I remember that if $f,g \in \mathcal{S}(\mathbb{R}^n)$ , then it is well-defined \begin{align*} \displaystyle (f \ast g)(x)= \int_{\mathbb{R}^n} g(x-y)f(y)dy=\int_{\mathbb{R}^n} (\tau_x \widetilde{g})(y)f(y)dy \end{align*} where $\widetilde{g}(x)=g(-x)$ e $(\tau_a g)(x)=g(x-a)$, obviously $\tau_x \widetilde{\varphi} \in \mathcal{S}(\mathbb{R}^n)$, and if $u \in \mathcal{S}'(\mathbb{R}^n)$, $\varphi \in \mathcal{S}(\mathbb{R}^n)$, define \begin{align*} \displaystyle (u \ast \varphi)(x)=\langle \varphi(x-y), u(y) \rangle = \langle (\tau_x \widetilde{\varphi})(y), u(y) \rangle , \forall \varphi \in \mathcal{S}(\mathbb{R}^n) \end{align*} The problem is to prove that if $\psi \in \mathcal{S}(\mathbb{R}^n)$ or $\psi \in \mathcal{D}(\mathbb{R}^n)$, and $u \in \mathcal{S}'(\mathbb{R}^n)$, then $\displaystyle \langle \psi , u \ast \varphi \rangle = \langle \psi \ast \widetilde{\varphi} , u \rangle$.

Proof. we assume initially $u \ast \varphi, \psi, \varphi \in \mathcal{S}(\mathbb{R}^n)$, in particular $u \ast \varphi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ and by definition $\langle \psi , u \ast \varphi \rangle$ it defines the distribution determined by $u \ast \varphi$ $\forall \psi \in \mathcal{D}(\mathbb{R}^n)$ or $\forall \psi \in \mathcal{S}(\mathbb{R}^n)$: \begin{align*} \langle \psi , u \ast \varphi \rangle &= \int_{\mathbb{R}^n} (\varphi \ast u)(x) \psi(x)dx \\ &=\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \varphi(x-y)u(y)dy \psi(x) dx \\ &= \int_{\mathbb{R}^n} u(y) \int_{\mathbb{R}^n} \varphi(x-y) \psi(x) dx dy \\ &= \int u(y) dy \int_{\mathbb{R}^n} \widetilde{\varphi}(y-x)\psi(x) dx \\ &= \langle \widetilde{\varphi} \ast \psi , u \rangle \end{align*} Now if $u \in \mathcal{S}'(\mathbb{R}^n)$ is fixed, we consider the inclusions $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$. Since $\mathcal{D}(\mathbb{R}^n)$ is dense $\mathcal{D}'(\mathbb{R}^n)$, it is dense also in $\mathcal{S}'(\mathbb{R}^n)$, i.e. if $\lbrace \varphi_\epsilon \rbrace_{\epsilon >0}$ is standard mollifier, then $u \ast \varphi_\epsilon \rightarrow u$ in $\mathcal{S}'(\mathbb{R}^n)$ and that is the limit (i) $\lim_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n} (u \ast \varphi_\epsilon) \psi(x) dx =\langle \psi, u \rangle$ $\forall \psi \in \mathcal{D}(\mathbb{R}^n)$, Therefore we can apply the previous case. Similarly it is proved that the thesis is true $\forall \psi \in \mathcal{S}(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$, since (i) worth for the approximation theorem in Lebesgue spaces, in other words $\mathcal{S}(\mathbb{R}^n)$ is dense in $\mathcal{S}'(\mathbb{R}^n)$.

I would be grateful if you are confirmed that this proof is correct

  • 1
    $\begingroup$ Hmmm, in your equation series you have assumed only $u\in\mathcal{S}'$, but you treat $u$ as a function $u(y)$, formally this needs clarification. But you are on the right track. You should also be careful in the mollifier approximation, $\mathcal{D}$ is dense in any of your mentioned spaces - but each of these spaces carries a different topology (i.e. the topology on $\mathcal{S}'$ is different as the one inherited from $\mathcal{D}'$). $\endgroup$
    – Vobo
    Mar 22, 2016 at 20:37
  • $\begingroup$ @Vobo What do you think of the answer? $\endgroup$
    – user288972
    Apr 1, 2016 at 18:39

1 Answer 1


In reference to the comments, that is, of the various topologies. It is known that the Schwartz class $\mathcal{S}(\mathbb{R}^n)$ is a Fréchet space and also that the space of test functions $\mathcal{D}(\mathbb{R}^n)$ is dense in $\mathcal{S}(\mathbb{R}^n)$. In particular $\mathcal{D}_K(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow L^1(\mathbb{R}^n)$ are continuous inclusions. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space $\mathcal{S}(\mathbb{R}^n)$. Now 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

  1. $\displaystyle |u(\varphi)| \leq C_N q_N(\varphi)=C_N \sup_{x \in \mathbb{R}^n; |\alpha| \leq N} (1+|x|^2)^N |D^\alpha \varphi(x)|$ for all $\varphi \in \mathcal{D}(\mathbb{R}^n)$

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. On $\mathcal{S}'(\mathbb{R}^n)$ we consider the topology $\sigma(\mathcal{S}'(\mathbb{R}^n), \mathcal{S}(\mathbb{R}^n))$, so that $u_k \rightarrow u$ in $\mathcal{S}'(\mathbb{R}^n)$ means that

  1. $\displaystyle \langle \varphi, u_k \rangle\rightarrow \langle \varphi, u \rangle$ for all $\varphi \in \mathcal{S}(\mathbb{R}^n)$.

In other words, according to this definition of tempered distributions, the limit in (i) applies in any case: if $\psi \in \mathcal{D}(\mathbb{R}^n)$ or if $\psi \in \mathcal{S}(\mathbb{R}^n)$, because they are precisely restrictions of distributions.

I think this should formalize the last part of the proof.


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