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I'm reading a paper that made a certain assumption as being trivial (I believe), but which I set out to prove. And now I'm kind of stuck.

Let $I=[a,b]$ and $\bf{x}$ $= \{ x_1, x_2, \ldots, x_N \} \in I$. In other words, $\bf{x}$ is a finite collection of points on interval $I$. Now let $L=[\text{inf}(\bf{x}$) $,\text{sup}(\bf{x}$ $)]$. Now let $\bf{\lambda } = \{\lambda _1, \lambda _2, \ldots, \lambda _N\}$; $\lambda _i \geq 0$, s.t. $\sum_{i=1}^{N} \lambda_i = 1$.

Is there a way to show that $\sum_{i=1}^{N} \lambda_i x_i \in L$, i.e. $\text{inf}(\bf{x}$$) \leq \sum_{i=1}^{N} \lambda_i x_i \leq \text{sup}(\bf{x}$$)$?

Is it even true?

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Intuitively true: $\sum_{i=1}^{N} \lambda_i x_i$ is a weighted average of ${\bf x}$, hence its value lies in L –  leonbloy Aug 31 '11 at 21:53
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FYI, $\sum_{i=1}^N \lambda_i x_i$ is called a convex combination of the $x_i$. –  Mike Spivey Aug 31 '11 at 21:56
    
@leonbloy yes, intuitively it makes perfect sense, but I'm a beginner in higher mathematics and I like to prove even the most intuitive things. –  Phonon Sep 1 '11 at 13:11
    
Mike, I'm actually using this fact to learn about convex functions, so you argument, though valid, isn't of help in this particular case. = ) –  Phonon Sep 1 '11 at 13:12
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2 Answers

up vote 4 down vote accepted

$\text{inf}(\bf{x})= \sum_{i=1}^{N} \lambda_i \text{inf}(\bf{x})\leq \sum_{i=1}^{N} \lambda_i x_i$ and the same for the other side. For the equality we just split $1$ to the sum of the $\lambda$s and for the inequality we just increased $\text{inf}(\bf{x})$ to $x_i$

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Hint: Let $X= \sup({\bf x})$, so $x_i \le X$.

Hence $\displaystyle \sum_{i=1}^{N} \lambda_i x_i \le \sum_{i=1}^{N} \lambda_i X = X$.

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