Characterisation of Gaussian measure on $\mathbb{R}$ The following Proposition is from Da Prato and Zabczyk's "Stochastic Equations in Infinite Dimensions":

Proposition 2.5. A measure $\mu$ on $\mathbb{R}$ is of the form $\mathscr{N}(0,q)$ for some $q \geq 0$, if and only if, for arbitrary independent real valued random variables $\xi$ and $\eta$ with law $\mu_\xi = \mu_\eta = \mu$ and numbers $\alpha, \beta$ such that  $\alpha^2+\beta^2=1$, we have the law $\mu_{\alpha \xi + \beta \eta} = \mu$.

($\Rightarrow$) direction is easy as we already know $\xi$ and $\eta$ follow normal distribution independently. How should I proceed the other direction? The book suggests to use characteristic functions. Take the characteristic functions of $\alpha \xi + \beta \eta$ we have
\begin{equation}
\varphi_\mu(t) = \varphi_{\alpha \xi + \beta \eta}(t) = \varphi_{\xi}(\alpha t) \varphi_\eta (\beta t) = \varphi_\mu(\alpha t)\varphi_\mu(\beta t)
\end{equation}
How should we deduce that $\varphi_\mu$ has the form of Gaussian? Thank you for any hint or suggestion!
 A: A posteriori, you know that you want to prove that $\varphi_\mu(t)=e^{-ct^2}$ for some $c\ge 0$ (either because you can somehow predict it, or because it is explicitly written in your statement). It motivates looking at $g(t)=\log \varphi_\mu(t)$ instead. Now, your observation can be written
$$g(t)=g(\alpha t)+g(\beta t),$$
for any $t$ and $\alpha^2+\beta^2=1$. In particular, for $\alpha=1/\sqrt{2}$, this gives for any $t$,
$$g(t)=2 g(t/\sqrt{2}).$$
Do you see what kind of function $g$ can satisfy this type of equality? (You can take derivatives.)
A: $\def\ph{\varphi}$I will use $\ph(t)=\int e^{itx}\mu(dx)$ to denote the characteristic function of $\mu$. We are given that
$$
\forall \alpha,\beta\in \mathbb R:\quad \alpha^2+\beta^2=1\implies \phi(t)=\phi(\alpha t)\phi(\beta t)
$$
The following statements hold for all  $t\in \mathbb R$.
Step 1: $\phi(-t)=\phi(t)$.

Proof: Take $\alpha=-1,\beta=0$.

Step 2: $\phi(t)$ is real.

Proof: $\overline{\phi(t)}=\overline{\int e^{itx}\mu(dx)}=\int \overline{e^{itx}}\mu(dx)=\int e^{i(-t)x}\mu(dx)=\ph(-t)=\ph(t)$.

Step 3: $\phi(t)> 0.$

Proof: Taking $\alpha=\beta=1/\sqrt{2}$, we get $\ph(t)=\ph(t/\sqrt{2})^2$, immediately implying $\ph(t)\ge 0$. To get strict inequality, we apply this relation $n$ times, yielding $$\phi(t)=\phi(t/(\sqrt 2)^n)^{2^n}.$$ Since $\lim_{t\to 0}\phi(t)=1$ (true for all characteristic functions), there exists $n$ large enough so $\phi(t/(\sqrt 2)^n)>0$. For that $n$, the above implies $\phi(t)>0$.

Step 4: $\phi(t)=\exp(-ct^2)$ for some $c\ge 0$.

Define
$$
f:[0,\infty)\to \mathbb C\\
f(t)=\log \ph(\sqrt{t})
$$
Because of steps $2$ and $3$, the range of $\ph$ over the domain $[0,\infty)$ is in the domain of $\log$, so this is well-defined. The functional equation for $\phi$ translates into this equation for $f$:
$$
\alpha^2+\beta^2=1\implies f(t)=f(\alpha^2t) + f(\beta^2 t)
$$
Now, for any $v,w\in [0,\infty)$, we can let $t=v+w$, $\alpha=\sqrt{\frac{v}{v+w}}$, and $\beta=\sqrt{w\over v+w}$, and the above implies
$$
\forall v,w\ge 0: f(v+w)=f(v)+f(w)
$$
Therefore, $f$ satisfies a one-sided version of the Cauchy functional equation. Since we also know that $f$ is continuous, being a composition of the continuous functions $\sqrt{\cdot}, \ph$ and $\log$, we conclude that $f(t)=ct$ for some $c\in \mathbb R$. This immediately implies
$$
\forall t>0:\log \ph(\sqrt{t})=ct\implies \forall t>0:\ph(t)=\exp(ct^2)
$$
Finally, the fact that $\ph(t)=\ph(-t)$ allows us to extend the above to conclude $\ph(t)=\exp(ct^2)$ for all $t\in \mathbb R$. In order for this to be a valid characteristic function, we need $c<0$.

Step 5: Conclude $\mu$ is normal.

Proof: This follows from the bijective correspondence between characteristic functions and distributions, and the fact that a normal distribution with mean zero, variance $\sigma^2$, has characteristic function $\exp(-(\sigma^2/2)t^2)$.

