Let $X$, $Y$ be i.i.d, that $X+Y$ and $X-Y$ are independent, and that $\varphi_{X}(2u)=(\varphi_{X}(u))^{3}\varphi_{X}(-u)$. Also, let $E\{X\}=0$ and $E\{X^{2}\}=1$. Show that $X$ is Normal $N(0,1)$.
We are given a hint that reads as follows: "Show that for some $\alpha>0$, we have $\phi(u)\neq 0$ for all $u$ with $|u|\leq a$. Let $\psi (u)= \frac{\varphi(u)}{\varphi(-u)}$ for $|u|\leq a$, and show $\psi(u)=\{\psi(u/2^{n})\}^{2^{n}}$; then, show this tends to 1 as $n \to \infty$. Deduce that $\varphi(t)=\{\varphi(t/2^{n})\}^{4^{n}}$, and let $n \to \infty$.
Now, I've done all this, and you wind up with $\varphi(t)\to 1$ as $n \to 0$ by definition of characteristic functions.
WHAT I WANT TO KNOW IS THIS: how does doing all of this tell us that $X$ is Normal $N(0,1)$? At this point, this is the only part of the proof that I have left.
I have to disagree that $\varphi(t) = 1$ for any $t$, since this would be the characteristic function for $X = 0$ a.e. However, suppose that you can conclude that for every $n$, $\varphi(t) = \big( \varphi(t/2^n) \big)^{4^n}$. Perhaps it might be helpful to realize that $\varphi'(0) = 0$ and $\varphi''(0) = -1$. Then $$ \varphi(t/2^n) = \varphi(0)+ \varphi'(0)\frac{t}{2^n} +\frac{1}{2}\varphi''(0) \frac{t^2}{4^n} + O(8^{-n}) = 1 - \frac{t^2}{2\cdot 4^n} + O(8^{-n}) $$ and therefore $$ \varphi(t) = \lim_{n \to \infty} \left( 1 - \frac{t^2}{2\cdot 4^n} + O(8^{-n}) \right)^{4^n} $$ This limit should result in the characteristic function for $N(0,1)$.