Timeline for Approximation of smooth real functions with an integer integral by a sum of $\delta$-functions with coefficients $\pm 1$
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18 at 21:52 | vote | accept | akhmeteli | ||
| Aug 16 at 2:37 | comment | added | fedja | @akhmeteli "However, I may try to write up your proof" Sure. Everything I put in the public domain is free for everyone to use and to spread without any special permission :-) That is, IMHO, the whole point of communication, scientific or otherwise. | |
| Aug 15 at 12:05 | comment | added | akhmeteli | As for "understandable"... Your proof is fabulous, and "daryonomu konyu..." merriam-webster.com/dictionary/… . However, I may try to write up your proof (of course, without any claim to originality) to customize it for my application and make it more accessible to us, lowly physicists:-) | |
| Aug 15 at 12:00 | comment | added | akhmeteli | Yes, I had taken note of jmerry's comment, but jmerry only mentioned $n$, and I worried more about $K_i$, although these issues are closely related. Until recently, I was pretty sure that the statement from your previous comment was incorrect, but when I tried to prove that...:-) (cntd.) | |
| Aug 14 at 8:15 | comment | added | fedja | @akhmeteli Yep, it wasn't a writing of exemplary clarity, indeed :lol: Jmerry had the same problem with it as you. But I hope the proof was still understandable without too much effort. | |
| Aug 14 at 6:23 | comment | added | akhmeteli | Thank you so much! Now it makes sense. The statement in your AoPS post looked somewhat confusing: $K_i$ and $n$ were mentioned first, and then there was this phrase “Show that if $\delta>0$ is small enough”, so it looked like $\delta$ depends on $K_i$ and $n$, whereas it seems to only depend on $d$. And I was wrong saying “But if they change arbitrarily, the statement does not seem correct in a general case.” | |
| Aug 13 at 11:33 | comment | added | fedja | @akhmeteli It means "For every $d$, there exists $\delta=\delta(d)>0$ such that for every set of compact sets $K_i$ ($i=1,\dots n$) such that the diameters of all $K_i$ are less than $\delta$, $K_i$ are connected, and the convex hull of $K_1+\dots+K_n$ contains the ball of radius $2$ centered at the origin, the sum itself contains the ball of radius $1$ centered at the origin." Now all quantifiers have been spelled out. Feel free to ask more questions if something is still unclear :-) | |
| Aug 13 at 5:09 | comment | added | akhmeteli | Are they intersections of the old $K_i$ with the ball having the radius of the new $\delta$? So I don’t understand this statement, however if we move in the opposite direction, keeping $\delta$ constant and increasing “the radius of the ball contained in the convex hull of $K$", the statement seems to make sense. I am still moving towards better understanding of your proof. | |
| Aug 13 at 5:09 | comment | added | akhmeteli | scalar product. Show that if $\delta>0$ is small enough, then this sum must contain the unit ball centered at the origin." I am afraid I don't quite understand what it means. I would think “if $\delta>0$ is small enough” assumes some process of decreasing $\delta$. But do $K_i$ change as $\delta$ decreases? They probably should, otherwise they eventually will not be contained in the ball of radius $\delta$. But if they change arbitrarily, the statement does not seem correct in a general case. Then how are the new $K_i$ related to the old ones? (cntd.) | |
| Aug 13 at 5:08 | comment | added | akhmeteli | Thank you very much indeed! Actually, I am glad I asked this trivial question because a related question on your AoPS post arose and your answer seems relevant to this question as well. “Suppose that $K_i$ ($i=1,\ldots,n$) are compact connected sets in $R^d$ such that each set is contained in the ball of radius $\delta$ centered at the origin and the sum $K_1+\ldots+K_n=\{y_1+\ldots+y_n:y_i\in K_i\}$ is not contained in any half-space $\{ y: (y,e)\leq 2\}$ where $e$ is any unit vector and $(\cdot,\cdot)$ is the usual (cntd.) | |
| Aug 10 at 12:19 | comment | added | fedja | @akhmeteli Then if we take $R/r$ copies of $K'$, the convex hull of the sum contains the ball of radius $2$, so the sum contains the ball of radius $1$, which is equivalent to the sum of $K$ containing the ball of radius $R$. We just need to choose $R>\frac {2\operatorname{diam}K}{\delta}$ to start with. | |
| Aug 10 at 12:16 | comment | added | fedja | @akhmeteli It is just trivial scaling. We need $\delta$ to be small compared to $1$ in the AoPS post, while here we need the fixed quantity (the diameter of $K$) to be small compared to the radius of the ball contained in the convex hull of $K+\dots+K$, which can be made arbitrarily large (it is just the number of terms times the radius of the ball contained in the convex hull of $K$). In other words, if the convex hull of $K$ contains a ball of radius $r$, we apply the AoPS lemma to $K'=\frac 2RK$ where $R$ is huge. Continued | |
| Aug 10 at 9:27 | comment | added | akhmeteli | do I miss an easy fix? I appreciate that "the diameter of $K$ in the corresponding finite dimensional Euclidean space is some fixed quantity", but it cannot be arbitrarily small, while we need $\delta$ "to be able to be" small. The following comment may be irrelevant or trivial, but I'd like to note that we cannot use fractions of dipoles (again, I am not sure it is relevant to "convex hull" of your posts). | |
| Aug 10 at 9:15 | comment | added | akhmeteli | Thank you very much. Now I am trying to trace the details of the connection between your AoPS post and the post here, namely, the connection between $K$ here and $K_i$ in AoPS. So $K_i$ "is contained in the ball of radius $ \delta$ centered at the origin", while "the convex hull of $K$ contains a small ball". I am not sure at the moment if $K$ itself contains a small ball, but let us assume it does. We can limit $K$ to this small ball to be able to use the AoPS statement, but it is not obvious that such a subset of $K$ will be connected. Is this a problem, or (cntd.) | |
| Aug 9 at 12:58 | comment | added | fedja | @akhmeteli Yep, though my idea was rather to choose $a$ at the maximum of the polynomial (truncated Fourier series) and $b$ at the minimum, but what you said works perfectly well too. | |
| Aug 9 at 6:47 | comment | added | akhmeteli | So one of the arguments ($a$ or $b$) can be chosen to provide the maximum square of the Fourier series, which is positive. The other argument ($a$ or $b$) can be chosen in such a way that the Fourier series vanishes. So I guess I have an idea how to prove the quoted statement. Will be sifting through your proof further. | |
| Aug 9 at 6:38 | comment | added | akhmeteli | Yes, I understand that. "if you have any trigonometric polynomial including only non-zero frequencies and of $L^2$ norm 1, then we can find a dipole with respect to which its integral is at least 1" Without worrying about the specific value of the constant, the scalar product of a vector built from arbitrary coefficients of a truncated real Fourier series $a_n,b_n$ and Fourier coefficients of a dipole will be a difference of Fourier series with arguments $a$ and $b$ instead of $x$. The lower bound for maximum of the square of one such series can be estimated using Parseval theorem. (cntd.) | |
| Aug 9 at 0:19 | comment | added | fedja | @akhmeteli Yep, though I'm not really concerned too much with the exact bound here as long as it depends on the dimension only, so any other bound and proof of it are fine for my purposes too. | |
| Aug 8 at 4:33 | comment | added | akhmeteli | Thank you very much. So I am slowly going through the proof (busy with other things). I guess the boundary from your comment for the "classical permutation lemma" was proven in mathnet.ru/links/de5273b00c5f18d3129ba8b3a5918530/faa1805.pdf . | |
| Aug 4 at 9:57 | comment | added | fedja | @akhmeteli It will take me some time to understand your proof. Feel free to ask as many questions as you need. Note that the proof is not entirely self-contained: it relies upon the classical permutation lemma saying that if you have a set of vectors of norm $\le 1$ in $\mathbb R^D$ with sum $0$, then you can arrange them in such an order that all partial sums will be bounded by $D$ in norm. The initial application was to construct the Dirichlet series with some fancy properties (I'm not sure I even remember the details now: it was another question someone asked somewhere). | |
| Aug 4 at 5:14 | comment | added | akhmeteli | Fantastic... Thank you very much indeed. It will take me some time to understand your proof. By the way, what was your initial application? | |
| Aug 4 at 0:58 | history | answered | fedja | CC BY-SA 4.0 |