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(My motivation for the following question is to understand the distribution theory)

The space of test functions:

$\mathcal{D}(\mathbb R)= \{\phi:\mathbb R \to \mathbb R : \phi \in C^{\infty}(\mathbb R), \ \text{and support of }\ \phi \ \text{is compact} \}.$ Let us introduce the norms, $$\|\phi\|_{N}=\text{max} \{|D^{\alpha}\phi(x)|: x\in \mathbb R, |\alpha| \leq N \}$$ for $\phi \in \mathcal{D}(\mathbb R)$ and $N=0,1,2,...$.

My Question: (1) How to use these norms to define locally convex metrizable topology on $\mathcal{D}(\mathbb R)$ ? (2) This topology is not complete; but I don't understand the reason; Pick $\phi \in \mathcal{D}(\mathbb R)$ with support in $[0,1], \phi>0$ in $(0,1),$ and define, $$\psi_{m}(x)=\phi(x-1)+\frac{1}{2}\phi(x-2) +...+\frac{1}{m}\phi(x-m);$$ so, my sub-question is: (a) How to verify $\{\psi_{m}\}$ is a Cauchy sequence in $\mathcal{D}(\mathbb R)$ ? (b) How to verify $\lim \psi_{m}$ does not have a compact support ?

I want to add one more question to it i.e. If $\Omega\subset_{open} \mathbb R^n$ and the norms are defined by $$\|\phi\|_{N}=\text{max} \{|D^{\alpha}\phi(x)|: x\in \mathbb \Omega, |\alpha| \leq N \}$$ for $\phi\in\text D(\Omega) \ \text{instead of}\ \text D(\mathbb R)$ then how to show that the metrizable topology on $\text D(\Omega)$ is not complete.

Thanks.

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    $\begingroup$ Any countable separating family of seminorms defines a metrisable locally convex topology on a vector space, that is a standard construction one should be acquainted with. It's similar to defining a metric defining the product topology on a countable product of (nonempty) metric spaces. The topology on $\mathcal{D}(\mathbb{R})$ one usually considers is strictly finer than the topology induced by these seminorms. Re (2)(a), consider $\lVert \psi_n - \psi_m\rVert_N$, concerning (b), the support of $\psi_m$ is $[1,m+1]$, and $\psi_{m+1}$ coincides with $\psi_m$ there. $\endgroup$ – Daniel Fischer Jun 23 '14 at 12:37
  • $\begingroup$ your sequence is cauchy because for any $x\in \mathbb R\ \psi_m(x) $contains atmost one term in $ \phi$ $\endgroup$ – bunny Sep 3 '17 at 6:52
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The standard (or not so standard, as analysis books tend to give messy constructions instead) conceptual way to define the topology is very well explained in Bourbaki's topological vectors spaces.

In general: Let $(E_i,f_{ij})$ be an inductive system of vector spaces (over $\mathbf{C}$, say) such that every $E_i$ is endowed with an LC topology making the maps $f_{ij}:E_i\rightarrow E_j$ continuous. If $E:=\varinjlim E_i$ denotes its inductive limit in $\mathbf{C}-\mathbf{Vect}$, then $E$ is said to be the locally convex inductive limit of $(E_i,f_{ij})$ (inductive limit in $\mathbf{LCS}$) if it is given the final locally convex topology with respect to the family of canonical maps $f_i:E_i\rightarrow E$.

The following is a special case: Consider a countable increasing family $E_1\subset E_2\subset\cdots$ of vector subspaces of a vector space $E$ such that $E=\bigcup_{i\in\mathbf{N}} E_i$ and every $E_i$ be endowed with a locally convex topology such that for every $i\in\mathbf{N}$ the subspace topology induced by $E_{i+1}$ on $E_i$ coincides with the topology of $E_i$. Then $(E_i,f_{ij})$ is an inductive system with respect to the inclusion maps $f_{ij}:E_i\hookrightarrow E_j$ ($i\leqslant j$), and $E$ is its strict locally convex inductive limit. If every $E_i$ is a Fréchet space, then $E$ is said to be an LF space.

In particular, let $\Omega$ be an open subset of $\mathbf{R}^n$; if $K$ is a compact subset of $\Omega$, let $\mathcal{D}_K(\Omega)$ be the subset of $C^\infty_0(\Omega)$ of maps with support contained in $K$. $\mathcal{D}_K(\Omega)$ is a Fréchet space with the topology given by uniform convergence of all derivatives; a generating family of seminorms is given by $\rho_k(\phi):=\sup_{x\in\Omega, |\alpha|\leqslant k}|(\partial^\alpha\phi)(x)|$. Let $K_1\subset K_2\subset\cdots$ be an exhaustion of $\Omega$ by compact subsets. Then $C^\infty_0(\Omega)=\bigcup_{i\in\mathbf{N}}\mathcal{D}_{K_{i}}(\Omega)$ and let $\mathcal{D}(\Omega):=\varinjlim\mathcal{D}_{K_{i}}(\Omega)$ be its strict locally convex inductive limit.

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1) You can use your norms to define a neighborhood base of zero, i.e. the sets $U_{N,\epsilon}:=\{\varphi \in D(\mathbb{R}) : \|\varphi\|_N<\epsilon\}$ as a local base. This topology is obviously metrizeable via a Frechét metric of the form $\rho(\varphi, \psi) = \sum_{N\in \mathbb{N}} \frac{\|\varphi - \psi\|_N}{2^N(1+\|\varphi - \psi\|_N)} $

2)I dont see your series to be a Cauchy sequence, but you can for example take

$\psi_n=\sum_{i=1}^n\frac{\phi(x-i)}{i^2}$ for the same functions $\phi$, then you have (wlog $m>n$)

$\|\psi_m-\psi_n\|_N=\|\sum_{i=n}^m\frac{\psi(x-i)}{i^2}\|_N\leq\sum_{i=n}^m\|\frac{\psi(x-i)}{i^2}\|_N=\| \phi\|_N \sum_{i=n}^m\frac{1}{i^2}$ which is obviously Cauchy for every $N$ as $\phi$ is bounded for every $N$ and the sum gets small. Therefore this sequence is Cauchy in your topology. The last step works because shifting of course does not change your maximum. The limit cannot have compact support because you shift it arbitrarily wide to the outside. Formally:

Because $\phi>0$ by assumption, supp($\psi)\subset$ supp$(\psi_n)$. Assume the support of $\lim_{n\rightarrow \infty}\psi_n$ lies in $[a,b]$, but there obviously exists an $k\in \mathbb{N}$ with $k>b$ and therefore $[k,k+1]$ is contained in the support of $\psi_k$ (because supp$(\phi)=[0,1]$, which is an obvious contradiction.

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you can just use the "Baire Category Theorem" as $\mathbb R=\cup_{i=1}^\infty K_i$ where each $K_i$ is compact and $K_i^o\subset K_{i+1}\forall i\in \mathbb N$ . Now each $D_K$ is a closed subspace of $D(\mathbb R)$ and {$D_K^o=\phi\ \forall\ K\subset_{compact}\mathbb R$} and $D(\mathbb R)=\cup_{K=1}^\infty D_K$ and $D(\mathbb R)$ is metrizable hence by Baire Category theorem it cannot be complete.

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