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I'm trying to understand how the linear mappings on the spaces of distributions $\mathcal{D}'(\Omega)$ and $\mathcal{E}'(\Omega)$(dist. with compact support in $\Omega$) are characterized ($\Omega$ open subbset of $\mathbb{R}^n$). If I get it right the topologies that one usually uses (the polar topologies?) are generated by semi-norms of the forms (on $\mathcal{D}'(\Omega)$ for example): $p_{\phi, B}(u)=\sup_{\phi \in B } |(u,\phi(x))|$, where $B$ is a bounded subset of $\mathcal{D}(\Omega)$. So if I want to show that a linear map from $T: \mathcal{D}'(\Omega) \to \mathcal{D}'(\Omega)$ to be continuous is it enough to find two semi-norms and a constant such that $p_{\phi, B}(Tu)\leq C p_{\psi, A}(u)$?

Also I wonder about sequential continuity in these spaces; is in general a sequentially continuous linear map $T: \mathcal{D}'(\Omega) \to \mathcal{D}'(\Omega)$ continuous?

I hope this didn't get messy, pleas ask if you want me to clarify anything!

user202542
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  • I do not think you have to consider limited sets $A$ and $B$. You call "sequential continuity", but it is a continuity with respect to weak* topology in a locally convex space. – user288972 Jul 09 '17 at 17:31
  • Sorry if that was unclear; with sequential continuity I mean $\phi_n \to \phi \implies T(\phi_n) \to T(\phi)$. Is this equivalent to continuity w.r.t w* topology in this case? – user202542 Jul 09 '17 at 17:45
  • Sorry but the operator is $T: \mathcal{D}'(\Omega) \to \mathcal{D}'(\Omega)$, and convergence must be compared to a sequence of distributions $u_k$ (and consider weak* topology on $\mathcal{D}'(\Omega)$). What you say is the sequential continuity of a distribution $u : \mathcal{D}(\Omega) \to \mathbb{C}$. – user288972 Jul 09 '17 at 18:08
  • Yes - you're perfectly right it should be $u_n \to u \implies T(u_n) \to T(u)$. – user202542 Jul 09 '17 at 18:15
  • Yes, and for all $\varphi \in \mathcal{D}(\Omega)$. – user288972 Jul 09 '17 at 18:22
  • @spynes It seems the topology described in the post is not the weak* topology (but the answer is the same). – Pedro Jul 10 '17 at 12:17
  • Yes, infact I have see your answer and upvoted – user288972 Jul 10 '17 at 12:20

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...If I get it right the topologies that one usually uses...

There are two topologies we usually use in $\mathcal{D}'(\Omega)$: the strong topology and the weak star topology.

The "strong topology" is the topology you have described:

  • Let $\mathcal{F}$ be the collection of all bounded subsets of $\mathcal{D}(\Omega)$. For each $B\in\mathcal{F}$, the application $p_{B}:\mathcal{D}'(\Omega)\to\mathbb R$ given by $$p_B(u)=\sup_{\varphi\in B}|\langle u,\varphi\rangle|$$ is a semi-norm on $\mathcal{D}'(\Omega)$. The strong topology on $\mathcal{D}'(\Omega)$ is the topology defined by the family $(p_{B})_{B\in\mathcal{F}}$.

This topology is used, for example, by Schwartz (p. 71) and Horvath (p. 313). (Actually, in Horvath's definition $\mathcal{F}$ is replaced by the collection $\mathcal{G}$ of all weak-bounded subsets of $\mathcal{D}(\Omega)$. But this changes nothing because $\mathcal{D}(\Omega)$ is a locally convex Hausdorff space and thus $\mathcal{G}=\mathcal{F}$ (see Rudin, p. 70).)

The "weak* topology" is the usual one:

  • The weak star topology on $\mathcal{D}'(\Omega)$ is the weakest topology on $\mathcal{D}'(\Omega)$ that makes every evaluation map $\text{ev}_\varphi:\mathcal{D}'(\Omega)\ni u\mapsto \text{ev}_\varphi(u)=\langle u,\varphi\rangle\in \mathbb K$ continuous.

This topology is used, for example, by Rudin (p. 160) and Grubb (p. 27). (Actually, Grubb uses the equivalent definition via seminorms.)

...is in general a sequentially continuous linear map $T: \mathcal{D}'(\Omega) \to \mathcal{D}'(\Omega)$ continuous?

If $\mathcal{D}'(\Omega)$ is equipped with the weak* topology, the answer is yes. In fact, it is a particular case of the following result, whose proof can be found here.

Let $E$ be a separable complete locally convex space. Let $E'$ be equipped with the weak*-topology. Then, any sequentially continuous linear map $T:E'\to E'$ is continuous.

If $\mathcal{D}'(\Omega)$ is equipped with the strong topology, the answer is also yes (because such a space is bornological (see Horvath, p. 314)). In fact, it is a particular case of the following result.

If $X$ is bornological and $Y$ is locally convex, then any sequentially continuous linear operator $T:X\to Y$ is continuous (Narici & Beckenstein, p. 452).

Remark. The topologies defined above have exactly the same convergent sequences (see p. 74 in Schwartz, or p. 358 in Treves). This explains why, in some contexts (like in the applications to partial differential equations), one usually

  • defines only one topology in $\mathcal{D}'(\Omega)$: the simplest one, which is the weak* topology (as Le Dret and Hörmander).

or

Pedro
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    Awesome answer! Thank you so much! – user202542 Jul 10 '17 at 16:47
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    @user202542 You're welcome. There is a thing about the two topologies that maybe I should have written in my original post (namely, that in the distribution space weak star convergence is the same as strong convergence. This explains why some approaches don't care about the different topologies). I've added it in the post as a final remark. – Pedro Jul 11 '17 at 17:35
  • @Pedro It's an aspect I did not know, thanks also for the references. – user288972 Jul 11 '17 at 21:26