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I was thinking a bit about distribution theory the last weeks and stumbled across the following question:

There are two natural locally convex topologies on the space of smooth functions of moderate growth $\mathcal{O}_M(\mathbb{R}^n) := \{f\in C^\infty(\mathbb{R}^n) \mid \forall\alpha: \partial^\alpha f \,\text{is bounded by a polynomial}\}$. One is generated by the seminorms $f\mapsto\|\phi \cdot \partial^\alpha f\|_{L^\infty}$ for all $\phi\in\mathcal{S}(\mathbb{R}^n)$ and all multiindices $\alpha$. Let's call that one $\tau_S$.

The other one is the colimit of the subspaces $\mathcal{O}_M^N := \{f \mid \forall\alpha: \partial^\alpha f(x) \in O(x^{N(\alpha)})\}$ where $N$ ranges over all functions $\mathbb{N}^n\to\mathbb{N}, \alpha\mapsto N(\alpha)$ (with the pointwise partial ordering this becomes a filtered colimit). Here $\mathcal{O}_M^N$ is endowed with the seminorms $f\mapsto \|\frac{\partial^\alpha f(x)}{1+\|x\|^{N(\alpha)}}\|_{L^\infty}$ for all multiindices $\alpha$. Let call this topology on $\mathcal{O}_M$ by the name of $\tau_G$.

My question is:

Are these topologies equal?

It is easy to see that $id: (\mathcal{O}_M,\tau_G)\to(\mathcal{O}_M,\tau_S)$ is continuous and I can show that the inverse map is bounded, that is: it maps bounded sets to bounded sets.

Because $\tau_G$ is the colimit of metric topologies, it is bornological. This means that $\tau_G$ is the "bornologification" of $\tau_S$. Therefore it is equivalent to ask:

Is $(\mathcal{O}_M,\tau_S)$ a bornological space?

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Yes, $\mathscr O_M$ is bornological. Grothendieck calls this "not trivial" and proves it in his thesis via a tensor product representation and quite general results about tensor products. Together with Julian Larcher, I recently gave an alternative proof which is still in the refereeing process. If you are interested I can send you the preprint by email.


The preprint is now in the arXiv.

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    $\begingroup$ Yes, I'm definitely interested. $\endgroup$ – Johannes Hahn May 31 '14 at 22:26

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