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Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{p^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$. Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) (1 - \beta_f(p)^2 p^{-s})}.\] Then for $\Re(s) > 1$, \[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{p^{ks}}.\] Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through.

Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{kp^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$ (in fact, this only requires the weaker bound $|\alpha_f(p)|, |\beta_f(p)| \leq p^{1/2 - \delta}$ for some $\delta > 0$). Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) (1 - \beta_f(p)^2 p^{-s})}.\] Then for $\Re(s) > 1$, \[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{kp^{ks}}.\] Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through (though this time, we need bounds towards the generalised Ramanujan conjecture of the form $|\alpha_f(p)|, |\beta_f(p)| \leq p^{1/4 - \delta}$ for some $\delta > 0$, which is certainly known).

Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 1}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{p^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$. Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) (1 - \beta_f(p)^2 p^{-s})}.\] Then for $\Re(s) > 1$, \[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 1}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{p^{ks}}.\] Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through.

Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{p^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$. Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) (1 - \beta_f(p)^2 p^{-s})}.\] Then for $\Re(s) > 1$, \[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 2}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{p^{ks}}.\] Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through.

Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 1}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{p^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$. Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Let $L(s,f)$ denote the $L$-function of a Hecke eigenform $f$, normalised so that the critical line is $\Re(s) = 1/2$. Then for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p) p^{-s} + p^{-2s}} = \prod_p \frac{1}{(1 - \alpha_f(p) p^{-s}) (1 - \beta_f(p) p^{-s})},\] where $\lambda_f(p)$ is the normalised $p$-th Hecke eigenvalue and $\alpha_f(p) + \beta_f(p) = \lambda_f(p)$, $\alpha_f(p) \beta_f(p) = 1$; by Deligne's proof of the Ramanujan conjecture, $|\alpha_f(p)| = |\beta_f(p)| = 1$. It follows that for $\Re(s) > 1$, \[\log L(s,f) = \sum_{p} \frac{\lambda_f(p)}{p^s} + \sum_{p} \sum_{k = 1}^{\infty} \frac{\alpha_f(p)^k + \beta_f(p)^k}{p^{ks}},\] and the second term is uniformly bounded as $s \searrow 1$. Moreover, the $L$-function $L(s,f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (by the usual proof of the prime number theorem; see chapter 5 of Iwaniec and Kowalski), so $\log L(s,f) \to \log L(1,f)$ as $s \searrow 1$.

In particular, it follows that $\sum_{p} \frac{\lambda_f(p)}{p^s}$ is bounded as $s \searrow 1$, and the same is obviously true for $\sum_{p} \frac{\lambda_f(p) - \lambda_g(p)}{p^s} = \sum_{p} \frac{\lambda_f(p)}{p^s} - \sum_{p} \frac{\lambda_g(p)}{p^s}$.

Similarly, let $L(s,\mathrm{sym}^2 f)$ denote the symmetric square $L$-function, so that for $\Re(s) > 1$, \[L(s,f) = \prod_p \frac{1}{1 - \lambda_f(p^2) p^{-s} + \lambda_f(p^2) p^{-2s} - p^{-3s}} = \prod_p \frac{1}{(1 - \alpha_f(p)^2 p^{-s}) (1 - p^{-s}) (1 - \beta_f(p)^2 p^{-s})}.\] Then for $\Re(s) > 1$, \[\log L(s,\mathrm{sym}^2 f) = \sum_{p} \frac{\lambda_f(p^2)}{p^s} + \sum_{p} \sum_{k = 1}^{\infty} \frac{\alpha_f(p)^{2k} + 1 + \beta_f(p)^{2k}}{p^{ks}}.\] Again, $L(s,\mathrm{sym}^2 f)$ is nonvanishing and has no poles on the line $\Re(s) = 1$ (though when the nebentypus is nonprincipal, there may be a pole; see this question). So the same argument goes through.