I'm having a hard time visualizing this function:
$$f(x) = \min_{\alpha_1,\alpha_2,\alpha_3\in[0,1]}\{\alpha_1x, \alpha_2x,\alpha_3x\}$$
such that $\alpha_1+\alpha_2+\alpha_3=1$.
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1If $x\ge 0$ it is $0$, if $x <0$ it is $x$, so $f(x) = \min(0,x)$. – copper.hat Feb 29 '20 at 06:51
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@copper.hat please write it as answer explaining the details – Aven Desta Feb 29 '20 at 06:57
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In the following $\alpha_k \ge 0$ and $\sum_k \alpha_k = 1$.
If $x \ge 0$, then $\alpha_k x \ge 0$ for all $k$ and choosing $\alpha = (1,0,0)$ we get $f(x) = 0$.
If $x < 0$, then $x \le \alpha_k x$ for all $k$ and choosing $\alpha = (1,0,0)$ we get $f(x) = x$.
Hence $f(x) = \min(0,x)$.
copper.hat
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$\alpha_1,\alpha_2,\alpha_3\in[0,1]$ I think $\alpha$ might have a range of values between 0 and 1. are you considering that in your proof? – Aven Desta Feb 29 '20 at 07:02
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If $\alpha_k \ge 0$ and $\sum_k \alpha_k = 1$ then we must have $\alpha_k \le 1$ since this case $\alpha_i \le \sum_k \alpha_k = 1$. – copper.hat Feb 29 '20 at 07:03