Timeline for Non-continuous higher differentiability
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2014 at 11:20 | comment | added | Pietro Majer | Even for n=1, it does not even imply continuity in other points than x, so that the second order derivability does not even make sense. Standard example: $f(t):=e^{-1/t}\chi_{\mathbb{Q}}$, that has polynomial expansion of any order at $0$, and is discontinuous at any other point. | |
| May 12, 2014 at 22:16 | comment | added | Mike Shulman | In case anyone is still watching this question, I am puzzled. On p183 of Strichartz' The Way of Analysis, after proving the second-order Taylor approximation to a $C^2$ function, he says "We will not discuss the problem of obtaining a converse kind of statement, deducing the existence of the second derivative from the existence of quadratic polynomial approximations, because such theorems are extremely difficult to prove and have few applications." But that implies that such theorems exist, which the answers to this question seem to deny. Can anyone guess what he had in mind? | |
| May 10, 2014 at 1:21 | comment | added | Tom Goodwillie | I don't know the Lang book. I was just going by the comment above. | |
| May 10, 2014 at 0:54 | comment | added | Mike Shulman | @TomGoodwillie I'm looking at the Lang chapters now, and I only see the second derivative of $f:E\to F$ defined as the derivative of $D f : E \to L(E,F)$ (for $E$ and $F$ vector spaces). He does of course use the expression $f(x+v+w)-f(x+w)-f(x+v)+f(x)$ in proving symmetry of a continuous second derivative, but I don't see it in a characterization of second differentiability. | |
| May 7, 2014 at 17:34 | comment | added | Tom Goodwillie | (1) Maybe the Lang chapters mentioned above? (2) I don't think so. (3) Yes: $f(a+h1+h2)−f(a+h1)−f(a+h2)+f(a)$ is a symmetric function of $h_1$ and $h_2$ and therefore so is its best bilinear approximation. | |
| May 7, 2014 at 15:53 | comment | added | Mike Shulman | @TomGoodwillie (1) is there a reference that develops higher derivatives using that point of view? (2) can you rephrase it in terms of the quadratic form? (3) does second-differentiability in that sense imply equality of mixed partials? | |
| May 7, 2014 at 15:22 | comment | added | Mike Shulman | @StevenGubkin Making formal sense of this point of view is still kind of a work in progress, but you can see what we've got at ncatlab.org/nlab/show/cogerm+differential+form and the nForum discussions linked at the bottom, plus nforum.mathforge.org/discussion/5817/cojet-differential-forms. One issue is that so far (related to this question) it makes the most sense for things that are already known to be smooth; the cogerm differential is a generalization of "all the directional derivatives", not necessarily implying true differentiability. | |
| May 7, 2014 at 15:01 | comment | added | Steven Gubkin | Maybe you could get Toby to write an answer to his question here mathoverflow.net/questions/60474/…, if he has worked something out? | |
| May 7, 2014 at 15:00 | comment | added | Steven Gubkin | @MikeShulman Do you have any references for this point of view? Of course, the first derivatives must be present because this is how second derivatives transform. So, invariantly, jet bundles should come into the picture somehow. | |
| May 7, 2014 at 12:13 | comment | added | Tom Goodwillie | The first derivative of $f$ at $a$ is the linear function of $h$ that best approximates $f(a+h)-f(a)$. The second derivative of $f$ at $a$ is the bilinear function of $(h_1,h_2)$ that best approximates $f(a+h_1+h_2)-f(a+h_1)-f(a+h_2)+f(a)$. Etc. | |
| May 7, 2014 at 10:15 | comment | added | Noah Stein | Fedor Petrov posted this on MO's big list of common false beliefs in mathematics: mathoverflow.net/a/25899/5963. | |
| May 7, 2014 at 5:09 | answer | added | Mike Shulman | timeline score: 3 | |
| May 7, 2014 at 3:09 | comment | added | Mike Shulman | @StevenGubkin: Toby Bartels has convinced me that the best point of view on the second derivative is as the second differential $\mathrm{d}^2f = \sum_{i,j}\partial_{i,j}f \, \mathrm{d}x_i \,\mathrm{d}x_j + \sum_i \partial_i f\,\mathrm{d}^2x_i$. E.g. this has the advantage of yielding the correct chain rule when you "calculate by substitution". But I don't really know how to express that as any sort of approximation. | |
| May 7, 2014 at 0:18 | vote | accept | Mike Shulman | ||
| May 6, 2014 at 22:26 | answer | added | Michal R. Przybylek | timeline score: 8 | |
| May 6, 2014 at 21:54 | comment | added | Ryan Reich | I sort of doubt this even implies that $f$ is differentiable except at $x$. For example, let $n = 1$, let $w(t)$ be your typical nowhere-differentiable continuous (i.e. Weierstrass) function, and let $f(t) = t^2 + t^3 w(t)$. Then this is "twice differentiable" according to you, but only at $t = 0$. | |
| May 6, 2014 at 21:52 | answer | added | Tom Goodwillie | timeline score: 5 | |
| May 6, 2014 at 21:47 | comment | added | user76758 | See Chapter XIII, section 3--6 of Lang's "Real and Functional Analysis" for a very elegant treatment under which higher derivatives are defined via multilinear maps and moreover the $p$th higher derivative is genuinely the "derivative" of the $(p-1)$th. One virtue of this approach is that it permits both a formulation and proof of the higher-dimensional Taylor formula which looks and feels exactly like the 1-dimensional case (with the mess of factorials hidden away within a clean formalism); of course, one can bust out coordinates and recover the usual messier explicit version from that. | |
| May 6, 2014 at 21:36 | comment | added | Steven Gubkin | You may also be interested in my online course with Jim Fowler here: ximera.osu.edu/course/kisonecat/m2o2c2/course/activity/week1. It is a little buggy, and a little sparse in terms of examples, but it covers this perspective on higher order derivatives in a kind of "interactive textbook" format. | |
| May 6, 2014 at 21:31 | comment | added | Steven Gubkin | I am not sure. I will think some about this. My definition of higher order differentiability would be the following: First $f$ needs to be differentiable in a nbhd of $p$. Then $f$ is twice differentiable at $p$ if there is a bilinear form $D^2f(p)$ such that $Df(p+v_1)(v_2) = Df(p)(v_2)+D^2f(v_1,v_2)+\epsilon |v_1||v_2|$. For some reason it is very hard to find sources where people talk about the second derivative as a bilinear form rather than just a quadratic form. | |
| May 6, 2014 at 21:26 | history | asked | Mike Shulman | CC BY-SA 3.0 |