Results & Lemmas (2)
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Lemma 3.5
Lemma 3.5 (ii) we can choose 1 2 open and relatively compact in ! such that there is no a 2 S! such that Ffag intersects 1 and !n 2. As…
Lemma 3.5 (ii) we can choose
1
2 open and relatively compact in ! such that there is no a 2 S! such that Ffag intersects
1 and !n
2. As in [25], Prop. 2.2, we consider L
z : sup w2@Qn
2 Rehw; zi max w2@Qn
2 Rehw; zi; z 2 CN: Obviously L H on SnS
2. If L
a H
a, then Rehw; ai H
a for some w 2 @Qn
2, hence a 62 S
1, since otherwise we would get a contradiction with 3.5 (ii). This proves L < H on S
1. Since SnS
2 SnS1, by (2) (for large jzj), there is some R > 1 such that the plu
Theorem 3.9
Theorem 3.9, we see that condition U fag; Q is fulfilled. Hence for all n ua z H z ÿ Hn a; z 2 CN: Since limn!1 Hn a 1, this…
Theorem 3.9, we see that condition U
fag; Q is fulfilled. Hence for all n ua
z H
z ÿ Hn
a; z 2 CN: Since limn!1 Hn
a 1, this implies that the nonzero plurisubharmonic function ua equals ÿ1 on fz j H
z < 1g. This contradicts the definition of Q0. We show that Q has nonempty interior: Assume that the interior is empty. Then there is b 2 S such that Q fz 2 CN j Rehz; bi 0g. Put a : ib. Choose La a to be the complex hyperplane