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.

Given the following:

  • $\sum_{i=1}^{n} a_i = 1, a_i \geq 0 \forall i$
  • $\sum_{i=1}^{n} b_i = 1, b_i \geq 0 \forall i$
  • $\sum_{i=1}^{n} |a_i - b_i| \leq e$ where $e \ll 1$

What is the upper bound on:

  • $\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$.

I am almost sure it is bounded by $2e$ but am not able to prove it.

share|improve this question
2  
Try to write $a_ia_j−b_ib_j as $a_ia_j−b_ia_j+b_ia_j - b_ib_j$ and use the triangle inequality. –  Soarer Aug 8 '10 at 13:18
2  
Got it. Thanks a lot. –  Manan Aug 8 '10 at 14:13
    
@Soarer: Please make this an answer so OP can accept it. –  KennyTM Aug 8 '10 at 16:00
    
@Kenny: I'm not sure about the social protocol on stack overflow. Is it necessary? Since I didn't give a fully-fledged answer (intended not to), I left it as a comment. –  Soarer Aug 8 '10 at 16:53
    
I am sorry. Am i supposed to do it? i am new here. anyway, i am writing the answer for it. let me know if anything is wrong. –  Manan Aug 8 '10 at 17:18

1 Answer 1

$\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|$ = $\sum_{i,j=1}^{n} |a_i\cdot a_j - b_i.a_j + b_i.a_j - b_i\cdot b_j|$

$= \sum_{i,j=1}^{n} |(a_i - b_i).a_j + (a_j - b_j).b_i|$

$\leq \sum_{i,j=1}^{n} [|(a_i - b_i).a_j| + |(a_j - b_j).b_i|]$ ........ (using the triangle inequality ,i.e., $|x+y| \leq |x|+|y|$)

$= (\sum_{i,j=1}^{n} |(a_i - b_i).a_j|) + (\sum_{i,j=1}^{n} |(a_j - b_j).b_i|)$

$\leq (\sum_{i}^{n} |(a_i - b_i).1|) + (\sum_{i}^{n} |e.b_i|)$ .... (By solving the inner summation over j and using the given (in)equations).

$\leq e + e$
$= 2.e$

Hence Proved.

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.