5
$\begingroup$

I am currently reading Iwaniec-Kowaleski's book on Analytic Number Theory.

My question is on page 431. Here is what it says:

For any $\chi \mod q$, we have the twisted series $K(s,\chi)=\sum_{n=1}^\infty \left( \sum_{d|n} \lambda_d \right) \left( \sum_{d|n} \theta_b \right) \chi(n) n^{-s}.$

Here, we have $\lambda(d)= \mu(d) \min \left(1, \frac{\log(z/d)}{\log(z/w)} \right)$ for $1 \leq d \leq z$ where $1<w<z$ and we set $\lambda_d=0$ if $d>z$.

Thus, we have $\sum_{d|n} \lambda_d=1$ if $n=1$ and $\sum_{d|n} \lambda_d=0$ if $1<n \leq w.$

Furthermore, we have $\theta_b = \frac{\mu(b) b}{G \phi(b)} \sum_{ab \leq y \atop (a,bq)=1} \frac{\mu^2 (a)}{\phi(a)}$ and $G$ is the normalization factor such that $\theta_1=1.$

Now, we can see that $K(s,\chi)$ factors into $L(s,\chi)M(s,\chi)$, where $M(s,\chi)=\sum_m \left(\sum_{[b,d]} \sum_{=m} \lambda_d \theta_b \right) \chi(m) m^{-s}$.

Now, if we take only a partial sum of $K(s,\chi)$, we can define $K_x(s,\chi)=\sum_{n=1}^x \left( \sum_{d|n} \lambda_d \right) \left( \sum_{d|n} \theta_b \right) \chi(n) n^{-s}$ with $x=(qT)^{23}.$ Let $m=[b,d] \leq bd \leq yz $ and $\chi \not= \chi_0.$

Then, why does this imply $\left|\sum_{n>x/m} \chi(n) n^{-s} \right| \leq 2q |s| (m/x)^{\sigma}$ where $s=\sigma +it$. I do not understand where the $2q|s|$ comes from.

Assuming this, the book states $|K(s,\chi)-K_x(s,\chi)| \leq 2q|s|yz x^{-\sigma}$. I do not understand where the yz term comes from.

Sorry for the length of this post. I wanted to give as much information as possible.

$\endgroup$
0

1 Answer 1

1
+50
$\begingroup$

Showing $\left|\sum_{n>x/m} \chi(n) n^{-s} \right| \leq 2q |s| (m/x)^{\sigma}$.

I will assume that $q > 1$ and that $\sigma \geq 1/2$. Then $$\sum_{n \geq N} \frac{\chi(n)}{n^s} = \sum_{n \geq N} S_\chi(n) \left( \frac{1}{n^s} - \frac{1}{(n+1)^s}\right) \tag{1},$$ where $S_\chi(n) = \sum_{N \leq m \leq n} \chi(m)$. This is an application of partial summation.

A trivial bound on $S_\chi(n)$ is $\lvert S_\chi(n) \rvert \leq q$. (It is here that I use that $q > 1$). In fact, this is a very weak bound, but that's ok.

I note that for $\mathrm{Re}(s) > 0$, and $0 < a < b$, we have $$ \left \lvert \frac{1}{a^s} - \frac{1}{b^s}\right \rvert \leq \frac{\lvert s \rvert}{\sigma} \left( \frac{1}{a^\sigma} - \frac{1}{b^\sigma}\right) \leq \frac{\lvert s \rvert(b - a)}{a^{\sigma + 1}}.$$ For a quick proof, look at $$ \int_a^b \frac{dx}{x^{s+1}} = \frac{1}{s} \left( \frac{1}{a^s} - \frac{1}{b^s}\right),$$ and thus $$ \left \lvert \frac{1}{a^s} - \frac{1}{b^s}\right \rvert \leq \lvert s \rvert \int_a^b \frac{dx}{\lvert x^{\sigma+1} \rvert},$$ giving the first inequality. For the second inequality, one can apply the mean value theorem to $1/x^{\sigma}$, or a trivial integral estimate.

In this case, this gives that $$ \left\lvert \frac{1}{n^s} - \frac{1}{(n+1)^s}\right\rvert \leq \frac{\lvert s \rvert}{n^{\sigma + 1}}.$$ Applying to $(1)$ shows that $$ \sum_{n \geq N} \frac{\chi(n)}{n^s} \leq \sum_{n \geq N} q \frac{\lvert s \rvert}{n^{\sigma + 1}} \leq 2 q \lvert s \rvert N^{-\sigma}.$$ In your case, you have $N = x/m$, and this proves the claim. $\diamondsuit$

$|K(s,\chi)-K_x(s,\chi)| \leq 2q|s|yz x^{-\sigma}$

I didn't work out the complete details, but I did work out part of the argument. In short, I didn't want to dive into any of the details concerning $x,y,z$ or $\theta_b, \lambda_d$, and so I use only trivial bounds for those parts.

From IK, we have the following properties:

  • $x = (qT)^{23}$
  • $y = (qT)^2$
  • $z = (qT)^8$
  • $w = (qt)^7$

And also $\lambda_d = 0$ if $d > z$, and $\theta_b = 0$ if $b > y$. Further, $\lvert \lambda_d \rvert \leq 1$ and $\lvert \theta_b \rvert \leq 1$. (IK has a whole section devoted to $\theta_b$ and the Selberg sieve, and restates some relevant bounds here for more precise bounds than I give below).

Then one can write $$ K_x(s, \chi) = \sum_{m \geq 1} \left( \sum_{[b, d] = m} \sum \lambda_d \theta_b \right) \frac{\chi(m)}{m^s} \sum_{n \leq x/m} \frac{\chi(n)}{n^s}$$ (in the same factoring sort-of argument that shows that $K(s, \chi) = M(s, \chi) L(s, \chi)$). Thus $$ \lvert K(s, \chi) - K_x(s, \chi) \rvert \leq \left \lvert \sum_{m \geq 1} \sum_{[b,d] = m} \lambda_d \theta_b \frac{\chi(m)}{m^s}\right \rvert \left \lvert \sum_{n \geq x/m} \chi(n)n^{-s}\right \rvert. \tag{2}$$

The latter part of $(2)$ is exactly what was considered above, and as $m \leq bd \leq yz$, we have that $$ \left \lvert \sum_{n > x/m} \chi(n) n^{-s} \right \rvert \leq 2 q \lvert s \rvert (yz)^\sigma x^{-\sigma}.$$

For the first sum of $(2)$, let us naively and lossfully note that $$ \sum_{[b,d] = n} \lambda_d \theta_b \leq \tau(n),$$ the number of divisors of $n$. Then the first $\lvert \cdot \rvert$ part can then be bounded above by $$ \sum_{n \leq yz} \tau(n) n^{-\sigma} \leq yz \log(yz) \left(1+ \frac{(yz)^{- \sigma}}{\sigma}\right) \leq 2(yz)^{1 - \sigma} \log(yz).$$

Putting these together, we find that $(2)$ is bounded above by $$2(yz)^{1 - \sigma} \log(yz) 2 q \lvert s \rvert (yz)^\sigma x^{-\sigma} = 4 q \lvert s \rvert yz \log(yz) x^{-\sigma}.$$ This is slightly weaker than the stated result in IK, but I think this describes how to get the sort of result as stated.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .