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There are three given lists of numbers in increasing order

$$\ell=\{\ell_1,\ell_2,...,\ell_n\}; \ell_i \geq\ell_{i+1}$$ $$\Lambda=\{\lambda_1,\lambda_2,...,\lambda_n\};\lambda_i \geq\lambda_{i+1}$$ $$\Psi=\{\psi_1,\psi_2,...,\psi_n\};\psi_i \geq\psi_{i+1}$$

We know that $\sum_{i=1}^{n}\lambda_i = \sum_{i=1}^{n}\psi_i$ is it true to claim that

(1) $$\sum_{i=1}^{n}(\ell_i-\lambda_i) = \sum_{i=1}^{n}(\ell_i-\psi_i)$$

(2) $$\sum_{i=1}^{n}|\ell_i-\lambda_i|= \sum_{i=1}^{n}|\ell_i-\psi_i|$$

If yes how to prove?

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    $\ell_i\ge\ell_{i+1}$ isn't what I would call **increasing order**. – 2017-01-01
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    @bof non decreasing ;) – 2017-01-01
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    Most people would call $\ell_i\le\ell_{i+1}$ "non decreasing order". – 2017-01-01

1 Answers 1

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For (1)

$\sum_{i=1}^{n}(\ell_i-\lambda_i) = \sum_{i=1}^{n}\ell_i-\sum_{i=1}^{n}\lambda_i$

Since given $\sum_{i=1}^{n}\lambda_i = \sum_{i=1}^{n}\psi_i$

Therefore $\sum_{i=1}^{n}\ell_i-\sum_{i=1}^{n}\lambda_i=\sum_{i=1}^{n}\ell_i-\sum_{i=1}^{n}\psi_i=\sum_{i=1}^{n}(\ell_i-\psi_i)$

Hence $$\sum_{i=1}^{n}(\ell_i-\lambda_i) = \sum_{i=1}^{n}(\ell_i-\psi_i)$$

For (2)

$$\sum_{i=1}^{n}|\ell_i-\lambda_i|= \sum_{i=1}^{n}|\ell_i-\psi_i|$$ will hold true iff

$$\ell_i\le\lambda_i,\psi_i\ \ \ or\ \ \ \ell_i\ge\lambda_i,\psi_i$$

for each and every $i$.