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This question is more of a check/validation of a concept.

Suppose I want to study $$\sum _{n\leq X}a_n$$ (e.g. $a_n=d(n)$, the divisor function). As is well-known it's standard practice to replace the sharp cut-off with a smooth function $$\sum _{n}a_nw_X(n),$$ where for some parameter $Y$ to be chosen according to our problem $w_X(t)=1$ for $t\in (1,X)$ and $w_X(t)=0$ for $t>X+Y$, and where $w_X(t)$ is smooth on the whole of $\mathbb R$ with derivatives $w^{(j)}_X(t)\ll _j1/Y^j$. This may lead to a "dual" sum $$\sum _{n}a_n\hat w_X(n)$$ which is more tractable, where $$\hat w_X(n)\approx \int _0^\infty w_X(t)\underbrace {f(nt)}_{\text {e.g. a Bessel function for $a_n=d(n)$}}dt$$ and where I'm being annoyingly imprecise but hopefully someone can find more sense than I in what I'm trying to say.

Suppose I want to study instead $$\sum _{n\leq X}a_ne(n\beta )$$ where $\beta \in (0,1/2)$. If I put in a weight function first and then try to remove the $e(n\beta )$ by partial summation, then I am doomed to fail (right?) because then I have a sharp cut-off again in the integral. So I have to do partial summation first and then put in the weight function, giving me something like \begin{eqnarray} e(X\beta )\sum _{n}a_nw_X(n)-2\pi i\beta \int _1^Xe(v\beta )\sum _na_nw_v(n)dv\\\ \approx \sum _na_n\int _0^\infty f(nt)\left (e(X\beta )w_X(t)-2\pi i\beta \int _1^Xe(v\beta )w_v(t)dv\right )dt. \end{eqnarray} The term in the brackets is I think something like $$\int _1^Xe(v\beta )\frac {d}{dv}w_v(t)dv$$ and it seems to me that the derivative is $\ll 1/v$ so essentially this term looks like $w_X(t)$. That would mean the problem is no different to the problem without the $e(n\beta )$ factor, which seems suspiciously simple and so almost certainly points to a conceptual misunderstanding on my part. (e.g. I'm not sure if I'm oversimplifying something with the two arguments of $w_X(t)$). Can anyone clarify me on this?

Thanks!

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  • $\begingroup$ maybe i'm just supposed to consider $e(n\beta )w(n)$ as the smooth weight? $\endgroup$
    – tomos
    Commented Feb 21, 2023 at 10:54
  • $\begingroup$ Do you mean $e(n\beta) = \exp(n\beta) = e^{n\beta}$ ? $\endgroup$ Commented Feb 22, 2023 at 12:48
  • $\begingroup$ i mean $e(n\beta )=e^{2\pi in\beta }$ $\endgroup$
    – tomos
    Commented Feb 22, 2023 at 17:33

1 Answer 1

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Notice that $$\Big|\sum_{n\leq x}a_n e(n\beta)-\sum_{n=1}^{\infty}a_n e(n\beta)w_X(n)\Big|\leq \sum_{X<n\leq X+Y}|a_n|.$$ So if you have strong control of $|a_n|$ in short intervals, then you can pass between the original sum and the smoothed sum with ease.

Let $0<\epsilon<1/2$ and $X\geq 3$. If $|a_n|$ is bounded by the $k$-divisor function $\tau_k(n)$ (with $\tau_2(n)$ the usual divisor function) and $X^{\varepsilon}\leq Y\leq X$, then Shiu's theorem implies that $$\sum_{X<n\leq X+Y}|a_n|\ll_{k,\epsilon} Y(\log X)^{k-1}.$$ This suffices for many applications.

Small side note: I think that one must in fact take $w_X$ to be 1 on $[Y,X]$, in $(0,1)$ in $(0,Y)\cup(X,X+Y)$, and $0$ otherwise. You want to give $w_X$ some time to transition from assuming the value 1 to assuming the value 0 so that the derivatives on the interval $[0,Y]$ are small. If $X^{\epsilon}\leq Y\leq X$, then we still have $$\sum_{n\leq Y}|a_n|\ll_{k,\epsilon}Y(\log X)^{k-1}.$$

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  • $\begingroup$ hi! thanks for looking at my question:) i actually can't remember what i was thinking/hoping for back then, apologies:D buuuuut: i think with your answer then i'm going to be using $e(n\beta )w(n)$ as a weight right? which means i'll get an extra $\beta $ factor in derivative bounds. but in my question, i'm saying i'm confused because it looks like the $e(n\beta )$ doens't give me any complications. however, i'm pretty sure that i'm just being too hand-wavey in the last part of my post and extra $\beta $ factors would probably fall out somewhere $\endgroup$
    – tomos
    Commented Apr 29, 2023 at 12:25
  • $\begingroup$ @tomos I don't know what you mean by "using $e(n\beta)w(n)$ as a weight". All I'm asserting is that unsmoothing is very easy if you can control $|a_n|$ in short intervals (I'm just using the fact that $|e(n\beta)w(n)|\leq 1$). Moreover, for the sequences you might consider (it seems that you want $|a_n|$ bounded by a generalized divisor function), you can in fact control $|a_n|$ in short intervals. So in your setting, it seems that unsmoothing is much less complicated than you let on, unless you want $Y$ smaller than $X^{\epsilon}$. $\endgroup$
    – 2734364041
    Commented Apr 29, 2023 at 23:26
  • $\begingroup$ ok, then i will leave it for now, because i can't remember well enough the problems i had so not sure how much sense i'll talk if i try to discuss it. at some point i'll return to it, and to your post, and i'll vote up in the meantime:) $\endgroup$
    – tomos
    Commented Apr 30, 2023 at 10:44
  • $\begingroup$ i think i'm assuming i don't know how to study $a_ne(\beta n)$ i.e. that the dual sum is just for $a_n$ $\endgroup$
    – tomos
    Commented May 27, 2023 at 20:51

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