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Let $p$ be a prime. Let $\omega$ be a $ p $-th root of unity. We know that $ \chi_\alpha (x) = \omega^{\alpha\cdot x} $ are the additive characters of $ \mathbb{Z}_p $.

I have a question about bounding incomplete sums of these characters.

Let $ T $ be a subset of $ \mathbb{Z}_p $ of size $ (p-1)/2 $. I want statistics on sums of the form: $$ \sum_{t \in T} \chi_\alpha(t) $$

In the worst-case, we know that this sum can be as large as $ c\cdot p $ where $ c $ is some constant. (e.g. when $ \alpha = 0 $ or $ \alpha =1 $ and $ T = \{ 1, 2, \dots, (p-1)/2 \}$ )

Can we say something stronger about the expectation? That is, for every subset $ T $, the expectation: $$ \mathbb{E}_{\alpha \gets \mathbb{Z}_p} \sum_{t \in T} \chi_\alpha(t) $$

For example, is it bounded by $ p^{\frac{1}{2}} $?

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The trivial estimate here would be $$ \mathbb E\,\Bigg|\sum_{t\in T}\chi_\alpha(t)\Bigg| \le \Bigg( \mathbb E\,\Bigg| \sum_{t\in T} \chi_\alpha(t) \Bigg|^2 \Bigg )^{1/2} = \Bigg( \sum_{t_1,t_2\in T} \frac1p\sum_{\alpha\in\mathbb Z_p} \omega^{\alpha\cdot(t_1-t_2)} \Bigg)^{1/2} = \sqrt{|T|} \le p^{1/2}. $$ Does this answer your question?

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    $\begingroup$ One can see that this bound is sharp to within a constant factor, as long as $|T| \leq p/2$, by considering a random subset $T$. $\endgroup$
    – Will Sawin
    Commented Oct 14, 2017 at 11:21
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    $\begingroup$ Also, without the absolute values, one has ${\mathbb E} \sum_{t \in T} \chi_\alpha(t) = 1_{0 \in T}$. $\endgroup$
    – Terry Tao
    Commented Oct 14, 2017 at 16:36

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