We want to prove the following proposition:
Let $X$ and $Y$ be two random variables independent and identically distributed with variance $\sigma^2.$ Let $\alpha,\beta \in \mathbb{R}$ such that $\alpha\beta \neq0$ and $\alpha^2+\beta^2=1.$ Suppose that $\alpha X+\beta Y$ and $X$ have the same distribution. Prove that $X$ is normal $N(0,\sigma^2).$
Following these steps: let $\phi$ be the characteristic function of $X,$ and $\psi(x)=\frac{\phi'(x)}{\phi(x)}.$
1) Prove that $\psi$ is a $C^1$ function, compute $\psi(0)$ and $\psi'(0).$
2) If W is a random variable such that $P_W=\alpha^2\delta_{\alpha}+\beta^2\delta_{\beta}.$ Prove that $$\forall x \in \mathbb{R},\psi'(x)=E[\psi'(Wx)].$$
3) Deduce $\psi'=-\sigma^2.$
4) Conclude.
Attempt:
1) Since $E[X^2]<+\infty,$ then $\phi$ is a $C^2$ function and hence $\psi$ is a $C^1$ function, we have $(\alpha+\beta-1)E[X]=0$ then $E[X]=0$, because $\alpha+\beta \neq 1$, which means that $\psi(0)=0$ and $\psi'(0)=-\sigma^2.$
2) We have $\forall x \in \mathbb{R},\phi(\alpha x)\phi(\beta x)=\phi(x),$ then $\alpha\phi'(\alpha x)\phi(\beta x)+\beta \phi(\alpha x)\phi'(\beta x)=\phi'(x)$ and $\alpha^2\phi''(\alpha x)\phi(\beta x)+2\alpha \beta \phi'(\alpha x)\phi'(\beta x)+\beta^2\phi(\alpha x)\phi''(\beta x)=\phi''(x),$ in conclusion $$\alpha^2E[\psi'(\alpha x)]+\beta^2E[\psi'(\beta x)]=\frac{\alpha^2\phi''(\alpha x)\phi(\beta x)\phi(x)-\alpha^2(\phi'(\alpha x)\phi'(\beta x))^2+\beta^2\phi''(\beta x)\phi(\alpha x)\phi(x)-\beta^2(\phi'(\beta x)\phi'(\alpha x))^2}{(\phi(\alpha x)\phi(\beta x))^2} $$$$=\frac{\phi''(x)\phi(x)-(\phi'(x))^2}{(\phi(x))^2}=\psi'(x)$$
4) $\psi'(x)=-\sigma^2$ which means $\psi(x)=-\sigma^2x$ and then $\phi(x)=e^{-\frac{1}{2}\sigma^2x^2}.$
I am having a problem with part 3), how to deduce that $\psi'(x)=-\sigma^2$ and also he defined $\psi$ by dividing by $\phi$, so why do we have $\forall x \in \mathbb{R},\phi(x) \neq 0$.
Suppose for a contradiction that there exists $t_0$ such that $\phi(t_0)=0$. Since $\phi(t)=\phi(\alpha t)\phi(\beta t)$, there exists $t_1\in\{\alpha t_0,\beta t_0\}$ such that $\phi(t_1) = 0$. Inductively, we may then construct a sequence $\{t_n\}$ with $t_n\in\{\alpha t_{n-1},\beta t_{n-1}\}$ such that $\phi(t_n)=0$ for all $n$. Since $\alpha^2 + \beta^2 = 1$ and $\alpha,\beta\notin\{0,1\}$, it follows that $r:=\max\{|\alpha|,|\beta|\}<1$, and obviously $|t_n| \le r|t_{n-1}|$, so we therefore have $|t_n|\le r^n|t_0|\to0$ as $n\to\infty$. But since $\phi$ is continuous at zero, we then have $1=\phi(0)=\lim_{n\to\infty}\phi(t_n)=0$, a contradiction. Hence $\phi$ is never zero, so $\psi$ is continuously differentiable.
The proof that $\psi'=-\sigma^2$ is quite similar. Suppose for a contradiction that there exists $t_0$ such that $\psi'(t_0)\neq-\sigma^2$. Assume without loss of generality that $\operatorname{Re}\psi'(t_0) \ge -\sigma^2+\epsilon$ (we can repeat the exact same argument for $\operatorname{Re}\psi'(t_0) \le -\sigma^2-\epsilon$, $\operatorname{Im}\psi'(t_0) \ge \epsilon$ and $\operatorname{Im}\psi'(t_0) \le -\epsilon$). Observe that $\psi'(t_0)=\alpha^2\psi'(\alpha t_0) + \beta^2\psi'(\beta t_0)$ is a convex combination of $\psi'(\alpha t_0)$ and $\psi'(\beta t_0)$, and so at least one of $\psi'(\alpha t_0),\psi'(\beta t_0)$ must have real part no smaller than $-\sigma^2+\epsilon$. That is, we have found $t_1\in\{\alpha t_0,\beta t_0\}$ such that $\operatorname{Re}\psi'(t_1) \ge -\sigma^2+\epsilon$. So once again we have a sequence $\{t_n\}$ with $|t_n|\le r|t_{n-1}|$ satisfying $\operatorname{Re}\psi'(t_n)\ge-\sigma^2+\epsilon$ for all $n$. Since $t_n\to0$ as $n\to\infty$ and $\psi'$ is continuous, we find $-\sigma^2=\operatorname{Re}\psi(0)=\lim_{n\to\infty}\operatorname{Re}\psi'(t_n) \ge -\sigma^2+\epsilon$, a contradiction. Hence $\psi'(t)=-\sigma^2$ for all $t$, completing the proof.