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The definition of a quasiconvex function $f$ is this: $$\text{All $\alpha$-sublevel sets $S_\alpha$ of $f$ are convex.}$$ The modified Jensen's inequality as it applies to quasiconvex functions is this: $$\forall \theta \in [0,\,1],\,\forall x,\,y \in \operatorname{dom}(f),\, f(\theta x + (1 - \theta) y) \leq \max\{f(x),\,f(y)\}$$

Is it possible to show that these statements are equivalent? If so, how?

Nurmister
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1 Answers1

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Call your properties (a) and (b). To show (a) $\implies$ (b), take any $x,y$, and define $\alpha = \max\{f(x), f(y)\}$. Since $x,y$ belong to the $\alpha$-sublevel set, also $(1-t)x+ty$ belongs to the $\alpha$-sublevel set, i.e., $$ f((1-t)x+ty) \leq \alpha = \max\{f(x), f(y)\}. $$

On the converse, if (b) is true, take any $x,y$ in a $\beta$-sublevel set, so that $f(x) \leq \beta$, $f(y) \leq \beta$. Then, for every $t \in [0,1]$, $$ f((1-t)x + ty) \leq \max\{f(x),f(y)\} \leq \beta, $$ i.e., $(1-t)x+ty$ belongs to the $\beta$-sublevel set.

Hugo
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  • Thank you for your answer. The proof of the converse is clear, and surprisingly intuitive. For the $\implies$ direction, I did not think about defining $\alpha$ in such a way. I suppose it is valid to do so because no generality is lost (there are no restrictions on $x$ and $y$). – Nurmister Oct 03 '18 at 13:09
  • And also there is no restriction on $\alpha$, which may be any real number. – Hugo Oct 03 '18 at 13:11