Doubt in understanding Space $\mathscr D(\Omega)$ I was reading about distributions from Rudin. I had 2 doubts in understanding space $\mathscr D(\Omega)$. Here is the relevant section:

6.2 The space $\mathscr{D}(\Omega)$ Consider a nonempty open set $\Omega \subset R^{n}$. For each compact $K \subset \Omega$, the Fréchet space $\mathscr{D}_{K}$ was described in Section 1.46. The union of the spaces $\mathscr{D}_{K}$, as $K$ ranges over all compact subsets of $\Omega$, is the test function space $\mathscr{D}(\Omega)$. It is clear that $\mathscr{D}(\Omega)$ is a vector space, with respect to the usual definitions of addition and scalar multiplication of complex functions. Explicitly, $\phi \in \mathscr{D}(\Omega)$ if and only if $\phi \in C^{\infty}(\Omega)$ and the support of $\phi$ is a compact subset of $\Omega$.
Let us introduce the norms
$$
\|\phi\|_{N}=\max \left\{\left|D^{\alpha} \phi(x)\right|: x \in \Omega,|\alpha| \leq N\right\}\tag1
$$
for $\phi \in \mathscr{D}(\Omega)$ and $N=0,1,2, \ldots$; see Section $1.46$ for the notations $D^{\alpha}$ and $|\alpha|$.
The restrictions of these norms to any fixed $\mathscr{D}_{K} \subset \mathscr{D}(\Omega)$ induce the same topology on $\mathscr{D}_{K}$ as do the seminorms $p_{N}$ of Section $1.46$. To see this, note that to each $K$ corresponds an integer $N_{0}$ such that $K \subset K_{N}$ for all $N \geq N_{0}$. For these $N,\|\phi\|_{N}=p_{N}(\phi)$ if $\phi \in \mathscr{D}_{K} .$ Since
$$
\|\phi\|_{N} \leq\|\phi\|_{N+1} \quad \text { and } \quad p_{N}(\phi) \leq p_{N+1}(\phi)\tag2
$$
the topologies induced by either sequence of seminorms are unchanged if we let $N$ start at $N_{0}$ rather than at $1 .$ These two topologies of $\mathscr{D}_{K}$ coincide therefore; a local base is formed by the sets
$$
V_{N}=\left\{\phi \in \mathscr{D}_{K}:\|\phi\|_{N}<\frac{1}{N}\right\} \quad(N=1,2,3, \ldots)\tag3
$$
The same norms (1) can be used to define a locally convex metrizable topology on $\mathscr{D}(\Omega)$; see Theorem $1.37$ and $(b)$ of Section $1.38$. However, this topology has the disadvantage of not being complete. For example, take $n=1, \Omega=R$, pick $\phi \in \mathscr{D}(R)$ with support in $[0,1], \phi>0$ in $(0,1)$, and define
$$
\psi_{m}(x)=\phi(x-1)+\frac{1}{2} \phi(x-2)+\cdots+\frac{1}{m} \phi(x-m)
$$
Then $\left\{\psi_{m}\right\}$ is a Cauchy sequence in the suggested topology of $\mathscr{D}(R)$, but $\lim \psi_{m}$ does not have compact support, hence is not in $\mathscr{D}(R)$.

(Trascribed from screenshots 1,2,3.)
Doubts:

*

*Why are the topologies on $\mathscr D(\Omega)$ and $\mathscr D_k $
the same?


*Why is $\{\psi_m\}$ a Cauchy sequnce but its limit doesn't have compact support?
I am studying functional analysis on my own with only the help of Math Stackexchange. Any help will be appreciated.
 A: Here is a crash course on the topology of $\mathcal{D}(\Omega)$.
Let $V$ be a vector space over $\mathbb{R}$. I will restrict to real scalars but one can also treat in the same way vector spaces over $\mathbb{C}$.
$V$ is called a topological vector space if it is equipped with a topology $\mathscr{T}$ such that $+:V\times V\rightarrow V$ and $\cdot:\mathbb{R}\times V\rightarrow V$ are continuous. Here $V\times V$ is given the product topology coming from $\mathscr{T}$ for each factor. Likewise $\mathbb{R}\times V$ is given the product topology of the usual topology of $\mathbb{R}$ and the topology $\mathscr{T}$ on $V$.
A map $\rho:V\rightarrow \mathbb{R}$ is called a seminorm on $V$ iff it satisfies the three conditions:


*

*$\forall v\in V, \rho(v)\ge 0$

*$\forall v,w \in V, \rho(v+w)\le \rho(v)+\rho(w)$

*$\forall v\in V, \forall \lambda\in\mathbb{R}, \rho(\lambda v)=|\lambda|\rho(v)$
Let $s(V)$ denote the set of all seminorms on $V$.
Given a subset $A$ of $s(V)$, one can define a topology $\mathscr{T}_A$ on $V$ as follows. First for $v\in V$, $r>0$ and $\rho\in A$, define the "open ball"
$$
B(v,r,\rho)=\{w\in V\ |\ \rho(w-v)<r\}\ .
$$
Now let $\mathscr{T}_A$ be the smallest topology on $V$ which contains the set of all such open balls (i.e., use the collection of these balls as a subbasis for defining a topology). This makes $V$ into a topological vector space (TVS) [Exercise 1: prove this].
A TVS which can be obtained in this way is called a locally convex TVS (LCTVS) [Remark 1: you don't have to prove this, it's a definition]. 
A seminorm $\eta$ on a LCTVS $V$ is called a continuous seminorm iff it is continuous in the usual sense, i.e., as a map between the topological spaces $V$ and $\mathbb{R}$.
If $V$ is given as above, starting from a set of defining seminorms $A$, then the latter property is equivalent to
$$
\exists k\ge 0, \exists \rho_1,\ldots,\rho_k\in A, \exists c_1,\ldots,c_k\ge 0,
\forall v\in V, 
$$
$$
\eta(v)\le c_1\rho_1(v)+\cdots+c_k\rho_k(v)\ .
$$
[Exercise 2: prove this equivalence]
Let $V_1,\ldots,V_n,W$ be LCTVS's. Let $\phi:V_1\times\cdots\times V_n\rightarrow W$ be an $n$-linear map. Give $V_1\times\cdots\times V_n$ the product topology. Then $\phi$
is a continuous map iff for all continuous seminorm $\eta$ on $W$, there exist continuous seminorms $\rho_1,\ldots,\rho_n$ on $V_1,\ldots,V_n$ respectively, such that
$$
\forall v_1\in V_1,\ldots,\forall v_n\in V_n,\ \ \eta(\phi(v_1,\ldots,v_n))\le \rho_1(v_1)\cdots\rho_{n}(v_n)\ .
$$
[Exercise 3: prove this last equivalence too]
Clearly, if the topology of $W$ is given as $\mathscr{T}_A$ for some $A\subset s(W)$, it is enough to check the last condition for $\eta$'s in $A$ only.
Example 1: Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. Let $K$ be a compact subset of $\Omega$. Now take $V=\mathcal{D}_{K,\Omega}$, the space of $C^{\infty}$ functions $\Omega\rightarrow\mathbb{R}$ with support contained in $K$.
Take $A=\{||\cdot||_N\ |\ N=1,2,3\ldots\}$ as in the question. Then $\mathscr{T}_A$ gives $\mathcal{D}_{K,\Omega}$ a LCTVS structure.
Example 2:
Now take instead $V=\mathcal{D}(\Omega)$. Let $B\subset s(V)$ be the set of all seminorms $\rho$ on $\mathcal{D}(\Omega)$, such that for all compact $K\subset\Omega$, $\rho\circ \iota_{K,\Omega}:\mathcal{D}_{K,\Omega}\rightarrow\mathbb{R}$ is a continuous map. Here $\iota_{K,\Omega}$ is the inclusion map of $\mathcal{D}_{K,\Omega}$ into $\mathcal{D}(\Omega)$. Now equip $\mathcal{D}(\Omega)$ with the topology $\mathscr{T}_B$. This is the standard topology of $\mathcal{D}(\Omega)$.
Example 3:
Again take $V=\mathcal{D}(\Omega)$.
Let $\mathbb{N}=\{0,1,\ldots\}$, and denote the set of multiindices by $\mathbb{N}^d$. A locally finite family $\theta=(\theta_{\alpha})_{\alpha\in\mathbb{N}^d}$ of continous functions $\Omega\rightarrow \mathbb{R}$ is one such that for all $x\in\Omega$ there is a neighborhood $V\subset\Omega$, such that $V\cap {\rm Supp}\ \theta_{\alpha}=\varnothing$ for all but finitely many $\alpha$'s.
For $f\in\mathcal{D}(\Omega)$, let
$$
||f||_{\theta}=\sup_{\alpha\in\mathbb{N}^d}\sup_{x\in\Omega}
|\theta_{\alpha}(x)D^{\alpha}f(x)|\ .
$$
Let $C$ be the set
of seminorms $||\cdot||_{\theta}$ where $\theta$ runs over all such locally finite families. Then $\mathscr{T}_C$ is also the standard topology of
$\mathcal{D}(\Omega)$. Namely, $\mathscr{T}_C=\mathscr{T}_B$, where $B$ is the set of seminorms from the previous example [Exercise 4: prove this equality].
Remark 2:
One can prove the above equality of topologies by showing that the identity map is a homeomorphism from $\mathcal{D}(\Omega)$ with the topology $\mathscr{T}_B$ to $\mathcal{D}(\Omega)$ with the topology $\mathscr{T}_C$, using the above criterion of continuity for multilinear maps (for $n=1$).
And for some more practice, Exercise 5: Prove that pointwise multiplication is continuous from $\mathcal{D}(\Omega)\times \mathcal{D}(\Omega)$ with the product topology, to $\mathcal{D}(\Omega)$.
For the solution of the last exercise see: https://mathoverflow.net/questions/234025/why-is-multiplication-on-the-space-of-smooth-functions-with-compact-support-cont/234503#234503
A: Answer for 2): Let $N$ be any positive integer. Since $\phi$ is non-negative it follows that $\psi_m(N+\frac  1 2)\geq \frac  1 N \phi (N+\frac  1 2-N)=\frac  1 N \phi (\frac  12 )$ whenever $m \geq N$. If $\psi = \lim \psi_m $ we get $\psi (N+\frac  1 2) \geq \frac  1 N \phi (\frac  1 2) >0$ for all $N$ . Hence $\psi$ does not have compact support. 
