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Mar 27, 2017 at 20:02 history edited Peter Humphries CC BY-SA 3.0
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Mar 27, 2017 at 19:49 comment added GH from MO Instead of Deligne's theorem (which is very deep and specific for holomorphic cusp forms), it suffices for the first part that $\sum_{n\leq x}|\lambda_f(n)|^2\ll_f x$, which is quite easy to prove. For the second part a bit more is needed, but Deligne's theorem is still an overkill.
Mar 27, 2017 at 19:31 history edited GH from MO CC BY-SA 3.0
the k-sums now start at k=2 instead of k=1 (as they should)
Mar 27, 2017 at 19:03 history edited Peter Humphries CC BY-SA 3.0
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Mar 27, 2017 at 18:53 comment added Goro Thank you! If I may ask a related question: I know that there is an automorphic rep with the same $L$-function as $Sym^2 f$. Why does it follow that $$\sum_{p} \frac{\lambda_f(p^2)}{p^s}$$ is also bounded when $s\to 1^{+}$?
Mar 27, 2017 at 18:47 vote accept Goro
Mar 27, 2017 at 18:41 history answered Peter Humphries CC BY-SA 3.0