Timeline for Differentiability along hyperplanes for rational functions
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 24 at 16:05 | history | bounty ended | CommunityBot | ||
| S Jun 24 at 16:05 | history | notice removed | CommunityBot | ||
| Jun 21 at 11:40 | comment | added | Pietro Majer | If we knew that there exists at least one direction $v_0$ such that the directional derivatives $\frac{\partial f}{\partial v_0}(x,y,z)$ are continuous at the origin , one would conclude by the MVThm like in the Total Differential Thm. But I don’t see why they should be even locally bounded at $(0,0,0)$; not even in the simplest case $\{q=0\}=(0,0,0)$. | |
| Jun 21 at 9:01 | comment | added | Jan Bohr | I've included continuity of $f$ into the conditions, then being rational (as in 1) should be well-defined. What I was thinking of was essentially extensions of the 2-dimensional example $xy^2/(x^2+y^2)$, where both polynomials are homogeneous of adjacent degrees and the only real zero of the denominator is at $0$. | |
| Jun 21 at 8:59 | history | edited | Jan Bohr | CC BY-SA 4.0 |
added 15 characters in body
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| Jun 17 at 6:33 | comment | added | Pietro Majer | I think such $f$ is at least Gateaux differentiable at the origin (that is, there exist a linear form $L$ such that for every $v\in\mathbb R^3$ there exists $\frac d{dt}f(tv)\big|_{t=0}=Lv$ | |
| Jun 16 at 21:44 | comment | added | Pietro Majer | how is f defined on the zero set of q ? | |
| S Jun 16 at 14:08 | history | bounty started | Jan Bohr | ||
| S Jun 16 at 14:08 | history | notice added | Jan Bohr | Draw attention | |
| Jun 12 at 13:38 | history | asked | Jan Bohr | CC BY-SA 4.0 |