I've been doing a lot of research about functional half-iteration, and I posed the following question to myself:
Consider the function $q:\mathbb R\mapsto\mathbb R$ defined as $$q(x)=x^2+1$$ Does $q^{\circ 1/2}$ exist? Does a continuous $q^{\circ 1/2}$ exist? What about a differentiable $q^{\circ 1/2}$?
It seems to me that $q^{\circ 1/2}$ exists, but I don't know how to prove that it exists (I certainly can't find it, since it's probably not an elementary function). So far, I've proven that if it exists and is continuous, then it must be bounded between $x$ and $q(x)$. My intuition tells me that a differentiable solution probably exists... but it may not be differentiable at $x=0$. I've worked out an "almost-graph" of a possible solution, but it's far from a rigorous proof:
Any ideas about how to attack this problem rigorously?
I claim there is a continuous $f$ such that $f \circ f = q$.
Recursively define $a_0 = 0$, $a_1 = 1/2 < 1 = q(a_0)$, and $a_{n+2} = q(a_{n})$.
Now start by taking $f$ to be any continuous increasing function from $[0,a_1]$ onto $[a_1,a_2]$, and define $f$ on $[a_n,a_{n+1}]$ for $n \ge 1$ by $f(x) = t^2+1$ where $x = f(t)$. For negative $x$ we define $f(-x) = f(x)$.
If $f(x) \sim 1/2 + \alpha x^2$ near $x=0$, we'll want $f'(1/2) = 1/\alpha$ to make $f$ differentiable. Thus one possibility is $f(x) = 1/2 + (2-\sqrt{3}) x^2 + 4 \sqrt{3} x^4$ for $0 \le x \le 1/2$.
On the other hand, if I'm not mistaken an analytic $f$ seems not to be possible.