Characterizations of ball-covering of separable Banach space and application

1.   Introduction and preliminaries

Let $(X, \|\cdot\|)$ denote a real Banach space and $X^{*}$ denote the dual space of $X$. Let $S(X)$ and $B(X)$ denote the unit sphere and unit ball of $X$, respectively. Let the set $B(x, r)$ denote the closed ball centered at point $x$ and of radius $r > 0$. Let $x_n\xrightarrow{w} x$ denote that the sequence $\{x_{n}\}_{n = 1}^{\infty}$ is weakly convergent to point $x$.

The geometric and topological properties of unit ball and unit sphere in Banach spaces play an important role in the geometry of Banach spaces. The geometry of Banach space can be said to be related to the unit ball and unit sphere of Banach space. Almost all geometric concepts are defined by the unit sphere, such as convexity and smoothness of Banach spaces. Not only that, many other research topics are related to the spherical representation of Banach space subsets, such as Mazur intersection property, noncompact measure and spherical topology problem. These topics have attracted the attention of many mathematicians since they were put forward. Through the tireless efforts of predecessors, many important results have been achieved in the study of these issues. These results often play an indispensable role in the in-depth study of the geometric properties of Banach spaces. It can be seen that the charm of the behavior of the ball family is amazing.

Starting with a different viewpoint, a notion of ball-covering property is introduced by Cheng^[1].

Definition 1.1. (see ^[1]) We call that $\mathfrak{B } = \{B(x_i, r_i)\}_{i\in I}$ is a ball-covering of $X$ if $S(X)\subset \cup_{i\in I}B(x_i, r_i)$ and $0 \not \in \cup_{i\in I}B(x_i, r_i)$. Moreover, if $I$ is a countable set, we call that $X$ has the ball-covering property.

Definition 1.2. (see ^[1]) A ball-covering $\mathfrak{B } = \{B(x_i, r_i)\}_{i\in I}$ is said to be $r$-off the origin if $\inf_{ x \in \cup \mathfrak{B }}\left\| x \right\| \geq r$.

It is easy to see that if $X$ is separable, then $X$ has the ball-covering property. However, if $X$ has the ball-covering property, then $X$ is not necessarily a separable space. In ^[1], Cheng proved that $l^{\infty}$ has the ball-covering property, but $l^{\infty}$ is not a separable spaces. In ^[2], Shang and Cui proved that if $X$ is a separable space and has the Radon-Nikodym property, then $X^{*}$ has the ball-covering property. As a corollary, Shang and Cui proved that if $M\in\nabla_{2}$, then Orlicz function space $L_{M}$ has the ball-covering property. This is an example of the ball-covering property of nonseparable function space. In 2021, Shang ^[3] studied the ball-covering property in dual space and proved the following theorem.

Theorem 1.3. The following statements are equivalent:

(1) The space $(X, \|\cdot\|)$ is a separable space;

(2) for every $0 < \varepsilon < 1$, there is a norm $\|\cdot\|_{1}$ of $X^{*}$ with $(1 + \varepsilon)^{-1}{\left\| {{x^{*}}} \right\|_1}\leq {\left\| {{x^{*}}} \right\|} \le \left\| {{x^{*}}} \right\|_1$ so that $\|\cdot\|_{1}$ is G${\hat{a}}$teaux differentiable on a dense subset of $X^{*}$ and $(X^{*}, \|\cdot\|_{1})$ has the ball-covering property.

In Theorem 1.3, we gave the ball-covering characteristics of separable spaces. In this paper, we further study the ball-covering characteristics of separable spaces. We first prove that a Banach space $(X, \|\cdot\|)$ is a separable space if and only if for every $\varepsilon\in$ $(0, 1)$, there exists a dense subset $G$ of $X^{*}$ and a $w^{*}$-lower semicontinuous norm $\|\cdot\|_{0}$ of $X^{*}$ such that

(1) the norm $\|\cdot\|_{0}$ is Frechet differentiable at every point of $G$ and $d_{F}\|x^{*}\|_{0}\in X$ is a $w^{*}$-strongly exposed point of $B(X^{**}, \|\cdot\|_{0})$ whenever $x^{*}\in G$;

(2) $\left(1+\varepsilon^{2}\right) {\left\| {{x^{***}}} \right\|_0} \le \left\| {{x^{***}}} \right\| \le \left(1 + \varepsilon \right){\left\| {{x^{***}}} \right\|_0}$ for each $x^{***}\in X^{***}$;

(3) there exists $\{x_{i}^{*}\}_{i = 1}^{\infty}\subset G$ such that ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $(X^{*}, \|\cdot\|_{0})$ is $(1+\varepsilon)^{-1}$-off the origin and $S(X^{*}, \|\cdot\|)\subset \cup_{i = 1}^{\infty}B({x_{i}^{*}}, {r_i})$.

Compared with Theorem 1.3, the result has the following two advantages

(1) The norm constructed in this result has better differentiability and better geometric properties than the norm constructed in Theorem 1.3;

(2) the closed ball sequence constructed in this result can cover the unit sphere of the original norm.

Moreover, we also prove that if the space $X$ is a weakly locally uniform convex space, then the necessary and sufficient condition that $X$ is a separable space is that for any $\alpha\in(0, 1)$, there is a sequence $\{ x_{i}^{*}\} _{i = 1}^\infty \subset X^{*}$ so that

(1) the ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $X^{*}$ is $\alpha$-off the origin;

(2) the norm of $X^{*}$ is G$\mathrm{\hat{a}}$teaux differentiable at every point of $\{ x_{i}^{*}\} _{i = 1}^\infty$;

(3) the point $d\|x_i^{*}\|$ belongs to $X$ for each $i\in N$.

As an application, we obtain that Orlicz sequence space $l_{M}$ has the ball-covering property. Other studies on ball covering properties can be found in ^{[4,5,6,7,8,9,10,11,12]}. First let us recall some definitions and lemmas that will be used in the further part of this paper.

Definition 1.4. (see ^[13]) Suppose that $D$ is an open subset of Banach space $X$, a continuous function $f$ is called G$\mathrm{\hat{a}}$teaux (Frechet) differentiable at $x\in D$ if there exists a functional $d_{G}f(x)\in X^{*}$ $(d_{F}f(x)\in X^{*})$ such that

Definition 1.5. (see ^[13]) A Banach space $X$ is said to be a G$\mathrm{\hat{a}}$teaux differentiability space if for every continuous convex function $f$ is G$\mathrm{\hat{a}}$teaux differentiable on a dense subset of $X$.

Definition 1.6. (see ^[14]) A Banach space is said to be smooth if every point of $S(X)$ is G$\mathrm{\hat{a}}$teaux differentiable point of norm.

Definition 1.7. (see ^[14]) A Banach space $X$ is said to be locally uniform convex if for every $x\in X$ and $\{ {x_n}\} _{n = 1}^\infty \subset S(X)$ with $\|x_n + x\|\to 2$ as $n\to\infty$ we have $\|x_n - x\|\to 0$ as $n\to\infty$.

Definition 1.8. (see ^[14]) We call that $X$ is a weakly local uniform convex space if for every $x\in X$ and $\{ {x_n}\} _{n = 1}^\infty \subset S(X)$ with $\|x_n + x\|\to 2$ as $n\to\infty$ we have $x_n\xrightarrow{w} x$ as $n \to \infty$.

It is easy to see that if $X$ is a locally uniform convex space, then $X$ is a weakly locally uniform convex space. We also know that if $X$ is a weakly locally uniform convex space, then $X$ is not necessarily a locally uniform convex space. Moreover, separable spaces have the norm of equivalent local uniform convexity.

Definition 1.9. (see ^[9]) Let $C^{*}$ be a subset of $X^{*}$. A point $x_{0}^{*}\in C^{*}$ is called a $w^{*}$-strongly exposed point of $C^{*}$ if there is a point $x_0\in$ $S(X)$ so that if $\{x_n^{*}\}_{n = 1}^{\infty}\subset C^{*}$ and $x_n^{*}(x_0)\to \sup_{x^{*}\in C^{*}} x^{*}(x_0)$, then $\|x_n^{*}- x_{0}^{*} \| \to 0$.

Definition 1.10. (see ^[14]) Let $C$ be a subset of $X$. A point $x_{0}\in C$ is called a strongly exposed point of $C$ if there is a point $x_0^{*}\in S(X^{*})$ so that if $\{x_n\}_{n = 1}^{\infty}\subset C$ and $x_0^{*}(x_n)\to \sup_{x\in C} x_0^{*}(x)$, then $\|x_n - x_{0} \|\to0$.

Lemma 1.11. (see ^[9]) Suppose that $C^{*}$ is a bounded subset of $X^{*}$. Then $\sigma_{C^{*}}$ is Frechet differentiable at point $x_0$ and $d\sigma_{C^{*}}(x_0) = x_0^{*}$ if and only if the point $x_0^{*}$ is a $w^{*}$-strongly exposed point of $C^{*}$ and exposed by $x_0$.

Lemma 1.12. (see ^[15]) Suppose that $C$ is a bounded closed set in $X$ and $C^{**} = \overline{C}^{w^{*}}$. If $x \in C$ is a strongly exposed point of $C$ and strongly exposed by $x^{*}\in X^{*}$, then $x$ is a $w^*$-strongly exposed point of $C^{**}$ and $w^*$-strongly exposed by $x^*$.

2.   A Characterization of ball-covering on separable space

Theorem 2.1. The space $(X, \|\cdot\|)$ is separable iff for every $\varepsilon\in$ $(0, 1)$, there is a dense subset $G$ of $X^{*}$ and a $w^{*}$-lower semicontinuous norm $\|\cdot\|_{0}$ of $X^{*}$ so that

(1) the norm $\|\cdot\|_{0}$ is Frechet differentiable at every point of $G$ and $d_{F}\|x^{*}\|_{0}\in X$ is a $w^{*}$-strongly exposed point of $B(X^{**}, \|\cdot\|_{0})$ whenever $x^{*}\in G$;

(2) $\left(1+\varepsilon^{2}\right) {\left\| {{x^{***}}} \right\|_0} \le \left\| {{x^{***}}} \right\| \le \left(1 + \varepsilon \right){\left\| {{x^{***}}} \right\|_0}$ for each $x^{***}\in X^{***}$;

(3) there exists $\{x_{i}^{*}\}_{i = 1}^{\infty}\subset G$ such that ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $(X^{*}, \|\cdot\|_{0})$ is $(1+\varepsilon)^{-1}$-off the origin and $S(X^{*}, \|\cdot\|)\subset \cup_{i = 1}^{\infty}B({x_{i}^{*}}, {r_i})$.

Proof. Necessity. (a) We first prove the condition (2). Since the space $(X, \|\cdot\|)$ is a separable Banach space, there exists an equivalent norm $\|\cdot\|_1$ such that the space $(X, \|\cdot\|_1)$ is a locally uniformly convex space. This implies that $\|\cdot\|$ and $\|\cdot\|_{1}$ of $X^{*}$ are two equivalent norms. Since the norms $\|\cdot\|$ and $\|\cdot\|_{1}$ of $X^{*}$ are two equivalent norms, there exists a real number $a\in (1, +\infty)$ such that

We pick a real number $\varepsilon\in (0, 1)$. Then we get that $\eta = \varepsilon/a\in (0, +\infty)$. Hence we define the symmetric bounded convex set

Since $\left\{ {{x^*} \in {X^*}:\left\| {{x^*}} \right\| \le 1} \right\}$ and $\left\{ {{x^*} \in {X^*}:\left\| {{x^*}} \right\|_{1} \le {\eta}/{2}} \right\}$ are two weak$^{*}$ compact convex sets, we obtain that ${C_0^{*}}$ is weak$^{*}$ compact. This implies that ${C_0^{*}}$ is a weak$^{*}$ bounded closed convex subset of $X^{*}$. Define the norm

in $X^{*}$. Then $\|\cdot\|_{0}$ is a $w^{*}$-lower semicontinuous norm of $X^{*}$ and $\left\| x^{*} \right\|_{0} < \left\| x^{*} \right\|$ for every $x^{*}\in X^{*}\backslash \{0\}$. Pick a point $x_0^{*}\in X^{*}$ such that $\|x_0^{*}\| = 1$. Then, by the formula $\left\| x_0^{*} \right\|_{0} < \left\| x_0^{*} \right\| = 1$, there exists $\lambda_0\in (0, +\infty)$ such that $\|(1+\lambda_0)x_0^{*}\|_{0} = 1$. We claim that $\|\lambda_0x_0^{*}\|_{1}\geq \eta/2$. In fact, suppose that $\|\lambda_0x_0^{*}\|_{1} < \eta/2$. Then we get that $\lambda_0x_0^{*}\in \mathrm{int} \left\{ x^* \in X^*:\| x^* \|_{1} \le {\eta}/{2} \right\}$. Therefore, by the formula $\|x_0^{*}\| = 1$ and the definition of ${C_0^{*}}$, we get that $(1+\lambda_0)x_0^{*}\in \mathrm{int}{C_0^{*}}$. Therefore, by the formula (2.2), we have $\|(1+\lambda_0)x_0^{*}\|_{0} < 1$, which contradicts $\|(1+\lambda_0)x_0^{*}\|_{0} = 1$. Moreover, we can assume without loss of generality that $2\varepsilon < 1/a^{2}$. Then, by the formulas $\|\lambda_0x_0^{*}\|_{1}\geq \eta/2$ and $\|x_0^{*}\| = 1$, we get that

Therefore, by the above inequalities and the formula $\|(1+\lambda_0)x_0^{*}\|_{0} = 1$, we have the following inequalities

Therefore, from the above inequalities, we have the following inequalities

On the other hand, we define the two norms

and

Then $(X^{**}, \|\cdot\|_{_0})$ is the dual space of $(X^{*}, \|\cdot\|_{_0})$ and $(X^{**}, \|\cdot\|)$ is the dual space of $(X^{*}, \|\cdot\|)$. Hence, we get that $\left\| x^{**} \right\|_{0}\geq\left(1+\varepsilon^{2}\right)\left\| x^{**} \right\|$ for every $x^{**} \in X^{**}$. Pick a point $x_0^{**} \in X^{**}$ such that the point $x_0^{**}$ is norm attainable on sphere $S(X^{*}, {\left\| \cdot \right\|_0})$. Hence exists a point $x_0^{*} \in {C_0^{*}}$ such that $x_0^{**}(x_0^{*}) = \left\|x_0^{**} \right\|_0$. Then, by the definition of $C_0^{*}$ and the formula $x_0^{*} \in {C_0^{*}}$, there exist two points

such that $x_0^{*} = y_0^{*}+ z_0^{*}$. Therefore, by the formula $x_0^{*} = y_0^{*}+ z_0^{*}$ and the formula $\|x^{**}\|_{1}\leq a \|x^{**}\|$, we have the following inequalities

Therefore, by formula (2.3) and the above inequalities, we get that

Therefore, by the Bishop-Phelps Theorem, we have the following inequalities

for every ${x^{**}} \in {X^{**}}$. Therefore, by the above inequalities, we obtain that

for every ${x^{***}} \in {X^{***}}$. Therefore, by the formula $\eta = \varepsilon/a$, we obtain that

for each ${x^{***}} \in {X^{***}}$. Moreover, it is easy to see that the norm $\|\cdot\|_{0}$ is a $w^{*}$-lower semicontinuous norm of $X^{***}$. Hence we get that the condition (2) is true.

(b) We will prove that there exists a dense subset $G$ of $X^{*}$ such that the norm $\|\cdot\|_{0}$ is Frechet differentiable on $G$ and $d_{F}\|\cdot\|_{0}(G)\subset X$. Define the set

Then we get that ${C_0}$ is a nonempty bounded closed convex set. Since the set ${C_0^{*}}$ is a weak$^{*}$ bounded closed convex subset of $X^{*}$, we get that

Then, using the Bishop-Phelps Theorem, we get that $\overline{A_0^{*}} = X^{*}$, where

Therefore, from Theorem 2.1 of ^[6], we obtain the following formulas $D_0^{*}\neq \emptyset$ and $\overline{D_0^{*}} = \left\{ {x^*} \in X^{*}:\|{x^*}\|_{0} = 1 \right\}$, where

Let $G = A_0^{*}$. Then we get that $G$ is a dense subset of $X^{*}$. We pick a point

Hence, we get that there exists a point $x_0\in C_{0}$ such that $\langle{y_{0}^*}, x_0\rangle = \sup \{ \langle {z^*}, x_0\rangle :{z^*} \in C_0^* \}$. Therefore, by the definition of $y_{0}^*$, we get that $y_0^*\in {C_0^{*}}$. Moreover, by the definition of $C_{0}^{*}$, there exist two point

such that $y_0^* = u_0^* + v_0^*$. Hence we have the following inequalities

Therefore, by the above inequalities, we have the following formula

Therefore, by the formula $v_0^{*} \in\left\{ {{x^*} \in {X^*}:\left\| {{x^*}} \right\|_{1} \le {\eta}/{2}} \right\}$, we obtain that the point $v_0^*$ is norm attainable on set $\{ {x} \in {X}:{\left\| {{x}} \right\|_1} \le 1\}$.

We next prove that the point $y_0^*$ is a Frechet differentiable point of norm $\|\cdot\|_{0}$ in $X^{***}$, i.e, the point $d_{F}\|y_0^*\|$ is a $w^{*}$-strongly exposed point of $B\left(X^{**}, \|\cdot\|_{0}\right)$. Define the closed convex set

It is well known that $y_{0}^{*}(x_{0}) = \|y_{0}^{*}\|_{0} = \|x_{0}\|_{0} = 1$. Pick a sequence $\{ x_n\} _{n = 1}^\infty \subset {C_0}$ such that $x_n({y_0^{*}}) \to 1$ as $n\to \infty$. Then, from the definition of ${C_0}$, we have the following equations

Since the set ${C_0^{*}}$ is a bounded set and $\mathrm{int}{C_0^{*}}\neq \emptyset$, we obtain that ${C_0}$ is a bounded subset of $X$. Therefore, by the formula $\{ x_n\} _{n = 1}^\infty \subset {C_0}$, we obtain that $\{ x_n\} _{n = 1}^\infty$ is a bounded sequence. Hence we can assume that without loss of generality that $\{ x_n(u_0^{*})\} _{n = 1}^\infty$ is a Cauchy sequence. Moreover, by the formula $y_0^* = u_0^* + v_0^*$, we get the following formula

Since $\{ x_n\} _{n = 1}^\infty \subset {C_0}$, by the formulas (2.5), we have the following inequalities

Therefore, from the above inequalities and formula (2.6), we obtain the following inequalities

Therefore, by the above inequalities, we have the following inequalities

Therefore, by the formulas $y_0^* = u_0^* + v_0^*$ and $v_0^*\in B(0, \eta/2)$, we obtain that

Since the sequence $\{ x_n\} _{n = 1}^\infty$ is a bounded sequence, we may assume without loss of generality that $\{ \|x_n\|\} _{n = 1}^\infty$ is a Cauchy sequence. This implies that

Moreover, there exists a sequence $\{ {k_n}\} _{n = 1}^\infty \subset {R^ + }$ such that $\left\| {{k_n}x_n} \right\| = \left\| {x_0} \right\|$ for all $n\in N$. Hence we have the following equations

Therefore, by the formula $\langle x_0, v_0^*\rangle = (\eta\left\| {x_0} \right\|_{1})/2$ and the formula (2.8), we obtain the following equations

Therefore, by the above equations, we have the following equation

Moreover, since $\left\| {{k_n}x_n} \right\|_{1} = \left\| {x_0} \right\|_{1}$ and $\|v_0^{*}\|_{1} = \eta/2$, by the above equation, we get the following equations

Since the space $(X, \|\cdot\|_{1})$ is a locally uniformly convex space, we get that $k_n x_n\to x_0$ as $n\to \infty$. Moreover, by the formula $x_n(y_0^{*}) \to x_0(y_{0}^{*})$, we obtain that $k_n\to1$ as $n\to \infty$. This implies that $\|x_n -x_0\|\to 0$ as $n\to \infty$. Hence the point $d_{F}\|y_0^{*}\|_{0}$ $\in X$ is a strongly exposed point of $B(X, \|\cdot\|_{0})$. Therefore, by Lemma 1.12, we get that $d_{F}\|y_0^{*}\|_{0}\in X$ is a $w^{*}$-strongly exposed point of $B\left(X^{**}, \|\cdot\|_{0}\right)$. Then, by the definition of $G$, we get that the norm $\|\cdot\|_{0}$ is Frechet differentiable at every point of $G$ and the point $d_{F}\|x^{*}\|_{0}\in X$ is a $w^{*}$-strongly exposed point of $B\left(X^{**}, \|\cdot\|_{0}\right)$ whenever $x^{*}\in G$.

(c) Let $p_{0}(x^{***}) = \|x^{***}\|_0$ for each $x^{***}\in X^{***}$. Then, by the proof of (b), we know that if $x^{*}$ is norm attainable on $S(X, {\left\| \cdot \right\|_0})$, then $x^{*}$ is a Frechet differentiable point of $p_{0}$ and $d_{F} p_{0}({x^*}) \in {X}$. Then we get that

Since the space $X$ is separable, we get that every subset of $X$ is separable. Let

Then there exists a sequence $\{ {x_n}\} _{n = 1}^\infty \subset A_0$ such that $\overline{\{ {x_n}\} _{n = 1}^\infty} = A_0$. Therefore, by the formula $\overline{\{ {x_n}\} _{n = 1}^\infty} = A_0$, we have the following equations

Pick a point $x_{0}^{*}\in X^{*}$ such that $\|x_{0}^{*}\| = 1$. Then, from the proof of (a), we obtain that $\|x_{0}^{*}\|_{0}\leq \|x_{0}^{*}\|\leq (1+\varepsilon/2) \|x_{0}^{*}\|_{0}$. Then we get that $(1+\varepsilon/2)^{-1}\leq \|x_{0}^{*}\|_{0}$. This implies that $\|(1+\varepsilon/2)x_{0}^{*}\|_{0}\geq 1$. Define the set

Then we get that $S(X^{*}, \left\| \cdot \right\|)\subset W_{0}$ and $S(X^{*}, \left\| \cdot \right\|_{0})\subset W_{0}$. Pick a point $z^{*}\in X^{*}$ such that $\|z^{*}\| = 1$. Then $\|z^{*}\|_{0}\geq (1+\varepsilon/2)^{-1}$. Therefore, by the formulas (2.9) and (2.10), we have the following inequalities

Therefore, by the above inequalities and the definition of $W_0$, we obtain that

Therefore, from the proof of (b) and the definition of $A_{0}$, there exists $\{ x_n^{*} \} _{n = 1}^\infty \subset S(X^{*})$ so that $\|x_n^{*} \|_{0} = 1$ and $x_n = d_{F}p_{0}({x_n^*})$ for every $n\in N$. Hence we have $\{ x_n^{*} \} _{n = 1}^\infty \subset G$. Moreover, we can assume without loss of generality that $(1+\varepsilon) < 2$. Define the closed ball sequences

in $(X^{*}, \|\cdot\|_{0})$. Then, by the formula $\|x_i^{*}\|_{0} = 1$, we get that if $y^{*}\in B_{i, m}^{0}$, then

Hence we get that $B_{i, m}^{0}$ has a positive distance ${1}/\left(1+\varepsilon\right)$ from the origin. Suppose that there exists a point $y^{*}\in W_0$ such that for every $i\in N$ and $m\in N$, we have $y^{*}\notin B_{i, m}^{0}$. Moreover, by the formula (2.11), there exists a natural number $n_0\in N$ such that $\langle y^{*}, x_{n_0} \rangle = \eta > 1/(1+\varepsilon)$. Define the hyperplane

in $X^{*}$. Hence there exists a point $h_{n_0}^{*}\in H_{n_0}$ such that $y^{*} = \eta x_{n_0}^{*}+ h_{n_0}^{*}$. Therefore, by the formulas $y^{*}\notin B_{i, m}^{0}$ and $y^{*} = \eta x_{n_0}^{*}+ h_{n_0}^{*}$, we get that

Therefore, by the above inequalities, we have the following inequalities

where $t = 1/\left(m - \eta \right)$. Moreover, since the point $x_{n_0}^{*}$ is a G$\mathrm{\hat{a}}$teaux differentiable point of norm $\|\cdot\|_{0}$ in $X^{*}$, form the above inequalities and the formula $h_{n_0}^{*}\in H_{n_0}$, we have the following inequalities

this is a contradiction. This implies that

Hence, there exists a sequence $\{x_{i}^{*}\}_{i = 1}^{\infty}\subset G$ such that ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $(X^{*}, \|\cdot\|_{0})$ is $(1+\varepsilon)^{-1}$-off the origin and $S(X^{*}, \|\cdot\|)\subset \cup_{i = 1}^{\infty}B({x_{i}^{*}}, {r_i})$.

Sufficiency. Since the space $(X^{*}, \|\cdot\|_{0})$ has the ball-covering property, we get that the space $X$ is separable, which finishes the proof.

Next, we will study what conditions can guarantee that the dual space of a separable space has the ball-covering property.

Theorem 2.2. Suppose that the space $X$ is weakly locally uniform convex. Then $X$ is a separable space if and only if for every $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}^{*}\} _{i = 1}^\infty$ such that

(1) the ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $X^{*}$ is $\alpha$-off the origin;

(2) the norm of $X^{*}$ is G${\hat{a}}$teaux differentiable at every point of $\{ x_{i}^{*}\} _{i = 1}^\infty$;

(3) the point $d_{G}\|x_i^{*}\|$ belongs to $X$ for each $i\in N$.

Proof. Necessity. (a) We first will prove that the norm of $X^{*}$ is G$\mathrm{\hat{a}}$teaux differentiable on a dense subset of $S(X^{*})$. In fact, by the Bishop-Phelps Theorem, we get that $A_0^{*}\subset S(X^{*})$ and $\overline{A_0^{*}} = S(X^{*})$, where

We claim that every point of $A_0^{*}$ is a G$\mathrm{\hat{a}}$teaux differentiable point of $X^{*}$. In fact, pick a point $x_0^{*}\in A_0^{*}$. Since the space $X$ is weakly locally uniform convex, there exists an unique point $x_0\in S(X)$ such that $x_0^{*}(x_0) = 1$. Suppose that there exists a functional $x_0^{**}\in S(X^{**})$ such that $x_0^{**}(x_0^{*}) = 1$ and $x_0^{**}\neq x_0$. Then there exists a weak$^{*}$ neighbourhood $V$ of origin in $X^{**}$. such that

Moreover, for every natural number $n\in N$, we define the weak$^{*}$ neighbourhood

of $x_{0}^{**}$ in $X^{**}$. Therefore, from the Goldstine Theorem, there is a point $x_n\in B(X)$ such that $x_n\in \left(x_0^{**}+V \right)\cap V_n$ for all $n\in N$. Hence we have $x_0^{*}(x_n)\to 1$ as $n \to \infty$. Therefore, by the formula $x_0^{*}(x_0) = 1$, we have

Since the space $X$ is weakly locally uniform convex, we obtain that $x_n\xrightarrow{w} x_0$ as $n \to \infty$. Since $V$ is a weak$^{*}$ neighbourhood of origin in $X^{**}$, we can assume that $x_n \in x_0+V$. However $x_n\in \left(x_0^{**}+V \right)\cap V_n$ for each $n\in N$, which contradicts the formula (2.12). This implies that every point of $A_0^{*}$ is G$\mathrm{\hat{a}}$teaux differentiable point of $X^{*}$.

(b) Let $\alpha\in\left(0, 1\right)$. Pick $\varepsilon\in\left(0, 1\right)$. Moreover, since the space $X$ is a separable space, there exists a sequence ${\{ {x_n}\} _{n = 1}^\infty} \subset S(X)$ such that $\overline{{\{ {x_n}\} _{n = 1}^\infty}} = S(X)$. Then we have the following equations

Therefore, from the previous proof, we define the following the set

Therefore, from the previous proof, we obtain that the norm of $X^{*}$ is G$\mathrm{\hat{a}}$teaux differentiable at every point of $G$ and every point of $G$ is norm attainable on set $S(X)$. Hence, for every natural number $n\in N$, there exists a point $x_n^{*}\in G$ with $\|x_n^{*}\| = 1$ such that $x_n^{*}(x_n) = 1$. Hence we define a sequence ${\{ {x_n^{*}}\} _{n = 1}^\infty}\subset S(X^{*})$.

Pick a point $y^{*}\in$ $S(X^{*})$. Then, by the formula (2.14), it is easy to see that there exists a natural number $n_0\in N$ such that $1\geq \langle y^{*}, x_{n_0} \rangle > 1/\left(1+2\varepsilon\right)$. We define the following hyperplane

in $X^{*}$. Then, by formula (2.13), there exists a point $h_{n_0}^{*}\in H_{n_0}$ and a real number $\eta\in(0, +\infty)$ such that $y^{*} = \eta x_{n_0}^{*}+ h_{n_0}^{*}$. Then, by the inequalities $1\geq \langle y^{*}, x_{n_0} \rangle >$ $1/\left(1+2\varepsilon\right)$ and $y^{*} = \eta x_{n_0}^{*}+ h_{n_0}^{*}$, we have the following inequalities

Moreover, since $\langle x_{n_0}^{*}, x_{n_0} \rangle = 1$, by formula (2.15) and the formula $y^{*} = \eta x_{n_0}^{*}+ h_{n_0}^{*}$, we have the following inequalities

Hence, for each natural numbers $i$ and $m$, we define the closed ball sequences

in $X^{*}$. Suppose that $y^*\notin B_{i, m}$ for all $i\in N$ and $m\in N$. Then for every natural numbers $m\in N$, we obtain that

Therefore, by the above inequalities, we have the following inequalities

where $t = 1/\left(m - \eta \right)$. Moreover, since the point $x_{n_0}^{*}$ is a G$\mathrm{\hat{a}}$teaux differentiable point of norm of $X^{*}$ and $\langle x_{n_0}^{*}, x_{n_0} \rangle = 1$, we have $d_{G}\|x_{n_0}^{*}\| = x_{n_0}\in X$. Therefore, by the formula (2.16), we have the following inequalities

this is a contradiction. Hence we have the following formula

Since $\varepsilon\in (0, 1)$, we can assume without loss of generality that $(1+2\varepsilon)^{-1}-\alpha > 0$. Pick a point $y^*\in B_{i, m}$. Since $(1+2\varepsilon)^{-1} > \alpha$, by the formula $\|x_i^{*}\| = 1$ and the triangle inequality, we have the following inequalities

Therefore, by the formula $x_{i}^{*} \in G$, we obtain that for every $0 < \alpha < 1$, there is a sequence $\{ x_{i}^{*}\} _{i = 1}^\infty$ of norm G$\mathrm{\hat{a}}$teaux differentiable points such that the ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $X^{*}$ is $\alpha$-off the origin. Hence, we get that the conditions (1) and (2) are true. Moreover, from the previous proof, it is that $d_{G}\|x_i^{*}\| = x_i\in X$ for every $i\in N$. The condition (3) is true.

Sufficiency. Since the space $X^{*}$ has the ball-covering property, we get that the space $X$ is separable, which finishes the proof.

Corollary 2.3. Suppose that the space $X$ is locally uniform convex. Then $X$ is a separable space iff for every $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}^{*}\} _{i = 1}^\infty$ such that

(1) the ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $X^{*}$ is $\alpha$-off the origin;

(2) the norm of $X^{*}$ is Frechet differentiable at every point of $\{ x_{i}^{*}\} _{i = 1}^\infty$;

(3) the point $d_{F}\|x_i^{*}\|$ belongs to $X$ for each $i\in N$.

Proof. By the proof of Theorem 2.1 and Theorem 2.2, we get that Corollary 2.3 is true, which finishes the proof.

3.   Application to Orlicz sequence spaces

In this section, we use the results of the ball-covering of Banach space to study the ball-covering theory of Orlicz sequence space. On the other hand, since the Orlicz sequence space is a kind of specific Banach space, we can get a more perfect conclusion on the Orlicz sequence space than the general Banach space.

Definition 3.1. (see ^[14]) A function $M: R\to R$ is said to be an Orlicz function if it has the following properties:

(1) $M$ is even, continuous, convex and $M(0) = 0$;

(2) $M(u) > 0$ for all $u > 0$;

(3) $\lim_{u\to 0}M(u)/u = 0$ and $\lim_{u\to \infty}M(u)/u = \infty$.

Definition 3.2. (see ^[14]) Let $M$ be an Orlicz function, $p$ be the right derivative of $M$, and $q(s) = \sup \{t:p(t)\leq s\}$. Then, we call that

is the complementary function of $M$.

By ^[14], we know that if the function $M$ is an Orlicz function, then the complementary function $N$ of $M$ is an Orlicz function. Moreover, by ^[14], we know that the complementary function of $N$ is $M$. Hence we say that $M$ and $N$ are complementary to each other (see ^[14]).

Definition 3.3. (see ^[14]) An Orlicz function $M$ is said to be satisfies condition $\Delta_{2}$ if there exist $K > 2$ and $u_0\geq0$ such that

In this case, we write $M\in \Delta_{2}$ or $N\in \nabla_{2}$, where $N$ is the complementary function of $M$.

For any sequence $x = (x(1), x(2), ...)$, we define its modular by

Then the Orlicz sequence space $l_{M}$ and its subspace $h_{M}$ are defined as follows:

For each $x\in l_{M}$, we define the Luxemburg norm

or the Orlicz norm

in $l_{M}$ (see ^[14]). It is well known that $l_{M}$ and $l_{M}^{0}$ are two Banach spaces (see ^[14]). Moreover, we know that $h_{M}$ is a closed subspace of $l_{M}$ and $h_{M}^{0}$ is a closed subspace of $l_{M}^{0}$ (see ^[14]). It is well known that $h_{M}$ and $h_{M}^{0}$ are separable spaces (see ^[14]). Moreover, it is well known that $(h_{N}^{0})^{*} = l_{M}$ and $(h_{N})^{*} = l_{M}^{0}$(see ^[14]). It is well known that $l_{M}(l_{M}^{0})$ is separable if and only if $M\in \Delta_{2}$ (see ^[14]).

Theorem 3.4. Suppose that $M\in \Delta_{2}$ or $M\in \nabla_{2}$. Then for any $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}\} _{i = 1}^\infty \subset l_{M}$ of norm G${\hat{a}}$teaux differentiable points such that

(1) the ball-covering $\{ B({x_{i}}, {r_i})\} _{i = 1}^\infty$ of $l_{M}$ is $\alpha$-off the origin;

(2) the norm of $l_{M}$ is G${\hat{a}}$teaux differentiable at every point of $\{ x_{i}\} _{i = 1}^\infty$.

Proof. Suppose that $M\in \Delta_{2}$. Then the space $l_{M}$ is a separable space. Hence $l_{M}$ is a weak Asplund space. It is easy to see that for any $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}\} _{i = 1}^\infty \subset l_{M}$ of norm G$\mathrm{\hat{a}}$teaux differentiable points such that (1) the ball-covering $\{ B({x_{i}}, {r_i})\} _{i = 1}^\infty$ of $l_{M}$ is $\alpha$-off the origin; (2) the norm of $l_{M}$ is G$\mathrm{\hat{a}}$teaux differentiable at every point of $\{ x_{i}\} _{i = 1}^\infty$.

Suppose that $M\in \nabla_{2}$. Then we get that $N \in\Delta_{2}$. This implies that $h_{N}^{0}$ has the Radon-Nikodym property. Therefore, by $(h_{N}^{0})^{*} = l_{M}$, we get that the norm of $l_{M}$ are G$\mathrm{\hat{a}}$teaux differentiable on a dense subset of $X^{*}$. Hence, the norm of $l_{M}$ are G$\mathrm{\hat{a}}$teaux differentiable on a dense subset of $S(X^{*})$. Therefore, from the proof of Theorem 2.2, we get that for every $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}^{*}\} _{i = 1}^\infty$ such that (1) the ball-covering $\{ B({x_{i}^{*}}, {r_i})\} _{i = 1}^\infty$ of $X^{*}$ is $\alpha$-off the origin; (2) the norm of $X^{*}$ is G$\mathrm{\hat{a}}$teaux differentiable at every point of $\{ x_{i}^{*}\} _{i = 1}^\infty$, which finishes the proof.

Theorem 3.5. Suppose that $M\in \Delta_{2}$ or $M\in \nabla_{2}$. Then for any $\alpha\in(0, 1)$, there exists a sequence $\{ x_{i}\} _{i = 1}^\infty \subset l_{M}^{0}$ of norm G${\hat{a}}$teaux differentiable points such that

(1) the ball-covering $\{ B({x_{i}}, {r_i})\} _{i = 1}^\infty$ of $l_{M}^{0}$ is $\alpha$-off the origin;

(2) the norm of $l_{M}^{0}$ is G${\hat{a}}$teaux differentiable at every point of $\{ x_{i}\} _{i = 1}^\infty$.

Proof. Similar to the proof of Theorem 3.4, we obtain that Theorem 3.5 is true, which finishes the proof.

Theorem 3.6. Let $l_{M}$ be an Orlicz sequence space. Then $l_{M}$ has a ball-covering $\{ B({x_{i}}, {r_i})\} _{i = 1}^\infty$ such that $\sup_{i\in N}r_{i} < +\infty$.

Proof. Let $Q$ denote rational number set. Then, for every natural number $n\in N$, we define the set

Define the set $Q_{0} = \cup_{n\in N}Q_{n}$. Then the set $Q_{0}$ is a countable set. Let $A = \{x\in Q_{0}: \|x\|\in [2,4] \}$. Since the set $A$ is a countable set, we can order it as a sequence $A = \{ x_{i}\} _{i = 1}^\infty$. Then we define the closed ball sequences

Pick a point $x\in S(l_{M})$. Then we obtain that $\rho_{M}(x)\leq1$. Let $x = (x(1), x(2), ...)$. Then there exists a natural number $i_{0}\in N$ such that

Therefore, by the above inequalities, we have the following inequalities

Since $\rho_{M}(x/3) > 0$, by the above inequalities, we obtain that

Moreover, we can assume without loss of generality that $x(i_0+1)\in[0, +\infty)$. We pick a real number $u_0\in (0, +\infty)$ such that $\|x_0\| = 3$, where

Therefore, by the formula (3.1) and $\rho_{M}(x)\leq 1$, we obtain that $u_0-x(i_0+1) > 0$. Therefore, by the formulas $x(i_0+1)\in[0, +\infty)$ and $u_0\in (0, +\infty)$, we obtain the following inequality

Since $\rho_{M}(x)\leq1$, by the formula $\|x_0\| = 3$ and the definition of $x_0$, we get that $\rho_{M}\left({x_0}/{3}\right) = 1$. Therefore, by the formulas (3.1)-(3.2) and the definition of $x_0$, we have the following inequalities

Since $\rho_{M}(x)\leq1$ and $\rho_{M}\left({x_0-x}\right) < +\infty$, there is a real number $\lambda_{0}\in (0, 3)$ such that $\rho_{M}\left({(x_0-x)}/{\lambda_{0}}\right)$ $= 1$. Hence we obtain that $\|x_0-x\| = \lambda_{0}$. Moreover, by the definition of $\{ x_{i}\} _{i = 1}^\infty$ and the formula $\|x_0\| = 3$, there exists a point $x_{j}\in \{ x_{i}\} _{i = 1}^\infty$ such that $\|x_{j}\|\geq 3$ and $\|x_{j}-x_0\| < (3-\lambda_{0})/2$. Then, by $\lambda_{0}\in (0, 3)$, we get that

Therefore, by the above inequalities, there is a natural number $m_0\in N$ so that

This implies that $l_{M}$ has a ball-covering $\{ B_{i, m}\} _{i = 1, \; m = 1}^\infty$ such that $\sup_{i\in N}r_{i} < 4 < +\infty$, which finishes the proof.

Corollary 3.7. Orlicz sequence space $l_{M}$ has the ball-covering property.

Acknowledgments

This research is supported by "China Natural Science Fund under grant 12271121" and "China Natural Science Fund under grant 11561053".

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The authors declare there is no conflict of interest.