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)