7
$\begingroup$

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 $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)) \subset \mathcal{T}$ ?

Improve this question
$\endgroup$
8
  • $\begingroup$ What would you like to know about $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)) \subset \mathcal{T}$ ? $\endgroup$
    – Mikhail Katz
    Commented Feb 9, 2016 at 13:45
  • $\begingroup$ If this inclusion is true. Basically, $\mathcal{T}$. It should be understood as the strong topology on $\mathcal{D}(\Omega)$ .At least I think so $\endgroup$
    – Andrew
    Commented Feb 9, 2016 at 14:12
  • 1
    $\begingroup$ "is a dual pair" is not an entirely standard usage. The space of test functions is indeed a "strict inductive limit", or "strict colimit", of Frechet spaces, and such things are called LF-spaces (for "limit of Frechet"). This completely determines the locally convex topology. The weak-dual topology on the space of continuous linear functionals (distributions) is completely determined. Issues about comparison with sequential continuity, or about explicit constructions of topologies by products and so on are secondary... are those the real questions, though? $\endgroup$
    – paul garrett
    Commented Feb 9, 2016 at 14:23
  • 1
    $\begingroup$ The fact that a Cauchy sequence (or bounded Cauchy net) of test functions provably lies in some limitand in the colimit is an example of the fact that bounded subsets of LF spaces must lie in some limitand. Then the topology on the limitands here is the subspace-of-Frechet topology, indeed. (... which is a projective limit of Banach spaces). Sure, there's the weak topology, too. What is the question, though? $\endgroup$
    – paul garrett
    Commented Feb 9, 2016 at 14:27
  • $\begingroup$ Yes, the point is that in the test function space there are two topologies: $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)$ and $\mathcal{T}$ whose respective convergence makes continuous distributions. I wanted to know so if these two topologies are comparable. $\endgroup$
    – Andrew
    Commented Feb 9, 2016 at 14:35

1 Answer 1

Reset to default
6
$\begingroup$

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 $\mathcal D_K(\Omega)$.

Improve this answer
$\endgroup$
0

You must log in to answer this question.