Can you define a real-valued function $f$ using standard arthimetical operations such that $2^{x} < f(f(f(x))) < 2^{2^x}$ for sufficiently large $x\in \mathbb{R}$?
I know that the rule $f(f(x))=2^{x}$ can't be established with standard arthimetical operations, but is it possible to find a function using standard arthimetical operations such that $2^{x}<f(f(f(x)))<2^{2^x}$ for sufficiently large $x\in \mathbb{R}$?
Following half exponential function does the trick. It only uses logarithms, exponentials and a linear function. (And an indicator function if you like.) For convenience I write $\exp_2(x) := 2^x = \exp(x\ln(2))$.
$$ f(x) = \begin{cases} x + 1/2 & \text{if } x \in [0, 1/2] \\ \exp_2(x- 1/2) & \text{if } x \in (1/2, 1] \\ \exp_2(f(\log_2(x))) & \text{if } x \in (1, \infty) \\ \log_2(f(\exp_2(x))) & \text{if } x \in (-\infty, 0) \end{cases} $$
By construction we have $f(f(x)) = \exp_2(x) = 2^x$. Furthermore it is easy to see that $\exp_2(x) > f(x) > x$. Therefore we get
$$2^{2^x} = \exp_2(\exp_2(x)) = f(f(f(f(x)))) > f(f(f(x))) > f(f(x)) = \exp_2(x) = 2^x.$$
The function $f$ is continuous but only piecewise differentiable.