05/22/2008, 11:25 AM

andydude Wrote:So by "bounded" are you referring to the fact that Kouznetsov's extension requires that the limit towards is finite? Is this the same as saying that this limit exists?

His assumption is that the limit is the fixed point in the upper complex halfplane, not only finite. I mentioned somewhere already that for with regular tetration this is no more true (i.e. has no limit), however it is still bounded. So perhaps this is a good generalization.

Quote:Where are these proven? How are these proven? By whom?

Ok, lets start with the uniqueness of (I dont know whether is somewhere written):

Proposition.Let then is the only holomorphic solution, defined on the right halfplane [this condition is not necessary, I just include it to emphasize on non-entire functions], of the equations , which is bounded on the strip given by .

Proof

We know that every other solution must be of the form where is a 1-periodic holomorphic function (this can roughly be seen by showing periodicity of ).

In this case this means:

where is also a 1-periodic function. As (and ) is bounded on , must be bounded too.

As is periodic it can be continued from to the whole plane and is hence an entire holomorphic function, which is still bounded. By Liouville must be constant:

. And now we apply and see that .

I will describe the proof for the Gamma function in another post, I found it in a German complex analysis book: Reinhold Remmert, "Funktionnentheorie", Springer, 1995. As reference is given: H. Wielandt 1939.

And the proof for the Fibonacci function is given in the sci.math.research thread by Waldek Hebisch (key points), and a bit fleshed out by me.