Timeline for Three optimization problems of uncertainty principle/Paley-Wiener type
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8 at 9:58 | comment | added | H A Helfgott | For (a), it would be reasonable to consider candidates such that $F$ is non-negative but $\widehat{\varphi}$ isn't (as then one is taking advantage of cancellation). | |
| Nov 26 at 10:58 | comment | added | H A Helfgott | Right, that works for (b). Now, for (c), the problem becomes: given those constraints on $g$, choose $g$ such that $-\int_{\mathbb{R}} (g\ast g)'(x)/x dx$ is minimal. Or, equivalently, so that $\int_{\mathbb{R}} (1-(g\ast g)(x)) dx/x^2$ is minimal. Or, equivalently, so that $\int_{\mathbb{R}} g(t) (Ag)(t) dt$ is minimal, where $Ag(t) = \int_{\mathbb{R}} \frac{g(t) - \frac{1}{2} (g(t+x) + g(t-x))}{x^2} dx$. I feel a solution is one step away... | |
| Nov 25 at 23:21 | comment | added | H A Helfgott | It may not be hard to solve (b). I have a not terribly rigorous argument that can probably be made rigorous - if you restrict yourself to $\phi(x)$ decreasing for $x>0$, the integral in part (b) can be made as close as you want to $1/2\pi^2$ - a function $\phi(x)$ that is $1$ for $|x|<1-\epsilon$ and then plunges to $0$ for $|x|>=1$ will do the trick. If you don't restrict yourself to $\phi(x)$ decreasing for $x>0$, the integral can be as close to $-\infty$ as you want, I think. | |
| Nov 25 at 21:44 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Nov 25 at 21:07 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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| Nov 25 at 16:09 | history | edited | H A Helfgott | CC BY-SA 4.0 |
added 203 characters in body; edited title
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| Nov 25 at 0:30 | comment | added | H A Helfgott | @KevinCasto I think you are forgetting "support on $[-1,1]$"? | |
| Nov 24 at 14:38 | history | asked | H A Helfgott | CC BY-SA 4.0 |