Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that:

$ \rightarrow\sum_{k=1}^n f(\frac{k}{n})\sum_{k=1}^n k{f(\frac{k}{n})}^2\le\sum_{k=1}^n kf(\frac{k}{n})\sum_{k=1}^n{f(\frac{k}{n})}^2 $

Given $f(x)$ is a positive function and also monotonic decreasing function.

share|improve this question
    
divide through by $(\sum_{k=1}^n f(\frac{k}{n}))^2$ and it looks like a correlation inequality, you want to show that the 2 increasing functions k and $f(\frac{k}{n})$ are positively correlated under etc. which is true and easy –  mike Oct 19 '12 at 17:29
    
I didnt follow it ? Can you elaborate it please –  Sai Krishna Deep Oct 22 '12 at 13:35
    
I see you have an answer , en.wikipedia.org/wiki/FKG_inequality is slightly abstract, but mentions the case you need. your case can also be done by the very intuitive rearrangement inequality, which is duscussed here en.wikipedia.org/wiki/Rearrangement_inequality –  mike Oct 22 '12 at 14:38
    
Hmmm... I dont see a trivial way by rearrangement inequality.Already tried that method! But i guesss FKG will work –  Sai Krishna Deep Oct 23 '12 at 14:23
    
write it down for all permutations and sum over the permuations, and it gives the correaltion inequality. –  mike Oct 23 '12 at 14:31

1 Answer 1

up vote 3 down vote accepted

Denote $a_k=f\left(\frac{k}{n}\right).$
(Proof by induction). Let $P(n)$ be the statement $$\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)\sum\limits_{k=1}^n k\left(f\left(\frac{k}{n}\right)\right)^2\leqslant\sum\limits_{k=1}^n kf\left(\frac{k}{n}\right)\sum\limits_{k=1}^n\left(f\left(\frac{k}{n}\right)\right)^2$$ or, in shorter form $$\sum\limits_{k=1}^n a_k\sum\limits_{k=1}^n ka_k^2\leqslant\sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2.$$

  1. (Basis) For $n=1$ we have: $a_1\cdot a_1= a_1\cdot a_1$
  2. (Induction step). We assume that $P(n)$ is true and prove that $P(n+1)$ is true. For LHS of $P(n+1)$: \begin{equation} \sum\limits_{k=1}^{n+1} a_k\sum\limits_{k=1}^{n+1} ka_k^2 = \left(\sum\limits_{k=1}^{n} a_k +a_{n+1} \right) \left(\sum\limits_{k=1}^{n} k a_k^2 + (n+1) a_{n+1}^2 \right)=\\ \sum\limits_{k=1}^{n} a_k\sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^3 \leqslant \\ \sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2 + (n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 + (n+1)a_{n+1}^3\overset{def}{=} A. \end{equation} RHS of $P(n+1)$ equals: \begin{equation}\sum\limits_{k=1}^{n+1} ka_k\sum\limits_{k=1}^{n+1} a_k^2 = \left(\sum\limits_{k=1}^{n} ka_k + (n+1)a_{n+1}\right)\left(\sum\limits_{k=1}^{n} a_k^2 + a_{n+1}^2 \right) = \\ \sum\limits_{k=1}^n ka_k\sum\limits_{k=1}^na_k^2 + (n+1)a_{n+1}\sum\limits_{k=1}^{n} a_k^2 +a_{n+1}^2\sum\limits_{k=1}^{n} ka_k + (n+1)a_{n+1}^3\overset{def}{=} B. \end{equation} Next, \begin{equation} A-B= \left((n+1)a_{n+1}^2 \sum\limits_{k=1}^{n} a_k + a_{n+1} \sum\limits_{k=1}^{n} ka_k^2 \right) - \\ \left( (n+1)a_{n+1}\sum\limits_{k=1}^{n} a_k^2 +a_{n+1}^2\sum\limits_{k=1}^{n} ka_k \right) = \\ (n+1)a_{n+1} \left(a_{n+1}\sum\limits_{k=1}^{n} a_k -\sum\limits_{k=1}^{n} a_k^2 \right) + a_{n+1} \left(\sum\limits_{k=1}^{n} ka_k^2 - a_{n+1}\sum\limits_{k=1}^{n} k a_k \right)=\\ a_{n+1}^2 \sum\limits_{k=1}^{n} \left((n+1)a_k -k a_k \right) + a_{n+1}\sum\limits_{k=1}^{n} \left(k a_k^2 - (n+1)a_k^2 \right)=\\ a_{n+1}^2 \sum\limits_{k=1}^{n} {a_k(n+1-k)} - a_{n+1} \sum\limits_{k=1}^{n} {a_k^2(n+1-k)}=\\ a_{n+1} \sum\limits_{k=1}^{n} \left(a_{n+1}a_k (n+1-k) - {a_k^2(n+1-k)} \right)=\\ a_{n+1} \sum\limits_{k=1}^{n}{(n+1-k)(a_{n+1}a_k - a_k^2)}=\\ a_{n+1} \sum\limits_{k=1}^{n}{(n+1-k)a_k(a_{n+1} - a_k)}<0 \end{equation} since $\{a_k \}$ decreases.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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