Timeline for Reference for a lemma on the asymptotic upper density of special sets with large gaps and intervals
Current License: CC BY-SA 3.0
9 events
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| May 27, 2015 at 15:53 | comment | added | Martin Sleziak | You are probably right. As not to digress from the original question, we can continue this discussion in chat, if needed. | |
| May 27, 2015 at 15:39 | comment | added | Salvo Tringali | @MartinSleziak. I'm not so convinced that would make the proof shorter, since you first need to show that ${\sf d}^\ast(X)=\limsup_n\frac{|X \cap [\![1, b_n]\!]|}{b_n}$. AFAICS, this can't avoid a rough estimate of the counting function of $X$, which is possible, e.g., by letting $N$ be as in Anthony Quas' comment and considering that $\frac{x}{y} \le \frac{x+z}{y+z}$ for all $x,y,z \in \mathbf R^+$ with $x \le y$. In any case, thank you for your comment! | |
| May 27, 2015 at 15:00 | comment | added | Martin Sleziak | It seems that you are essentially asking about $d^*(A)=\limsup\frac{(b_1-a_1)+\dots+(b_n-a_n)}{b_n}$. From Stolz-Cesaro Theorem you get that this is in fact equal to $\lim\frac{b_n-a_n}{b_n-b_{n-1}}=1-\ell$. (I know that this does not answer the question, but still it makes the proof a bit shorter, I guess.) | |
| May 25, 2015 at 9:32 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Updated to take into account Anthony Quas' comment
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| May 25, 2015 at 8:32 | comment | added | Salvo Tringali | Because "about half a page" is not "half a page" (but an approximation in excess, which takes into account 5 lines for the statement and two more lines for a second point, which I didn't mention in the OP), and because we were working with $b_n \le N < b_{n+1}$ rather than $a_n<N \le a_{n+1}$, which is, indeed, much better (thank you!). Anyway, we would still prefer a reference, if any. | |
| May 25, 2015 at 7:37 | comment | added | Anthony Quas | So I don't see why this would take 1/2 page: If $a_n<N\le a_{n+1}$, then $|X\cap [1,N]|/N \le |X\cap [1,b_n]|/b_n$ so the $\limsup$ is attained along the sequence of $b_n$'s. And $b_n-a_n \le |X\cap [1,b_n]|\le (b_n-a_n)+b_{n-1}$. Dividing by $b_n$ and taking the limit gives the result. | |
| May 25, 2015 at 7:30 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Improved formatting
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| May 25, 2015 at 6:29 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Fixed a typo
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| May 25, 2015 at 6:24 | history | asked | Salvo Tringali | CC BY-SA 3.0 |