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I could solve the problems: We show that $\pi$ is open onto its image $\pi(S)$. Let $\{s\} \subseteq S$, then we have $\{\delta_s\} = \pi(s) \cap (\operatorname{ev}_{1_{\{s\}}})^{-1}(\mathbb C \setminus \{0\})$, where $\operatorname{ev}_{1_{\{s\}}}(\varphi) = \varphi(1_{\{s\}})$ for all $\varphi \in G_S$. Hence $\{\delta_s\}$ is open in $\pi(S)$ and $\pi(S)... 0 The functorial property of the inductive limit makes life much easier: The inclusion$\mathscr D(\Omega) \hookrightarrow C(\Omega)$where the latter space is endowed with the topology of convergence on all compact sets is continuous because so are all$\mathscr D(K)\hookrightarrow C(\Omega)$. Since$C(\Omega)$is Hausdorff (trivially: if$f\neq g$there is$...

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As for the first question, the singleton $\{\varphi_1\}$ is closed because you have just shown that the complement $\mathcal{D}(\Omega)\setminus\{\varphi_1\}$ is open. I suspect that the Hausdorffness argument is not correct, as the equivalent of this in $\mathbb{R}$ would be that $\varphi_1$ and $\varphi_2$ are points, and the set $\varphi_2+U$ would be the ...

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It is correct provided that one understand the notation $\int_{-\infty}^\infty\delta(x)\phi(x)\ dx$ correctly. This is by no means an integral and $\delta$ is not a function with real variables. Also, one should specify in what function space is the function $\phi$ so that your identities would make sense. Any distribution $T$ is infinitely ...

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The first question follows by definition, $H_m(\mathbb{R}^n)$ is the completion of $C_{c}^\infty(\mathbb{R}^n)$ iff (by definition) $H_m(\mathbb{R}^n) = \overline{C_{c}^\infty(\mathbb{R}^n)}^{|| \cdot ||_m}$, i.e. $\forall u \in H_m(\mathbb{R}^n)$ exists $\lbrace u_k \rbrace \subset C_{c}^\infty(\mathbb{R}^n)$ such that $u_k \rightarrow u$ in the $H_m$-norm. ...

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