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Here is an assertion I have read from these lecture notes:

Let $f(x)$ be a convex function, then the set $I_\beta= \{f(x)\leq \beta\}$ is convex for every $\beta$

This is not hard to prove. we simply pick $x$ and $y$ from $I_\beta$, then we see by convexity, we have

$f(ax + (1-a)y) \leq af(x) +(1-a) f(y) \leq a\beta + (1-a)\beta$ for $0<a<1$.

Hence the set is convex. Then it asked below if the converse true? I tried to prove this or find a counter-example. I think it is very simple question but it is eluding me, help anyone?

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Such a function is called a quasiconvex function. It need not always be a convex function: for example, the floor function is quasiconvex, but not convex (or concave). Another example is the following curve (courtesy: Wikipedia).

Quasiconvex

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  • $\begingroup$ thanks... that is actually a really obvious counter-example... $\endgroup$ – Lost1 Mar 26 '13 at 15:33

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