Timeline for Tempered distributions at non-coinciding points and density of Schwartz functions
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| May 8 at 18:47 | comment | added | Pedro Lauridsen Ribeiro | Regarding the use of resolution of singularities in (perturbative) renormalization, although the idea seems natural at first, it's somewhat limited in practice as far as I know (one can only renormalize logarithmic singularities at worst), see e.g. arxiv.org/abs/0908.0633 | |
| May 8 at 18:43 | comment | added | Abdelmalek Abdesselam | indeed......... | |
| May 8 at 18:42 | comment | added | Pedro Lauridsen Ribeiro | @AbdelmalekAbdesselam well, the whole subject of renormalization of distributions (as done in perturbative QFT) is a constructive form of the Hahn-Banach theorem, as Klaus Hepp put it long ago.... | |
| May 8 at 18:36 | comment | added | Abdelmalek Abdesselam | ...the big diagonal is as hard as it gets from the point of view of resolution of singularities. BTW see mathoverflow.net/questions/470858/… for a new question about resolution of singularities for the big diagonal. | |
| May 8 at 18:34 | comment | added | Abdelmalek Abdesselam | I was going to mention the strategy of extension, followed by the usual mollification and truncation to get the approximation in the usual $S'$ but Pedro beat me to it. It's better to say sequentially dense for the strong dual topology, which is a stronger statement anyway, instead of weak-$\ast$ sequentially dense. Note that Hahn-Banach is a cheap solution, the alternative being to note that $T$ has at most algebraic singularities on the big diagonal and therefore admits an extension using resolution of singularities, or Bernstein polynomial techniques. My understanding is that... | |
| May 8 at 18:28 | comment | added | Pedro Lauridsen Ribeiro | It should be remarked that if ${}^0\mathcal{S}(\mathbb{R}^{mN})$ is endowed with the topology induced from $\mathcal{S}(\mathbb{R}^{mN})$, then any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ admits a(n usually non-unique) continuous linear extension to the whole of $\mathcal{S}(\mathbb{R}^{mN})$ by the Hahn-Banach theorem. In other words, any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ is the restriction to ${}^0\mathcal{S}(\mathbb{R}^{mN})$ of some $\widetilde{T}\in\mathcal{S}(\mathbb{R}^{mN})'$. Since $\mathcal{S}$ is weak-* sequentially dense in $\mathcal{S}'$, the result should follow. | |
| May 8 at 14:28 | vote | accept | Isaac | ||
| May 8 at 12:17 | comment | added | Isaac | Thank you. I will think about the answer myself a bit more and will upload another post if I fail. | |
| May 8 at 12:09 | comment | added | Iosif Pinelis | @Isaac : I think this should also be true, but do not have a complete proof at this point. Since there should be only one question in one post you may want to consider posting the sequential density separately. | |
| May 8 at 11:55 | comment | added | Isaac | Thank you as always! How about sequential density in the weak$^*$ topology? | |
| May 8 at 11:47 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |