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6 Made clearer the argument in the case p doesn't divide N

The Weil conjectures proved by Deligne imply that every root $\alpha$ of the polynomial $T^2-a_p T +\psi(p) p^{k-1}$ satisfies $|\alpha| \leq p^{(k-1)/2}$.

Now there are two main cases :

1) $p$ divides $N$. Then $\psi(p)=0$ and $u_m = a_p^m$ for every $m \geq 0$. Since It can be shown, using purely analytical arguments, that $|a_p| \leq p^{(k-1)/2}$ (see Theorem 3 in Li's article "Newforms and functional equations", Math. Ann., 1975). Thus we get $|u_m| \leq p^{m(k-1)/2}$ which gives the desired inequality (and in fact, a stronger one).

[EDIT : In the case $p$ divides $N$, it seems that the bound $|a_p| \leq p^{(k-1)/2}$ can be obtained by purely analytical arguments, and was known before Deligne. See Theorem 3 in Li's article "Newforms and functional equations", Math.Ann., 1975 and the discussion in Mazur-Tate-Teitelbaum's article "On p-adic analogues of BSD" (section 12).]

2) $p$ doesn't divide $N$. By Deligne, we have Put $1-a_p X +\psi(p) p^{k-1} X^2 = (1-\alpha X)(1-\beta X)$with . In the case $|\alpha|=|\beta|=p^{(k-1)/2}$. k \geq 2$, the Weil conjectures proved by Deligne imply that$|\alpha|=|\beta|=p^{(k-1)/2}$(this still holds if$k=1$, by a theorem of Deligne and Serre). There are two subcases : Remark Remarks : It is conjectured that case 2b never happens if$k \geq 2$(semi-simplicity of crystalline Frobenius). Note also that in case 2a, we can write$u_m = \beta^m (1+\lambda+ \ldots + \lambda^m)$, where$\lambda := \alpha/\beta$satisfies$|\lambda|=1$and$\lambda \neq 1$. Thus in fact we get a bound of the form$|u_m| \leq C \cdot p^{m(k-1)/2}$, where the constant$C$depends only on the argument of$\alpha/\beta$. Finally, note that we also have the strict bound$|a_p|<2p^{(k-1)/2}$in case 2a. 5 added 13 characters in body Let me focus on how to deduce a bound for all coefficients$a_n$assuming the Deligne bound on$a_p$(this is a standard argument). Let$f$be a newform of weight$k$, level$N$and Nebentypus character$\psi$modulo$N$. Assume the Deligne bound$|a_p| \leq 2p^{(k-1)/2}$, and let us prove the inequality$|a_n| \leq d(n) n^{(k-1)/2}$for every$n \geq 1$(here$d(n)$is the sum of the positive divisors of$n$). Since both sides are multiplicative in$n$, it suffices to consider the case$n=p^m$, where$p$is prime. Put$u_m=a_{p^m}$. By looking at the Euler factor of$f$at$p$, we know the following formal identity :$\sum_{m \geq 0} u_m X^m = \frac{1}{1-a_p X +\psi(p) p^{k-1} X^2}$. The Weil conjectures proved by Deligne imply that every root$\alpha$of the polynomial$T^2-a_p T +\psi(p) p^{k-1}$satisfies$|\alpha| \leq p^{(k-1)/2}$. Now there are two cases : 1)$p$divides$N$. Then$\psi(p)=0$and$u_m = a_p^m$for every$m \geq 0$. Since$|a_p| \leq p^{(k-1)/2}$, we get$|u_m| \leq p^{m(k-1)/2}$which gives the desired inequality (in fact, a stronger one). [EDIT : In the case$p$divides$N$, it seems that the bound$|a_p| \leq p^{(k-1)/2}$can be obtained by purely analytical arguments, and was known before Deligne. See Theorem 3 in Li's article "Newforms and functional equations", Math. Ann., 1975 and the discussion in Mazur-Tate-Teitelbaum's article "On p-adic analogues of BSD" (section 12).] 2)$p$doesn't divide$N$. By Deligne, we have$1-a_p X +\psi(p) p^{k-1} X^2 = (1-\alpha X)(1-\beta X)$with$|\alpha|=|\beta|=p^{(k-1)/2}$. There are two subcases : 2a)$\alpha \neq \beta$. Solving the linear recurrence relation satisfied by$u_m$gives$u_m = \frac{\alpha^{m+1}-\beta^{m+1}}{\alpha-\beta} = \alpha^m+\alpha^{m-1} \beta + \ldots + \beta^m$. In particular$|u_m| \leq (m+1) p^{m(k-1)/2}$which is what we want. 2b)$\alpha = \beta = \frac{a_p}{2}$. In this case$u_m = (m+1) \alpha^m$for every$m \geq 0$, and the inequality also follows. Remark : It is conjectured that case 2b never happens if$k \geq 2$(semi-simplicity of crystalline Frobenius). Note also that in case 2a, we can write$u_m = \beta^m (1+\lambda+ \ldots + \lambda^m)$, where$\lambda := \alpha/\beta$satisfies$|\lambda|=1$and$\lambda \neq 1$. Thus in fact we get a bound of the form$|u_m| \leq C \cdot p^{m(k-1)/2}$, where the constant$C$depends only on the argument of$\alpha/\beta$. Finally, note that we also have the strict bound$|a_p|<2p^{(k-1)/2}$in case 2a. EDIT : case 2b can indeed happen in the case$k=1$(thanks to David for pointing this out). 4 added 336 characters in body Let me focus on how to deduce a bound for all coefficients$a_n$assuming the Deligne bound on$a_p$(this is a standard argument). Let$f$be a newform of weight$k$, level$N$and Nebentypus character$\psi$modulo$N$. Assume the Deligne bound$|a_p| \leq 2p^{(k-1)/2}$, and let us prove the inequality$|a_n| \leq d(n) n^{(k-1)/2}$for every$n \geq 1$(here$d(n)$is the sum of the positive divisors of$n$). Since both sides are multiplicative in$n$, it suffices to consider the case$n=p^m$, where$p$is prime. Put$u_m=a_{p^m}$. By looking at the Euler factor of$f$at$p$, we know the following formal identity :$\sum_{m \geq 0} u_m X^m = \frac{1}{1-a_p X +\psi(p) p^{k-1} X^2}$. The Weil conjectures proved by Deligne imply that every root$\alpha$of the polynomial$T^2-a_p T +\psi(p) p^{k-1}$satisfies$|\alpha| \leq p^{(k-1)/2}$. Now there are two cases : 1)$p$divides$N$. Then$\psi(p)=0$and$u_m = a_p^m$for every$m \geq 0$. Since$|a_p| \leq p^{(k-1)/2}$, we get$|u_m| \leq p^{m(k-1)/2}$which gives the desired inequality (in fact, a stronger one). [EDIT : In the case$p$divides$N$, it seems that the bound$|a_p| \leq p^{(k-1)/2}$can be obtained by purely analytical arguments, and was known before Deligne. See Li's article "Newforms and functional equations", Math. Ann., 1975 and the discussion in Mazur-Tate-Teitelbaum's article "On p-adic analogues of BSD" (section 12).] 2)$p$doesn't divide$N$. By Deligne, we have$1-a_p X +\psi(p) p^{k-1} X^2 = (1-\alpha X)(1-\beta X)$with$|\alpha|=|\beta|=p^{(k-1)/2}$. There are two subcases : 2a)$\alpha \neq \beta$. Solving the linear recurrence relation satisfied by$u_m$gives$u_m = \frac{\alpha^{m+1}-\beta^{m+1}}{\alpha-\beta} = \alpha^m+\alpha^{m-1} \beta + \ldots + \beta^m$. In particular$|u_m| \leq (m+1) p^{m(k-1)/2}$which is what we want. 2b)$\alpha = \beta = \frac{a_p}{2}$. In this case$u_m = (m+1) \alpha^m$for every$m \geq 0$, and the inequality also follows. Remark : It is conjectured that case 2b never happens if$k \geq 2$(semi-simplicity of crystalline Frobenius). Note also that in case 2a, we can write$u_m = \beta^m (1+\lambda+ \ldots + \lambda^m)$, where$\lambda := \alpha/\beta$satisfies$|\lambda|=1$and$\lambda \neq 1$. Thus in fact we get a bound of the form$|u_m| \leq C \cdot p^{m(k-1)/2}$, where the constant$C$depends only on the argument of$\alpha/\beta$. Finally, note that we also have the strict bound$|a_p|<2p^{(k-1)/2}$in case 2a. EDIT : case 2b can indeed happen in the case$k=1\$ (thanks to David for pointing this out).

3 edited body
2 Corrected : case 2b can indeed happen in the case k=1.
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