In the book where I'm studying there is the following exercise.

If $x \in \mathbb{R}^n$, $\varphi \in \mathcal{S}(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$ we define $(u \ast \varphi)(x)=\langle \tau_x \widetilde{\varphi} , u \rangle$, where we place $(\tau_x \widetilde{\varphi})(y):=\widetilde{\varphi}(y-x):=\varphi(x-y)$. Then

(a) $(u \ast \varphi)(x)$ is continuous with respect to $u \in \mathcal{S}'(\mathbb{R}^n)$, with respect to $\varphi \in \mathcal{S}(\mathbb{R}^n)$, and with respect to $x \in \mathbb{R}^n$

(b) $u \ast \varphi$ is a tempered distribution.

(c) If $\psi \in \mathcal{D}(\mathbb{R}^n)$, we have \begin{align*} \langle \psi, u \ast \varphi \rangle = u \left ( \int_{\mathbb{R}^n} \psi(x) (\tau_x \widetilde{\varphi})(\cdot) dx \right ) \end{align*} and extend this identity to case $\psi \in \mathcal{S}(\mathbb{R}^n)$.

HINT: To prove (b), check that $|(u \ast \varphi)(x)| \leq C(1+|x|^2)^N$ by proving an estimate $q_N(\tau_x \varphi) \leq 2^N(1+|x|^2)^N q_N(\varphi)$.

Note that for me there are these definitions. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space of $\mathcal{S}(\mathbb{R}^n)$. We have that the mapping $u \in \mathcal{S}'(\mathbb{R}^n) \longmapsto v=u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one because convergence in $\mathcal{D}(\mathbb{R}^n)$ implies convergence in $\mathcal{S}(\mathbb{R}^n)$, and $u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ determines uniquely $u \in \mathcal{S}'(\mathbb{R}^n)$. Then a distribution $v \in \mathcal{D}'(\mathbb{R}^n)$ is the restriction of an element $u \in \mathcal{S}'(\mathbb{R}^n)$ if and only if there exist $N \in \mathbb{N}$ and a constant $C_N>0$ such that

\begin{align*} |u(\varphi)| \leq C_N q_N(\varphi)=C_N \sup_{x \in \mathbb{R}^n; |\alpha| \leq N} (1+|x|^2)^N |D^\alpha \varphi(x)| , \forall \varphi \in \mathcal{D}(\mathbb{R}^n) \end{align*}

where $q_N(\varphi)$ are seminorm that make $\mathcal{S}(\mathbb{R}^n)$ a Fréchet space. The elements of $\mathcal{S}'(\mathbb{R}^n)$ or their restriction to $\mathcal{D}(\mathbb{R}^n)$ are called tempered distributions.

To prove (a). I thought can be done with an application of the closed graph theorem, proving that

$\tau_a \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \tau_a(\varphi)=\varphi(x-a) \in \mathcal{S}(\mathbb{R}^n)$

$\widetilde{\varphi} \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \widetilde{\varphi}=\varphi(-x) \in \mathcal{S}(\mathbb{R}^n)$

are continuous with respect to convergence in $\mathcal{S}(\mathbb{R}^n)$. Is it correct, no?

Do you have any idea to the point (b) and (c)?

Note that (b) and (c) I tried to show in a different way, as here Tempered distributions and convolution

Thanks for any help

  • $\begingroup$ if $\varphi$ is Schwartz and $u$ is a tempered distribution, then $\varphi \ast u$ is $C^\infty$ and a tempered distribution, and $(\varphi \ast u) . \phi$ (with $.$ the pointwise multiplication) is a Schwartz function. This is how you define the Fourier transform of tempered distributions explicitely : $$FT[u] = \lim_{\varphi \to \delta} \lim_{\phi \to 1} FT[(\varphi \ast u) . \phi]$$ where the limits are in the sense of distributions $\endgroup$
    – reuns
    Jun 11 '16 at 0:42
  • 1
    $\begingroup$ @user1952009 I do not understand why use the Fourier transform, this is not the case. The main problem is to prove that $u \ast \varphi$ is a tempered distribution using estimates in "hint", and poin (c). $\endgroup$
    – user288972
    Jun 11 '16 at 9:50
  • $\begingroup$ (let $n =1$) I was just saying that $h(x) = u \ast \varphi(x)$ is $C^\infty(\mathbb{R})$, since $h'(x) = u \ast \varphi'(x), \ h^{(k)}(x) =u \ast \varphi^{(k)}(x)$, and those are continuous (since $\varphi^{(k)}-\tau_\epsilon \varphi^{(k)} \to 0$ in the Schwartz topology, so that $u(\tau_x \varphi^{(k)}) - u(\tau_{x+\epsilon} \varphi^{(k)}) \to 0$). And $h(x)$ is also clearly tempered, by the definition of $u$ is tempered. Hence, $g(x) = \phi(x) h(x)$ is Schwartz. That's all, and from this we can define the Fourier transform of $u$ explicitely. $\endgroup$
    – reuns
    Jun 11 '16 at 13:03
  • $\begingroup$ @user1952009 I know that there are other ways to prove this result, as in Classical Fourier Analysis by Grafakos, whose proof is similar to that where you try that $u \ast \varphi \in C^\infty$ when $u \in \mathcal{E}'$ is a distribution with compact support and $\varphi$ is test function, using Riemann sums. But my problem is to prove that $u \ast \varphi$ is tempered distribution using estimate in "hint". $\endgroup$
    – user288972
    Jun 11 '16 at 13:05
  • $\begingroup$ I didn't mean to help you with your exercice, just with the concept behind. (and me neither I don't like the closed graph theorem so much :) ). $\ \ \ $ and also I wanted you to remember : regularization of distributions is easy, just consider $\phi . (\varphi \ast u)$. (if $u$ is not tempered, $\varphi$ has to be compact support) $\endgroup$
    – reuns
    Jun 11 '16 at 13:14

Let us start by proving the inequality in the hint: For fixed $x\in\mathbb{R}^{n}$, we have $$ \partial_y^\alpha(\tau_x\tilde{\psi})(y) = \partial_y^\alpha(\psi(x-y)) = (-1)^{|\alpha|}(\partial^\alpha\psi)(x-y) $$ and $$ (1+|x-z|^2)^N \leq (1+2|x|^2+2|z|^2)^N \leq 2^N(1+|x|^2+|z|^2)^N \leq 2^N (1+|x|^2)^N(1+|z|^2)^N. $$ Hence $$ \sup_{|\alpha|\leq N,y\in\mathbb{R}^{n}} (1+|y|^2)^{N} |\partial^\alpha (\tau_x\tilde{\psi})(y)| = \sup_{|\alpha|\leq N,z\in\mathbb{R}^{n}} (1+|x-z|^2)^{N}|\partial^\alpha \psi(z)| \leq 2^N(1+|x|^2)^N q_N(\psi). $$ Now observe that since $\tau_x\tilde{\psi}\in\mathcal{S}(\mathbb{R}^{n})$, for all $u\in\mathcal{S}'(\mathbb{R}^{n})$ and all $N\in\mathbb{N}$, there is $C_N>0$ such that $$ |\psi*u(x)| = |\langle \tau_x\tilde{\psi},u\rangle |\leq C_n q_N(\tau_x\tilde{\psi}) \leq 2^N (1+|x|^2)^N C_N q_N(\psi). $$ Since $u\ast\psi$ is continuous and therefore locally integrable, we get that $u\ast\psi$ is a distribution since, for $\varphi\in\mathcal{D}(\mathbb{R}^{n})$ $$ \langle \varphi, u\ast\psi\rangle = \int_{\mathbb{R}^{n}} \varphi(x)(u\ast\psi)(x)\;{\rm d}x. $$ Using the above estimate for $N=1$, we get $$ |\langle \varphi, u\ast\psi\rangle| \leq \int_{\mathbb{R}^{n}} |\varphi(x)||(u\ast\psi)(x)|\;{\rm d}x \leq 2 C_1 \int_{\mathbb{R}^{n}} (1+|x|^2) \frac{(1+|x|^2)^{n+2}}{(1+|x|^2)^{n+2}} |\varphi(x)|\;{\rm d}x \leq 2 C_1 q_1(\psi) q_{n+2}(\varphi) \int_{\mathbb{R}^{n}} \frac{{\rm d}x}{(1+|x|^2)^{n+1}} \leq C q_{n+2}(\varphi), $$ where $C:=2 C_1 q_1(\psi)\int_{\mathbb{R}^{n}}\frac{{\rm d}x}{(1+|x|^2)^{n+1}}$.

Now the claim should follow from the fact that $\mathcal{D}(\mathbb{R}^{n})\subset\mathcal{S}(\mathbb{R}^{n})$ is dense.


I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either $\mathcal{S}=\mathcal{S}(\mathbb{R}^d)$ to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied ($\mathcal{S}$ is normable or compact, at least I don't think so). Instead, it might just be easier to prove (a) directly. Say you want to prove (a) with respect to $x$. Just take a sequence $x_n$ in $\mathbb R^n$ that converges to $x$, say. Then

$$\lim_{n\to \infty} \langle\tau_{x_n} \tilde{\varphi},u\rangle= \langle\lim_{n\to\to \infty}\tau_{x_n} \tilde{\varphi},u\rangle=\langle \tau_x \tilde{\varphi},u\rangle$$ where the last inequality holds once you prove $\tau_{x_n}\varphi\to \tau_{x}\varphi$ in $\mathcal{S}$, which shouldn't be too hard to prove. For continuity with respect to $u$, you want to prove that the function $u\mapsto u*\varphi(x)$ is continuous for each pair $(x,\varphi)$. This follows by definition of convergence of tempered distribution since $u_n\to u$ as tempered distribution implies that $$\lim_{n\to \infty} \langle\tau_{x} \tilde{\varphi},u_n\rangle= \langle\tau_{x} \tilde{\varphi},u\rangle.$$ I'm too lazy to do (a) w.r.t $\varphi$. But I would imagine it's the same sort of argument.

For (b) and (c), there is a proof in Classical Fourier Analysis by Grafakos, and it's Theorem 2.3.20. in the 3rd edition. Unfortunately I don't completely understand his proof, especially the Riemann sum part.

  • 1
    $\begingroup$ The closed graph theorem is also true in Fréchet spaces, and $\mathcal{S}(\mathbb{R}^n)$ is a Fréchet space. For (b) and (c) I know the proof you say, but here the problem is finding the estimate (as in HINT) $\endgroup$
    – user288972
    Jun 10 '16 at 23:37

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