# Uniqueness of $L$-series of cusp forms

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?

• $(2\pi)^{-s} \Gamma(s) L(s,f)$ 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) – reuns Nov 18 '18 at 15:36

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).