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If $\mathbb{E}^*(\mathbb{R}^n)$ is the topological dual of the space o smooth function on $\mathbb{R}^n$ and $\mathbb{S}^*(\mathbb{R}^n)$ is the topological dual of the Schwartz space $\mathbb{S}(\mathbb{R}^n)$ then do we know of the any of the following is true?

1)$\mathbb{E}^*(\mathbb{R}^n)$ endowed with the weak* topology is metrizable
2) $\mathbb{S}^*(\mathbb{R}^n)$endowed with the weak* topology is metrizable

we know that $\mathbb{S}(\mathbb{R}^n)$ and $\mathbb{E}(\mathbb{R}^n)$ are Frechet spaces and separable.

I am asking because in Dieudonne's Treatise in Analysis Volume VI there is a function $F:\mathbb{S}^*(\mathbb{R}^n)\rightarrow\mathbb{D}^*(\mathbb{R}^n)$ that we know that is linear and we want to prove that is continuous and it proves it using sequences! The same to prove that a linear function $F:\mathbb{E}^*(\mathbb{R}^n)\rightarrow\mathbb{S}^*(\mathbb{R}^n)$ is continuous.

We know that convergence on the topological dual of a space X is pointwise convergence.But why continuity is equivalent with sequential continuity?

Except if in general when the topological dual of a space X is endowed with the weak* topology then the notion of continuity means sequential continuity.I mean that we use the word continuity but we actually mean sequential continuity!

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  • $\begingroup$ As shown here math.stackexchange.com/questions/144475/…, the weak star topology is retrievable only if the underlying space has an at most countable Handel dimension. In your case, this is not fulfilled. The "proof" given there actually shows that we do not have first countability in your case. What kind of map is Dieudonne talking about? $\endgroup$
    – PhoemueX
    Commented Oct 10, 2015 at 7:09
  • $\begingroup$ they are restrictions.If for instance f is a function on $\mathbb{E}^*(\mathbb{R}^n)$ then F(f) is the restriction of f from $\mathbb{E}(\mathbb{R}^n)$ to $\mathbb{S}(\mathbb{R}^n)$ $\endgroup$ Commented Oct 10, 2015 at 7:42

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