We begin this section from the following useful lemmas.
Proof.
The implication (1)⇒(2) is trivial. Let x be a complex extreme point of the unit ball B(LΦ,p) and there exists k0∈Kp(x) such that μ{t∈T: k0|x(t)|<e(t)}>0. Then we can find d>0 such that μ{t∈T:k0|x(t)|+d<e(t)}>0. Indeed, if μ{t∈T: k0|x(t)|+d<e(t)}=0 for any d>0, then we set dn=1/n and define the subsets
(16)Tn={t∈T:k0|x(t)|+dn<e(t)}
for each n∈ℕ. Notice that T1⊆T2⊆⋯⊆Tn⊆⋯ and {t∈T:k0|x(t)|≤e(t)}=⋃n=1∞Tn; we can find that
(17)μ{t∈T:k0|x(t)|≤e(t)}=0
which is a contradiction.
Let T0={t∈T:k0|x(t)|+d<e(t)} and define y=(d/k0)χT0; we have y≠0 and for any λ∈ℂ with |λ|≤1,
(18)∥x+λy∥Φ,p≤1k0{1+IΦp(k0(x+λy))}1/p=1k0{+IΦ(k0xχT0+λdχT0)]p1+[IΦ(k0xχT∖T0)+IΦ(k0xχT0+λdχT0)]phhhhhhhhhh+IΦ(k0xχT0+λdχT0)]p}1/p≤1k0{1+[IΦ(k0xχT∖T0)((k0|x|+d)χT0)hhhhhhhhh+IΦ((k0|x|+d)χT0)]p}1/p=1k0{1+[IΦ(k0xχT∖T0)]p}1/p=1k0(1+IΦp(k0x))1/p=1,
which shows that x is not a complex extreme point of the unit ball B(LΦ,p).
(
3
)
⇒
(
1
)
. Suppose that x∈S(LΦ,p) is not a complex strongly extreme point of the unit ball B(LΦ,p),; then there exists ε0>0 such that Δc(x,ε0)=0. That is, there exist λn∈ℂ with |λn|→1 and yn∈LΦ,p satisfying ∥yn∥Φ,p≥ε0 such that
(19)∥λnx0±yn∥Φ,p≤1, ∥λnx0±iyn∥Φ,p≤1,
which gives
(20)∥x0±ynλn∥Φ,p≤1|λn|, ∥x0±iynλn∥Φ,p≤1|λn|.
Setting zn=yn/λn, we have
(21)∥zn∥Φ,p≥∥yn∥Φ,p≥ε0,∥x0±zn∥Φ,p≤1|λn|,∥x0±izn∥Φ,p≤1|λn|.
For the above ε0>0, by Lemma 1, there exists δ0∈(0,1/2) such that if u,v∈ℂ and
(22)|v|≥ε08maxe|u+ev|,
then
(23)|u|≤1-2δ04Σe|u+ev|.
For each n∈ℕ, let
(24)An={t∈T:|zn(t)|≥ε08maxe|x0(t)+ezn(t)|},zn(1):zn(1)(t)=zn(t)(t∉An), zn(1)(t)=0(t∈An),zn(2):zn(2)(t)=0(t∉An), zn(2)(t)=zn(t)(t∈An).
It is easy to see that zn=zn(1)+zn(2), ∀n∈ℕ. Since |λn|→1 when n→∞, the following inequalities
(25)∥zn(1)∥Φ,p =infk>01k{1+[∫t∉AnΦ(t,k|zn(t)|)dt]p}1/p ≤infk>01k{[∫t∉AnΦ(t,kε08maxe|x0(t)+ezn(t)|)dt]p}1/p1+hhhhhhhhhh[∫t∉AnΦ(t,kε08maxe|x0(t)+ezn(t)|)dt]p{[∫t∉AnΦ(t,kε08maxe|x0(t)+ezn(t)|)dt]p}1/p1+}1/p ≤ε08∥maxe|x0+ezn|∥Φ,p ≤ε08∥Σe|x0+ezn|∥Φ,p ≤ε02|λn|<3ε04
hold for n large enough.
Therefore, ∥zn(2)∥Φ,p>ε0/4 which shows that μ(An)>0 for n large enough. Furthermore, for any t∈An, we have
(26)|x0(t)|≤1-2δ04Σe|x0(t)+ezn(t)|.
To complete the proof, we consider the following two cases.
(I) One has kn→∞ (n→∞), where kn∈Kp((1/4)Σe|x0+ezn|).
For each n∈ℕ, we get
(27)1=∥x0∥Φ,p≤1kn{1+IΦp(knx0)}1/p=1kn{∫1+[∫t∈AnΦ(t,kn|x0(t)|)dthhhhhhhhhh+∫t∉AnΦ(t,kn|x0(t)|)dt]p}1/p≤1kn{1+[∫t∈AnΦ(t,kn(1-2δ04hhhhhhhhhhhhhhhhh×Σe|x0(t)+ezn(t)|(1-2δ04))dthhhhhhhhhhhhhhhhhh+∫t∉AnΦ((14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt[∫t∈AnΦ(t,kn(1-2δ04]p}1/p≤1kn{+∫t∉AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt]p1 +[+∫t∉AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt]p(1-2δ0)hhhhhhhh×∫t∈AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dthhhhhhhh+∫t∉AnΦ(×(14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt[+∫t∉AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt]p(1-2δ0)]p}1/p=1kn{×(14Σe|x0(t)+ezn(t)|))dt]p1 +[IΦ(kn(14Σe|x0+ezn|))-2δ0hhhhhhhh×∫t∈AnΦ(×(14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt]p}1/p.
Furthermore, we notice that
(28)∫t∈AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt ≥∫t∈AnΦ(t,kn|zn(t)|)dt ≥[knp∥zn(2)∥Φ,pp-1]1/p ≥[(ε04kn)p-1]1/p.
Since kn→∞ when n→∞, we can see that the inequality ((ε0/4)kn)p-1>0 holds for n large enough. Moreover, we find that
(29)IΦ(kn(14Σe|x0+ezn|)) ≥IΦ(knx0) ≥(knp-1)1/p≥2δ0((ε04kn)p-1)1/p.
Then we deduce
(30)1=∥x0∥Φ,p≤1kn{14]p1+[IΦ(kn(14Σe|x0+ezn|))-2δ0hhhhhhhhhh×∫t∈AnΦ(×(14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt]p}1/p≤1kn{1+[-2δ0((ε04kn)p-1)1/p]pIΦ(kn(14Σe|x0+ezn|))hhhhhhhhhh-2δ0((ε04kn)p-1)1/p]p}1/p<1kn{1+IΦp(kn(14Σe|x0+ezn|))}1/p=1,
a contradiction.
(II) One has kn→k0(n→∞), where kn∈Kp((1/4)Σe|x0+ezn|), k0∈ℝ.
Then for each n∈ℕ,
(31)1λn≥∥14Σe|x0+ezn|∥Φ,p=1kn{1+IΦp(kn4Σe|x0+ezn|)}1/p≥1kn{1+IΦp(knx0)}1/p≥∥x0∥Φ,p=1.
Let n→∞, we deduce that 1=(1/k0){1+IΦp(k0x0)}1/p=∥x0∥Φ,p.
From (I), we obtain that
(32)1=∥x0∥Φ,p≤1kn{∫t∈AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt]p}1/p1+[IΦ(kn(14Σe|x0+ezn|))-2δ0hhhhhhhhhh×∫t∈AnΦ(×(14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt]p{∫t∈AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt]p}1/p1+}1/p.
Now we consider the following two subcases.
(a) Consider ∫t∈AnΦ(t,kn((1/4)Σe|x0(t)+ezn(t)|))dt>0 for some n∈ℕ.
Then we obtain
(33)1=∥x0∥Φ,p≤1kn{×(14Σe|x0(t)+ezn(t)|))dt]p}1/p1+[IΦ(kn(14Σe|x0+ezn|))-2δ0hhhhhhhhhh×∫t∈AnΦ(×(14Σe|x0(t)+ezn(t)|))t,knhhhhhhhhhhhhhhh×(14Σe|x0(t)+ezn(t)|))dt]p}1/p×(14Σe|x0(t)+ezn(t)|))dt]p}1/p<1kn{1+IΦp(kn(14Σe|x0+ezn|))}1/p=1,
a contradiction.
(b) Consider ∫t∈AnΦ(t,kn((1/4)Σe|x0(t)+ezn(t)|)) dt=0 for any n∈ℕ.
For n large enough, we observe that
(34)kn>1-2δ01-δ0k0.
Hence, we get
(35)0=∫t∈AnΦ(t,kn(14Σe|x0(t)+ezn(t)|))dt≥∫t∈AnΦ(t,|kn1-2δ0x0(t)|)dt≥∫t∈AnΦ(t,|k01-δ0x0(t)|)dt≥0.
Thus, we see the equality ∫t∈AnΦ(t,|(k0/(1-δ0))x0(t)|) dt=0 holds for n large enough. It follows that
(36)|k01-δ0x0(t)|≤e(t), μ-a.e.t∈An
since μ(An)>0 for n large enough.
On the other hand, by (3), we deduce that
(37)|k0x0(t)|≥e(t), μ-a.e. t∈An.
Hence, we get a contradiction:
(38)|k01-δ0x0(t)|≥e(t)1-δ0>e(t), μ-a.e. t∈An.
Proof.
The implication (1)⇒(2) is trivial. Now assume that LΦ,p is complex strictly convex. If μ{t∈T:e(t)>0}>0, let T0={t∈T:e(t)>0} and it is not difficult to find an element x∈S(LΦ,p) satisfying supp x=T∖T0. Take k∈Kp(x), and define
(39)y(t)={e(t)2kfor t∈T0,x(t)for t∈T∖T0.
Obviously, ∥y∥Φ,p≥∥x∥Φ,p=1. On the other hand,
(40)∥y∥Φ,p≤1k{1+[∫t∈T0Φ(t,e(t)2)dthhhhhhhhhhh+∫T∖T0Φ(t,kx(t))dt]p}1/p =1k(1+IΦp(kx))1/p=1.
Thus, ∥y∥Φ,p= (1/k)(1+IΦp(ky))1/p=1. However, for t∈T0, we find k|y(t)|=e(t)/2<e(t), which implies y∈S(LΦ,p) is not a complex extreme point of B(LΦ,p) from Theorem 3.
(
3
)
⇒
(
1
)
. Suppose that x∈S(LΦ,p) is not a complex strongly extreme point of B(LΦ,p). It follows from Theorem 3 that μ{t∈T:k0|x(t)|<e(t)}>0 for some k0∈Kp(x), consequently μ{t∈T:e(t)>0}>0 which is a contradiction.