proof of pseudo convexity implies strict quasi convexity

Question

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!

Answer 1

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.