# proof of pseudo convexity implies strict quasi convexity

I am reading the proof of the theorem stated in the title from this paper: Mangasarian, Olvi L. "Pseudo-convex functions." Journal of the Society for Industrial and Applied Mathematics, Series A: Control 3.2 (1965): 281-290.

Definition. $$f(x)$$ is said to be strictly quasi-convex on C,if C is convex and if for every $$x_1$$ and $$x_2$$ in C, $$x_1 \neq x_2$$, $$f(x_2) < f(x_1) \implies f(\lambda x_1 + (1-\lambda)x_2) < f(x_1)$$ for every $$\lambda$$ such that $$0 < \lambda < 1$$

Theorem. Let C be convex. If $$f(x)$$ is pseudo-convex on C, then f(x) is strictly quasi-convex (and hence quasi-convex) on C, but not conversely.

Proof. Let $$f(x)$$ be pseudo-convex on C. We shall assume that f(x) is not strictly quasi-convex on C and show that this leads to a contradiction. If $$f(x)$$ is not strictly quasi-convex on C then it follows from the definition that there exist $$x_1 \neq x_2$$ in C such that $$f(x_2) < f(x_1)$$ and $$f(x) \geq f(x_1)$$ for some $$x \in L$$, where $$L = \{x | x = \lambda x_1 + (1-\lambda)x_2 , 0<\lambda<1\}$$ Hence there exists an $$\bar{x} \in L$$, such that $$f(\bar{x}) = \max\limits_{x \in L}f(x)$$

My question: I don't understand why there should exist such a $$\bar{x}$$ given the fact that L is an open set, if I am not missing anything else.

Any help is appreciated, thank you in advance!

The thing is that you can just consider the maximum over the closed set $$\overline{L}:=L\cup\{x_1,x_2\}$$. By the above assumptions (i.e. $$f(x) \geq f(x_1) > f(x_2)$$) it holds that $$\max_{x \in \overline{L}} f(x) = \max_{x \in L} f(x).$$ More specifically, if $$\max_{x \in \overline{L}} f(x) > f(x_1)$$, then we are settled. If $$\max_{x \in \overline{L}} f(x) = f(x_1)$$, then take the $$x \in L$$ from the proof.