# Prove that a given distribution is tempered

Suppose I have a distribution $$E$$ such that $$\phi \ast E$$ is square-integrable for all $$\phi \in C_c^\infty \left( \mathbb{R}^d \right)$$. Is it possible to prove that $$E$$ is tempered? It seems plausible to me, but I only get so far:

For brevity, define

$$\begin{equation} G_\phi = \phi \ast E \end{equation}$$

for any $$\phi \in C_c^\infty$$. Now, because convolution is commutative, for all $$\phi, \psi \in C_c^\infty$$ we have

$$\begin{equation} \mathcal{F} \left( \phi \ast \psi \ast E \right) = \tilde{G}_{\phi \ast \psi} = \left( 2 \pi \right)^{d/2} \tilde{\phi} \cdot \tilde{G_\psi} = \left( 2 \pi \right)^{d/2} \tilde{\psi} \cdot \tilde{G_\phi} \end{equation}$$

where the tildes denote Fourier transformed quantities. Then, we get

$$\begin{equation} \frac{\tilde{G}_\phi}{\tilde{\phi}} = \frac{\tilde{G}_\psi}{\tilde{\psi}} =: F \end{equation}$$

which is of course what we expect, since we would like to interpret $$F$$ as the Fourier transform of our distribution $$E$$. We can now deduce that $$F \in L^1_{\mathrm{loc}} \left( \mathbb{R}^d \right) \cap L^2_{\mathrm{loc}} \left( \mathbb{R}^d \right)$$ by

$$\begin{equation} \left \Vert F \right \Vert_{L^1 \left( K \right)} \le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)} \le \left \Vert \frac{1}{\tilde{\phi}} \right \Vert_{L^2 \left( K \right)} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)} < \infty \end{equation}$$

and

$$\begin{equation} \left \Vert F \right \Vert_{L^2 \left( K \right)} \le \left \Vert \frac{1}{\tilde{\phi}} \tilde{\phi} F \right \Vert_{L^1 \left( K \right)} \le \sqrt{\left \Vert \frac{1}{\left \vert \tilde{\phi} \right \vert^2} \right \Vert_{L^\infty \left( K \right)}} \left \Vert \tilde{\phi} F \right \Vert_{L^2 \left( K \right)} < \infty \end{equation}$$

for any compact $$K$$ by using some $$\phi \in C_c^\infty \left( \mathbb{R}^d \right)$$ with strictly positive Fourier transform (These do exist).

I seem to be unable to get any further though. It is now evident, that $$F$$ is in fact a distribution (as it is locally integrable) and it remains to show that we can approximate any Schwartz function under the integral in a (Schwartz)-continuous way.

• Thank you! I removed that statement. – iolo Jun 4 '20 at 12:26
• Got it, thanks. – Nate Eldredge Jun 8 '20 at 14:46

## 2 Answers

In Laurent Schwartz's Théorie des Distributions (page 245, chap. VII, §5) you can find something similar: A distribution $$T\in \mathscr D'(\mathbb R^d)$$ is tempered if and only if all regularizations $$T \ast \varphi\in \mathscr O_M$$ for $$\varphi\in\mathscr D(\mathbb R^d)$$, where $$\mathscr O_M$$ is the space of slowly increasing $$C^\infty$$-functions $$f$$, i.e., for every $$\alpha\in\mathbb N_0^d$$ there is $$k\in\mathbb N$$ such that $$\partial^\alpha f(x)/(1+|x|)^k$$ is bounded.

• Sweet! I will check that reference tomorrow! – iolo Jun 4 '20 at 18:18
• As you accepted this hint to the certainly highly relevant work of Schwartz as an answer: Is it clear that this solves your problem as stated? I guess that the answer is negative but I don't have an example. – Jochen Wengenroth Jun 6 '20 at 11:06
• Hm, I may be wrong (it is getting late again), but since $\phi \ast E \in L^2$ and all its derivatives are as well, we should be able to invoke a Sobolev statement? Thus $\phi \ast E$ as well as its derivatives are in $L^\infty$. Then they are also trivially in $\mathscr O_M$. – iolo Jun 6 '20 at 18:21
• You are right. Schwartz's result indeed implies your statement. – Jochen Wengenroth Jun 7 '20 at 7:42

Edit: even if another answer has been accepted, I edited mine in order to correct (hopefully) the issues raised in the comments, and

• possibly list more easily readable references to Łojasiewicz's solution of the division problem, and
• prove that the tempered distribution $$S$$, found by using Łojasiewicz's division theorem and such that $$\phi\ast S=\phi\ast E$$, is equal to $$E$$.
• prove a stronger property that the one required by the asker: namely, if $$\phi\ast E\in\mathscr{S}^\prime$$ for a single function $$\phi\in C^\infty_c(\Bbb R^n)$$, then $$E\in\mathscr{S}^\prime$$.

The result can be proved by using Stanisław Łojasiewicz's solution of the division problem in $$\mathscr{S}^\prime(\Bbb R^n)$$ (see  and  or  pp. 99-101 or  chapter VI, §VI.1): the equation $$\Phi S=T\label{div}\tag{DIV}$$ has a tempered distribution solution $$S\in\mathscr{S}^\prime(\Bbb R^n)$$ for every non-null real analytic function $$\Phi\in \mathscr{A}(\Bbb R^n)$$ and every datum $$T\in\mathscr{S}^\prime(\Bbb R^n)$$. Indeed, since $$G_\phi=\phi \ast E\in L^2(\Bbb R^n)\qquad \forall \phi\in C_c^\infty(\Bbb R^n), \label{1}\tag{1}$$ we have also that $$G_\phi\in\mathscr{S}^\prime(\Bbb R^n)$$ as a distribution, and thus $$\hat{G}_\phi\in\mathscr{S}^\prime(\Bbb R^n)$$ by the isomorphism theorem for the Fourier transform in $$\mathscr{S}^\prime$$ (see , chapter VII, §7.1, theorem 7.1.10, p. 164). Then we can choose a test function $$\phi\not\equiv 0$$ and, by using the division theorem, find a tempered distribution $$S$$ such that $$\hat{\phi}\hat{S}=\hat{G}_\phi\label{2}\tag{2}$$ since $$\phi\in C_c^\infty(\Bbb R^n) \implies\phi\in \mathscr{E}^\prime(\Bbb R^n)$$ as a distribution, and therefore $$\hat{\phi}\in \mathscr{A}(\Bbb R^n)$$ i.e. $$\hat{\phi}$$ is a complex valued real analytic function (see, for example, , chapter VII, §7.1, theorem 7.1.14 pp. 165-166). Now applying the inverse Fourier transform to both sides of equation \eqref{2} and considering equation \eqref{1} we have $$\phi\ast S=\phi\ast E\iff \phi\ast(S-E)=(S-E)\ast\phi =0 \label{3}\tag{3}$$ Lemma. Equation \eqref{3} implies $$S=E$$.
Proof for a given $$\phi\in C^\infty_c(\Bbb R^n)$$, consider the following convolution equation: $$\DeclareMathOperator{\invs}{\small{inv}}$$ $$\phi\ast\psi(x)=-\varphi(-x)=-\varphi\circ\invs(x) \quad \forall \varphi(x)\in C^\infty_c\Bbb R^N \label{4}\tag{4}$$ where $$\Bbb R^n \ni x\mapsto \invs(x)=-x$$ is the point reflection map. Again by the division theorem, this equation is solvable and its solution, i.e. $$\psi(x)= -\mathscr{F}^{-1} \left[\hat{\phi}^{-1}\right] \ast \varphi\circ\invs(x)$$ apart from being tempered as a distribution, is a $$C^\infty$$ function since it is equal to the convolution of a tempered distribution whit a compactly supported and $$C^\infty$$-smooth function. Now define $$\eta_r\in C_c^\infty(\Bbb R^n)$$, $$r>0$$, as $$\eta_r(x) = \begin{cases} 1 & |x|r+1, \end{cases}$$ Then the family $$\{\psi_r(x)\}_{r>0}=\{\eta_r(x)\cdot\psi(x)\}_{r>0}$$ is a family of compactly supported $$C^\infty$$ functions converging to $$\psi$$.
Now consider the structure of the convolution on the left side of \eqref{3}: we have that $$(S-E)\ast \phi = \big\langle S-E ,\phi(x-\cdot)\big\rangle$$ and thus $$\begin{split} \langle (S-E)\ast \phi, \varphi\rangle & = \int_{\Bbb R^n}\langle S-E ,\phi(x-\cdot)\rangle\varphi(x)\mathrm{d} x\\ & = \left\langle S-E,\int_{\Bbb R^n}\phi(x-y)\varphi(x)\mathrm{d} x\right\rangle\\ & = \big\langle E- S,\phi\ast\varphi\circ\invs\big\rangle \end{split}\label{5}\tag{5}$$ so, again considering the relation \eqref{3} and the definition of the family of compactly supported functions $$\{\psi_r(y)\}_{r>0}$$ we have $$\begin{split} \lim_{r\to+\infty}\big\langle (S-E)\ast \phi,\psi_r\big\rangle=\langle S-E,\varphi\rangle=0\quad \forall \varphi\in C_c^\infty(\Bbb R^n) \end{split}$$ and thus $$E-S=0\iff E=S\;\blacksquare$$.

Finally, the above lemma implies $$E=S\iff E\in\mathscr{S}^\prime$$.

Notes.

• The expositions for Łojasiewicz's solution of the division problem, apart from the original works  and  are the books by Bernhard Malgrange  and Jean-Claude Tougeron  (the latter two deal with the work of Malgrange which generalizes Łojasiewicz's solution to systems and even to $$C^\infty$$ division in some special cases): however, neither of them is particularly readable to the accustomed to the "ordinary" theory of distributions and its application, since the techniques used are more from the theory of analytic sets (varieties) and from the (however related) theory of ideals of smooth functions than from functional analysis. Nevertheless I like the work of Malgrange , partly because of its generality and partly because of its newly improved digital version produced by the Tata Institute of Fundalmental research. However, as stated above, they are not easy read.
• Jochen Bruning and iolo point out that the solution of \eqref{4} is unique if we chose $$\phi$$ in such a way that $$\hat{\phi}(\xi)>0$$ for all $$\xi\in\Bbb R^n$$: this is alway possible by the Paley-Wiener theorem.
• The relation \eqref{5} can also be proved directly by using the standard definition of convolution of distributions: however, using the fact that $$\varphi\in C^\infty_c$$ simplifies bit the formal development.

References

 Lars Hörmander (1990), The analysis of linear partial differential operators I, Grundlehren der Mathematischen Wissenschaft, 256 (2nd ed.), Berlin-Heidelberg-New York: Springer-Verlag, ISBN 0-387-52343-X/ 3-540-52343-X, MR1065136, Zbl 0712.35001.

 Stanisław Łojasiewicz (1959), "Sur le problème de la division" (French),
Studia Mathematica 18, 87-136, DOI: 10.4064/sm-18-1-87-136, MR0107168, Zbl 0115.10203.

 Stanisław Łojasiewicz, Sur le problème de la division, (French), Rozprawy Matematyczne 22, pp. 57 (1961), MR0126072, Zbl 0096.32102.

 Malgrange, Bernard, Ideals of differentiable functions, (English) Studies in Mathematics. Tata Institute of Fundamental Research 3. London: Oxford University Press, pp. 106 (1966), MR0212575, Zbl 0177.17902.

 Jean-Claude Tougeron, Ideaux de fonctions différentiables (French) Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 71. Berlin-Heidelberg-New York: Springer-Verlag. pp. VII+219 (1972), MR0440598, Zbl 0251.58001.

• I do not agree with your first point since $\hat{E}$ is not à priori defined because we do not know from the start whether $E$ is tempered. But since we can perform the Fourier transform (in L^2) and the result is in fact a tempered distribution, this is not an issue. I admit I have trouble reading your refs: First, i fail to see that Łojasiewicz actually writes about tempered distributions, but that might be to a language barrier. Second, it seems we need in fact to use a $\hat{\phi} > 0$ to obtain a unique solution - they exists, so this is again a non-issue. – iolo Jun 5 '20 at 7:15
• @iolo, of course you do not know a priori if $E$ is tempered or not. However, you know that its convolution with any $\phi$ is tempered, and thus you can solve a division problem like \eqref{2} and determine a tempered solution distribution which is unique except for a solution of the homogeneous equation, i.e. $$\phi\ast E_o=0$$ Now, are solutions of this equation tempered? I’ll have a look this evening to this problem. – Daniele Tampieri Jun 5 '20 at 7:54
• I don't see how your edit solves the problem that, a priori, $E$ cannot be Fourier transformed. Anyway, Lojasiewicz's theorem is a tremendous sledge hammer but I am somehow missing the point why it is relevant, here: For all test functions $\varphi$ there is a a tempered distribution $S$ with $\hat S \hat \varphi = \widehat{E\ast \varphi}$ and thus $S\ast \varphi=E\ast\varphi$. Why does this imply $E\in\mathscr S'$? – Jochen Wengenroth Jun 7 '20 at 9:02
• Since $\phi \ast E \in L^2$ it is a tempered distribution and its Fourier transform $T$ is a tempered distribution as well. Now we can pose the division problem $\tilde{\phi} S = T$ for a $\phi \in \mathcal{D}$ with a strictly positive Fourier transform $\tilde{\phi}$, which incidentally is also analytic by Paley-Wiener. Thus the above theorem applies and there exists a (unique, because $\tilde{\phi} > 0$) solution $S$ to the above equation. However, I do not see where Lojasiewicz's theorem states that this solution is tempered. If so, however, it must be the Fourier transform of $E$. – iolo Jun 7 '20 at 11:14
• Okay, eventually I agree: We know $S\ast\varphi=E\ast\varphi$ for some $\varphi\in\mathscr D$ with $\hat\varphi>0$. To conclude $S=E$ is is enough to prove $S\ast\psi=E\ast\psi$ for all test functions $\psi$ (because $E(\psi)=(E\ast \check{\psi})(0)$) and this follows from $\widehat{S\ast\psi}=\widehat{S\ast\varphi\ast\psi}/\hat\varphi=\widehat{E\ast\varphi\ast\psi}/\hat\varphi=\widehat{E\ast\psi}$. – Jochen Wengenroth Jun 8 '20 at 7:45