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Given $$\left| \sum_{1\leq n \leq N} a_f (n)e^{2\pi i n \alpha}\right| \leq c_f N^{k\over 2}\log N$$ for any $f\in S_k$ where $f(\tau) = \sum\limits_{n=1}^\infty a_f (n)q^n$, any real $\alpha$, and any $N\geq 10$.

Show that $$\left| \sum_{1 \leq n \leq N , n \equiv a \pmod q} a_f (n)\right| \leq c_f N^{k \over 2} \log N $$ for any positive integers $q$ and $a$.

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$$\sum_{n \equiv b \bmod q} a_n(f) e^{2i \pi n z}= \frac1q\sum_{m=1}^q e^{-2i \pi bm/q} f(z+m/q)$$

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  • $\begingroup$ Replace $f(z) = \sum_n a_n(f) e^{2i \pi nz}$ by $f(z) = e^{2i \pi nz}$ to see why it is true. The sum of LHS is over every $n$ in the congruence class. It is obvious it implies your claim. $\endgroup$ – reuns Sep 29 at 22:08
  • $\begingroup$ $\sum_{m=1}^q e^{2i \pi cm/q} = 0$ when $q\nmid c$ $\endgroup$ – reuns Sep 30 at 0:15

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