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Given a test function space, in particular $\mathcal{S}=\mathcal{S}(\mathbb{R}^n)$ (the Schwartz space) or $\mathcal{D}=\mathcal{D}(\mathbb{R}^n)$ (the space of compactly supported smooth test functions with its usual topology, as defined for instance here), I understand that generalised functions may be defined as elements of the topological dual space, in our examples resp. $\mathcal{S}'$ or $\mathcal{D}'$.

$\mathcal{S}(\mathbb{R}^n)$ is a metrisable space, hence sequential. Therefore its topological dual is the same as its sequential dual, by which I mean the space of sequentially continuous functionals on $\mathcal{S}$. $\mathcal{D}$, on the other hand, is not metrisable. I recall having seen somewhere that it is not even first-countable (I would welcome verification). Nevertheless, I have a vague notion that for a functional $f$ to belong in $\mathcal{D}'$, it is sufficient that it be sequentially continuous on $\mathcal{D}$. Hence my following questions:

  1. Is it true that sequentially continuous functionals on $\mathcal{D}$ are the same as the continuous ones? Put differently, do the sequential and continuous duals of $\mathcal{D}$ coincide?

  2. Assuming 1 is true, does it follow that $\mathcal{D}$—in spite of not being first-countable—is a sequential space? In other words, do the notions of continuity and sequential continuity coincide for general mappings from $\mathcal{D}$ to an arbitrary topological space $X$?

  3. For general test function spaces which may not be sequential, which is more appropriate: To define generalised functions as elements of their continuous dual space, or of the sequential dual?

  4. Is 3 even relevant (i.e. can such test function spaces be reasonably conceived), given the many requirements that are normally placed on a test function space, such as nuclearity?

Thanks very much in advance.

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    Pouya, unfortunately since MathOverflow is not (yet) a member of the StackExchange 2.0 network, we cannot merge your account there with your one here.2011-12-29
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    Would you care to add definitions of $\mathcal{S}$ and $\mathcal{D}$ as you are using them? It would make your question more self-contained and eliminate any ambiguity.2011-12-29
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    Pouya, for 2. probably you mean $\mathcal{D}$ instead of $\mathcal{D}'$? That's what I assumed in my answer.2011-12-29
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    @nate-eldredge, I have linked to the definitions, thanks.2011-12-29
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    @Vobo, thanks for the answer, and yes, I did mean $\mathcal{D}$ (corrected).2011-12-29
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    Okay, maybe I am stupid, but why is $\mathcal{D}$ not metrizable? I always sort of assumed in the back of my mind that it was, and if I had to guess at a metric, I'd try $$d(f,g) = \sum_{m,k=0}^\infty 2^{-k-m} \sup_{[-m,m]^n} |f^{(k)} - g^{(k)}|.$$ What am I missing?2011-12-30
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    @NateEldredge This is proved for instance in remark 6.9 of Rudin's Functional Analysis. In brief, $\mathcal{D}$ is sequentially complete (Cauchy sequences converge in it), and can be written as a countable union of subspaces $\mathcal{D}(K_i)$ where $K_i$ are increasing compacts such that $\bigcup K_i =\mathbb{R}^n$. The latter subspaces are closed and have empty interior in $\mathcal{D}(\mathbb{R}^n)$. Baire's category theorem then implies that $\mathcal{D}$ is not metrisable.2011-12-30
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    @PouyaTafti: Thanks, I'll look that up. Maybe I am a candidate for [this question](http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics-closed).2011-12-31

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A first general remark: A topological vector space is metrizable, if and only if it is first countable. To answer your questions:

  • Yes, Q1 and Q2 are true. Let $K_i$ be a defining sequence of compact sets for the space $\mathcal{D}$. As the topology on $\mathcal{D}$ is the final topology (i.e. the finest such that all injections $\mathcal{D}_{K_i} \to \mathcal{D}$ are continuous), a map $T$ on $\mathcal{D}$ is continuous, iff its restriction to each $\mathcal{D}_{K_i}$ is continuous. And for this, sequential continuity of $T$ is sufficient (assuming the known fact, that a sequence of test functions converges iff the support of the functions are contained in one $K_i$ and the sequence converges in $\mathcal{D}_{K_i}$.
  • Q3 and Q4: Don't know if there exist relevant examples. I would always require continuity.
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    I believe Q2 is actually false. $\mathcal{D}$ gets its topology from the inductive limit in the _category of locally convex tvs_. The space $\mathcal{D}$ is _not_ sequential. However, sequential continuity is equivalent to continuity when restricted to _linear_ functions. See [here](http://math.stackexchange.com/questions/706061/topologies-of-test-functions-and-distributions#comment1483406_706463).2015-02-05
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    @federico Following your links, I see that $\mathcal{D}$ is not sequential. However, sequential continuity and continuity for (not necessarily linear) maps from $\mathcal{D}$ to another toological space conincide.2015-02-07