$\mathcal{E}'(\mathbb{R^n}) \subset \mathcal{S}'(\mathbb{R^n})$?

Question

Can I have pls the proof of why is the space of distribution with compact support $\mathcal{E}'(\mathbb{R^d})$ subset in the set of tempered distribution $\mathcal{S}'(\mathbb{R^n})$?

Answer 1

I'm assuming here that your $\mathcal E$ is $C^\infty$. Say $\mathcal S$ is the Schwarz space. Define $i:\mathcal S\to\mathcal E$ by $if=f$.

Show $i$ is bounded.

Calculate $i^*:\mathcal E'\to\mathcal S'$.

Answer 2

By definition of $\mathcal{S}$ we have $\mathcal{S} \subset C^\infty = \mathcal{E}$.

Next, if $\varphi_k \to 0$ in $\mathcal{S}$ then by definition of convergence in $\mathcal{S}$, for every multi-index $\alpha$ and $\beta$, $\sup |x^\alpha \partial^\beta \varphi_k|\to 0.$ Especially, taking $\alpha=(0,\ldots,0),$ one has $\sup |\partial^\beta\varphi_k|\to 0$ for all multi-index $\beta$, which means that $\varphi_k\to 0$ in $\mathcal{E}.$ Thus, if $\varphi_k \to 0$ in $\mathcal{S}$ then $\varphi_k\to 0$ in $\mathcal{E}.$

Take $u\in\mathcal{E}'$ and $\varphi_k\to0$ in $\mathcal{S}$. Since $\mathcal{S}\subset\mathcal{E}=C^\infty,$ the pairing $\langle u, \varphi_k \rangle$ is defined. And since $\varphi_k\to 0$ in $\mathcal{S}$ implies that $\varphi_k\to 0$ in $\mathcal{E}$ we have $\langle u, \varphi_k \rangle \to 0$. Thus, $u$ is continuous also over $\mathcal{S}$. This shows that $\mathcal{E}' \subset \mathcal{S}'.$