Let $L(s, \chi)$ denote the Dirichlet $L$-function associated to the character $\chi$.

In his paper A Mean Value Estimate for Real Character Sums, Heath-Brown proves a mean value bound for $L(s,\chi)$ when $\chi$ is a real character. In particular, he shows that $$ \sum_{\substack{\chi \leq Q \\ \chi \text{ real}}} \lvert L(\tfrac{1}{2} + it, \chi) \rvert^4 \ll (Q (1 + \lvert t \rvert) + 1)^{1 + \epsilon}, \tag{1} $$ where we interpret the sum to be over those real characters with conductors less than $Q$.

I am looking for a proof of $(1)$ (or perhaps a second moment, if that's more readily available) when $\chi$ is allowed to be complex. For instance, suppose that $\chi$ is allowed to run over $n$th-power characters of conductor less than $Q$ (by $n$th-power, I mean that $\chi^n(a) = 1$ for $a$ relatively prime to the conductor of $\chi$).

From what I understand, the key reason why Heath-Brown's work doesn't extend immediately is because he employs a large sieve involving quadratic characters in the form $$ \sum_{m \leq M} \lvert \sum_{n \leq N} a(n) \chi_m(n) \rvert^2 \ll (MN)^\epsilon (M + N) \sum_{n \leq N} \lvert a(n)\rvert^2 $$ where the $a(n)$ are general complex numbers. I am unfamiliar with most work involving such sieves, but I would not be surprised if one had managed to produce good bounds for sieves with $n$-th power characters; and from there, one might produce good bounds for moments of $L$-functions with $n$-th power characters as well.

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    $\begingroup$ Have you looked at this paper? arxiv.org/abs/1101.3646 $\endgroup$ – Peter Humphries Aug 25 '16 at 21:31
  • $\begingroup$ @PeterHumphries Thank you for the reference. I hadn't seen that before. Looking into papers that cite it led me to this paper of Blomer, which has a result exactly analogous to the one I want, but for Hecke characters. $\endgroup$ – davidlowryduda Aug 25 '16 at 21:57
  • $\begingroup$ I don't think you're going to get a result for larger values of $n$ without some serious effort. Blomer's result works because, as stated in his introduction, he's assuming the base field $K$ over which he works contains the $n$-th roots of unity. When $K = \mathbb{Q}$, we only can work with $n = 2$ via this method. $\endgroup$ – Peter Humphries Aug 25 '16 at 22:00
  • $\begingroup$ Take a look at arxiv.org/pdf/1112.1650v1.pdf in particular Theorem 1.3 and Corollary 1.4. $\endgroup$ – user97698 Aug 25 '16 at 23:55
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    $\begingroup$ Stephan Baier and I looked at some problems in this vein for $n=3$. See arxiv.org/pdf/0804.2233v4.pdf $\endgroup$ – Matt Young Aug 26 '16 at 2:58

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