11/18/2008, 06:18 PM
When I continued to find a suitable domain of definition for the slog such that the it is unique there by the universal uniqueness criterion II, I followed a line of thoughts I will describe later and came up with the following condition:
Proposition. There is at most one holomorphic super logarithm
that has a convergence radius of at least
when developed at 0 and that maps an open set
containing
- which is defined below - (or
) biholomorphically to some set
, that contains for each real value
a horizontal line of length
with imaginary part
.
Here
is the first fixed point of
and
can be roughly seen on the following picture:
The idea behind is the following. If we look at the straight line between
and
which can be given by
for
, then the area
bounded by
and
can be considered as an initial area from which you can derive for example the values on
or on
by
, or
.
You can see very well on the picture that
lies on the circle with radius
(red dashed line). This can be easily derived:
By
we know that
and hence
which is an arc with radius |L| around 0.
To be more precise we define exactly what we mean:
Let
be the set enclosed by
and
for
,
included but
excluded. This set is not open and hence not a domain, but if we move
slightly to the left we get a domain
containing
.
We call a function
defined on the domain
a super exponential iff it satisfies
and
for all
such that
.
We call a function
defined on the domain
a super logarithm iff
for each
(but not necessarily
for each
).
With those specifications we can come to the proof.
Proof. Assume there are two holomorphic super logarithms
and
.
Then both are defined on the domain
and map it bihomorphically, say
and
.
is holomorphic on the domain
. By the condition on
and by
it can be continued to an entire function. The same is true for
, which is the inverse of
and hence must
as it was shown in the proof in universal uniqueness criterion II.
Proposition. There is at most one holomorphic super logarithm
Here
The idea behind is the following. If we look at the straight line between
You can see very well on the picture that
By
To be more precise we define exactly what we mean:
Let
We call a function
We call a function
With those specifications we can come to the proof.
Proof. Assume there are two holomorphic super logarithms
Then both are defined on the domain