Today's problem originates in this conversation with Willie Wong about the Fourier transform of a Gaussian function

$$g_{\sigma}(x)=e^{-\sigma \lvert x \rvert^2},\quad x \in \mathbb{R}^n;$$

where $\sigma$ is a complex parameter. When $\Re (\sigma) \ge 0$, $g_\sigma$ is a tempered distribution$^{[1]}$ and so it is Fourier transformable.

On the contrary, it appears obvious that if $\Re(\sigma) <0$ then $g_\sigma$ is *not* tempered.

Question 1. What is the fastest way to prove this?

My guess is that one should exploit the fact that the pairing $$\int_{\mathbb{R}^n} g_\sigma(x)\varphi(x)\, dx$$ makes no sense for some $\varphi \in \mathcal{S}(\mathbb{R}^n)$. But is it enough? I am afraid that this argument is incomplete.

Question 2. More generally, is there some characterization oftempered functions, that is, functions which belong to the space $L^1_{\text{loc}}(\mathbb{R})\cap \mathcal{S}'(\mathbb{R})$?

The only tempered functions that I know are *polynomially growing functions*. By this I mean the functions of the form $Pu$, where $P$ is a polynomial and $u \in L^p(\mathbb{R}^n)$ for some $p\in[1, \infty]$.

Question 3. Is it true that all tempered functions are polynomially growing functions?

$^{[1]}$ The definition of *tempered distribution* I refer to is the following.

A distribution $T \in \mathcal{D}'(\mathbb{R}^n)$ is called *tempered* if for every sequence $\varphi_n \in \mathcal{D}(\mathbb{R}^n)$ such that $\varphi_n \to 0$ in the Schwartz class sense, it happens that $\langle T, \varphi_n \rangle \to 0$. If this is the case then $T$ uniquely extends to a continuous linear functional on $\mathcal{S}(\mathbb{R}^n)$ and we write $T \in \mathcal{S}'(\mathbb{R}^n)$.