Clearly Sxpj(u;o) —•
and Stpj(dF0) —•
uniformly on com-
pact subsets of D \ {0}. The lemma now follows by taking the limit as j —• oo
of (2.8) in the sense of currents.
Let us use (2.7) to prove that dpz(Sl) is a complex subspace for all z £
Bn \ {0}. By definitions,
Si = {veCn: (w0 | J (t;, v') = 0 for all v' £ S 2 },
and hence
dp,(Sl) = {v€Cn:(w \M) (v, v') = 0 for all v' e T%t)}.
This is a complex subspace because T^,^ is, and because a; is a (1,1) form. This
finishes the proof of Theorem 2.2.
The main step in the proof of Theorem 2.3 is the following.
2.9. Suppose that p satisfies (1.1)—(1.4). Then for each z £ D\ {0},
dzd&z ^( z ) IS a positive-definite matrix.
Notice that
F(z) is continuous in z on D \ {0}, because of Lemma 2.6.
Let A(z) denote the self-adjoint complex-linear transformation on C n associ-
ated to this matrix. For each z in D \ {0} A(z) has trivial kernel, because of
(2.7) and the fact that cuo is nondegenerate. Hence A{z) must have a negative
eigenvalue for all z if it has one for some z, since A(z) is continuous in z. Let us
assume that this is true, and get a contradiction using the minimum principle.
We have to be a little careful, because of the singularity at the origin.
For each e 0 set
fe(z) = F(z) - t\og\z\
on D\{0} , and let At{z) be the linear transformation associated to
Because log|z| is plurisubharmonic A€(z) has at least one negative eigenvalue
for each e 0 a n d z £
D \ { 0 } . This implies that
(2.10) inf/(*)= jnf fc(z)
z£U z£oU
for each open set U with U C D \ {0}. This is actually a little tricky; we cannot
use the standard minimum principle directly, because f€ may not be C2, even
fe is continuous. However, this difficulty is easily overcome by a
standard approximation argument.
It is easy to see that (2.10) is impossible if e is very small and U is almost all
of D \ {0}, because F = 1 on 3D, F 1 on D, and /c(0) = oo for all e 0.
This contradiction establishes Lemma 2.9.
Let us now prove Theorem 2.3. Suppose that p satisfies (1.1)—(1.4). Then
F is plurisubharmonic on D, because of Lemma 2.9, and we have F 1 on D}
F(z) —• 1 as z dD, which implies that D is pseudoconvex.
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