Let
be a function on
with
and suppose there exists
such that
for![]()
Then
for![]()
maps the open disc
into the the open disc![]()
Proof: Since
is analytic on
we can write
as a Taylor series about 0.
for
(
since
)
Then
for![]()
Thus the function
provides an analytic extension of
from
to![]()
Now apply the maximum principle to
on the open disc
where
(we cannot allow
since
– and so
– is not known to be continuous on![]()
This gives
where![]()
for![]()
This inequality holds for all
such that
we deduce on letting
that
for![]()
Hence
for
so that
for
but since this inequality obviously holds for z=0 we have
for![]()