bio | website | cornellmath.wordpress.com |
---|---|---|
location | Ithaca, NY | |
age | 31 | |
visits | member for | 4 years, 11 months |
seen | Mar 23 at 5:33 | |
stats | profile views | 933 |
I was a graduate student at Cornell University studying harmonic analysis with Camil Muscalu. Now I'm a postdoc at Washington University in St. Louis.
Aug 12 |
awarded | Nice Answer |
Jul 31 |
awarded | Scholar |
Jul 31 |
accepted | Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform |
Jul 31 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
I'm embarrassed I didn't make that observation myself about the finite dimensionality of $V$. It's a nice little argument. It also highlights that the $2$ in the $L^2$ condition isn't key as you wind up in a finite dimensional setting. Showing that the inner product, $\langle f(x-k_0), f(x-k)\rangle$, tends to zero is equivalent to proving the Riemann-Lebesgue lemma, but if you just approximate $f$ with a $C_c^\infty$ function, I think all you need is the dominated convergence theorem and Cauchy-Schwarz, so trigonometric polynomials never enter the picture. Thanks! |
Jul 29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
Sorry, maybe I'm just being dense, but why does linear dependence imply the space is finite dimensional? Certainly if you have one dependence relation, you will have infinitely many, but it doesn't seem obvious at first glance that I should get a finite dimensional space. |
Jul 29 |
awarded | Student |
Jul 29 |
asked | Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform |
Apr 30 |
awarded | Yearling |
Dec 20 |
comment |
When is a collection of exponentials dense in $L^2(K), |K|<\infty$
Alexei Poltoratski gave a very nice series of lectures with a number of comments in this direction this summer at Clemson. I don't think he really discussed the R^d case, but notes similar to the lectures he gave are available here: internetanalysisseminar.gatech.edu/lectures_uncp Some of the references may point you in the right direction. |
Sep 14 |
awarded | Informed |
Jun 25 |
awarded | Citizen Patrol |
May 25 |
awarded | Necromancer |
Apr 30 |
awarded | Yearling |
May 8 |
comment |
Square roots of the Laplace operator
Now that I think about it, the analytic continuation might need some additional work... there might be some poles at certain integers $\alpha$, so you would need to add an additional term to balance things out. This computation has to be written down someplace. Sorry that I don't know of a good place. |
May 8 |
comment |
Square roots of the Laplace operator
Yes, much of this stuff is fairly disjointed, as I recall, which is unfortunate. Many of the basic computations are not terribly difficult (just tricky), so I think they are not written down in many places. The issue you are having in the computation in your comment comes from having to use some analytic continuation technique to extend the range of possible $\alpha$s in the inverse transform of $|\xi|^\alpha$. $|x|^{-n-1}$ is not locally integrable, as Tom pointed out, so its a bit more delicate. |
May 8 |
answered | Square roots of the Laplace operator |
Apr 30 |
awarded | Yearling |
Nov 23 |
comment |
Carleson's Theorem (on the Adeles and other exotic groups)
I don't think you can accept more than one answer, but that's ok. I just hope what I wrote helped in some way! |
Oct 19 |
comment |
Can the supremum of continuous functions be discontinuous on a set of positive measure?
Hm, I suppose you could also have just made it 1 on $K_n$ and 0 on $E$ and avoided that $\sup(-f_n)$ business... |
Oct 19 |
answered | Can the supremum of continuous functions be discontinuous on a set of positive measure? |