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 =\Re(L)+i\Im(L)t)
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 )=\text{slog}(z)+1)
, or )=\text{slog}(z)-1)
.
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 }=|e^L|=|L|)
and hence
}=e^{\Re(L)+i\Im(L)t}=|L|e^{i\Im(L)t})
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 =1)
and =\exp(\text{sexp}(z)))
for all 
such that 
.
We call a function defined on the domain 
a super logarithm iff )=z)
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 =\delta(z)+1)
it can be continued to an entire function. The same is true for 
, which is the inverse of 
and hence must =z)
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
