Connection to zeroes on the critical line
Suppose (for fictitious expositional simplicity)
= 0 for a cusp form on
It is not difficult to handle arbitrary
Let H(t) = M
Then H(t) = H(-t) is real.
be an approximate identity such that M
If L(s) has only a finite number of zeroes on the critical line, then the
following integral must also be of order T:
But it cannot if
(x) vanishes to infinite order at x=1 (
near a point where
behaves as if it is zero).
In that case this integral decays as O(T
) for any N > 0!
The above was for a cusp form on SL(2,
For congruence groups, the
point x=1 changes to
q, q = level.
The bound S(T,x) = O
that the last integral is still o(T) with room to spare.
: numerically that integral decays only like T