Let $F_{1},...,F_{n}$ be Fréchet spaces. Is the cartesian product $F_{1}\times\cdots \times F_{n}$ again a Fréchet space? In particular, is the cartesian product $\mathcal{S}(\mathbb{R}^{d})\times\mathcal{S}(\mathbb{R}^{d})$ of Schwartz spaces (of rapid decrease functions) again a Fréchet space?
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With the product topology? – md2perpe May 10 '20 at 13:08
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Yes! With the product topology! – JustWannaKnow May 10 '20 at 15:11
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Yes. An equivalent definition of $V$ being Fréchet is the existence of a countable collection of seminorms $||\cdot||_n$, $n\in\mathbb{N}=\{0,1,\ldots\}$, on $V$ such that:
- The topology of $V$ is the locally convex topology defined by this collection of seminorms.
- The distance $$ d(x,y)=\sum_{n=0}^{\infty}\ 2^{-n}\ \min(1,||x-y||_n) $$ makes $V$ a complete metric space.
So if you have two Fréchet spaces $V,W$ with respective collections of seminorms as above, $||\cdot||_{V,n}$ and $||\cdot||_{W,n}$, just take the collection $$ ||(v,w)||_{m,n}=\max(||v||_{V,m},||w||_{W,n})\ . $$ This gives a countable collection indexed by $\mathbb{N}^2$ on $V\times W$ which shows it is also Fréchet.
For a quick review of the definition of seminorms, locally convex topologies etc. see my answer Doubt in understanding Space $D(\Omega)$
Abdelmalek Abdesselam
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1Thanks. Look at the answer I linked to if you want to learn more about distributions. – Abdelmalek Abdesselam May 11 '20 at 16:45