Timeline for Surjectivity of curl
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2015 at 21:17 | comment | added | Bazin | @Denis Serre: I do not understand your Borel-type argument for the smoothness of $w^0$ at the origin. The size of $w_k$ on the sphere could be anything and your argument ("differs from…") would provide smoothness for $\sum_{k\ge 0}\phi(kx)a_k x^k$ for any sequence. Given a sequence $(a_k)_{k\ge 0}$, it is possible to choose a sequence $(\mu_k)_{k\ge 0}$ (depending heavily on the $a_k$) such that $\sum_{k\ge 0}\phi(\mu_k x)a_k x^k$ is smooth (i.e. $C^\infty$). | |
| Feb 1, 2015 at 13:19 | comment | added | Igor Khavkine | As Denis wrote, the series defining $w^0(x)$ is locally finite, in the sense that $\phi(kx)w_k(x)$ is non-zero only for finitely many $k$ at any $x\ne 0$. | |
| Feb 1, 2015 at 9:14 | comment | added | Hachino | @IgorKhavkine : This is convergence at $0$, not around $0$. Assume that the dominant coefficient of $v_k$ grows like $(k!^k)$ (which may indeed happen), how come that $w^0$ makes sense ? I trust Denis Serre when he says so, but still, I would like to understand why this holds. | |
| Jan 31, 2015 at 15:16 | comment | added | Igor Khavkine | @Hachino, by virtue of being homogeneous polynomials, all $w_k(0) = 0$, except for $k=1$. | |
| Jan 31, 2015 at 6:51 | comment | added | Hachino | Nice one ! Could you please elaborate a bit on why the series defining $w^0$ converges in a neighborhood of the origin ? I'm probably missing something simple, but since the Taylor coefficients at $0$ of $v$ could be anything, convergence is not completely obvious. | |
| Jan 30, 2015 at 14:41 | history | answered | Denis Serre | CC BY-SA 3.0 |