# Continuous distribution-valued function induces distribution

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 $$$$\langle E,\psi\rangle = \int_{\mathbb{R}^n} \langle E_\eta, \psi \rangle d\mu(\eta),$$$$ 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.

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 $$\sup_{\psi\in A}||\psi||<\infty$$.
• 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? – Joseph Expo Jan 24 at 9:23
• 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 $\mathbb{R}^n$. – Abdelmalek Abdesselam Jan 24 at 15:37