Does strict quasiconvexity imply continuity?

Question

Strict quasiconvexity definition:

$f(tx + (1-t)y) < \max \{f(x), f(y)\}$ for any $x,y$ in the domain and $t \in (0,1)$

Answer 1

Theorem 8 of the document mentioned in the comment is not true.

Consider $$f(x) = -(x-1)^2, \qquad g_1(x) = \begin{cases} -1 & \text{if } x = 0, \\ |x| & \text{if } x \ne 0. \end{cases}$$ Then, $f$ is strictly concave and $g_1$ is strictly quasi-convex. Moreover, $x^* = 0$ is the unique solution of $$\text{Maximize}\; f(x) \quad \text{subject to } g_1(x) \le 0$$ and the constraint is non-binding. However, $\hat x = 1$ is the unique solution of the relaxed problem $$\text{Maximize}\; f(x).$$