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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:

\begin{equation*} \sum_{i,j=1}^{n} |a_i\cdot a_j - b_i\cdot b_j|? \end{equation*}

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

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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
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

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

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

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

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