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I am looking for another solution to the problem below.

Prove that the set of real-valued functions $\{|x - \lambda_1|, |x - \lambda_2|,\ldots, |x - \lambda_n|\}$ is linearly independent over $\mathbb R$ for $\lambda_i \neq \lambda_j, i \neq j$ ($1 \le i, j \le n$).

I have been able to come up with a proof:

  • Reorder the set $\{\lambda_1, \lambda_2, \ldots, \lambda_n\}$ such that $\lambda_1 \lt \lambda_2 \lt \cdots \lt \lambda_n$.
  • Differentiate both sides of the equation: $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$ on intervals $(\lambda_1, \lambda_2)$, $\ldots$, $(\lambda_{n-1}, \lambda_n)$, $(\lambda_n, +\infty)$.
  • Obtain the system of $n$ equations, from which we can conclude $\alpha_1 = \ldots = \alpha_n = 0$:$$ \left\{ \begin{array}{c} \alpha_1-\alpha_2+\cdots-\alpha_n=0 \\ \alpha_1+\alpha_2+\cdots-\alpha_n=0 \\ \ldots\\ \alpha_1+\alpha_2+\cdots+\alpha_n = 0 \end{array} \right. $$

Are there anything wrong with my proof and can you folks come up with another?

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    $\begingroup$ Your proof is (correct and) nice! $\endgroup$ Mar 1, 2021 at 5:03

2 Answers 2

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Here's a different proof, also starting with reordering the $\lambda_j$ in increasing order. Suppose that $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$. If $\alpha_j\ne0$, then near $x=\lambda_j$ we have $$ |x - \lambda_j| = \frac1{\alpha_j} \bigg( \sum_{i<j} \alpha_i(x-\lambda_i) + \sum_{i>j} \alpha_i(\lambda_i-x) \bigg); $$ but this is impossible since the right-hand side is a linear function of $x$ while the left-hand side is not near $x=\lambda_j$.

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If $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$ then $\alpha_i$ must be $0$ for each $i$ simply because RHS is differentiable at $\lambda_i$ but LHS is not unless $\alpha_i=0$ : Note that all terms except the $i-$th one are differentiable at $\lambda_i$.

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  • $\begingroup$ @BrianMoehring Thank you! $\endgroup$ Mar 1, 2021 at 6:00

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