We can concluded that $\mathcal{D}(\Omega):=\bigcup_{K \in \mathcal{K}(\Omega)} \mathcal{D}_K(\Omega)$ (where $\mathcal{K}(\Omega)$ denotes the union of all compacts set content in a open subset $\Omega \subset \mathbb{R}^n$) it's a countable union of Fréchet spaces, and $\mathcal{D}(\Omega)$ is a locally convex space not metrizable.
We can also conclude that space of distributions $\mathcal{D}'(\Omega)$ is a locally convex space.
More specifically, for the second question, it's correct to say that $(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ is a dual pair, since $\langle \varphi , u \rangle := u(\varphi)=0$ $\forall u \in \mathcal{D}'(\Omega)$ implies $\varphi=0$ in the sense that since $\mathcal{D}(\Omega) \subset L^1_{loc}(\Omega) \subset \mathcal{D}'(\Omega)$ we have $\int_{\Omega} |\varphi|^2 dx = \langle \varphi , \overline{\varphi} \rangle =0$.
Therefore, $\mathcal{D}'(\Omega)$ is equipped with the weak* topology $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ making it a locally convex space, with topology defined by sufficient (separable) family of seminorm $\mathcal{F}:=\lbrace p_{\varphi}(u):=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace $ which it is the restriction ad $\mathcal{D}'(\Omega) \subset \mathbb{K}^{\mathcal{D}(\Omega)}:=\prod_{x \in \mathcal{D}(\Omega)} \mathbb{K}$ of the product topology or the topology of pointwise convergence. In particular, it is the minimal topology that makes continuous distributions: okay, $\mathcal{D}'(\Omega)$ is a locally convex space, it's correct?
Therefore we say that $u_k \rightarrow u$ if $\langle \varphi , u_k \rangle \rightarrow \langle \varphi , u \rangle$ $\forall \varphi \in \mathcal{D}(\Omega)$. Many authors, call this property "sequential continuity" of a distributions sequence $\lbrace u_k \rbrace \subset \mathcal{D}'(\Omega)$, but it would be more appropriate to call it weak* convergence. I do not know if you agree.
I'm getting another doubt, that in the space of Fréchet $\mathcal{D}_K(\Omega)$ is defined topology $\mathcal{T}_K$, through seminorm
$\displaystyle p_N(\varphi)=\sup_{|\alpha| \leq N} \left \| D^\alpha \varphi \right \|_{K_{N}}$ , $\displaystyle \left \| D^\alpha \varphi \right \|_{K_{N}}:=\sup_{x \in K} |D^\alpha \varphi|$
and $\varphi_k \rightarrow \varphi$ in $\mathcal{D}(\Omega)$ if and only if
(1) $\exists K \in \mathcal{K}(\Omega): \mathrm{supp}(\varphi_k), \mathrm{supp}(\varphi) \subset K$ $\forall k \in \mathbb{N}$.
(2) $D^\alpha \varphi_k \rightarrow D^\alpha \varphi$ uniformly on $K$ $\forall \alpha \in \mathbb{N}^n$
By definition, a distribution is a continuous linear functional than the previous convergence.
In particular, this convergence has compared a locally convex topology $\mathcal{T}$ on $\mathcal{D}(\Omega)$, where it's local base $\mathcal{U}$ is formed by the family of all convex balanced subsets $U$ such that $U \cap \mathcal{D}_K(\Omega)$ is an open subset in $\mathcal{D}_K(\Omega)$.
Now, on $\mathcal{D}(\Omega)$, there is also the weak topology $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega))$ defined by sufficient (separable) family of seminorm $\mathcal{F}=\lbrace p_u(\varphi)=|u(\varphi)| : u \in \mathcal{D}'(\Omega)\rbrace$ and $\varphi_k \rightharpoonup \varphi$ if $u(\varphi_k) \rightarrow u(\varphi)$ $\forall u \in \mathcal{D}'(\Omega)$.
then we can say that $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)) \subset \mathcal{T}$ ?
Best Answer
In Schwartz's Théorie des distributions, chapter III, Théorème VII : $\mathcal D$ is a Montel space, where bounded sets are relatively compact. Then the weak and strong topologies, restricted to bounded sets, coincide, and convergent sequences are the same in these two topologies (and also in weaker Hausdorff topologies). In more concrete terms: whenever a sequence $\varphi_k$ is such that $u(\varphi_k)\to u(\varphi)$ $\forall u\in \mathcal D'$, there does exist a compact $K\subset\Omega$ such that $\varphi_k\to\varphi$ in the Fréchet space $\mathcal D_K(\Omega)$.