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Let $a$ be an integer, $p$ a prime (much) greater than $a$, and $\chi$ a Dirichlet character. There is an abundant literature on the sums $$S(\chi,a)=\sum_{i=1}^a \chi(i),$$ called short (or incomplete) character sums, and their average in various sense. Yet I have not found in this literature an answer to the following specific question (perhaps because of its very abundance):

Let us form the average of the modulus of $S(\chi,a)$ over $p$; that is, let's define: $$AS(p,a) = \frac{1}{p-1} \sum_{\chi} |S(\chi,a)|,$$ where the sum is over the $p-1$ Dirichlet characters mod $p$. An easy upper bound for $AS(p,a)$, namely $$AS(p,a) \leq \sqrt{a}$$ can be obtained by applying Cauchy-Schwarz.

Is this upper bound essentially the best possible? Precisely, is it true or false that $AS(p,a)=o(\sqrt{a}),$ uniformly in $p >a$? (And what if one restricts to the domain where $p$ is not too large, say $p < a^K$ for some fixed constant $K$ -- so that the character sums are not too short?)

I have done some numerical experimenta, which are not really conclusive in what sense or another. I might very well have overlooked some paper in the literature treating this question. Thanks for any idea, guess, conjecture, proof or reference...

PS: here is the proof of the trivial upper bound: $AS(p,a) \leq \frac{1}{p-1} \sqrt{(p-1) \sum_{\chi} |S(\chi,a)|^2}$ by Cauchy-Schwarz, and $\sum_\chi |S(\chi,a)|^2$ can be expanded as $\sum_{\chi,i,j} \chi(i) \overline{\chi(j)}=\sum_{i,j} \sum_\chi \chi(ij^{-1})$ where $i,j$ run from $1$ to $a$ and $ij^{-1}$ is computed modulo $p$; then the sum over $\chi$ is $0$ unless $i=j$ which happens $a$ times, and it this case it is $p-1$, so $\sum_\chi |S(\chi,a)|^2=(p-1)a$, hence $AS(a,p)\leq \sqrt{a}$.

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Two comments: you may find it interesting to look at arxiv.org/abs/1011.4917 which is a paper of Chatterjee and Soundararajan. Also, your sums are slightly sensitive to the principal character. For instance, take $a=p-2$ and compare what happens when you do not include the principal character in your definition of $AS(p,a)$. –  Matt Young May 1 '13 at 16:25
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1 Answer

up vote 8 down vote accepted

Updated answer: Here's an unconditional lower bound of exactly the right order $\sqrt{a}$, however I need $a$ to be comparable to $p$. It is known that (for any $a$)

$ \frac{1}{p-1} \sum_{\chi} |S(\chi, a)|^{2k} \ll_{k} p^{k} $

[See: H. Montgomery, R. Vaughan, Mean values of character sums. Canad. J. Math. 31 (1979), no. 3, 476–487]

Thus $( \frac{1}{p-1} \sum_{\chi} |S(\chi,a)|^{4})^{1/4} \ll p^{1/2} $. Now by Hölder, we have

$ \frac{ ( \frac{1}{p-1} \sum_{\chi} |S(\chi,a)|^{2})^{3/2} } { ( \frac{1}{p-1} \sum_{\chi} |S(\chi,a)|^{4})^{2/4} } \leq \frac{1}{p-1} \sum_{\chi} |S(\chi, a)|.$

This gives:

$\frac{a^{3/2 }}{p} \ll \frac{1}{p-1} \sum_{\chi} |S(\chi, a)| $

So in the range $a \sim p$ this gives a sharp lower bound.

Original answer: Here's an argument that shows one can't replace $a^{1/2}$ with $a^{1/2-\delta}$ for any $\delta>0$, assuming GRH. This gives a better estimate than the argument above for small $a$.

From orthogonality (as you have shown) we have that $\frac{1}{p-1}\sum_{\chi} |S(\chi, a)|^2 = a$. On the other hand, assuming GRH, one has that $|S(\chi,a)| \leq \sqrt{a}p^{\epsilon} $ (for non-principal characters).

So ignoring principal characters (which are minor order) we have: $\frac{1}{p-1}\sum_{\chi} |S(\chi, a)| \geq \frac{1}{p-1}\sum_{\chi} |S(\chi, a)|^2 /(\sqrt{a} p^{\epsilon}) \geq a / (\sqrt{a} p^{\epsilon}) $. So you can't hope for an estimate better than $\sqrt{a} p^{-\epsilon}$.

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Thank you, Mark. –  Joël May 1 '13 at 13:11
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