In this section, our main aim is to calculate the weakly convergent sequence coefficientfor an Orlicz sequence space and further discuss the fixed point property of thisspace.
Let X denote a reflexive infinite dimensional Banach space, without Schurproperty automatically. For each sequence in X, we define the asymptoticdiameter and asymptotic radius respectively by
The weakly convergent sequence coefficient concerned with normal structure is animportant geometric parameter. It was introduced by Bynum  as follows.
is the supremum of the set of all numbersM with the property that for each weakly convergent sequence with asymptotic diameter A, thereis some y in the closed convex hull of the sequence such that
A sequence in X is said to be asymptoticequidistant if
In this paper, we use the following equivalent definition of . This definition was introduced in , where it was proved that
It is obvious that (see). A Banach space X is saidto have weakly uniform normal structure provided . See for further information about thiscoefficient.
For ,and . A formula for calculatingthe weakly convergent sequences of reflexive Orlicz and Musielak-Orlicz sequence spacesequipped with the Luxemburg or Amemiya norm is found in [21, 23], respectively.
Theorem 3.1 If, ,then
Proof Let . For any , thereexists such that
Then there exist andsuch that .Define
where have pairwise disjoint supports. Then() for all and all.According to the same method as in the proof of Theorem 2.2, we have weakly,
Then ,so ,since ε is arbitrary, we have .
On the other hand, let in be an arbitrary asymptoticequidistant sequence such that weakly.
Since , then by Lemma 1.5 for,there exists such that
Let and pick such that and choosesuch that . By as for ,we can find with such that . And so on,by induction, we find the sequence and of natural numbers with ,satisfying
If , then , so .
If , set , then
(3) If , then
hence we get again. Consequently . By thearbitrariness of in , it follows that. □
Theorem 3.2 Letand.hasweakly uniform normal structure if and only if.
Proof Necessity. If , by Theorem 3.1,
Assume , then for anythere exist and satisfying.By and , there exists such that. Hence,
This is a contradiction.
Sufficiency. If not, , then for any thereexists such that
We can find such that
Take satisfying . Then. Set
Then and is an asymptoticequidistant sequence. And because ,we have weakly. Finally,,then
which implies that
so . □
According to the above proof, we have the following.
Corollary 3.3 Let,and, then.
Remark In the case that , the situationdegrades to the case of classical Orlicz norm.
If , then .
If , then
Next, we discuss the fixed point property of .
Theorem 3.5and,,thencontainsan asymptotically isometric copy of.
Proof If , then there exist the sequence and such that
for .It is obvious that P is linear.
For any ,since , then there exists such that for all,hence
which implies .Moreover, for any ,we have
On the other hand, for any , there exists such that
where (), then
which implies that contains an asymptotically isometric copy of . □
Theorem 3.6 Let,,thenhasthe fixed point property if and only if it is reflexive.
Proof Since a reflexive Banach space X with has the fixedpoint property, we only need to prove the necessity.
Suppose , then by Theorem 3.5,contains an asymptotically isometric copy of . Hencedoes not have the fixed point property.
Suppose , then there exists such that for any , and for everysequence decreasing to 0, thereexist such that
Set ,then there exists such that.Hence, for any ,we have
Hence contains an asymptotically isometric copy of . ByTheorem 2 of , does not have the fixed point property. □
Theorem 3.7 If,,thendoesnot have the fixed point property.
Proof If , forany , define , then
Therefore, for any .
Moreover, we can prove that .Hence, define by
for all . Thenthe operator P is obviously linear, since
so we have , then P is an orderisometry of ontoa closed subspace of . □