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Let $ f_{1}, f_{2},..., f_{n} $ convex functions in the interval $[0,1]$

such that $ max(f_{1},f_{2},...,f_{n}) \geq 0 $

show that there exist positive real numbers $a_{1}, a_{2},...,a_{n} $ not all equal to 0 such that $ a_1 \times f_1 + a_2 \times f_2 + ... + \ a_n \times f_n \geq 0 $

This problem seems to be not easy, I tried many ways but no results please help

Terminator
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1 Answers1

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Note that $g = \max(f_1,f_2,\ldots,f_n)$ is convex. Let its minimum on $[0,1]$ be at $x=p$. There are three cases to consider: $p=0$, $p=1$ and $0 < p < 1$. I'll do the third case.

Note that $g$ has left and right one-sided derivatives $D_-g(p) \le 0$ and $D_+g(p) \ge 0$ at $p$. There exist $J$ such that $g(p) = f_{J}(p)$ and $D_-g(p) = D_- f_{J}(p)$, and similarly $K$ such that $g(p) = f_{K}(p)$ and $D_+g(p) = D_+ f_{K}(p)$. We have $f_{J}(x) \ge g(p) + (x-p) D_- g{}(p)$ and $f_{K}(x) \ge g(p) + (x-p) D_+ g(p)$. If $D_- g(p) = D_+g(p) = 0$ you can take $a_{J} = 1$ and all others $0$. Otherwise take $a_{J} = D_+ g(p)$, $a_{K} = -D_- g(p)$, and all the others $0$.

Robert Israel
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  • Should it be $D_+g(p) = D_+ f_{K}(p)$ ? – Martin R Jan 01 '15 at 22:15
  • Thanks for catching that. Edited. – Robert Israel Jan 01 '15 at 22:29
  • One can probably fix this, but a convex (not necessarily continuous) function does not have to attain its minimum. Consider e.g. $f(x) = x$ for $0<x\leq 1$ and $f(0) = 5$ on $[0,1]$. – PhoemueX Jan 01 '15 at 22:58
  • If necessary, redefine the functions at the endpoints to make them continuous. – Robert Israel Jan 01 '15 at 23:00
  • sorry, The functions aren't necessarily differenciable – Terminator Jan 01 '15 at 23:15
  • @Terminator: That is no problem. For convex functions, the one-sided derivatives $D_+$ and $D_-$ always exist (with values in $[-\infty, \infty]$). See this question http://math.stackexchange.com/questions/946311/a-convex-function-is-differentiable-at-all-but-countably-many-points. – PhoemueX Jan 01 '15 at 23:30