On Friday 28 September 2007 12:38:44 pm you wrote:
> [2] The L_infinity norm of a function over given interval (say of
> size 1) is always BIGGER than (or equal to) the L_2 norm. (For
> simplicity, suppose that the function f is positive, since it doesn't
> matter. Let M be its max (i.e., its L_infinity norm) so f(x) \le M and
> so taking L_2 norms of both of these functions f(x) and M we get that
> the L_2-nomrm of f in \le that of M, i.e. M. SO, any conjecture that
> bounds L_infinity norms from above is STRONGER (a priori) than a
> conjecture that bounds L_2 norms.
One thing I was confused about at first with A-T versus us, is that
the L_2 norm is the integral of f^2, whereas the L_infinity norm is
the max of f. Thus the 1/sqrt(x) bound on the L_oo norm roughly
corresponds to a 1/x bound on L_2.
By the way, if f is e.g., the constant function f(x) = 2 on the
interval [0,1], then the L_oo norm is 2 and the L_2 norm is
integral(4, 0, 1) = 4 > 2. So your remarks above I guess
apply only when f <= 1, which is of course exacty the situation
of interest.
> [3] I'm going to check that I am writing the A-T conjecture right,
> and---although I hope not--- it is possible that I have written it
> incorrectly, or a bit strangely, so let me work on this...i.e. more
> soon.
I finally just tried adapting my code to compute the L_oo norm
instead of the L_2 norm. The graphs are much like before, except
the jaggedy line rises to 1/2 instead of 1. And, like before,
if when curves have large rank the graph starts near 1/4 instead
of 1/2.
-------------
So here are some questions we could think about next week:
1. Is our L_2 conjecture implied by the A-T conjecture?
The answer to this should be "trivially yes".
2. Does our L_2 conjecture imply A-T?
3. Does our L_2 conjecture imply GRH for L(E,s)?
This is not trivial, since our conjecture might not
imply the A-T conjecture.
4. Compute data both for the L_2 conjecture and the A-T conjecture
for curves of ranks 0,1,...,8, and all primes p < 10^7.
This will take a lot of time (a day), but the code is
already written.
5. Formulate a conjecture that incorporates the rank. This
is just begging to be done, since it connects GRH and BSD.
In light of the A-T conjecture, I think our L2 conjecture
is probably that for all eps > 0 we have
Delta(X) <= 1/X^(1-eps) for X >> 0.
(There might be a limsup subtlety, too.)
However, in the data if the curve has large rank
then X must be taken much much much larger to get a
bound like above. The problem is thus to figure
out how large.
Let r be the rank of our elliptic curve E. Maybe there is a "nice" function
f(r) such that
Delta(X) <= 1/X^(1 - f(r)/log(X))
for *all* X?
For example, for my example rank 6 curve it's *striking* how close
the graphs of
Delta(X) and 1/X^(1 - 4/log(X))
arr...
Eyeballing graphs for some r gives:
r | f(r) that seems to work very well
| for x < 10^5 with 80 sample points
----------------------------------------
0 | 0
1 | 1
2 | 2
3 | 3
4 | 3.5 (and 3 doesn't work)
5 | 3.5
6 | 4
7 | 4.25 = 17/4
8 | 4.25 = 17/4
28 | 6.25 = 25/4
Delta(X) <= 1/X^(1 - f(r)/log(X)) for *all* X,
where f(r) = r/log(r)