Suppose that the map $\mathbb{R}^n \to \mathcal{D}'(\mathbb{R}^n), \hspace{3mm}\eta\mapsto E_\eta$ is continuous. Furthermore let $\mu$ be a Radon-measure with compact support.

I'm having trouble showing that the functional $E$ on $\mathcal{D}(\mathbb{R}^n)$, whose action on a test function $\psi$ is defined by \begin{equation} \langle E,\psi\rangle = \int_{\mathbb{R}^n} \langle E_\eta, \psi \rangle d\mu(\eta), \end{equation} is in fact a distribution.


The question is a bit ambiguous because the topology on $\mathcal{D}'(\mathbb{R}^n)$ is not specified when mentioning that $\eta\rightarrow E_{\eta}$ is continuous. I don't see immediately if $E$ is indeed continuous when the chosen topology is the weak-$\ast$ topology. However, if the topology is the strong one (a better choice anyway) then the answer is yes as is shown below.

Firstly, by writing $\mu$ as a difference of positive measures one can reduce the problem to the case where $\mu$ is positive. Moreover, by assumption there is a compact set $K$ on which it is supported so $$ \langle E,\psi\rangle=\int_K \langle E_\eta,\psi\rangle d\mu(\eta)\ . $$ Let $A$ be a bounded set in $\mathcal{D}(\mathbb{R}^n)$ and for a distribution $T$ let us use the notation $$ ||T||_A=\sup_{\psi\in A} |\langle T,\psi\rangle|\ . $$ These seminorms define the strong topology on $\mathcal{D}'(\mathbb{R}^n)$ and are thus continuous. By composition, the map $K\rightarrow [0,\infty)$, $\eta\mapsto ||E_\eta||_A$ is continuous and therefore, by compactness, is bounded by some constant $M$.

Now let $(\psi_n)$ be a sequence that converges to $\psi$ in $\mathcal{D}(\mathbb{R}^n)$. Let $A=\{\psi_n\ :\ n\in\mathbb{N}\}$ which is bounded. We have $|\langle E_{\eta},\psi_n\rangle|\le M$ for all $n\in\mathbb{N}$ and $\eta\in K$. Moreover, for fixed $\eta$, $\langle E_\eta,\psi_n\rangle\rightarrow \langle E_\eta,\psi\rangle$. Therefore, by dominated convergence $$ \langle E,\psi_n\rangle\rightarrow \langle E,\psi\rangle $$ and $E$ is a distribution.

Addendum recalling some prerequisites:

Here the space of smooth compactly supported functions $\mathcal{D}(\mathbb{R}^n)$ is a topological vector space. Its topology is defined by an uncountable collection of seminorms which are described explicitly here. As for the notion of bounded set in $\mathcal{D}(\mathbb{R}^n)$, the following are equivalent, for $A\subset\mathcal{D}(\mathbb{R}^n)$:

  1. $A$ is bounded
  2. For every open set $V\subset\mathcal{D}(\mathbb{R}^n)$ which contains the origin, there exists $\lambda>0$ such that $\lambda A\subset V$.
  3. For every continuous seminorm $||\cdot||$ on $\mathcal{D}(\mathbb{R}^n)$, we have $\sup_{\psi\in A}||\psi||<\infty$.
  4. For every seminorm $||\cdot||$ in a defining set of seminorms (e.g., the ones in the MO link I gave above), we have $\sup_{\psi\in A}||\psi||<\infty$.
  • $\begingroup$ Thanks for the reply. Could you specify what you mean when a set is bounded in $\mathcal{D}(\mathbb{R}^n)$? Are you using the supremum norm or is the topology also induced by a family of seminorms? $\endgroup$ – Joseph Expo Jan 24 at 9:23
  • $\begingroup$ I edited my answer to recall what a bounded set is. If by supremum norm you mean $\sup_{x\in\mathbb{R}^n}|\psi(x)|$, that's not what I was using. The notion of bounded set here involves a supremum over a set of test functions, not over a set of points in $\mathbb{R}^n$. $\endgroup$ – Abdelmalek Abdesselam Jan 24 at 15:37

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