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Let us consider the Euclidean space $\mathbb{R}^n$ of any $n \in \mathbb{N}$.

Then, I somehow vaguely recall that the Schwartz space $\mathcal{S}(\mathbb{R}^n)$ is dense in the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$.

Here, two questions arise, on which I have not been able to find satisfactory answers.

I am aware that two seemingly different topologies are possible $\mathcal{S}'(\mathbb{R}^n)$ : namely, the weak$^*$ topology and strong dual topology, cf. Wiki. With respect to which of these two topologies is $\mathcal{S}(\mathbb{R}^n)$ dense in $\mathcal{S}'(\mathbb{R}^n)$?

The following post ME says that test functions may NOT be sequentially dense in the distribution space. In the case of $\mathcal{S}(\mathbb{R}^n)$ and $\mathcal{S}'(\mathbb{R}^n)$, do we have sequential density at least w.r.t one of the two topologieis on $\mathcal{S}'(\mathbb{R}^n)$?

Could anyone please clarify for me?

Keith
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    Are you familiar with the idea of a mollifier, i.e. a sequence of functions in $\mathcal{S}(\mathbb{R}^n)$ that, interpreted in $\mathcal{S}'(\mathbb{R}^n)$ converges weakly to a Dirac delta? You can convolve a mollifier with an arbitrary tempered distribution to obtain a sequence of Schwartz functions converging to that distribution. (In fact I now see that this was given in a comment to the question you linked). You can even use compactly supported mollifiers to prove that compactly supported functions are sequentially dense in (general) distributions. – Robert Furber Feb 21 '24 at 11:15
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    I also know an abstract argument deriving from the fact that $\mathcal{S}(\mathbb{R}^n)$ is a separable nuclear Fréchet space and therefore the strong dual topology is the sequentialization of the weak-* topology, but I'd never use this here because the mollifier argument is much better, it just does what you want and gives you the understanding of how to approximate a distribution by a smooth function. – Robert Furber Feb 21 '24 at 11:22
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    So anyway, for clarity: Yes, $\mathcal{S}(\mathbb{R}^n)$ is both dense and sequentially dense in $\mathcal{S}'(\mathbb{R}^n)$ in both the weak-* and strong dual topologies. – Robert Furber Feb 21 '24 at 11:30
  • @RobertFurber Thank you. I thought convolution of a tempered distribution with a smooth function is just a smooth and slowly increasing function. How do you ensure that convolution with a mollifier is actually a Schwartz function? – Keith Feb 21 '24 at 11:37
  • @RobertFurber what do you mean by "sequentiization"? It would be nice if the strong dual topology and weak$^*$ one coincide for Schwartz spaces.. – Keith Feb 21 '24 at 11:38
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    The topologies don't coincide, but they have the same convergent sequences. The sequentialization of a topology is the finest topology having the same convergent sequences. – Robert Furber Feb 21 '24 at 11:44
  • @RobertFurber I see. I remember the strong dual topology to be finer than weak$^*$ one in general. Plus, neither of them is sequential in general. – Keith Feb 21 '24 at 11:50
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    In that case I wasn't speaking in general, but specifically for $\mathcal{S}(\mathbb{R}^n)$, in that case the strong dual is strictly finer than the weak-* topology, and it is a sequential topology. But this is not a very good direction to go in - I shouldn't have mentioned it. To get a sequence of Schwartz functions, convolve by a mollifier and multiply by a sequence of compactly supported smooth functions converging to the constant function 1 as distributions. – Robert Furber Feb 21 '24 at 11:56
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    "In case of the Schwartz space, the strong dual topology is strictly finer than the weak-* one and not sequential. However, it turns out that we can always find a common convergent" sequence in both topologies, and the weak-* topology is sequential. <= Am I correct?" No. The correct statement is that in $\mathcal{S}'(\mathbb{R}^n)$, a sequence $(\phi_i)_{i \in \mathbb{N}}$ converges to $\phi$ in the weak-* topology if and only if it does so in the strong dual topology, the strong dual topology is sequential and strictly finer than the weak-* topology. – Robert Furber Feb 21 '24 at 12:01
  • @RobertFurber I am confused...you said it is strictly coarser than the strong dual one.. Plus, you just said that $\mathcal{S}(\mathbb{R}^n)$ is sequentially dense in $\mathcal{S}'(\mathbb{R}^n)$ w.r.t to the weak$^*$ topology. – Keith Feb 21 '24 at 12:04
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    That's exactly why it's not sequential. The sequentialization of a topology $\mathcal{T}_1$ is the finest topology $\mathcal{T}_2$ having the same convergent sequences as $\mathcal{T}_1$. We get it by adding all the sequentially open sets to $\mathcal{T}_1$. So if $\mathcal{T}_1$ is strictly coarser than $\mathcal{T}_2$, there is a sequentially open set for $\mathcal{T}_1$ that is not open for $\mathcal{T}_1$, i.e. $\mathcal{T}_1$ is not sequential. – Robert Furber Feb 21 '24 at 12:09
  • @RobertFurber Ooops....I am sorry for lack of knowledge...and grateful for your clarifications..... I understand the stuff now.. – Keith Feb 21 '24 at 12:12
  • @RobertFurber If you are still merciful with me, I would be even more grateful for any reason why convolution with a mollifier is a Schwartz function rather than just a slowly increasing smooth function. – Keith Feb 21 '24 at 12:14
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    I omitted the need to multiply the mollified function by a sequence of compactly supported smooth functions that converges to the constant function $1$. This gives you a sequence of compactly supported smooth functions. As the mollifier converges to a delta function and the compactly supported function converge to 1, the sequence converges to the distribution. This is what reuns suggests in the post you linked to. – Robert Furber Feb 21 '24 at 12:17
  • @RobertFurber Ok, I missed your remark on "multiplying by a sequence of compactly supported smooth functions converging to the constant $1$ as distributions". That settles everything for me. – Keith Feb 21 '24 at 12:21

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