1
$\begingroup$

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 $b$).

$\endgroup$
1
  • 1
    $\begingroup$ Context? Motivation? Indication of the cases or similar versions that you have already tried? $\endgroup$
    – Yemon Choi
    Mar 20, 2011 at 21:23

1 Answer 1

7
$\begingroup$

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).

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – ogn
    Mar 21, 2011 at 17:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – fedja
    Mar 21, 2011 at 18:35
  • $\begingroup$ Thank you very much! I understand now; I learn a new trick, that's nice. $\endgroup$
    – ogn
    Mar 22, 2011 at 10:45
  • $\begingroup$ 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. $\endgroup$
    – ogn
    Mar 30, 2011 at 17:50
  • $\begingroup$ See the edit :). $\endgroup$
    – fedja
    Mar 31, 2011 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.