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If $g_i$ are concave functions where $i = 1,2,\ldots,m$ and $b_i$ are constants where $i = 1,2,\ldots,m$. Why is the set :

$$ S = {x : g_i(x) \ge b_i i = 1,2,...,m} $$ a convex set?

I know by definition if $f$ is a concave function then $$ f( tx_1 + (1-t)x_2) \ge tf( x_1) + (1-t)f(x_2)$$

and $-f(x)$ is a convex function.

moli
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    Typo? It should be $ f( tx_1 + (1-t)x_2) \ge tf( tx_1) + (1-t)f(x_2)$ – Maksim Dec 12 '20 at 13:53
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    First show that the set $S_i = { x \mid g_i(x) \ge b_i }$ is convex for each $i$. Then show that the intersection of convex sets is again convex. – Martin R Dec 12 '20 at 14:00
  • @MartinR I got stuck showing $$Si={x∣gi(x)≥bi}$$ I think it is related to epigraph or hypograph properties. But could not figure out how. – moli Dec 12 '20 at 14:05
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    You have to show that if $g_i(x_1) \ge b_i$ and $g_i(x_2) \ge b_i$ then also $g_i(tx_1 + (1-t)x_2) \ge b_i$ for $0 < t < 1$, and that follows immediately form the concavity formula. – Martin R Dec 12 '20 at 14:11

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