Proper Way To Compute An Upper Bound

Question

I regard to the proof of Lemma 10 in "A remark on a conjecture of Chowla" by M. R. Murty, A. Vatwani, J. Ramanujan Math. Soc., 33, No. 2, 2018, 111-123,

the authors used the average value $(\log x)^c$, $c$ constant, of the number of divisors function $\tau(d)=\sum_{d|n}1$ as an upper bound for $\tau(d)^2$, where $d \leq x$. To be specific, they claim that $$\sum_{q \leq x^{2\delta}}\tau(q)^2 \left | \sum_{\substack{m \leq x+2\\ m \equiv a \bmod q}} \mu(m)\right | \ll x (\log x)^{2c},$$

where $2 \delta <1/2$.

The questions are these:

1. Is the main result invalid? The upper bound should be $$\sum_{q \leq x^{2\delta}}\tau(q)^2 \left | \sum_{\substack{m \leq x+2\\ m \equiv a \bmod q}} \mu(m)\right | \ll x ^{1+2\delta}.$$ This is the best unconditional upper bound, under any known result, including Proposition 3.

2. It is true that the proper upper bound $\tau(d)^2 \ll x^{2\epsilon}$, $\epsilon >0$, is not required here?

3. Can we use this as a precedent to prove other upper bounds in mathematics?

Answer 1

Good question, and I agree that the authors should have been more explicit here. However, I think I can reconstruct their argument: note that \begin{align*} \sum_{q \leq x^{2\delta}}\tau(q)^2 \bigg | \sum_{\substack{m \leq x+2\\ m \equiv a \bmod q}} \mu(m)\bigg | &\le \sum_{q \leq x^{2\delta}}\tau(q)^2 \sum_{\substack{m \leq x+2\\ m \equiv a \bmod q}} |\mu(m)| \\ &\le \sum_{q \leq x^{2\delta}}\tau(q)^2 \sum_{\substack{m \leq x+2\\ m \equiv a \bmod q}} 1 \\ &\ll \sum_{q \leq x^{2\delta}}\tau(q)^2 \frac xq = x \sum_{q \leq x^{2\delta}} \frac{\tau(q)^2}q. \end{align*} And this remaining sum is indeed $\ll_\delta (\log x)^{2c}$ for some constant $c$; indeed, it's not hard to show that $$ \sum_{q \leq y} \frac{\tau(q)^2}q \sim \frac{(\log y)^4}{4\pi^2}. $$