If for every $\alpha$ the set $A_{\alpha}= \{x : f(x) < \alpha\}$ is convex, then $f$ is convex. Is it true?

Question

I was asked to prove or disprove this statement:

If for every $\alpha$ the set $A_{\alpha} = \{x : f(x) < \alpha\}$ is convex, then $f$ is convex.

I tried by definition to solve it, but I got stuck with the inequality with $\alpha$:

$f(cx+(1-c)y) < cf(x)+(1-c)f(y) < a.$

What should I do next to prove or disprove it?

Answer 1

This is a false statement. A counterexample is $f: \Bbb R \to \Bbb R$ $$f(x)=x^3$$

Answer 2

It is not true.

Example: Let $f\colon\mathbb{R}\to\mathbb{R}$ be a strictly increasing function. Then $A_\alpha$s are intervals $(-\infty,b)$ but not every strictly increasing function is convex.