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If $x = \sum_{i = 1}^n a_i y_i$ with $a_i > 0$, $\sum_{i = 1}^n a_i = 1$ and $|x| \geq |y_i|$ why is it true that $x = y_i$ for all $i = 1, \ldots, n$?

I can see that $|x| \leq \sum_{i = 1}^n a_i |y_i| \leq \sum_{i = 1}^n a_i |x| = |x|$ which means that $|x| = \sum_{i = 1}^n a_i |y_i|$. But how to I conclude?

1 Answers 1

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$|x| \leq \sum_{i = 1}^n a_i |y_i| \leq \sum_{i = 1}^n a_i |x|$

Therefore

$0 \leq \sum_{i = 1}^n a_i (|y_i|-|x|) \leq 0$

Thus: $0 \leq \sum_{i = 1}^n a_i (|y_i-x|)\leq \sum_{i = 1}^n a_i (|y_i|-|x|) \leq 0$

Hence:

$0 = \sum_{i = 1}^n a_i (|y_i-x|)$

Since all terms in this sum are nonnegative, therefore we conclude that $|y_i-x|=0$