How do I Solve $\varphi(t)=(\varphi(t /2^n))^{4^n}$? I am trying to solve the equation $\varphi(t)=(\varphi(t /2^n))^{4^n}$ for $\varphi$. By guessing I see that $\varphi(t)=e^{- \sigma^2t^2/2}$  for all $\sigma>0$ satisfies the equation above. How could I show it in a more precise way? The hint in the book is to compute the limit of the $R.H.S$ of the equation above  as $n \to \infty$. I also know that $\varphi$ is real valued and bounded above by $1$.
Any hints?

If $y(t) = \ln \varphi(t)$, the equation becomes
$$ y(t) = 4^n y(t/2^n) $$
Assume $y$ is twice differentiable at $0$.
By Taylor's theorem the
 right side is $4^n (y(0) + 2^{-n} t y'(0) + 2^{-2n} t^2 y''(0)/2 + o(2^{-2n})$, so for the limit to exist we need $y(0)=0$ and $y'(0)=0$,
and then $y(t) = c t^2$ where $c = y''(0)/2$.  Thus
$$ \varphi(t) = \exp(ct^2)$$ 
