Timeline for smooth families of analytic functions
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 2, 2012 at 12:51 | comment | added | Florian | Just to be sure: So under mildly stronger assumptions than the ones I stated one can indeed conclude analyticity of $\frac{\partial f}{\partial x^i}$? Is there an easy example where one sees that my conditions don't suffice? | |
| Aug 24, 2012 at 20:40 | comment | added | Bazin | @Jochen I need indeed that for each $y$, $f(⋅,y)$ is a distribution on $\mathbb R^m_x$, then I can differentiate with the above rule. Using the $\overline{\partial}$ operator as suggested in my answer, you can prove a weak analyticity result. I doubt that much more could be proven: think about a smooth non-analytic function $f$ depending only on $x$. Sorry for the wrong reference for the variation on this topic: it is Theorem 2.1.3. – Bazin 4 hours ago | |
| Aug 24, 2012 at 11:47 | comment | added | Jochen Wengenroth | Doesn't one need something about the function $x\mapsto f(x,y)$ for $y\ in \mathbb C$ (or at least some open complex sets) to differentiate under the integral? Moreover, is it obvious that the analyticity of the lhs for all test functions $\phi$ implies the analyticity of $\frac{\partial f}{\partial x}$? Finally, are you sure about the reference to 2.3.1 ? (In my edition this is states that $\mathscr E'$ is the dual of $\mathscr C^\infty$.) | |
| Aug 24, 2012 at 9:56 | history | edited | Bazin | CC BY-SA 3.0 |
edited body
|
| Aug 23, 2012 at 16:32 | history | answered | Bazin | CC BY-SA 3.0 |