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For $f: \mathbb R^n \to \mathbb R$ prove that the two statements are equal:

  1. For all $x,y \in \mathbb R^n$ and for all $t \in [0,1]$, $f(tx+(1−t)y)\geq \min (f(x),f(y))$

  2. For all $k \in \mathbb R$, $\{ x : f(x) \geq k\}$ is a convex set.

Alex M.
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Eddie Yu
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1 Answers1

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For $\implies$: Fix $k \in \mathbb{R}$. Then if $x,y \in \{s : f(s) \geq k\}$, must have for any $t \in [0,1]$, $f(tx + (1-t)y) \geq \min(f(x),f(y)) \geq k$, so must have $tx + (1-t)y \in \{s : f(s) \geq k\}$, i.e. it is convex.

For $\impliedby$: Let $\min(f(x),f(y)) = k$. Then by convexity, must have $tx + (1-t)y \in \{s : f(s) \geq k\}$. So must have $f(tx + (1-t)y) \geq k =\min(f(x),f(y))$.

E.Lim
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