϵ\right)\ge 1-2{\left[1-{\left(\frac{ϵ}{2}\right)}^p\right]}^{\frac{2}{p}}.$ This completes the proof. □

In the following, we shall state the relationship between the angle modulus of convexity ${\rho }_X^a\left(ϵ\right)$ and the classical modulus ${\rho }_X\left(ϵ\right)$.

Proposition 7.

Let ${\rho }_X\left(ϵ\right)$ be the modulus of smoothness. Then ${\rho }_X^a\left(ϵ\right)=-2{(1-{\rho }_X\left(ϵ\right))}^2+1$.

Proof. 

It is easy for us to obtain

$$\begin{array}{cc}\hfill {\rho }_X^a\left(ϵ\right)& =sup\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =ϵ\right\}\hfill \\ & =1-2inf{\left\{\frac{1}{2}\parallel x+y\parallel :x,y\in S_X,\parallel x-y\parallel =ϵ\right\}}^2\hfill \\ & =-2{(1-\rho \left(ϵ\right))}^2+1.\hfill \end{array}$$

This completes the proof. □

Corollary 10.

If X is an inner product space, then ${\rho }_X^a\left(ϵ\right)=-1+\frac{{ϵ}^2}{2}$.

Proof. 

From [6], we can obtain ${\rho }_X\left(ϵ\right)=1-\sqrt{1-\frac{{ϵ}^2}{4}}$ if X is an inner product space. Then we can obtain ${\rho }_X^a\left(ϵ\right)=-1+\frac{{ϵ}^2}{2}$ by Proposition 7. □

Corollary 11.

If X is a real normed linear space, then ${\rho }_X^a\left(ϵ\right)$ is not decreasing on $(0,2)$.

Proof. 

From [16], ${\rho }_X\left(ϵ\right)$ is not decreasing on $(0,2)$. Then we can obtain that ${\rho }_X^a\left(ϵ\right)$ is not decreasing on $(0,2)$ by Proposition 7. □

Now, we will recall some facts concerning the geometry of the two-dimensional real-normed linear space. Assume that $x,y$ are linearly independent elements. The set $L=L(x,y)$ is defined in the way

$$L(x,y)=\{\alpha x+\beta y:\alpha \in \mathbb{R},\beta \ge 0\}$$

will be called a two-dimensional half-plane in space X. In this case, x is said to be the diametral element of the half-plane L.

In what follows, we shall need the following lemma:

Lemma 2.

Let ${ϵ}_1,{ϵ}_2$ be fixed positive numbers, such that ${ϵ}_1<{ϵ}_2<2$. Further assume that $y_1,y_2$ are linearly independent elements of the unit sphere $S_X$, such that $\parallel y_1-y_2\parallel ={ϵ}_2$. Then, in the half-plane $L(y_1,y_2)$, there exist elements $z_1,z_2\in S_X$, such that $\parallel z_1-z_2\parallel ={ϵ}_1$, then

$$\left(1-\frac{\parallel y_1+y_2{\parallel }^2}{2}\right)-\left(1-\frac{\parallel z_1+z_2{\parallel }^2}{2}\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}.$$

Proof. 

From Theorem 2 in [18], we have

$$\left(1-\frac{\parallel y_1+y_2\parallel }{2}\right)-\left(1-\frac{\parallel z_1+z_2\parallel }{2}\right)\le \frac{2\sqrt{5}+1}{2}·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}.$$

Thus, we can obtain

$$\begin{array}{cc}\hfill \left(1-\frac{\parallel y_1+y_2{\parallel }^2}{2}\right)-\left(1-\frac{\parallel z_1+z_2{\parallel }^2}{2}\right)& =\frac{\parallel z_1+z_2{\parallel }^2}{2}-\frac{\parallel y_1+y_2{\parallel }^2}{2}\hfill \\ & =\frac{1}{2}(\parallel z_1+z_2\parallel -\parallel y_1+y_2\parallel \left)\right(\parallel z_1+z_2\parallel +\parallel y_1+y_2\parallel )\hfill \\ & \le 2(\parallel z_1+z_2\parallel -\parallel y_1+y_2\parallel )\hfill \\ & =4\left[\left(1-\frac{\parallel y_1+y_2\parallel }{2}\right)-\left(1-\frac{\parallel z_1+z_2\parallel }{2}\right)\right]\hfill \\ & \le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}.\hfill \end{array}$$

This completes the proof. □

Next, we give an important property of ${\rho }_X^a\left(ϵ\right)$ as a function.

Proposition 8.

The modulus ${\rho }_X^a\left(ϵ\right)$ is continuous on the interval $(0,2)$.

Proof. 

Assume first that ${ϵ}_1$, ${ϵ}_2$ are fixed arbitrarily and $0<{ϵ}_1<{ϵ}_2<2$. Further, let $\eta >0$ be small enough. According to the definition of ${\rho }_X^a\left(ϵ\right)$, we can find $x,y\in S_X$, $\parallel x-y\parallel ={ϵ}_2$, such that

$${\rho }_X^a\left({ϵ}_2\right)-\eta \le 1-\frac{{\parallel x+y\parallel }^2}{2}\le {\rho }_X^a\left({ϵ}_2\right).$$

Next, in view of Lemma 2, we can find two elements $x_1$, $y_1$ in the half-plane $L(x,y)$, such that $x_1,y_1\in S_X$, $\parallel x_1-y_1\parallel ={ϵ}_1$ and

$$\left(1-\frac{{\parallel x+y\parallel }^2}{2}\right)-\left(1-\frac{\parallel x_1+y_1{\parallel }^2}{2}\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}.$$

Hence, we can obtain

$${\rho }_X^a\left({ϵ}_2\right)-\eta -\left(1-\frac{\parallel x_1+y_1{\parallel }^2}{2}\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}$$

or equivalently

$${\rho }_X^a\left({ϵ}_2\right)-\left(1-\frac{\parallel x_1+y_1{\parallel }^2}{2}\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}+\eta .$$

Thus, we can obtain

$${\rho }_X^a\left({ϵ}_2\right)-{\rho }_X^a\left({ϵ}_1\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}+\eta .$$

Since $\eta >0$ is arbitrarily small, then we can obtain

$${\rho }_X^a\left({ϵ}_2\right)-{\rho }_X^a\left({ϵ}_1\right)\le 2(2\sqrt{5}+1)·\frac{{ϵ}_2-{ϵ}_1}{2-{ϵ}_1}.$$

Hence, the function $ϵ\to {\rho }_X^a\left(ϵ\right)$ is continuous on the interval $(0,2)$ by Corollary 11 and the above inequality. □

The constants

$$J\left(X\right)=sup\{min\{\parallel x+y\parallel ,\parallel x-y\parallel \}:x,y\in S_X\}$$

$$S\left(X\right)=inf\{max\{\parallel x+y\parallel ,\parallel x-y\parallel \}:x,y\in S_X\}$$

are defined by Gao [29] in order to measure the degree of uniform non-squareness; they are usually called the James constant and Schäffer constant, respectively.

Later, the equivalent definitions of the James constant and Schäffer constant

$$J\left(X\right)=sup\{\parallel x+y\parallel :x,y\in S_X,\parallel x+y\parallel =\parallel x-y\parallel \}$$

$$S\left(X\right)=inf\{\parallel x+y\parallel :x,y\in S_X,\parallel x+y\parallel =\parallel x-y\parallel \}$$

are introduced by He and Cui [30].

Now, we will study the relationship between ${\rho }_X^a\left(1\right)$ and $S\left(X\right)$.

Proposition 9.

Let X be a real normed linear space. Then, ${\rho }_X^a\left(1\right)\le 1-\frac{1}{2}{S}^2\left(X\right)$.

Proof. 

For any $x,y\in S_X$ and $\parallel x-y\parallel =1$, then we can obtain

$$2=\parallel 2x\parallel =\parallel (x+y)+(x-y)\parallel \le \parallel x+y\parallel +\parallel x-y\parallel =\parallel x+y\parallel +1,$$

so we have $\parallel x+y\parallel \ge 1$. Thus, we can obtain

$$\begin{array}{cc}\hfill {\rho }_X^a\left(1\right)& =sup\left\{1-\frac{1}{2}{\parallel x+y\parallel }^2:x,y\in S_X,\parallel x-y\parallel =1\right\}\hfill \\ & =1-\frac{1}{2}inf\{\parallel x+y\parallel :x,y\in S_X{,\parallel x-y\parallel =1\}}^2\hfill \\ & =1-\frac{1}{2}inf\{max\{\parallel x+y\parallel ,\parallel x-y\parallel \}:x,y\in S_X{,\parallel x-y\parallel =1\}}^2\hfill \\ & \le 1-\frac{1}{2}inf\{max\{\parallel x+y\parallel ,\parallel x-y\parallel \}:x,y\in S_X{\}}^2\hfill \\ & =1-\frac{1}{2}{S}^2\left(X\right).\hfill \end{array}$$

This completes the proof. □

Corollary 12.

Let X be a real normed linear space and ${\rho }_X^a\left(ϵ\right)$ be the angle modulus of smoothness of X. Then, ${\rho }_X^a\left(1\right)\le \frac{1}{2}$.

Proof. 

Since $1\le S\left(X\right)\le \sqrt{2}$ (see [29,30]), then from Proposition 9, we can obtain

$${\rho }_X^a\left(1\right)\le 1-\frac{1}{2}{S}^2\left(X\right)\le 1-\frac{1}{2}=\frac{1}{2},$$

which completes the proof. □

In [24], James proposed the concept of uniform non-squareness and showed that the uniformly convex space is a uniformly non-square space, and a uniformly non-square space is reflexive. In order to give a characterization of uniform non-squareness in terms of ${\rho }_X^a\left(1\right)$, we first recall the equivalent form of a uniformly non-square space, namely, there exist some $ϵ>0$, such that $max\{\parallel x-y\parallel ,\parallel x+y\parallel \}\ge 1+ϵ$ for all $x,y\in S_X$ (see [7]).

Theorem 3.

Let X be a real normed linear space. Then X is not uniformly non-square if and only if ${\rho }_X^a\left(1\right)=\frac{1}{2}$.

Proof. 

By Proposition 9, we can obtain $S\left(X\right)\le \sqrt{2(1-{\rho }_X^a\left(1\right)})$. If ${\rho }_X^a\left(1\right)=\frac{1}{2}$, then we can obtain $S\left(X\right)\le 1$. Thus, we can obtain $S\left(X\right)=1$ from $1\le S\left(X\right)\le \sqrt{2}$. Hence, X is not uniformly non-square (see Corollary 15 in [30]).

Conversely, if X is not uniformly non-square, then for every $ϵ>0$, there exist a pair $x,y\in S_X$, such that $\parallel x-y\parallel <1+ϵ$ and $\parallel x+y\parallel <1+ϵ$, which implies that ${\rho }_X^a(1+ϵ)>1-\frac{{(1+ϵ)}^2}{2}=\frac{1-2ϵ-{ϵ}^2}{2}$. Since $ϵ$ is arbitrary and ${\rho }_X^a$ is continuous on the interval $(0,2)$ by Proposition 8, then we can obtain ${\rho }_X^a\left(1\right)\ge \frac{1}{2}$. Thus, we can obtain ${\rho }_X^a\left(1\right)=\frac{1}{2}$ by Corollary 12. □

From Proposition 7, we can obtain ${\rho }_X^a\left(1\right)=\frac{1}{2}$ if and only if ${\rho }_X\left(1\right)=\frac{1}{2}$. Hence, from Theorem 3 and Proposition 7, we can obtain Corollary 13, which implies Theorem 3 in [7].

Corollary 13.

For a real normed linear space X, the following statements are equivalent:

(1) X is not uniformly non-square;

(2) ${\rho }_X^a\left(1\right)=\frac{1}{2}$;

(3) ${\rho }_X\left(1\right)=\frac{1}{2}$.

The geometric constant

$$c_{NJ}^{\prime }\left(X\right)=inf\left\{\frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}:x,y\in S_X\right\}$$

has been studied by Gao [31] and Tanaka [32]. They also obtained that X is an inner product space if and only if $c_{NJ}^{\prime }\left(X\right)=1$. Now, we give the relationship between ${\rho }_X^a\left(ϵ\right)$ and $c_{NJ}^{\prime }\left(X\right)$.

Proposition 10.

Let X be a real normed linear space. Then,

$c_{NJ}^{\prime }\left(X\right)=inf\left\{\frac{1}{2}(1-{\rho }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\}$.

Proof. 

For any $x,y\in S_X$, ${\rho }_X^a(\parallel x-y\parallel )\ge 1-\frac{{\parallel x+y\parallel }^2}{2}$. Then, we can obtain

$$\begin{array}{cc}\hfill \frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}& \ge \frac{{\parallel x-y\parallel }^2}{4}+\frac{1}{2}(1-{\rho }_X^a(\parallel x-y\parallel \left)\right),\hfill \end{array}$$

which implies that

$$c_{NJ}^{\prime }\left(X\right)\ge inf\left\{\frac{1}{2}(1-{\rho }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\}.$$

On the other hand, let $0\le ϵ\le 2$. Then for any $\alpha >0$, there exist $x,y\in S_X$, such that $\parallel x-y\parallel =ϵ$ and

$${\rho }_X^a\left(ϵ\right)-\alpha \le 1-\frac{{\parallel x+y\parallel }^2}{2}.$$

Thus, we can obtain

$$\begin{array}{cc}\hfill c_{NJ}^{{}^{\prime }}\left(X\right)& \le \frac{{\parallel x+y\parallel }^2+{\parallel x-y\parallel }^2}{4}\hfill \\ & \le \frac{1}{2}(1-{\rho }_X^a\left(ϵ\right)+\alpha )+\frac{{\parallel x-y\parallel }^2}{4}.\hfill \end{array}$$

Since $\alpha$ is arbitrary, we can obtain

$$c_{NJ}^{\prime }\left(X\right)\le inf\left\{\frac{1}{2}(1-{\rho }_X^a\left(ϵ\right))+\frac{{ϵ}^2}{4}:0\le ϵ\le 2\right\},$$

which completes the proof. □

Corollary 14.

Let X be a real normed linear space, then ${\rho }_X^a\left(ϵ\right)\le (1+\frac{{ϵ}^2}{2})-2c_{NJ}^{{}^{\prime }}\left(X\right)$ for any $0\le ϵ\le 2$, and the inequality attains equality if X is an inner product space.

4. Conclusions

Inspired by the concept of P-angle, we introduced the angle modulus of convexity ${\delta }_X^a\left(ϵ\right)$ and the angle modulus of smoothness ${\rho }_X^a\left(ϵ\right)$ based on the classical modulus of convexity and modulus of smoothness, and established their relations. Certain related geometrical properties of the two moduli were analyzed. In addition, we described some relationships between the well-known geometric constants and the two parameters through some equalities and inequalities. Besides the geometric constants mentioned in the paper, what other important geometric constants are closely related to ${\delta }_X^a\left(ϵ\right)$ and ${\rho }_X^a\left(ϵ\right)$?