A character sum $\sum_{0<n\leq Y}\chi_4(n)\chi_0^{(k)}(n)$ estimate

I'm reading the paper 'Jutila, Matti. "On the Mean Value of $$L (1/2, χ)$$ FW Real Characters." Analysis 1.2 (1981): 149-161.'

Let $$\chi_4(n)$$ be the real primitive nonprincipal character of modulo 4, that is,

$$\chi_4(n)=\begin{cases} 1, & n \equiv 1 \quad(4) \\ -1, & n \equiv 3 \quad(4) \\ 0, & \text{otherwise} \end{cases}$$,

and $$\chi_0^{(k)}$$ be a principal character mod $$k$$. We denote the function that gives the number of positive divisor and the mobius function by $$\tau(n)$$ and $$\mu(n)$$ respectively.

In the paper p.152, the author estimates a character sum (stated below) as the following:

$$\frac{1}{2}\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a) \sum_{0 $$=\frac{1}{2}X(\phi(2k)/2k)\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a)a^{-2}+O(X^{1/2}\tau(k)).$$

It is easy to see that $$\frac{1}{2}\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a) \sum_{0 $$\approx \frac{1}{2}X(\phi(2k)/2k)\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a)a^{-2},$$

so I think that the term $$O(X^{1/2}\tau(k))$$ is contributed by $$\frac{1}{2}\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a)\sum_{0.

However I cannot derive this equation. The author says that he used the 'elementary estimates for character sums', but if I use the 'elementary estimates', for example, $$\sum \chi(n) \leq q$$ or the Polya-Vinogradov inequality $$\sum \chi(n) \ll q^{1/2}\log q$$ (here $$q$$ is the modulus of character $$\chi$$), it is not really effective in my view. (I'll explain soon)

I think the $$X^{1/2}$$ part in the big $$O$$ is coming from the term $$\frac{1}{2}\sum_{a\leq X^{1/2}}\mu(a)\chi_0^{(2k)}(a)$$.

Meanwhile, the sum of principal character $$\chi_0^{(k)}$$ only depends on $$p(k)$$, the product of distinct prime divisors of $$k$$, so I didn't catch why $$\tau(k)$$ is presented.

My approach using the Polya-Vinogradov estimates:

If I use the Polya-Vinogradov estimates, since the modulus of the character $$\chi_4\chi_0^{(k)}$$ is smaller than $$4p(k)$$, we have

$$\sum \chi_4(n)\chi_0^{(k)}(n) \ll (4p(k))^{1/2}\log(4p(k)).$$

Thus we want to that $$(4p(k))^{1/2}\log(4p(k)) \ll \tau(k)$$, but there is a possibility that $$\tau(k)=2^{\omega(k)}$$, where $$\omega$$ is the prime omega function (that gives the number of prime divisors of $$k$$), so I believe that this estimate does not hold.

How I derive the estimate in the paper?

• Please correct the LaTeX errors. I tried, but I am not sure how some of the formulas should be corrected so I gave up. – Piotr Hajlasz Mar 27 at 1:26
• I'm sorry, I didn't check the error. Now the erros are corrected. – LWW Mar 27 at 1:29