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 of tempered 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 $T \in \mathcal{S}'(\mathbb{R}^n)$.


2 Answers 2


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 $k$.

Question 2 You have the structure theorem of tempered distributions. (See Theorem 8.3.1 in Friedlander and Joshi).

Theorem Every tempered distribution is a (distributional) derivative of finite order of some continuous function of polynomial growth.

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.

  • $\begingroup$ 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! $\endgroup$ Commented Apr 23, 2011 at 17:37

Thinking $L^1_{loc}\cap\mathcal{S}'$ as a subset of $\mathcal{D}'$, it is not true that every element of $L^1_{loc}\cap\mathcal{S}'$ is a polynomially growing function.

For example, in $\mathbb{R}$, define

$$f:\mathbb{R}\to\mathbb{R}, t\mapsto \cos(e^t) e^t.$$

Then $f\in L^1_{loc}(\mathbb{R})$, so it represents by integral pairing an element of $\mathcal{D}'(\mathbb{R})$.

Also, define:

$$g:\mathbb{R}\to\mathbb{R}, t\mapsto \sin(e^t).$$

Then $g\in L^\infty(\mathbb{R})$, so it represents by integral pairing an element of $\mathcal{S}'(\mathbb{R})$.

Also, denoting the distributional derivative with the symbol $D$, we have that:

$$\forall\varphi\in\mathcal{D}(\mathbb{R}), f(\varphi)= \int_\mathbb{R}f(t)\varphi(t)\operatorname{d}t =\int_\mathbb{R} \cos(e^t) e^t\varphi(t)\operatorname{d}t \\ = \int_\mathbb{R} \left(\frac{\operatorname{d}}{\operatorname{d}t}\sin(e^t)\right)\varphi(t)\operatorname{d}t =- \int_\mathbb{R} \sin(e^t)\varphi'(t)\operatorname{d}t = -g(\varphi')=Dg(\varphi).$$ So, being $\mathcal{S}'(\mathbb{R})$ closed with respect to distributional derivative, we get that $f=Dg\in\mathcal{S}'(\mathbb{R})$.

However $f$ is not a polynomially growing function, so we have got an example of $f\in L^1_{loc}(\mathbb{R})\cap\mathcal{S}'(\mathbb{R})$ that is not of polynomial growth.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .