Let $f=\sum\_{n\geq 1} a\_nq^n$ and $g=\sum\_{n\geq 1} b\_nq^n$ be two distinct cusp eigenforms of same weight $k\geq 2$ with real Fourier coefficients. We normalize the coefficients by defining: $\widetilde {a\_n}:=\frac{a\_n}{n^{(k-1)/2}}$ and similarly $\widetilde{b\_n}:=\frac{b\_n}{n^{(k-1)/2}}$. I was wondering if there is a quick way to see that $$\sum\_{p-prime}\frac {\widetilde{a\_p}-\widetilde{b\_p}}{p^s}$$ is bounded as $s\to 1^{+}$. 



L-series of difference of two cusp forms
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 $\sum\_{p} \frac{\lambda\_f(p) - \lambda\_g(p)}{p^s} = \sum\_{p} \frac{\lambda\_f(p)}{p^s} - \sum\_{p} \frac{\lambda\_g(p)}{p^s}$.




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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](https://mathoverflow.net/questions/178654/characterizing-the-newforms-s-t-the-associated-symmetric-square-l-function-ha)). 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).


