8 STEPHEN SEMMES

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 =

8{p-l){dduQ),

which could also have been proved exactly as in Lemma 2.6. Because

(dduo)n

=

0,

(ddu0)n"1

^ 0 on Bn \ {0}, the corresponding results hold for u, which gives

(c).

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 h£ 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.