Lets summarize what we have so far:
Proposition. Let
be a vertical strip somewhat wider than 
, i.e. <x_1+1\})
for some 
and 
.
Let
for some 
and let 
be two domains (open and connected) for values, and let 
be holomorphic on 
. Then there is at most one function 
that satisifies
(1) is holomorphic on 
and \subseteq G\subseteq f(D)=G')
(2) is real and strictly increasing on 
(3)=F(f(z)))
for all 
and =y_1)
(4) There exists an inverse holomorphic function
on 
, i.e. a holomorphic function such that )=z=f^{-1}(f(z)))
for all 
.
Proof. Let
be two function that satisfy the above conditions. Then the function =g^{-1}(h(z)))
is holomorphic on (because \subseteq G)
and (4)) and satisfies =g(\delta(z)))
. By (3) and (4)
=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1)
and =0)
.
So
can be continued from to an entire function and is real and strictly increasing on the real axis.
=\mathbb{C})
by our previous considerations.
By Big Picard every real value of is taken on infinitely often if is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still =\mathbb{C})
. But this is in contradiction to )
. So must be a polynomial that takes on every real value at most once. This is only possible for =x+c)
with 
because .
.
In the case of tetration one surely would chose
and 
or 
. However I am not sure about the domain which must contain )
and hence give some bijection 
, with some 
.
Of course in the simplest case one just chooses)
if one has some function in mind already. However then we can have a different function 
with \neq f_2(S))
but our intention was to have a criterion that singles out other solutions.
So we need an area on which every slog should be defined at least and satisfy =\text{id}=\text{slog}(\text{sexp}))
as well as \subseteq G)
.
Proposition. Let
Let
(1)
(2)
(3)
(4) There exists an inverse holomorphic function
Proof. Let
So
By Big Picard every real value of
In the case of tetration one surely would chose
Of course in the simplest case one just chooses
So we need an area
