Timeline for generalized mean inequality extension
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2015 at 14:15 | comment | added | Sergei | Above are norm inequalities. Devide by n with proper powers and they will reduce to means inequalities with best constants. Not so? | |
| Apr 6, 2015 at 21:39 | comment | added | behrad mahboobi | @Sergei, would you please make an answer based on your claim so i can accept you answer ? | |
| Apr 6, 2015 at 12:40 | comment | added | Sergei | $||x||_p \le ||x||_r \le n^{1/r-1/p} ||x||_p$, | |
| Apr 6, 2015 at 12:37 | comment | added | Sergei | In Wiki we find exact inequalities: for $p>r>0$ it follows | |
| Apr 6, 2015 at 10:50 | comment | added | behrad mahboobi | @sergei , would please give us a link or reference on embedding space theorems constants? i have looked over net and couldn't find anything yet. | |
| Apr 6, 2015 at 8:08 | vote | accept | behrad mahboobi | ||
| Apr 6, 2015 at 8:06 | comment | added | behrad mahboobi | in general not all power means are norm, only for p>1. the provided bound looks very fascinating, however the dependency of $C$ on $\mathbf{x}$ violates the requirement of the question. but because it is yet very useful and application, I would accept that. | |
| Apr 6, 2015 at 4:21 | comment | added | Sergei | In fact power means are norms on finite dimensional spaces. All such norms are equivalent, so the inequalities you asked for exist and constants in them are embedding space theorems constants. | |
| Apr 6, 2015 at 0:00 | history | answered | kodlu | CC BY-SA 3.0 |