Smoothing in analytic number theory

Question

Often we are interested in the sum \[ \sum _{n\leq x}a_n\] for some number theoretic sequence $a_n$ and often the study of the "smooth" sum \[ \sum _{n=1}^\infty a_n\phi _x(n)\] if simpler (so here $\phi _x(n)$ is a nice "smooth" function).

Can anyone provide me with any material that helped them understand "smoothing" better? In particular examples from number theory would be great. This resource on smoothing sums is already pretty good but I'd love to have some more material. I do get the idea, but I'd like to see a few more examples. I often see it used in papers but with all the details left out, and I'm aware that I can't really fill in the gaps in some papers, see e.g. the proof of Theorem 1 here. They don't explicitly say anything about their weight function so I don't really know how to calculate with it - in particular I don't understand the bounds on page 277. (I'm not sure what goes on in the "iterated integration by parts" bit.)

I think Kowalski's book Un cours de théorie analytique des nombres has some stuff on smoothing done with some details, but only in French... does anyone know if there's a translation?

Answer 1

Smoothing can lead to an elegant analytic proof of $L(1,\chi)\gg q^{-1/2}$ for primitive quadratic $\chi$ modulo $q$. Now, we define

$$ f(n)=\sum_{d|n}\chi(d) $$

This definition suggests that its Dirichlet series has a very neat connection with L functions:

$$ \sum_{n\ge1}{f(n)\over n^s}=\zeta(s)L(s,\chi)\tag1 $$

Then by the multiplicative property of $\chi$ it can be shown that $f(n)\ge0$ holds for all $n\ge1$. This allows us to obtain a nice upper pound for the following partial sum:

$$ F(x)=\sum_{n\le x}f(n)\le\sum_{n\ge1}f(n)e^{1-n/x}\tag2 $$

Moreover, it can be verified that $f(n)\ge1$ whenever $n$ is a perfect square, so $F(x)$ also has a very elementary lower bound:

$$ F(x)\ge\sum_{\substack{n\le x\\n\text{ is square}}}1\ge x^{1/2}-1\tag3 $$

By Mellin's inversion formula, we know the exponential function can be expressed as follows.

$$ e^{-y}={1\over2\pi i}\int_{2-i\infty}^{2+i\infty}\Gamma(s)y^{-s}\mathrm ds\tag4 $$

Thus, combining (1), (2), (3) and (4) gives

$$ x^{1/2}\ll{1\over2\pi i}\int_{2-i\infty}^{2+i\infty}x^s\Gamma(s)\zeta(s)L(s,\chi)\mathrm ds\tag5 $$

To estimate the integral on the right hand side, we move the path of integration to the line segment $\Re(s)=c<1$, so a calculation of the residue at $s=1$ gives:

$$ {1\over2\pi i}\int_{2-i\infty}^{2+i\infty}x^s\Gamma(s)\zeta(s)L(s,\chi)\mathrm ds=xL(1,\chi)+\mathcal O\left(\int_{c-i\infty}^{c+i\infty}\right)\tag6 $$

When $s=c+it$, it follows from Stirling's approximation that as $t\to\pm\infty$:

$$ |\Gamma(s)|\ll|t|^{c-1/2}e^{-t\pi/2}\tag7 $$

Moreover, by the functional equation for $L(s,\chi)$, it can be shown that

$$ L(c+it,\chi)\ll(q|t|)^{1/2-c}|L(1-c-it,\overline\chi)| $$

As a result, if we impose $c<0$ then the above bound becomes

$$ L(c+it,\chi)\ll(q|t|)^{1/2-c}\tag8 $$

Combining (7) and (8) gives

$$ \int_{c-i\infty}^{c+i\infty}\ll q^{1/2-c}x^c $$

As a result, (6) gets simplified into

$$ L(1,\chi)\gg x^{-1/2}-q^{1/2-c}x^{c-1}\tag9 $$

Finally, plugging in $x=rq$, we get

$$ L(1,\chi)\gg(1-r^{c-1})q^{-1/2} $$

Now, picking some negative $c$ and some large $r$ will make the right hand side positive, obtaining the desired result.

This suggests that Siegel's theorem $L(1,\chi)\gg q^{-\varepsilon}$ is effective whenever $\varepsilon\ge1/2$.

Answer 2

A common tool to study partial sums like $$ S(X) := \sum_{n \leq X} a(n) $$ is to study the associated Dirichlet series $$ D(s) := \sum_{n \geq 1} \frac{a(n)}{n^s}. $$ Mellin inversion (a form of Fourier inversion) then relates analytic properties of $D(s)$ to growth properties of $S(X)$. At its most basic, we have Perron's formula, which states that $$ \sum_{n \leq X} a(n) = \lim_{T \to \infty} \frac{1}{2 \pi i} \int_{\sigma - iT}^{\sigma + iT} D(s) X^s \frac{ds}{s}, $$ assuming $D(s)$ is uniformly convergent for $\mathrm{Re} s > \sigma - \epsilon$ for some $\epsilon > 0$ and that $X$ is not exactly an integer. Thus if $D(s)$ is nice enough for $\mathrm{Re} s > \sigma$, then we can hope to get good understanding of $S(X)$.

This continues a bit further. If $D(s)$ is very nice, then our understanding of the associated sum is better. It can be convenient to force $D(s)$ to be very nice by multiplying by something that decays rapidly (such as $\Gamma(s)$, which is done in the previous answer) or which cancels poles or accomplishes some other analytic goal. Broadly, one considers Mellin pairs $(v, V)$ satisfying

$$ v(x) = \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} V(s) x^{-s} ds. $$

These give summatory relations

$$ \sum_{n \geq 1} a(n) v(n/x) = \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} D(s) V(s) X^s ds, $$

proved in the same way as for Perron's formula. In practice, only a few families of Mellin pairs $(v, V)$ are used frequently. These include the classic Perron pair, the exponential pair $(\exp(-x), \Gamma(s))$, and the Cesaro/Riesz weights $((1 - x)^k, \Gamma(k + 1) \Gamma(s) / \Gamma(s + k + 1))$, and perhaps a few others.

Alternately, it can be convenient to specify exactly one of $v$ or $V$. One common choice is to choose $v(x) = v_y(x)$ to be a smooth function that is identically $1$ from $0$ to $1$, bounded between $0$ and $1$ from $1$ to $1 + 1/y$, and identically $0$ above $1 + 1/y$. As $v$ is smooth and compactly supported, $V$ has arbitrary polynomial decay (analogous to properties of typical Fourier transforms), which makes many convergence problems in the integral disappear. (Then one can optimize choices of $y$ later). This is approximately the same in spirit as the weights used in the OP-linked paper by Iwaniec and Fouvry --- compact support guarantees nice enough integration.

A good reference to see more about this is Montgomery and Vaughan's book on Analytic Number Theory. These techniques are used extremely often and sometimes are called "Tauberian techniques".