Can the characteristic function of $\phi(t)$ satisfying $\phi(t)=\phi^2\left(\frac{t}{\sqrt{2}}\right)$ conclude it as C.F. of normal distribution? Suppose $\phi(t)$ is a characteristic function of a random variable $X$. It satisfies $$\phi(t)=\phi^2\left(\frac{t}{\sqrt{2}}\right)$$
Further we know $\phi \in C^2(R)$.
Can we derive $\phi(t)$ is a characteristic function of a Gaussian random variable?
First since $\phi'(t) = \sqrt{2} \phi\left(\frac{t}{\sqrt{2}}\right) \phi'\left(\frac{t}{\sqrt{2}}\right)$, $\phi'(0)=0$, thus $E[X]=0$. For $X \sim N(0, \sigma^2)$, its characteristic function exactly satisfies that equation. But I'm stuck to prove the other side. Thanks for any help!
 A: $$
\phi(t)=\phi^2\left(\frac{t}{\sqrt{2}}\right)=...
=\phi^{2^n}\left(\frac{t}{(\sqrt{2})^{n}}\right)=\left(1+\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)^{\frac{1}{\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1} 2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)}
$$
As we know:
$$
\lim_{n\to \infty} \left( 1+\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1 \right)^{\frac{1}{\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1}}=e
$$
So
$$
\phi(t)=e^{\lim_{n\to \infty}2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)}
$$
Since $\phi(t)=1+\frac{1}{2}\phi''(0)t^2+o(t^2)$, set $\frac{t}{(\sqrt{2})^{n}}=x$, then $n=\frac{2\ln(\frac{t}{x})}{\ln2}$. We have:
$$
\lim_{n\to \infty}2^n\left(\phi\left(\frac{t}{(\sqrt{2})^{n}}\right)-1\right)=\lim_{x\to 0} (\frac{t}{x})^2(\phi(x)-1)
=\frac{1}{2}\phi'’(0)t^2
$$
A: Let $\left(Y_i\right)_{i\geqslant 1}$ be an i.i.d. sequence such that $Y_1$ has the same distribution as $X$. Let 
$$
Z_n:=2^{-n/2}\sum_{i=1}^{2^n}Y_i.
$$
By the functional equation, $Z_n$ has the same distribution as $X$. Since $\phi\in C^2$, $X$ has a finite variance. Combining this with the already shown fact that $X$ is centered gives, by the central limit theorem, that $X$ has a normal distribution. 
