How are two definitions of quasi-convexity equivalent? [duplicate]

Question

Why are these two definitions equivalent? (Assuming $C$ is convex)

• $f: C \rightarrow \mathbb{R} $ is quasi-convex if for any $x, y \in C$ and $\lambda \in [0,1]$, $$ f((1-\lambda)x + \lambda y) \leq \max\{f(x), f(y)\} .$$

• $f: C \rightarrow \mathbb{R} $ is quasi-convex if for any $\alpha \in \mathbb{R}$, $\operatorname{Lev} (f, \alpha) $ is convex.

Where $\operatorname{Lev} (f, \alpha) $ is defined as: $$ \{x \in S: f(x) < \alpha \}$$

I was successful to show that if we have the first definition, it implies the second one, but got stuck on the reverse side.

assume $ x, y \in \operatorname{Lev} (f, \alpha) $ so: $$f(x) < \alpha,$$ $$f(y) < \alpha.$$

Then we have: $$ max\{f(x), f(y)\} < \alpha $$ So by the first definition, we have: $$ f((1-\lambda)x + \lambda y) < \alpha $$ Which implies that:

$$ ((1-\lambda)x + \lambda y) \in \operatorname{Lev}(f, \alpha) .$$

Answer 1

In order to show that the second condition implies the first, assume that $\operatorname{Lev}(f, \alpha) $ is convex for all $\alpha \in \Bbb R$.

Now let $x, y \in C$ and $0 \le \lambda \le 1$.

For arbitrary $\epsilon > 0$ set $\alpha = \max(f(x), f(y)) + \epsilon$. Then $$ f(x) < \alpha \implies x \in \operatorname{Lev}(f, \alpha) $$ and $$ f(y) < \alpha \implies y \in \operatorname{Lev}(f, \alpha) $$ and therefore, since $\operatorname{Lev}(f, \alpha)$ is convex, $$ (1-\lambda)x + \lambda y \in \operatorname{Lev}(f, \alpha) \, . $$ So $$ f((1-\lambda)x + \lambda y) < \alpha = \max(f(x), f(y)) + \epsilon $$ for all $\epsilon > 0$, which implies $$ f((1-\lambda)x + \lambda y) \le \max(f(x), f(y)) \, . $$ This proves that $f$ is quasi-convex.