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Assume $\chi\neq\chi_{0}^q$ and $\chi$ is a character modulo $q$. At the lecture the following result was introduced:

$\vert \sum_{n\leq x} \chi(n) \vert \leq \varphi(q) -1$

I'm not very happy about the $-1$ part. I can see why the following is true:

$\vert \sum_{n\leq x} \chi(n) \vert\leq \varphi(q)$

because the sum can be "at most" $\vert\chi(1)\vert+\dots + \vert\chi(q-1)\vert$ and from the elements $1,\dots, q-1$ there are $\varphi(q)$ elements coprime to $q$.

Question(s)

Is $\vert \sum_{n\leq x} \chi(n) \vert \leq \varphi(q) -1$ true? And in this case how do I prove it?

guestguest
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1 Answers1

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Since $\chi$ is not a trivial character (one which only takes the values $1$ and $0$ on $\mathbb{Z}$) we have $$ \sum_{n=0}^{q-1}\chi(n)=0. $$ This means that $$ F(x)=\sum_{n\leq x}^{q-1}\chi(n)=\sum_{kq\leq n\leq x}\chi(n)=\sum_{0\leq n \leq x-kq}\chi(n) $$ where $kq$ is the largest multiple of $q$ that is smaller than $x$ (so $0\leq x-kq< q$). Now there are two cases: If $x-kq\geq q-1$ then $F(x)$ is $0$ by the first equation. If $x-kq<q-1$ then at most $\phi(q)-1$ terms are non-zero in the sum and since those terms have absolute value $1$ we find $|F(x)|\leq \phi(q)-1$.

MichalisN
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