# modular form -Petersson inner product

my question is about Petersson inner product.
i need to prove that $(E_k,f) =0$ $\forall f \in S_k(SL_2(\mathbb{Z}))$

the only thing that i think that should help me is that the space of cusp form has a basis of Hecke eigenform.
i cant use in the Hermitian property of the Hecke operator (i dont think that help) because is work only in the case both function are cusp but $E_k$ isnt a cusp form.

i know that $M_k(SL_2(\mathbb{Z})=<E_k> \oplus S_k(SL_2(\mathbb{Z}))$

thank

here is the prove :

Lemma 1 : $E_k$ is eigenfunction $\forall \space T_n$
Lemma 2 : $(T_nE_k,g) = (E_k,T_ng)$ $\space \forall g\in S_k$
Lemma 3: there is a basis $\{f_1,...f_m\}$ of $S_k$ such that is simulation eigenfunction $\forall T_n$
please note that $\{f1 ,... f_m,E_k\}$ is basis for $M_k$ and by multiplication we can assusme that$a_1(f'_1)=...=a_1(f'_m)=a_1(E'_k)=1$.

so it will be enough to show that $(f'_i,E_k)= 0$ $\space 0 < i <m+1$
now we know that if $a_1(f)=1 => T_n(f) =a_n(f)f.$
so if there exist n such that $a_n(f'_i) \neq a_n(E'_k)$ we finish.
assume not then $\forall n>0 \space a_n(f'_i) = a_n(E'_k)$ then $g=f'_i-E_k \in M_k$ however when we look at fourier series we see that g=1. when we know the only constant function in $M_k$ are zero. contradiction

i think it should work what do you think?