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For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have $$ \tilde{f}(s) = (2\pi)^{-s} \Gamma(s) L(s,f). $$ My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.

I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?

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    $\begingroup$ $(2\pi)^{-s} \Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-\sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^\sigma, \sigma \in Gal(\overline{\mathbb{Q}},\mathbb{Q})$ is a modular form if $f$ is an eigenform) $\endgroup$ – reuns Nov 18 '18 at 15:36
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If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i \infty$.

By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n \geq 1$. This implies that $f=g$ (having the same Laurent series).

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