Average of Fourier coefficients of a cusp form of half integral weight

Suppose $f$ is a cusp form of half integral weight $k$ w.r.t. the group $\Gamma_0(4)$ ($k$ is not very low, can assume $k \ge 11/2$), and $a_n$ is its Fourier coefficient. The Linnik bound says that if $f$ is square-free then we have the estimate $$|a_n| \ll_\epsilon n^{k/2-2/7+\epsilon}$$

My question is, is there a nontrivial bound for $\left| \sum_{n=N}^{N+H} a_n \right|$? Here nontrivial means better than simply adding up the Linnik bound individually. Also will averaging take care of the square-free issue in the Linnik bound?

An equivalent problem is to estimate $$\sum_{n=N}^{N+H} \sum_{4|c} \frac{K(m,n,c)}{c}$$

Note: $m$ is assumed to be fixed and is not averaged. To give some order of magnitude, $H$ is supposed to be around $N^{1/5}$, so it is a rather short sum.

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I think it is misleading to use the term "Linnik bound" as it was proved by Iwaniec in 1987, while Linnik died in 1972. This bound was improved several times, e.g. by Blomer-Harcos to $n^{k/2-5/16+\epsilon}$, see Corollary 2 in J. reine angew. Math. 621 (2008), 53-79. The bound is true as long as the primes dividing the level have bounded exponents in $n$ (which includes all square-free $n$'s, but many more). – GH from MO Jul 15 '13 at 9:07
I don't have time to work on this, but the convexity bound for the Dirichlet series $\sum_n a_n/n^s$ (which is entire and satisfies a functional equation) coupled with Mellin transform techniques should yield better cancellation for your sum than the "Linnik bound" coupled with the triangle inequality. On the other hand, this will only work for some $H>N^{1/2}$, I am afraid. – GH from MO Jul 15 '13 at 9:28
Thanks to the above. Just FYI, I'm actually balancing this estimate with another estimate, with $H$ an adjustable parameter. As per the bound in Blomer-Harcos, $H$ is now $N^{11/52}$. The better the bound is, the larger the $H$, so I'm still curious whether averaging in $n$ could take advantage of that. – Fan Zheng Jul 15 '13 at 15:32
I am glad I could help. If you have a preprint (later in the future), please post the link here, as the topic is interesting. – GH from MO Jul 15 '13 at 15:41
So, if we assume $H \gg N^{1/2}$, what kind of result will we get? I'm trying to decide if it will actually pushes $H$ that large, so that it is self-fulfilling. – Fan Zheng Jul 18 '13 at 17:07