Universal uniqueness criterion? - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: Universal uniqueness criterion? ( /showthread.php?tid=165) |

Universal uniqueness criterion? - bo198214 - 05/21/2008
Inspired by Kouznetsov's consideration I investigated a bit more in this direction and found interesting results: 1. Let F be holomorphic on the right half plane, and let F(z+1)=zF(z), F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is the gamma function. 2. is also determined as the only function that satisfies , and is bounded on the strip 1<=R(z)<2 (or 0<=R(z)<1). 3. I started a thread in sci.math.research and it turned out that the same criterion also makes the Fibonacci function , the unique extension of , , . So I really would guess that this criterion (which is a slightly weaker demand than Kouznetsov's criterion) also implies uniqueness for tetration, at least for base . Maybe its not even difficult to prove. RE: Universal uniqueness criterion? - andydude - 05/22/2008
Where are these proven? How are these proven? By whom? 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? Andrew Robbins RE: Universal uniqueness criterion? - andydude - 05/22/2008
Also, I admire the criticism of the people in that thread, but given the right adjectives, there is really no problem with what was said. I think that if you replace "entire" with "holomorphic over domain D" (where D is the complex plane minus any singularities or branch cuts) and suffix "unique" with "unique up to branch choice" (or state some property that only the principal branch has) then we might get better responses, and certainly more clarity. Andrew Robbins RE: Universal uniqueness criterion? - bo198214 - 05/22/2008
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. RE: Universal uniqueness criterion? - andydude - 05/22/2008
Very nice. I followed your proof up until "By Liouville", what does this mean? Is this the name of a theorem? RE: Universal uniqueness criterion? - bo198214 - 05/22/2008
andydude Wrote:Very nice. I followed your proof up until "By Liouville", what does this mean? Is this the name of a theorem? Theorem of Liouville RE: Universal uniqueness criterion? - bo198214 - 05/23/2008
Ok, now the promised proof for the uniqueness of the gamma function, which was point 1. here. I nearly only translate it from the mentioned book: We consider the function on the right half plane, it also satisfies . Hence has a meromorphic continuation to . Poles can be at most at 0, -1, -2 , .... As we have , hence has a holomorphic continuation to 0 and also to to each , by . restricted to is bounded because is bounded there. Then is also restricted on given by . defined by is bounded on because and have the same values on . Now , hence is bounded on whole and by Liouville . Hence and . RE: Universal uniqueness criterion? - Gottfried - 06/25/2008
Perhaps the following two msgs in sci.math of David Libert are appropriately posted to this thread. I just copy&paste the whole msg to provide the context. Perhaps we move this post to another place - Henryk? Code: `(David Libert ah170@FreeNet.Carleton.CA )` and a correction, posted later on the day Code: `David Libert (ah170@FreeNet.Carleton.CA) writes:` Proof attempt for a uniqueness criterion - bo198214 - 09/18/2008
We have the usual conditions (1) is a holomorphic function defined on (2) (3) , . (4) is real on and strictly increasing. The following criterion should pick a unique function out of the possible solutions for the above conditions: (U1) is bounded on the strip . (U2) The values of fill the complex plane: , . Proof attempt: Let be some function that also satisfies (1), (2), (3),(4) and (U1), (U2). We put appropriate cuts on the domain of such that we have an inverse which always maps reals to reals. By some considerations previously made already on this forum is a holomorphic and 1-periodic function. We know that if a 1-periodic holomorphic function is bounded on then it is already a constant. So if we could show that has a bounded real part on then would be bounded and 1-periodic on , hence hence hence for some . But of course because . So we still need show that has bounded real part on . We know already that is bounded. By condition (U2) we divide the area into images of of , i.e. . Because is bounded, it is a subset of a compact set, which can be covered by merely finitely many : so which has bounded real part. RE: Proof attempt for a uniqueness criterion - bo198214 - 09/26/2008
I am not so convinced whether is invertible (it is usually not injective on the complex plane) without getting trouble. So here is the next revision/attempt: bo198214 Wrote:We have the usual conditionsThe following criterion should pick a unique function out of the possible solutions for the above conditions: (U1) is bounded on the strip . (U2) for all . ( is just an abbreviation for the more save .) Proof attempt: Let be some function that also satisfies (1), (2), (3),(4) and (U1), (U2). Then , where is a 1-periodic function (This is easily provable for bijective and but I am not 100% sure yet for arbitrary and , but it should go through.) As in the previous proof we just need to show that has bounded real part on . We know already that is bounded. We can write where is the next lower integer (floor) of the real part of and . hence is bounded by (U1) and for every by (U2). If the real part of would be unbounded then also would be unbounded on . We choose a sequence such that is strictly increasing (and hence unbounded). The sequence is bounded and hence has a converging subsequence. Let be a subsequence of such that . The sequence is bounded and hence has a converging subsequence. Let be a subsequence of such that converges, let . We still have , and is strictly increasing. is element of the closure . (U2) holds also on by (3). Thatswhy in contradiction to being finite. |