We put \[E(x; q, a) = \psi(x; q, a) - \frac{x}{\varphi(q)}\] for \((a,q) = 1\). We let \[E(x; q) = \max_{\substack{a \\ (a,q)=1}} |E(x; q, a)| \quad\text{and}\quad E^*(x,q) = \max_{y \le x} E(y; q).\]
Theorem. Let \(A > 0\) be fixed. Then \[\sum_{q \le Q} E^*(x, q) \ \ll\ x^{1/2} Q (\log x)^6,\] provided that \[x^{1/2} (\log x)^{-A} \ \le\ Q \ \le\ x^{1/2}.\]
Proof. First we show that the bound follows from the bound \[\sum_{q \le Q} \frac{q}{\varphi(q)} \sum_{\substack{\chi \bmod q \\ \chi \ \text{primitive}}} \max_{y \le x} \left| \psi(y, \chi) \right| \ \ll\ (x + x^{1/2} Q + x^{1/2} Q^2) (\log x)^5\] for all \(x \ge 1\), \(Q \ge 1\), where \[\psi(y, \chi) = \sum_{n \le y} \Lambda(n) \chi(n).\] Then we establish this bound by an estimation over the bilinear form.
By orthogonality of characters, we have \[\psi(y; q, a) = \frac{1}{\varphi(q)} \sum_{\chi \bmod q} \overline{\chi}(a) \, \psi(y, \chi).\] Then \[E(y; q, a) = \frac{1}{\varphi(q)} \sum_{\chi \bmod q} \overline{\chi}(a) \, \psi'(y, \chi),\] where \[\psi'(y, \chi) = \begin{cases} \psi(y, \chi), & \chi \ne \chi_0,\\ \psi(y, \chi_0) - y, & \chi = \chi_0. \end{cases}\] Hence, \[|E(y; q, a)| \le \frac{1}{\varphi(q)} \sum_{\chi} |\psi'(y, \chi)|.\] Note this bound also holds for \(E(y; q)\). If \(\chi \pmod{q}\) is induced by \(\chi_1 \pmod{q_1}\), then \[\psi'(y, \chi) - \psi'(y, \chi_1) = \sum_{\substack{p^k \le y \\ p \mid q}} \chi(p^k) \log p \ \ll\ \sum_{p \mid q} \left[ \frac{\log y}{\log p} \right] \log p \ \ll\ (\log qy)^2.\] Therefore, \[E^*(y, q) \ \ll\ (\log qx)^2 + \frac{1}{\varphi(q)} \sum_{\chi} \max_{y \le x} |\psi'(y, \chi)|.\] By summing \(q\) to \(Q\), and considering the right-hand side as a combination of primitive characters modulo \(q\) and its induced characters, we obtain \[\sum_{q \le Q} E^*(y, q) \ \ll\ Q (\log Qx)^2 + \sum_{q \le Q} \sum_{\chi}{}^* \max_{y \le x} |\psi'(y, \chi)| \left( \sum_{k \le Q/q} \frac{1}{\varphi(kq)} \right)\] \[\ \ll\ (\log x) \sum_{q \le Q} \frac{q}{\varphi(q)} \sum_{\chi}{}^* \max_{y \le x} |\psi'(y, \chi)|.\]
We then separate the summation over \(q\) into \([1, (\log x)^A]\) and \(((\log x)^A, Q]\). The first part we estimate with the bound we assumed is correct. Let \(R = (\log x)^A\), then we have \[\ \ll\ (\log x) \sum_{q \le R} \frac{q}{\varphi(q)} \sum_{\chi} {}^*\max_{y \le x} |\psi(y, \chi)| \left( \frac{1}{Q} + \int_{q}^{Q} t^{-2} \, \mathrm{d}t \right)\] \[\ \ll\ x^{1/2} Q (\log x)^6 + x^{1/2} R^{-1} (\log x)^6 \ \ll\ x^{1/2} Q (\log x)^6.\] for \[x^{1/2} (\log x)^{-A} \ \le\ Q \ \le\ x^{1/2},\] as for the summation of \(q\) over \((R, Q]\), we estimate by the Siegel–Walfisz theorem: \[\ll\ (\log x)^{A} \, x (\log x)^{-2A} \ \ll\ x (\log x)^{-A}.\] Therefore, we are left to show the bound is correct. The idea is to apply Vaughan’s identity of \(\sum_{n} \Lambda(n) n^{-s}\). First note, by using the large sieve of characters, it yields \[\sum_{q \le Q} \frac{q}{\varphi(q)} \sum_{\chi}{}^* \left| \sum_{m \le M} \sum_{n \le N} a_m b_n \chi(mn) \right|\] \[\ \ll\ (M + Q^2)^{1/2} (N + Q^2)^{1/2} \left( \sum_{m \le M} |a_m|^2 \right)^{1/2} \left( \sum_{n \le N} |b_n|^2 \right)^{1/2}.\] Then, we want to insert the maximal condition. Let \(\gamma > 0\) and define \[\delta(\gamma) = \begin{cases} 1, & 0 \le \beta \le \gamma, \\ 0, & \beta > \gamma. \end{cases}\]
Let \[C = \int_{-\infty}^{\infty} \frac{\sin \alpha}{\alpha} \, \mathrm{d}\alpha,\] then \(C\) exists and \(C > 0\). Then \[\int_{-\infty}^{\infty} \mathrm{e}^{-i\beta\alpha} \frac{\sin \gamma\alpha}{C\alpha} \, \mathrm{d}\alpha = \delta(\beta),\] and let \(A > 0\), by integration by parts, it follows \[\delta(\beta) = \int_{-A}^{A} \mathrm{e}^{-i\beta\alpha} \frac{\sin \gamma\alpha}{C\alpha} \, \mathrm{d}\alpha + O\!\left( \frac{1}{A \, |\log \beta|} \right).\]
Let \(y > 0\), by changing variables with \(\beta = \log mn\) and \[\gamma = \log\left( \left\lfloor y \right\rfloor + \tfrac12 \right),\] it yields that for \[\delta(\log mn) = \begin{cases} 1, & mn \le y, \\ 0, & mn \ge y. \end{cases}\] Since \[\min_{m,n} \left| \log\left( \left\lfloor y \right\rfloor + \tfrac12 \right) - \log mn \right| = \min\!\left( \log\frac{\left\lfloor y \right\rfloor + \tfrac12}{\left\lfloor y \right\rfloor}, \log\frac{\left\lfloor y \right\rfloor + 1}{\left\lfloor y \right\rfloor + \tfrac12} \right) \gg \frac1{y},\] it follows that \[\delta(\log mn) = \int_{-A}^{A} (mn)^{i\alpha} \frac{\sin\gamma\alpha}{C\alpha} \, \mathrm{d}\alpha + O\!\left( \frac{y}{A} \right).\] Therefore, by inserting the above as the maximal condition, we get for \(y \le x\): \[\sum_{m \le M} \sum_{n \le N} a_m b_n \chi(mn) = \sum_{m \le M} \sum_{n \le N} a_m b_n \chi(mn) \, \delta(\log mn)\] \[\ll \int_{-A}^{A} \left| \sum_{m \le M} a_m m^{i\alpha} \sum_{n \le N} b_n n^{i\alpha} \chi(mn) \frac{\sin\gamma\alpha}{C\alpha} \, \right |\mathrm{d}\alpha\] \[\ll \int_{-A}^{A} \left| \sum_{m \le M} a_m m^{i\alpha} \sum_{n \le N} b_n n^{i\alpha} \chi(mn) \right| \min\!\left( \log x, \frac{1}{|\alpha|} \right) \, \mathrm{d}\alpha,\] where we take \(A = x / (MN)\), and the error term is acceptable.
Therefore, it follows \[\sum_{q \le Q} \frac{q}{\varphi(q)} \sum_{\chi}{}^* \max_{y \le x} \left| \sum_{m \le M} \sum_{n \le N} a_m b_n \chi(mn) \right|\] \[\ll (\log x MN) (M+Q^2)^{1/2} (N+Q^2)^{1/2} \left( \sum_{m \le M} |a_m|^2 \right)^{1/2} \left( \sum_{n \le N} |b_n|^2 \right)^{1/2}.\]
If \(Q^2 > x\), we take \(M=1\), \(a_1 = 1\), \(N = x\), \(b_n = \Lambda(n)\), then the bound follows.
If \(Q^2 \le x\), let \(U = \nu = \min(Q^2, x^{2/3}, x/Q^2)\). If
the maximum is taken in \(\nu <
U^2\), then it also follows. Take \(u,
v\) above to be the ones in Vaughan’s identity, and take \[f(n) = \chi(n) \quad\text{if } n \le u, \qquad
f(n) = 0 \ \text{otherwise}.\] Then it allows us to write \[T(\chi, Q) = \sum_{j=1}^4 T_j
= \sum_{j=1}^4 \sum_{q \le Q} \frac{q}{\varphi(q)}
\sum_{\chi}{}^* \max_{y \le x} |S_j(\chi)|.\] Then,
since \(u \le U^2\), we can bound \(T_q\) trivially. For \(j=1\), by Pólya–Vinogradov inequality, we
can easily bound \(T_1\). For \(j=3\), since we have the bilinear bound for
\(\chi(mn)\), the bound follows. For
\(j=2\), we simply split up the sum by
terms with \(m \le u\) and \(u < m \le U^2\), and apply the method
for \(T_1\) and \(T_3\).
Then we obtain the bound, and hence, it completes the proof.