Consider the fractional iterations of the expontential function denoted
$$ \exp^{[t]}(x) $$
$$ \exp^{[0]}(x) = x $$
$$ \exp^{[t]}(x) = \exp(\exp^{[t-1]}(x) $$
$$ \exp^{[l + m]} = \exp^{[l]}(\exp^{[m]}(x))$$
Where $t,l,m,x$ are real and $t > 0$.
My friend Tommy proposed once a condition for an infinitely differentiable solution.
$t,x$ are real and $t > 0$:
$$ \frac{d^n}{d^n x} \exp^{[t]}(x) > 0 $$
It seems impossible to satisfy for analytic solutions because for $ 0 < t < 1$ we have singularities in the complex plane that force the radius being finite and the signs of the derivatives of the taylor series alternating ( because the singularities have taylor series with alternating signs and near the singularities we must have it too). For instance if you are close to a log singularity your taylor series must eventually copy the sign changes of the logarithm.
However if the bundle of functions $exp^{[t]}$ are nowhere analytic on the real line but infinitely differentiable, this might be true ??
CONJECTURE 1 (existance conjecture)
There exist a bundle of infinitely differentiable functions solution satisfying
$$ \frac{d^n}{d^n x} \exp^{[t]}(x) > 0 $$
CONJECTURE 2 (uniqueness conjecture)
The solution to conjecture 1 is unique (if it exists!) ; there is only one (non-analytic) solution for tetration satisfying the condition
$$ \frac{d^n}{d^n x} \exp^{[t]}(x) > 0 $$
Can we prove conjecture 1 or 2 ?
Is this a very general thing or exclusive for iterations of exponential functions ? I mean, do many entire functions $g(x)$ with all derivatives positive ( at any $x$) , no real fixpoint ( and thus above $id(x)$ btw ) (like exp) satisfy
$$ \frac{d^n}{d^n x} g^{[t]}(x) > 0 $$
??
If conjecture 1 or 2 fail for function $g(x)$ does it work for some real $v$ and function $g(x) + v$ ?? In particular the function $\exp(x) + v$ ?