Timeline for The hypercube: $|A {\stackrel2+} E| \ge |A|$?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2011 at 10:37 | history | edited | Seva | CC BY-SA 3.0 |
added 42 characters in body
|
| Nov 7, 2011 at 9:56 | vote | accept | Seva | ||
| Nov 7, 2011 at 9:53 | history | edited | Seva | CC BY-SA 3.0 |
added 73 characters in body
|
| Nov 5, 2011 at 12:15 | comment | added | fedja | The full Fourier system on the cube '$\{-1,1\}^n$' consists of products $\prod_{j\in J}x_j$ with '$J\subset\{1,2,\dots,n\}$'. They are eigenvectors of the averaging operator $Th(x)=\frac 1n\sum_{e\in E}h(x+e)$ with the eigenvalues $1-\frac{2|J|}{n}$. The two bad ones are $J=\varnothing$ (constant, eigenvalue 1) and '$J=\{1,\dots,n\}$' (alternating, eigenvalue -1). $g$ is just $f$ with these two Fourier components removed. Removing an eigenspace is my slang for "restricting the operator to the orthogonal complement of that eigenspace". | |
| Nov 5, 2011 at 7:37 | comment | added | Seva | I am impressed by the fact that there is a progress, but I am afraid I do not understand much in your solution. To begin with, what do you mean by "removing the constant and alternating components"? Exactly how is $g$ defined? And what is "removing an eigenspace"? | |
| Nov 5, 2011 at 2:29 | history | answered | fedja | CC BY-SA 3.0 |