# Jensen's inequality for quasiconvex functions

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?

• Maybe you mean the $\alpha$-sublevel sets. – Hugo Oct 3 '18 at 12:56
• Oh, sorry about that. I made the correction. Thanks. – Nurmister Oct 3 '18 at 12:58

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.
• 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 3 '18 at 13:09
• And also there is no restriction on $\alpha$, which may be any real number. – Hugo Oct 3 '18 at 13:11