Functional Equation $ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) $ I have the following functional equation:
$$ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) $$
Under the precondition that $ f ( n ) = \omega ( 1 ) $, monotonicity and an initial value $ f ( 1 ) = \Theta ( 1 ) $, one can show (by induction) that $ \Omega \big( ( \log n ) ^ k \big) = f ( n ) = O ( n ^ \alpha ) $ for arbitrary large $ k $ and small $ \alpha $ (see below).
It seems to me that these bounds are very tight so the question is:

Is there a (simple) function that satisfies these conditions?


Proof sketch:
$$ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) \le 2 \left( \frac n { f ( n ) } \right) ^ \alpha = \left( \frac 2 { f ( n ) ^ \alpha } \right) n ^ \alpha \le n ^ \alpha $$
for sufficiently large $ n $.
$$ f ( n ) = 2 f\left( \frac n { f ( n ) } \right) \ge 2 \left( \log \frac n { f ( n ) } \right) ^ k \ge 2 \left( \log \frac n { n ^ \alpha } \right) ^ k = 2 ( 1 - \alpha ) ^ k ( \log n ) ^ k \ge ( \log n ) ^ k $$
for sufficiently large $ n $ and small $ \alpha $.

Hint:
$f(n)=2f\left(\dfrac{n}{f(n)}\right)$
$\dfrac{f(n)}{f\left(\dfrac{n}{f(n)}\right)}=2$
$\dfrac{\dfrac{n}{f(n)}}{f\left(\dfrac{n}{f(n)}\right)}=\dfrac{2n}{(f(n))^2}$
Let $g(n)=\dfrac{n}{f(n)}$ ,
Then $g(g(n))=\dfrac{2(g(n))^2}{n}$
