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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    $\begingroup$ 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 $AS(p,a)$. $\endgroup$ – Matt Young May 1 '13 at 16:25

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 $\sqrt{a} p^{-\epsilon}$.


This is a very interesting question, and I want to point out a paper that just appeared on the arXiv today by Bondarenko and Seip which sheds light on the problem. The question that Bondarenko and Seip consider is slightly different (in the $t$-aspect) rather than characters, but the problem remains essentially the same. They attribute to H. Helson the analogous problem (in $t$-aspect) of whether the $L^1$ norm is $o(\sqrt{a})$. Their work doesn't give a definitive resolution of Helson's problem (or this question), but what it shows (or more precisely, what an adaptation of their method should show; I didn't check all the details of the adaptation) is the following:

Suppose $a\le p^{\frac 14}$ (and $a$ is large enough) then $$ \frac{1}{p-1} \sum_{\chi \pmod p} \Big|\sum_{n\le a} \chi(n)\Big| \gg \sqrt{a}(\log a)^{-\delta/4}, $$ where $$ \delta = 1- \frac{1+\log \log 2}{\log 2} = 0.086071\ldots $$ is the constant appearing in the multiplication table problem.

The main idea is to consider the character sums over integers with a specified number of prime factors, and to identify the range (of the number of prime factors) in which the $L^2$ and $L^4$ norms are comparable. Then one gets lower bounds on the $L^1$ norms by Holder. The comparison of the $L^2$ and $L^4$ norms leads to an analysis of the same spirit as the multiplication table problem. There is one more input from analysis to show that the restricted sums lead to lower bounds for the unrestricted sums; this is the step that I didn't check fully whether it extends to the question on characters. Alternatively, one could show that the moments for character sums match the moments in $t$-aspect in certain ranges, and then simply use the work of Bondarenko and Seip.


In his beautiful talk at MSRI today (1 May 2017), Adam Harper announced the following theorem which settles this problem. Let $p$ be a large prime, and let $x$ be less than $p$. Put $L=\min (x,p/x)$. Then $$ \frac{1}{p-1} \sum_{\chi \pmod p} \Big| \sum_{n\le x} \chi(n) \Big| \ll \frac{\sqrt{x}}{\min (({\log\log L})^{\frac 14}, ({\log \log \log p})^{\frac 14})}. $$ One might expect that the $(\log \log \log p)^{\frac 14}$ term above can be removed with more work. In particular, Harper's result shows that when $x=o(p)$, the $L^1$ norm is $o(x)$. MSRI should have the video of this lecture up in a short while.


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