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Feb
7
comment Stone-Weierstrass theorem for holomorphic functions?
Alex, you can contact me directly by e-mail: mathnet.ru/php/…
Feb
7
comment Stone-Weierstrass theorem for holomorphic functions?
Yes, it's obvious, if $M$ is $\sigma$-compact: then ${\mathcal C}^\infty(M)$ is a Fréchet space (I don't know, perhaps people consider now non-$\sigma$-compact manifolds as well). For the case when $M$ is an open subset in $\mathbb{R}^m$ this topology on ${\mathcal C}^\infty(M)$ is described in Rudin's "Functional analysis": amazon.com/Functional-Analysis-Walter-Rudin/dp/0070542368
Feb
6
comment Unifying (& “justifying”) the various definitions for differential operators
Ah, yes! $\phantom{*}$
Feb
6
comment Unifying (& “justifying”) the various definitions for differential operators
You can add another definition: a linear operator $D:C^\infty(X)\to C^\infty(X)$ is a differential operator of order $n$ if for any sequence of functions $f_0,f_1,...,f_n\in C^\infty(X)$ the commutator with the operators of multiplication by $f_i$ vanishes: $[...[[D,f_0],f_1]...,f_n]=0$.
Feb
6
comment Unifying (& “justifying”) the various definitions for differential operators
No, you don't need the topology on ${\mathcal O}_X$. You can look at the proof of J.Peetre's theorem in S.Helgason's book: amazon.com/Geometric-Invariant-Differential-Operators-Spherical/… He also gives an equivalent definition: the operator $D:C^\infty(X)\to C^\infty(X)$ must not extend the support of function: $\text{supp}Df\subseteq \text{supp}f$. And these operators indeed have locally finite order.
Feb
2
comment Stone-Weierstrass theorem for holomorphic functions?
@AlexM. I edited. There is no need to take into account the stereotype theory, when you define topology on ${\mathcal C}^\infty(M)$. This is equivalent to your condition with vector fields $X_i$.
Dec
11
comment Stone-Weierstrass theorem for holomorphic functions?
Simon, I don't understand you. Why meromorphic functions? You can consider a narrower class, the class of entire functions on ${\mathbb C}$, and even in this situation the problem is unsolved: if $A$ is a subalgebra in ${\mathcal O}({\mathbb C})$, nobody knows which conditions $A$ should satisfy for being dense in ${\mathcal O}({\mathbb C})$.
Dec
10
comment Stone-Weierstrass theorem for holomorphic functions?
Necessary, but I would be surprised if it were essential in the answer. I believe, there must be a much more strong condition instead of separating tangent vectors.
Dec
10
comment Stone-Weierstrass theorem for holomorphic functions?
David, actually, I would not expect that the criteria for ${\mathcal O}(M)$ and for ${\mathcal C}^\infty(M)$ will be similar. I would suppose that separating points will be their common part, but separating tangent vectors sounds strange for me.
Dec
10
comment Stone-Weierstrass theorem for holomorphic functions?
David, your example is interesting, but I think, you did not understand my question. First, I have no hypotheses on what the answer should be. Second, separating points does not mean the equality of spectra. For example, from the Stone-Weierstrass theorem it follows that the algebra $A$ of almost periodic functions is dense in ${\mathcal C}({\mathbb R})$, but $\text{Spec}(A)\cong\text{Bohr compactification of} \ {\mathbb R}\ne{\mathbb R}\cong\text{Spec}({\mathcal C}({\mathbb R}))$ (if $\text{Spec}$ is defined somewhat reasonably, say, if it means continuous involutive characters).
Dec
10
comment Stone-Weierstrass theorem for holomorphic functions?
Yes, of course!
Dec
10
comment Stone-Weierstrass theorem for holomorphic functions?
@SimonHenry, as far as I understand, even in the case of $M={\mathbb C}$ there is no answer.
Dec
9
comment Stone-Weierstrass theorem for holomorphic functions?
Ah, I see. Maybe...
Dec
9
comment Stone-Weierstrass theorem for holomorphic functions?
Runge's theorem is about a concrete subalgebra $A$, the algebra of rational functions, but I am asking about an arbitrary subalgebra $A\subseteq{\mathcal O}(M)$. I don't know, maybe there were generalizations...
Dec
7
comment Categorical description of dense homomorphisms of topological algebras
@Zhen Lin: yes, but in this approach the question arises, what are closed subobjects from the categorical point of view?
Oct
4
comment Traces of operators in nuclear spaces
Ah, yes, thank you.
Oct
4
comment Traces of operators in nuclear spaces
Something is strange for me here... For a locally convex space $X$ let us denote by $X^{\mathbb N}$ and $X_{\mathbb N}$ the direct product and the locally convex sum of contable copies of $X$. Then I would think that $({\mathbb R}^{\mathbb N})'_{\beta}\otimes_{\pi}{\mathbb R}^{\mathbb N}=({\mathbb R}^{\mathbb N})_{\mathbb N}\ne({\mathbb R}_{\mathbb N})^{\mathbb N}=L_{\beta}({\mathbb R}^{\mathbb N})$.
Oct
4
comment Traces of operators in nuclear spaces
In the stereotype world, en.wikipedia.org/wiki/Stereotype_space, this equality is not true: ${\mathbb R}_{\mathbb N}\circledast {\mathbb R}^{\mathbb N}\ne {\mathcal L}({\mathbb R}^{\mathbb N})$.
Sep
26
comment What are Reinert's reproaches to the Ricardo theory?
Your last example with scarcity and real numbers does not convince me, since I believe that there is nothing bad in discussing which phenomena should be reflected in the theory, but I agree that the question is too vague, that's not good.
Sep
26
comment What are Reinert's reproaches to the Ricardo theory?
Yes, I agree. If there is a possibility, I would prefer this question to be deleted.