Timeline for Regularity of functions everywhere approximable by $n$-th degree polynomials
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 8, 2022 at 20:58 | history | suggested | Dirk Werner | CC BY-SA 4.0 |
some language
|
| Jan 8, 2022 at 20:57 | review | Suggested edits | |||
| S Jan 8, 2022 at 20:58 | |||||
| Jan 8, 2022 at 14:56 | comment | added | NameNo | The problem might be that to define a good 2-approx. a single variable point $y$ is not enough. A line is given by two points (one could be your fixed $x$, but two variable points converging to $x$ give the better strong differentiability). But three points are needed for second order approximation (like the osculating circle to a curve). | |
| Jan 8, 2022 at 13:59 | history | edited | Kacper Kurowski | CC BY-SA 4.0 |
added 1 character in body
|
| Jan 8, 2022 at 13:47 | comment | added | Kacper Kurowski | I haven't done that yet, thank you for the recommendation! | |
| Jan 8, 2022 at 13:37 | history | edited | Kacper Kurowski | CC BY-SA 4.0 |
added a note and second question
|
| Jan 8, 2022 at 13:09 | comment | added | DCM | Have you looked in Federer's "Geometric Measure Theory"? I remember there being some discussion of this sort of thing shortly before the discussion of the Whitney extension theorem (which you may have come across already if you're thinking about things like this). | |
| Jan 8, 2022 at 4:52 | comment | added | NameNo | Does 2-approx. (in a point or on a open set) imply strong differentiabilty (on the same set)? In the sense of mathoverflow.net/questions/404397/… | |
| Jan 7, 2022 at 21:53 | history | edited | Kacper Kurowski | CC BY-SA 4.0 |
edited title
|
| Jan 7, 2022 at 21:45 | comment | added | Kacper Kurowski | $(n+1)$-approximability implies $n$-approximability for any $n\in \mathbb{N}$. If $f \colon U \to Y$ is $(n+1)$-approximable at $x\in X$, then $f(y) = P(y) + \omicron( \lVert y-x \rVert_X^{n+1})$, where $P$ is an $(n+1)$-th order polynomial. Therefore, there exists $(n+1)$-linear bounded map $T \colon X^{n+1} \to Y$ such that $X \ni y \mapsto P(y) - T(y, \ldots, y)$ is an $n$-th degree polynomial. By boundedness of $T$ we have $y \mapsto T(y) \in O( \lVert y-x \rVert_X^{n+1}) \subseteq \omicron(\lVert x-y \rVert_X^n)$ and $T(0, \ldots, 0) = 0$ and $P-T$ $n$-approximates $f$. | |
| Jan 7, 2022 at 21:17 | comment | added | NameNo | I have no idea (too old ...) but have you already tried some "Taylor inspired" obvious questions like: does $n+1$-approx. imply $n$-approx. ? For a differentiable $f$, are there implications between "$f$ is 2-approx." and "$f'$ is 1-approx. i.e. differentiable"? | |
| Jan 7, 2022 at 20:24 | history | edited | Kacper Kurowski | CC BY-SA 4.0 |
fix typo
|
| Jan 7, 2022 at 20:13 | history | asked | Kacper Kurowski | CC BY-SA 4.0 |