Question in Iwaniec-Kowaleski's book : bound for a twisted series

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 $$1 and we set $$\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

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.

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.