[FOM] Analyticity of half-exponentials

[joeshipman@aol.com]

>> Is there a monotonic real analytic function defined  on the
>> non-negative real numbers such that f(f(x)) = 2^x, or f(f(x))=e^x?

> "For the equation
>(*)              f^2(x) = e^x,
>a real analytic solution has been found by H. Kneser.
>This solution, however, is not single-valued (Baker)
>and, as pointed out by G. Szekeres, there is no
>uniqueness attached to the solution. It seems reasonable
>to admit f(x)=F^(1/2)(x), where F^u is the regular
>iteration group of  g(x)=e^x, as the "best" solution of
>the equation (*) (best behaved at infinity). However,
>we do not know whether this solution is analytic for
>x>0.

This is not very helpful; from this information I cannot tell
1) is Kneser's "real analytic solution" a function, or isn't it?
2) if it is a function, is it monotonic, or isn't it?
3) how is "F^(1/2)(x)" is defined?
4) is it monotonic?

If "F^(1/2)(x)" is uniquely defined then it is very surprising that 
whether it is analytic is open.

I'm trying to ask as precisely focused a question as possible, but I 
cannot tell from your reply whether the answer to my original question 
is "yes", "no", or "open".

If I look at a modification of the question and ask for a function f 
such that f(f(x)) = (e^x - 1) instead of e^x, then I can actually build 
a formal power series that uniquely solves the functional equation, but 
which has some negative coefficients (and is therefore probably 
non-monotonic) and whose radius of convergence I cannot calculate. This 
implies that the answer to my question is negative if I replace e^x by 
(e^x - 1) AND I strengthen the requirement "real analytic" to "real 
analytic with an everywhere-convergent power series"; but it's possible 
for a function to be real analytic and need different power series in 
different parts of the domain, and this example says nothing about the 
original question with f(f(x))=e^x or f(f(x))=2^x because the 
construction of the formal power series solution depends on the 
constant term being zero.

-- Joe Shipman
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