\input arma
\input mizenpaj

\centerline{\bf Two Counterexamples Concerning the}
\centerline{\bf Pluri-Complex Green Function in $\cb^n$}
\vskip10pt
~\hskip15mm{\sl Eric Bedford and Jean-Pierre Demailly}\footnote*
{This paper was written while the authors were visiting the Mittag-Leffler
Institute, and they wish to thank the Institute for its hospitality.}
\hfil\break\indent~\hskip15mm
{\sl Indiana University and Universit\'e de Grenoble I}
\vskip20pt

Given a domain $\Omega\subset\cb^n$ and a point $z\in\Omega$, we
consider the function
$$
u_z(\zeta)= \sup\{v(\zeta): v\hbox { is psh on } \Omega, v<0, \hbox{
and } v(\zeta)\le\log|\zeta-z|+O(1)\},
$$
which is the {\sl pluri-complex Green function} on $\Omega$ with
logarithmic pole at $z$\break 
(cf. [2,3,4]).  For a large class of domains
$\Omega$, (e.g. if $\partial\Omega$ is strongly pseudoconvex and bounded),
then $u_z\in\ci(\bar\Omega-\{z\})$, $u_z=0$ on $\partial\Omega$,
and $u_z(\zeta)=\log|z-\zeta|+O(1)$.  Several further properties of
$u_z$ were derived in [2], where the following questions were also
raised: 

1. Is $u_z\in C^2(\bar\Omega-\{z\})$ ?

2. Is $u_z$ symmetric, i.e. $u_z(\zeta)=u_\zeta(z)$ ?  

Lempert [4] has shown
that if $\Omega\subset\cb^n$ is strictly convex and smoothly bounded,
then the answer to these questions is ``Yes." The convex situation,
however, is quite special, and the point of this note is to show that
for strongly pseudoconvex domains the answer to both these questions
is ``No."

\theo Proposition| 1||Let $\Omega\subset\cb^2$ be a bounded, connected,
strongly pseudoconvex domain with $C^2$ boundary.  Let $\Omega$
be invariant under the transformation $(z,w)\mapsto(z,-w)$, and
suppose that $\Omega\cap\{w=0\}=D_1\cup\dots\cup D_s$, where
$\bar D_i\cap\bar D_j=\emptyset$, and $D_j$ is a smoothly
bounded disk.  Then if $p\in D_1$, the function $u_p$ is not $C^2$ 
in a neighborhood of the set $\bar D_j$ for any $2\le j\le s$.
\fintha

\remark Remark||For a specific example,  we may let $\Omega$ be a
small neighborhood of the unit 
circle $\{(x_1,x_2)\in\rb^2\subset\cb^2: x_1^2+x_2^2=1\}$.

\proof  Since both $p$ and the domain $\Omega$ are invariant
under the mapping $(z,w)\mapsto(z,-w)$, it follows that $u_p$ is also
invariant.  If $u_p$ is of class $C^2$ on a neighborhood of the
set $\bar D_2$, then we have an invariant solution of
$\det(\partial^2u_p/\partial z_i\partial\bar z_j)=0$ on $\bar D_2$, with 
$u_p=0$ on $\partial D_2$.  But this is a contradiction to the Proposition of
[1], which completes the proof. 

\theo Proposition| 2||There is a bounded, strongly pseudoconvex
domain $\Omega$ in $\cb^2$ with real analytic boundary for which the
Green function is not symmetric.
\fintha

\proof Let $u(z,w)=\max\left({1\over 2}
\log(|w^2-z^2(z-a)|/\varepsilon^2),\log|z|\right)$, where $|a|<1$. 
We will show that for $\varepsilon>0$ small, the Green function for the domain
$\Omega=\{u(z,w)<0\}$, is not symmetric.  To prove Proposition 2, we
consider an increasing sequence of strongly pseudoconvex domains
$\Omega_j$ with real analytic boundary, which increase to $\Omega$.
If the Green function $u_z(\zeta,\Omega_j)$ is symmetric for all $j$,
then so is $\lim_{j\to\infty}u_z(\zeta,\Omega_j)=u_z(\zeta)$.  Thus
one of the sets $\Omega_j$ must give us the set desired in
Proposition 2.

We note that $u$ is the Green function for the domain
$\Omega$ with pole at 
$(0,0)$. Thus $u_{(0,0)}(a,0)=u(a,0)=\log|a|$. On the other hand, the pole
of $u_{(a,0)}$ is estimated by the function
$$
v(z,w)=\left\{
\eqalign{
&\log|w|/2\hbox{~~if~~}|z-a|\le\varepsilon~,\cr
&\max\left(\log|w|/2,\left({1\over 2}+{1\over{\sqrt {\log
1/\varepsilon}}}\right)\log\left|{z-a\over 1-\bar az}\right|\right)
\hbox{~~if~~}|z-a|\ge\varepsilon~,\cr}\right.
$$
Indeed, we have $|w|<2$ on $\Omega$, and thus $v<0$.  Moreover the
inequalities
$$
|w|^2\ge|z^2(z-a)|-\varepsilon^2\ge(|a|-\varepsilon)\varepsilon-\varepsilon^2
$$
on $\Omega\cap\{|z-a|=\varepsilon\}$ show that $v$ is psh.  As
$v(z,w)\le\log(|w|+|z-a|)+O(1)$ near $(a,0)$, it follows that
$v(z,w)\le u_{(a,0)}(z,w)$.  Thus if $\varepsilon$ is small, we have
$$
u_{(a,0)}(0,0)\ge\left({1\over 2} + {1\over {\sqrt {\log
1/\varepsilon}}}\right)\log|a|>u_{(0,0)}(a,0),
$$
and so the Green function is not symmetric.

{\sl Remark.}  In general, the Green function $u_z(\zeta)$
cannot even be expected to be psh in $z$.  In fact, for any domain
$\Omega$, the symmetry property is equivalent to the fact that
$v(z)=u_z(\zeta)$ is psh in $z$ for every fixed $\zeta$.  To see
this, observe that $v<0$ and $v(z)=\log|z-\zeta|+O(1)$ near $\zeta$.
the plurisubharmonicity of $v$ implies therefore 
$u_z(\zeta)=v(z)\le u_\zeta(z)$, and similarly $u_\zeta(z)\le u_z(\zeta)$. 

\titre References|
\vskip15pt
\eightpoint
\article 1|E. Bedford {\rm and} J.-E. Forn\ae ss|Counterexamples to
regularity for the complex Monge-Amp\`ere equation|Inv. Math.|50|1979|129--134|
\article 2|J.-P. Demailly|Mesures de Monge-Amp\`ere et mesures
pluriharmoniques|Math. Z.|194|1987|519--564|
\article 3|M. Klimek|Extremal plurisubharmonic functions and invariant
pseudodistances|Bull. Soc. Math. France|113|1985|123--142|
\article 4|L. Lempert|La m\'etrique de Kobayashi et la repr\'esentation des 
domaines sur la boule|Bull. Soc. Math. France|109|1981|427--474|

\bye
