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Hello,

Given positive integers $k$ and $n$. Are there upper bounds on coefficients $A$ and $B$ such that they depends only on $k$ (eg., $2 k^k$) and for all non-negative integer sequences $(a_i)_{1}^n, (b_i)_{1}^n$ and non-negative increasing real sequence $(p_i)_{1}^n$, the following inequality holds?

$$ \sum_{i=1}^n b_i \left(\sum_{j=1}^i a_j p_j \right)^k \leq A \sum_{i=1}^n a_i \left(\sum_{j=1}^i a_j p_j \right)^k + B \sum_{i=1}^n b_i \left(\sum_{j=1}^i b_j p_j \right)^k $$

Do you know any result or reference related to the question?

Edit: 11/4 Due to the asymmetry of the left-hand side, we can prove the inequality for $A = k/(k+1)$ and $B = \Theta(k^k)$. Is it possible for the same kind of $A,B$ (up to a constant) such that

\begin{eqnarray*} \sum_{i=1}^n &b_i& (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \\ &\leq& A \sum_{i=1}^n a_i (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \\ &+& B \sum_{i=1}^n b_i(b_1p_1 + \ldots + b_{j-1}p_{j-1} + (b_j + \ldots + b_n) p_j)^k \end{eqnarray*}

The difficulty is due to the tail $(a_{j+1} + \ldots a_n)p_n$ (idem for $b$).

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Context? Motivation? Indication of the cases or similar versions that you have already tried? –  Yemon Choi Mar 20 '11 at 21:23
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up vote 7 down vote accepted

This is true.

I prefer to denote $q_i=p_i^{-1}$, $\alpha_i=a_ip_i$, $\beta_i=b_ip_i$, $A_i=\sum_{j=1}^i\alpha_j$, $B_i=\sum_{j=1}^i\beta_j$. Now we have to check that $$ \sum_i q_i\beta_iA_i^k\le C\sum_i q_i\alpha_iA_i^k+C\sum_i q_i\beta_iB_i^k $$ This is linear in $q_i$, so we just need to check that $$ \sum_{i=1}^n\beta_iA_i^k\le C\sum_{i=1}^n\alpha_iA_i^k+C\sum_{i=1}^n\beta_iB_i^k $$ for all $n\ge 1$. But $xX^k$ is comparable with $X^{k+1}-(X-x)^{k+1}$ for $0\le x\le X$, so the right hand side is essentially $A_n^{k+1}+B_n^{k+1}$ and the left hand side is dominated by $(A_n+B_n)^{k+1}$. The rest should be clear.

Edit: To cover your second inequality, let's show that the "missing part" $$ \sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n a_j\right)^k\le C\sum_{i=1}^n p_i^k a_i\left(\sum_{j=i}^n a_j\right)^k+ C\sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n b_j\right)^k $$ holds. By now it shouldn't be surprising that it will suffice to check it for the sequence $p_i$ consisting of several zeroes followed by several ones, in which case it is just exactly the same story as before but written backwards (with summations starting with $n$ and going down).

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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. –  ogn Mar 21 '11 at 17:59
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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. –  fedja Mar 21 '11 at 18:35
    
Thank you very much! I understand now; I learn a new trick, that's nice. –  ogn Mar 22 '11 at 10:45
    
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. –  ogn Mar 30 '11 at 17:50
    
See the edit :). –  fedja Mar 31 '11 at 4:09
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