# “Tempered” functions?

It happens more and more often that the answers I receive here to one question span another question!

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 $\Re e (\sigma) \ge 0$, $g_\sigma$ is a tempered distribution and so it is Fourier transformable.

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

Question 1: What is the fastest way to prove this? I would say, as a conditioned reflex, that is because $\int g_\sigma(x)\varphi(x)\, dx$ makes no sense for some $\varphi \in \mathcal{S}(\mathbb{R}^n)$. Is it enough? This looks somewhat incomplete as an argument. (*)

Question 2: More generally, is there some characterization of "tempered functions" $L^1_{\text{loc}}(\mathbb{R})\cap \mathcal{S}'(\mathbb{R})$? I know that slowly growing functions (of the form $Pu$ for $P$ polynomial and $u \in L^p$) are tempered. Is the converse true?

(*)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 $\varphi_n \in \mathcal{D}(\mathbb{R}^n)$ s.t. $\varphi_n \to 0$ in the Schwartz class sense, there 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 $T \in \mathcal{S}'(\mathbb{R}^n)$.

-

Question 1 What you have is almost enough. Assume $\Re\sigma \leq -\epsilon < 0$. Test $\exp (-\sigma |x|^2)$ "against" $\phi(x) = \exp( (\sigma+\epsilon/2)|x|^2)$ in the following way: you can construct a sequence of annular cut-off functions $\chi_k$ such that $\chi_k \phi \to 0$ in $\mathcal{S}$ (using the exponential decay of $\phi$) and $\langle g_\sigma, \chi_k\phi\rangle > c > 0$ for all $k$.
If you intersect against $L^1_{loc}$, this just guarantees that the distributional derivative is actually the weak derivative. From this you can conclude that an appropriate version of what you stated is true.
Ok for the annular cut-off argument. Very nice, too! I see that the question is not as trivial as I would have expected. In fact, I thought it was easy to prove something like "if there exists a $\varphi \in \mathcal{S}$ s.t. $f\varphi \notin L^1$, then $f$ is not tempered". I will have a look at that book you are recommending. Thank you for everything, you are helping me quite a bit in those days! –  Giuseppe Negro Apr 23 '11 at 17:37