Timeline for Coefficient bounds of an inequality
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2011 at 16:12 | comment | added | ogn | @fedja: see edit of question | |
| Mar 31, 2011 at 14:41 | comment | added | ogn | Yes, you are right! | |
| Mar 31, 2011 at 4:09 | comment | added | fedja | See the edit :). | |
| Mar 31, 2011 at 4:07 | history | edited | fedja | CC BY-SA 2.5 |
responded to the last comment
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| Mar 30, 2011 at 17:50 | comment | added | ogn | I play with this kind of inequality. I am wondering if the following inequality holds for some C depending only on k (always with the same assumptions on the sequences). \begin{align*} \sum_{i=1}^n & b_i(a_1p_1+ \ldots + a_ip_i+ (a_{i+1}+ \ldots + a_n)p_i)^k \\ ≤& C\sum_{i=1}^n a_i(a_1p_1+ \ldots + a_ip_i+(a_{i+1} +\ldots +a_n)p_i)^k \\ &+ C\sum_{i=1}^n b_i(b_1+…+b_ip_i+(b_{i+1}+…+b_n)p_i)^k \end{align*} I think that is true but the technique above does not give a proof. | |
| Mar 22, 2011 at 10:45 | vote | accept | ogn | ||
| Mar 22, 2011 at 10:45 | comment | added | ogn | Thank you very much! I understand now; I learn a new trick, that's nice. | |
| Mar 21, 2011 at 18:35 | comment | added | fedja | Sure. Any decreasing positive real sequence is a linear combination of "elementary" sequences 1,0,0,0... ; 1,1,0,0...; 1,1,1,0,... etc. with positive coefficients like (4,2,1,0.5)=0.5(1,1,1,1)+0.5(1,1,1,0)+1(1,1,0,0)+2(1,0,0,0). So, if you want to show that $\sum_j q_j Q_j\ge 0$ for all decreasing positive sequences $q_j$, it suffices to check that all partial sums of $Q_j$ are non-negative. | |
| Mar 21, 2011 at 17:59 | comment | added | ogn | Hello, I am slow so I don't understand why the first inequality is linear in $q_i$ so we just need to check the second inequality? Please explain. | |
| Mar 21, 2011 at 1:54 | history | answered | fedja | CC BY-SA 2.5 |