Timeline for Special Schwartz function on the positive interval
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 10 at 20:17 | vote | accept | SnowRabbit | ||
| Jan 4 at 16:19 | answer | added | fedja | timeline score: 6 | |
| Jan 4 at 14:40 | comment | added | Christian Remling | @ChristopheLeuridan: Yes, I know that this is not obvious (as I stated at the end of my long comment), but on the other hand the space of solutions is huge (Nevanlinna parametrization) and one can perhaps take advantage of this to produce a Schwartz solution. | |
| Jan 4 at 8:08 | comment | added | Christophe Leuridan | The known conditions on Stieltjes moment problem do not ensure the existence of a measure with given moments and having a density which belongs to the Schwartz space. | |
| Jan 4 at 4:43 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Jan 4 at 3:33 | comment | added | Christian Remling | One can also try to use moment problems, perhaps as follows: start out with a $g\ge 0$ with indeterminate moments $m_n=\int_0^{\infty} x^n g(x)\, dx$. Then the Stieltjes moment problem $m_0+1, m_1,m_2,\ldots$ still has solutions (obvious from the standard criteria) and should still be indeterminate (not clear to me right now how to prove it, but it feels right). In particular, there are absolutely continuous solutions $h\, dx$, and then we can take $\zeta=h-g$. This satisfies (1), (2), $\textrm{supp }\zeta\subseteq[0,\infty)$, but is perhaps not in $\mathcal S$. | |
| Jan 3 at 20:51 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 6 characters in body
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| Jan 3 at 20:23 | comment | added | Christian Remling | The slightly weaker assignment where we replace (3) by $\textrm{supp }\zeta\subseteq [0,\infty)$ is equivalent to finding a function $f=\widehat{\zeta}$ satisfying $f\in H^2\cap\mathcal S$, $f(0)=1$, $f^{(k)}(0)=0$. This looks potentially difficult to me, as $H^2$ functions are determined by their values on any positive measure subset of $\mathbb R$. For example, there is no such function that is $=1$ near zero. | |
| Jan 3 at 18:39 | comment | added | Aleksei Kulikov | $supp (\zeta) \subset (0, \infty)$ more or less means that $\zeta$ has an analytic continuation to the lower half-plane plus the convergence of the logarithmic integral. We can probably get an analytic function satisfying all these conditions, so the answer must be yes. | |
| Jan 3 at 17:42 | history | asked | SnowRabbit | CC BY-SA 4.0 |