I was studying theorem 6.2 of Apostol Modular functions and Dirichlet series in number theory and I am asking if my proof of this part of the theorem is right.
Part (d) - The only entire cusp form of weight $k<12$ is zero function.
What I think about it - If $f$ is a cusp form then constant coefficients in its Fourier expansion vanish. Using Theorem 6.1 , $k = 12N + 6 N(i) + 4N(\rho) +12N(i\infty)$, as $k <12$, so $N( i\infty) = 0 $. If I put $\tau= i \infty$ in $f$, then $f(i\infty) = 0$ as $c(0) =0$. So, this means $N(i\infty)$ greater than equal to $1$, which is a contradiction. So, f must be a zero function .
Can somebody please check whether I am right or not and give hints if my proof is wrong?
If you are still looking for an answer, here it is. If $f$ is a cusp form that is not identically zero then obviously from its Fourier expansion we can see that $$f(i\infty)=c(0)=0$$ because $e^{-2\pi \infty}=0$. Thus we know that $N(i\infty)>0$. This contradicts with the fact that $12N(i\infty)\leqslant k<12$. Therefore, we know that the weight formula is not applicable to $f$. Therefore the only exception is when $f$ is identically zero.
So yes you are right.