Timeline for Differentiability: Partially Defined Functions
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| May 19, 2015 at 8:16 | answer | added | Peter Michor | timeline score: 3 | |
| Jul 22, 2014 at 22:54 | history | edited | C-star-W-star | CC BY-SA 3.0 |
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| Jul 22, 2014 at 22:26 | history | edited | C-star-W-star | CC BY-SA 3.0 |
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| Jun 12, 2014 at 20:53 | comment | added | C-star-W-star | @ChristianRemling: Yes I'm aware that there might be no global smooth extension what is a simple consequence of the fact that holomorphic functions turn out to be analytic in fact. But my point is rather that there might be no global extension that makes the function at all points in the original domain differentiable - not continuously differentiable of course since this is too strong to build counterexamples... | |
| Jun 12, 2014 at 18:59 | comment | added | Christian Remling | @WillieWong : I think the OP is not totally unambiguous on this point, but I thought he was asking whether local smooth extension at all $x\in A$ will guarantee existence of a global smooth extension to the whole space $E$ ($=\mathbb R^2$ in my example). Both readings seem compatible with the current wording, so that could use clarification too. | |
| Jun 12, 2014 at 7:48 | comment | added | Willie Wong | @ChristianRemling: is it? Let the function $f(r,\theta)$ in polar coordinate be defined by $$f = \frac12(1 + \cos\theta) $$ when $r \neq 0$ and $0$ when $r = 0$. This function is differentiable away from the origin, and is an extension of your $f$ (I assume the notation $(-1,0)$ means the segment on the real axis, and not the single point.) Note that the condition the OP quoted did not require the extension to be differentiable on the whole of $E$, just at the point $x$ which we care about. | |
| Jun 11, 2014 at 19:41 | comment | added | Christian Remling | We definitely need to assume that $A$ is closed, otherwise this is totally hopeless: $A=(-1,0)\cup (0,1)\subset\mathbb R^2$ and $f=0$ on one half of $A$ and $f=1$ on the other. | |
| Jun 11, 2014 at 10:22 | comment | added | C-star-W-star | Yes holomorphic is quite a strong requirement that's true but I'm rather concerned with the definition merely differentiable not continuously differentiable and there about what exotic things might happen so then if we wanna do holomorphic functional calculus we really see we have to require that there is one! neighborhood that makes it differentiable simultaneously for all points in the subset rather then for every point a neighborhood separately | |
| Jun 11, 2014 at 10:03 | comment | added | C-star-W-star | @WillieWong: yeah right that was a 'typo' I actually meant a merely subset like $\{1\}\cup[2,3)$. | |
| Jun 11, 2014 at 10:00 | history | edited | C-star-W-star | CC BY-SA 3.0 |
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| Jun 11, 2014 at 7:14 | comment | added | Willie Wong | @Freeze_S: my point is that there may be stronger rigidity from holomorphy than would be implied by real differentiability. The simple example being that if two holomorphic functions on $U\subset \mathbb{C}$ agree on a 1 (real) dimension curve, then they agree on $U$. So if your real motivation is about holomorphic functional calculus, I am not really sure the question you actually asked is relevant. | |
| Jun 10, 2014 at 18:17 | answer | added | C-star-W-star | timeline score: 2 | |
| Jun 10, 2014 at 18:07 | comment | added | C-star-W-star | Heeeyy I guess I got an example :) but I'll need some help to rigorously prove it - I will post it as a answer... | |
| Jun 10, 2014 at 16:45 | comment | added | Pietro Majer | The complex motivation is maybe not strong enough, but the question seems interesting to me, and I really do not see why down-voting... | |
| Jun 10, 2014 at 15:47 | comment | added | C-star-W-star | Yeah I know but up to that point as merely being a definition this should not depend on the ground field... | |
| Jun 10, 2014 at 15:11 | comment | added | Willie Wong | And then there is the problem that real differentiability is quite a different beast from complex differentiability. So I don't really see the connection between the first and second parts of your question. (With real differentiability you are allowed to use partitions of unity and what not.) | |
| Jun 10, 2014 at 10:53 | history | asked | C-star-W-star | CC BY-SA 3.0 |