I was looking for a simple explanation of why the Gaussian random variable can be the only distribution appearing in the Central limit theorem.
From the statement of the Central limit theorem, it is pretty natural to expect that if $X$ is the limiting random variable in the Central limit theorem, then $X$ must be infinitely divisible, that is
\begin{equation}\tag{1} X\stackrel{d}{=}X^{(n)}_1+\dots+ X^{(n)}_n, \quad\text{for all } n\in\mathbb N, \end{equation}
where $(X^{(n)}_i)_{i\in\mathbb{N}}$ is a sequence of i.i.d. random variables for all $ n\in\mathbb N$.
We can also assume that $$\mathbb{E}[X]=0 \qquad\text{and}\qquad \mathbb{Var}[X]=1. \tag{2}$$
This and the infinite divisibility assumption, immediately implies that
$$\mathbb{E}[X^{(n)}_1]=0,\quad \text{for all } n\in\mathbb{N}.$$
Finally, it is natural to expect from the statement of the central limit theorem that the law of $X$ is the same as the law of $X^{(n)}_1$ up to a difference in the variance. In particular, since the variance of $X^{(n)}_1+\dots+ X^{(n)}_n$ is equalt to $n$ times the variance of $X^{(n)}_1$, we get that
$$\mathbb{Var}[X^{(n)}_1]=\frac 1 n,\quad \text{for all } n\in\mathbb{N}.$$
This discussion leads to the additional assumption that
$$X\stackrel{d}{=}\sqrt{n}\,X^{(n)}_1, \quad\text{for all } n\in\mathbb N. \tag{3}$$
Now my question is, if $X$ satisfies the assumptions in Equations $(1), (2),$ and $(3)$, is it true that $X$ must be a Gaussian random variable with zero mean and unit variance?
If not, under what additional assumption on $X$ we can conlude that $X$ must be a Gaussian random variable with zero mean and unit variance? For instance, I am scared that we also need to assume that all the moments of $X$ are finite, but is this enough?