What is this function's behavior? What is behavior of this $f(a)=(\sqrt a)^{f(\sqrt a)}$ at a given $a\in\Bbb R$?
Does it converge to some explicit value at $a=2$ and if so what is it?
 A: I'm also unsure about an analytic solution—however, there are some interesting observations we can make:


*

*If we write the functional relationship another way,$$f(b^2) = b^{f(b)}$$
we can see that if $f(b)= 2$, then $b^2$ is a fixed point of $f$, since then $f(b^2) = b^{f(b)} = b^2.$ (The case $b=1$ is special, but the conclusion still holds).

*Also for a fixed point $b^2$, we have that $f(f(b)) = f(2)$.

*We know that $f(1) = 1$. Hence because repeated applications of the square root function converge to 1, we can write:
$$f(x) \approx \begin{cases}1 & \text{if }x \approx 1\\ \sqrt{x}^{f(\sqrt{x})}&\text{otherwise}\end{cases}$$
and the result will converge.

*With such approximations, we find that:


*

*$f(2) \approx 1.52040060871$

*$f$ is strictly increasing; there's exactly one point $b$ for which $f(b)=2$. It occurs at $b \approx 2.80449202$

*This point $b$ is significant because it implies that $b^2$ is a fixed point of $f$. Indeed, $b^2 = f(b^2) \approx 7.8651755$  


*Finally, we can plot the function; it looks like this:


*There's a family of similar relations which define other functions:
$$f(a) = a^{k\cdot f(a^k)}$$
The function described here represents the case $k=\frac{1}{2}$.
When $k=1$, we can actually find a closed form for the solution. The functional relation becomes simply $f(a) = a^{f(a)}$, so 
$$\log(f(a)) = \log(a) \cdot f(a) $$
$$\frac{\log(f(a))}{f(a)} = \log(a)$$
$$f(a) = \frac{W[-\log(a)]}{-\log(a)}$$
where $W$ is the Lambert $W$ function (which may be useful for solving this problem more generally.)
A: It seems we have to take $a>0$ anyway. 
For $a<0$ it is undefined and $f(0)=0^{f(0)}$ is also full of contradictions.
$f(1)=1^{f(1)}=1$
So assuming $f$ is continuous and since $\sqrt[n]{a}\to 1$ we can expect $\Large f(a)=\sqrt{a}^{\sqrt[4]{a}^{\sqrt[8]{a}^{\cdots}}}$
However I don't know if we can rewrite this tetration in a simpler way.
