Suppose we have a quasiconvex function $f$. Since it is quasiconvex, we know that by definition all its $\alpha$-sublevel sets have to be convex.
Does dom($f$) have to be convex?
I suspect it does, given $f$ is real-valued. I say this by recognizing that $S_\infty \equiv$ dom($f$) is convex.
Yes, the domain has to be convex. Suppose on the contrary that $x_1$ and $x_2$ are in the domain of $f$, but that there exists $\lambda \in [0,1]$ such that $\lambda x_1 + (1-\lambda) x_2$ is not in the domain of $f$. Then the sublevel set $\{x : f(x) \leq \alpha\}$ for $\alpha = \max\{f(x_1),f(x_2)\}$ contains $x_1$, $x_2$ and should by convexity also contain $\lambda x_1 + (1-\lambda) x_2$, which is a contradiction.
