Abstract

The problem of packing spheres in Orlicz sequence space equipped with the p-Amemiya norm is studied, and a geometric characteristic about the reflexivity of is obtained, which contains the relevant work about and classical Orlicz spaces discussed by Rankin, Burlak, and Cleaver. Moreover the packing constant as well as Kottman constant in this kind of spaces is calculated.

1. Introduction and Preliminaries

The packing constant is an important and interesting geometric parameter for studying the geometric structure, isometric embedding, noncompactness, and reflexivity in Banach spaces [14]. Let be a Banach space. We denote by the unit ball of and by the unit sphere of . The packing constant of is the real number such that if , then an infinite number of spheres of radius can be packed in , and if , only a finite number of spheres can be done. It began in the 1950s studying the packing constant of special sequence spaces. Burlak et al. [1] proved that and for . Rankin found and in 1955 and 1958, respectively. In 1976, Cleaver discussed Orlicz sequence space equipped with the Orlicz norm under a strong condition, and he found upper and lower bounds of . In 1983, Ye investigated Orlicz sequence space equipped with the Luxemburg norm and obtained a formula for [5].

In this paper, an analogue for Orlicz sequence spaces equipped with the p-Amemiya norm is illustrated, and some useful definitions and lemmas are presented.

Definition 1 (see [1]). The packing constant of a Banach space is defined by It is obvious that , if .

Lemma 2 (see [2]). Let be an infinite-dimensional Banach space. Define which is called the Kottman constant of . Then

It is known that . Due to Riesz lemma, it can be summarised that for any infinite-dimensional Banach space . Finite-dimensional spaces have Kottman constant equal to zero. Furthermore, Elton and Odell in [6] proved that if is an infinite-dimensional Banach space, then there exists an such that . Consequently, . Hudzik proved that and for every nonreflexive Banach lattice [7].

Recall that a Banach space is said to be -convex (see [2]) if , for some , , where

Kottman [2] has proved that any -convex Banach space is reflexive.

The packing problem in Orlicz sequence spaces was investigated in [811]. The packing constant for Musielak-Orlicz sequence spaces and Cesaro sequence spaces have been calculated in [12, 13].

For any map , define A map is said to be an Orlicz function, if ; is not identically equal to zero; it is even and convex on the interval and left-continuous at .

For every Orlicz function , we define its complementary function by the formula The complementary function is also an Orlicz function. The convex modular is defined on (the space of all real sequences) by for any .

Definition 3 (see [1416]). The Orlicz sequence space is defined as the set The Luxemburg norm and the Orlicz norm are expressed as respectively. The Orlicz space equipped with the Luxemburg norm and the Orlicz norm are denoted by and , respectively.
For any and , define and define for all . Note that the functions and are convex. Moreover, the function is increasing on , for , but the function is increasing on the interval only.

Definition 4 (see [17, 18]). Let . For any , define the p-Amemiya norm by the formula The Orlicz space equipped with the p-Amemiya norm will be denoted by .
It is known that and . If ,  , then (See [17].)
Let be the right-hand side derivative of on and put . Define the function by and the functions and by It is obvious that for every and .
Set .

Definition 5 (see [14]). We say an Orlicz function satisfies the -condition (, for short) if there exist constants and such that and
For more details about Orlicz spaces, we refer the reader to [15, 16, 18, 19].

Lemma 6 (see [20]). Assume that , . Then, for any and , there exists such that for any there holds the implication

2. Main Results

Assume that , . Then, for any and , there exists a unique such that Set Then and . Denote

In the sequel, the packing constant is calculated, and the main results of this paper are proposed.

Theorem 7. If , , then and .

Proof. For any , there exists such that , so for all . Define Then have pairwise disjoint supports and   . For all and all , Then , so we have , since is arbitrary.
In the following, will be illustrated as an important part of our results.
For any sequence , which means that , , for any , then is bounded for all .
Since is bounded, there exists a subsequence such that is convergent, but is bounded, so there exists a subsequence such that is convergent. In a similar way, using the diagonal method, we can find a subsequence such that, for any , is convergent. Denoting and setting as , then as for all .
Let , , and . Then Since , then . For any , there exists , such that . Moreover, as for . So we have and, consequently, .
For the above , since is order continuous, there exists such that for . Take such that . And for , there exists such that . Then Take such that , and for , there exists such that . Then Analogously, we can find by induction a subsequence of and such that , such that, for any , Since , for all . Therefore, for any , Setting (for all ), then In this way, we get
For any , by the definition of , there exists such that , where satisfies the equality .
Setting and taking , we have Then Now we obtain that . If not, , we have whence This is a contradiction. Hence, So ; we get due to the arbitrariness of .

Theorem 8. If , , then .

Proof. Denote
Since , then ; so for , according to Riesz lemma, there exists satisfying . Then we have Since there exists a subsequence such that and Let Then for any , Due to the arbitrariness of , we have .

Lemma 9. If , , then

Proof. (1) Since and the norm convergence and the modular convergence are equivalent, there exists such that
For any and , we have so ; then
(2) If , then there exists such that For any and , we have ; then for any , we get whence . Moreover, according to the Young inequality since is arbitrary, we deduce that .

Ye et al. [21] have proved that Orlicz function space as well as Orlicz sequence space equipped with the Luxemburg norm is -convex if and only if it is reflexive; that is, satisfies the suitable -condition and -condition (i.e., the -condition at zero in the sequence case). We will prove now an analogous result for in terms of .

Theorem 10. If , , then .

Proof. If , then due to Lemma 6, for any , there exists such that where . According to Lemma 9, . Set . If , then there exists such that , so for all .
Since , we can find such that Let us notice that Thus, we have . Since is arbitrary, we obtain ; this is a contradiction. Therefore, and .

Corollary 11. If , then

Proof. For any , where and , .
In fact, since , then and . Set By , we get . Since , we have
Set . From the equation we deduce that . Therefore, We have . So for .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by the Provincial Education Department Fund (12531185) and supported by the National Natural Science Foundation of China (61203191).