If p is bilipschitz on Bni then the eigenvalues of A(z) can be bounded away
from 0 on D \ {0}, because of (2.7) and the nondegeneracy of LJQ. Up is also
smooth on a neighborhood of dBni then it follows easily that D is strongly
pseudoconvex. [Extend F smoothly to a neighborhood of 3D, and use F 1 as
a defining function.] This proves (a).
Let us check (c) before (b). Observe that
(2.11) ddu = dd(\ogF) = d(F~ldF) = F~lddF - F~20F AdF.
These equalities hold in the sense of currents on D\ {0}. They can be justified
rigorously using an approximation argument and the fact (from Lemma 2.6) that
ddF is continuous. This implies that ddu is continuous, and that
(2.12) ddu =
which could also have been proved exactly as in Lemma 2.6. Because
^ 0 on Bn \ {0}, the corresponding results hold for u, which gives
To prove (b) it suffices to show that
dz dT
u(z) is a nonnegative matrix for
all z G D \ {0}. Because of (2.11) and Lemma 2.9, it can have at most one
non-positive eigenvalue. On the other hand, (c) says that it must have a zero
eigenvalue, whence the desired nonnegativity.
It remains to verify (e), since (d) is immediate from the definitions. Clearly
u is a competitor for the supremum in (2.4), by (b) and (d), and so u is not
greater than (2.4). We need to show that u v for all v as in (2.4).
Fix v. Given z G dBn, define a real-valued function h on A by
h(X) = v(p(\z)).
Then h 0 on A, h is subharmonic, and h(X) log |A| -f C, where C depends
on v and p but not A. This implies that h(X) log |A| on A. [For each e 0 set
h€(X) = h(\) (l c) log |A|. Then he is subharmonic on A\{0} and h€(0) = —oo,
and so he is subharmonic on A. The maximum principle implies that 0 for
each e 0, and hence h(X) log |A|.]
On the other hand we have u(p(Xz)) = log |A|. Because z, A are arbitrary we
obtain u v. This completes the proof of Theorem 2.3.
Let us now turn to the proof of Theorem 2.4. Let v, / , #, and a be as in the
statement of Theorem 2.4, and assume that p satisfies (1.1)—(1.4). We may as
well require that g'(Q) ^ 0.
Consider the function
«(A) = «(0(A))-log|A|
defined on A \ {0}. By Theorem 2.3(b) 5(A) is subharmonic on A \ {0}. It is
bounded in a neighborhood of 0, because of (1.2) and the definition of w, and
so it has a subharmonic extension to A. [One way to check this uses the fact
5(A) -j- elog |A| is subharmonic on A for all e 0.] Hence
limsup 5(A) lim sup 5(A) = 0.
Previous Page Next Page