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:

[attachment=417]

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 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:

[attachment=417]

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.