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Given $\Omega\subseteq\mathbb{R}^n$, consider the Frechet space $C^{\infty}(\Omega)$ with the family of seminorms $$P_n(f)=\sup_{|\alpha|\leq n,\,x\in K_n}|D^{\alpha}f(x)|,\quad n\geq0,$$ where $\{K_j\}_{j=0}^{\infty}$ is a fixed family of compact subsets of $\Omega$ satisfying $\Omega=\cup_{j=0}^{\infty}K_j$ and $K_j\subseteq \text{Int}(K_{j+1})$ for all $j\geq0$.

Given a compact subset $K$ of $\Omega$, define $$D_K=\{f\in C^{\infty}(\Omega):\,\text{support}(f)\subseteq K\}.$$ With the topology induced by $C^{\infty}(\Omega)$, $D_K$ is a Frechet space. Observe that in this space the topology is also induced by the seminorms $$\Vert f\Vert_n=\sup_{|\alpha|\leq n,\,x\in \Omega}|D^{\alpha}f(x)|,\quad n\geq0.$$

Define $$D(\Omega)=\bigcup_{K\text{ compact in }\Omega}D_K.$$ I was told the following definition of convergence in $D(\Omega)$: we say that $\{\varphi_n\}_{n=1}^{\infty}\subseteq D(\Omega)$ converges to $\varphi\in D(\Omega)$ if there exist $K$ compact in $\Omega$ and $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ we have $\varphi_n\in D_K$, $\varphi\in D_K$ and $\varphi_n\rightarrow\varphi$ in $D_K$.

I would like to know if that definition is equivalent to the following one: we say that $\{\varphi_n\}_{n=1}^{\infty}\subseteq D(\Omega)$ converges to $\varphi\in D(\Omega)$ if $\varphi_n\rightarrow\varphi$ in $C^{\infty}(\Omega)$.

I would appreciate if your ideas use the concepts that I have developed above.

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Choose $\psi \in D(\Bbb R)$ with support $[-1,1]$ and let $\phi_n(x) = \frac1n \psi(x/n)$. Then $\phi_n$ is in $D(\Omega)$ and has support $[-n,n]$ and $\|\phi_n\| \le \frac{\|\psi\|}{n} \to 0$, also for each $k\in \Bbb N$, $\|\phi_n^{(k)}\| = \| \frac1{n^{k+1}} \psi^{(k)}(x/n)\| \le \frac{\|\psi^{(k)}\|}{n^{k+1}} \to 0$.

This shows that $\phi_n \to 0$ in $C^{\infty}(\Omega)$, although $\phi_n$ does not converge in $D(\Omega)$ because its support grows arbitrarily large.

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  • $\begingroup$ Thank you for your answer! I would like to ask you the following: I was defined a distribution as a linear map $\Delta:D(\Omega)\rightarrow\mathbb{C}$ such that, if $\{\varphi_n\}_{n=1}^{\infty}\subseteq D(\Omega)$ converges in $D(\Omega)$ to $\varphi$, then $\Delta(\varphi_n)\rightarrow \Delta(\varphi)$ (essentially, $\Delta|_{D_K}:D_K\rightarrow\mathbb{C}$ is continuous). We write $\Delta\in D'(\Omega)$. Do we have $C^{\infty}(\Omega)'\subseteq D'(\Omega)$? $\endgroup$ – user39756 Oct 31 '16 at 16:43
  • $\begingroup$ @user39756 my pleasure! I guess this boils down to whether or not convergence in $D(\Omega)$ implies convergence in $C^{\infty}(\Omega)$. Which is true if my brain is functioning well at the moment. $\endgroup$ – user384138 Oct 31 '16 at 20:46

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