As I now see, even if the uniqueness criterion is valid, the precondition:

for each

can not be true for any

for

(and probably for each

)!

Michał Misiurewicz [1] showed 1981 that the

Julia set of

is the whole

(that is not directly visible from the

fractal). The Julia set is the boundary of the set

of all

such that

. That means that the set

and its complement

is dense in

. In every neighborhood of any complex number

there is a complex number

such that

and also a

such that

!

And that implies that

, which contains an open non-empty set, always contains points

such that

.

[1] Michał Misiurewicz (1981). On iterates of e^z. Ergodic Theory and Dynamical Systems, 1 , pp 103-106, doi:10.1017/S014338570000119X

bo198214 Wrote:...

And that implies that , which contains an open non-empty set, always contains points such that .

...

Why do you think that this set is open? Wiki says:

In metric topology and related fields of mathematics, a set U is called open if, intuitively speaking, starting from any point x in U one can move by a small amount in any direction and still be in the set U. In other words, the distance between any point x in U and the edge of U is always greater than zero.

http://en.wikipedia.org/wiki/Open_set
The small deviation from a point where the limit is finite allows to get (in the limit) so high walues as you want, because the right hand side of the plot of tetration is pretty scratched. I would replace "open non-empty set" to "non-empty set".

Uniqueness of analytic tetration, scheme of the proof.

Notations.

Let

be set of complex numbers.

Let

be set of real numbers.

Let

be set of integer numbers.

Let

Let

be base of tetration.

Assume that there exist analytic tetration

on base

, id est,

(0)

is analytic at

(1) for all

, the relation

holds

(2)

(3)

is real increasing function at

.

Properties.

From assumption (1) and (2) it follows, that function

has singularity at -2, at -3 and so on.

Consider following Assumption:

There exist entire 1-periodic function

such that

is also analytic tetration on base

.

Then, function

is not allowed to take values -2, -3, ..

being evaluated at elements of

.

This means that function

is entire function.

(Weak statement which seems to be true)

Function

cannot grow faster than linear function at infinity in any direction.

Therefore, it is linear function. Therefore,

Therefore, there exist only one analytic tetration.

Kouznetsov Wrote:bo198214 Wrote:...

And that implies that , which contains an open non-empty set, always contains points such that .

...

Why do you think that this set is open?

Just take a point

of the interior of

such that

(if such a point does not exist then

is constant), then there is a neighborhood

of

such that

is bijective on

. That means

and

are continuous (because analytic). And the preimage of an open set is open under a continuous map. So the preimage

of

under

is open and so we find arbitrary many points

in

such that

.

Kouznetsov Wrote:Do you make any difference between and ?

Oh no its both the same here.

Kouznetsov Wrote:Consider following Assumption:

There exist entire 1-periodic function such that

is also analytic tetration on base .

Then, function is not allowed to take values -2, -3, ..

being evaluated at elements of .

But

does not need to be entire, it can also omit the values -2, -3 , ... as arguments.

For an entire function we would know that it can omit at most 1 value (

Little Picard).

Edit: Oh I see now

needs not to be entire, but can be continued being entire by

if

can be/is defined on the strip

.

Ok, I give it another try:

Proposition.

Let

,

,

and

being a domain (open and connected) of definition and

being a domain (open and connected) of values for a holomorphic function.

Let

be a holomorphic function on

,

, such that

(0)

.

(1)

(2)

(U)

has bounded real part.

Then

for every other on

holomorphic

satisfying (0), (1), (2), (U).

Proof.

has bounded real part on

. We consider

and

and so

to be holomorphic on the same Riemann surface

.

is a 1-periodic function, holomorphic on

. As

it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip

otherwise it is a constant. Now

has bounded real part and hence can not take on every value, so

and

.

bo198214 Wrote:Ok, I give it another try:

..

Bo, Congratulations! You seem to do well. I copypast your theorem into the latex source, but I shall not work on it until I collect all the pieces into the single file.

(I have not yet type the uniqueness of the "Children's factorial" you suggested). While, you generate results faster than I describe them. The draft becomes longer, but does not yet approach to the finish. If you want to work with the text, let me know; I send you the last version. Keep in touch, K.

"We call a region H an initial region of F iff F(z) {is not in} H for all z {in} H..."

does this mean

(there is SOME z such that F(z) is NOT in H) or

) (there is NO z such that F(z) IS in H)?