bio | website | maths.ox.ac.uk/~greenbj |
---|---|---|
location | Auckland, New Zealand | |
age | 37 | |
visits | member for | 4 years, 5 months |
seen | 17 hours ago | |
stats | profile views | 11,994 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
May 15 |
awarded | Nice Answer |
May 14 |
answered | Proof of the weak Goldbach Conjecture |
Apr 23 |
awarded | Yearling |
Apr 19 |
accepted | Additive Combinatorics - reference request |
Apr 19 |
comment |
Additive Combinatorics - reference request
I finally dug out the reference myself using a MathSciNet search. MR1931192 (2003f:11016) Reviewed Schoen, Tomasz(PL-POZN) The cardinality of restricted sumsets. (English summary) J. Number Theory 96 (2002), no. 1, 48–54. |
Apr 19 |
comment |
Additive Combinatorics - reference request
Thanks Seva - but I'm sure I've seen this exact argument somewhere. I guess not in one of your papers then? |
Apr 19 |
asked | Additive Combinatorics - reference request |
Apr 9 |
comment |
Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$
Is this exactly "coupon collecting analysis"? For each choice of B = {b_1,..,b_k} you can try and estimate the probability that AB is all of G, if A is chosen randomly. But then you have to sum over a large selection of B's, and the triangle inequality may be too crude. In fact, I posed as an open problem (Barbados workshop, March 2012) the question of deciding whether the log N should really be there. Can Z/NZ be covered by O(N^{1/2}) translates of a random set A of size N^{1/2}? |
Apr 8 |
awarded | Good Answer |
Mar 26 |
comment |
Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones
I think that from standard asymptotics on the number of zeros of height up to T you get a bound of O(log T). Apparently even on RH is is only known that the multiplicity of 1/2 + iT as a zero is O(log T/log log T). So I guess you shouldn't expect too much better than this O(log T) bound, |
Mar 11 |
comment |
On the $L^1$-norm of certain exponential sums.
This is nicely done. |
Mar 8 |
comment |
On the $L^1$-norm of certain exponential sums.
Yep, this issue (of whether a 1-dissociated set, or "subsum-distinct" set) has a positive density 2-dissociated subset seems a little unclear. It seems closely related to work of Pisier and Bourgain from the late 80's on Sidon sets. In particular I wonder whether Pisier's arithmetic characterisation of Sidon sets works with 2-dissociated in place of dissociated. Perhaps the proposer could comment on whether he needs the full generality? |
Mar 8 |
comment |
On the $L^1$-norm of certain exponential sums.
Noam, Very nice! To do the splitting, I was planning to use Holder: $\int |fg| \leq (\int |f|)^{1/3}(\int |g|)^{1/3}(\int |f g|^2)^{1/3}$. You have the trivial bound on each factor (note that fg is the FT of the set of subset sums of S), so you only need to win in one factor, say the first one. So actually, I think all you need is that any subsum-distinct set has a 2-dissociated subset of positive density. That's clear in the case of powers of 2 (just thin out every other power of two) but actually not so obvious in general. I'm looking at some literature on that right now. |
Mar 8 |
comment |
The polynomial Freiman-Ruzsa conjecture for the special case when $f$ is a bijection
It seems unlikely to me that the bijection assumption would help. If f is an arbitrary function then, provided f is not "really far" from a bijection I'd expect that some modification f'(x) = f(x) + eps(x) with eps varying in a small set would make f' pretty close to a bijection (perhaps use Hall's marriage theorem or something). PFR for f and for f' are basically the same problem. Being a bijection is not a property that is useful in connection with several of the existing techniques in the area, particularly Fourier analysis (cf. the work of Sanders). |
Mar 7 |
comment |
On the $L^1$-norm of certain exponential sums.
I should add, though, that I have to prepare an undergraduate lecture now - so I'll try and write down the details tomorrow. |
Mar 7 |
comment |
On the $L^1$-norm of certain exponential sums.
I chatted with Tom Sanders about this today in Oxford, and I think we can more-or-less solve it, at least the second case. The key idea is to write |1 + e(t)| as 2^{1/2} (1 + cos(2 pi t))^{1/2}, then use the inequality (1 + x)^{1/2} \leq 1 + x/2 - cx^2, valid for some $c > 0 (in fact for c = 3/2 - 2^{1/2}). Now expand everything out, and you get a bound for your integral of 2^{n/2}(1 - c/4)^n if S is "good": has no relations with coefficients <= 2. A bit of fiddling should give exactly what you want, with your weaker assumption on S; in the powers of two case you can split S into two good sets |
Mar 6 |
comment |
On the $L^1$-norm of certain exponential sums.
Joel, this is a nice question. I haven't thought about it yet, but the first thing that comes to my mind (especially in connection with your second condition) is the paper of Mauduit and Rivat on binary digits of primes. They have to estimate the L^1 norm of the exponential sum of some Riesz products quite similar to yours, and they do beat the trivial Cauchy-Schwarz bound by an expontial factor. |
Jan 22 |
awarded | Good Answer |
Dec 16 |
awarded | Popular Question |
Oct 19 |
comment |
Showing non-expansion for $x\rightarrow x+1, x\rightarrow 2x.$
Harald: I think his name is Gonzalo Fiz-Pontiveros. |