Prove the other side of this proposition.

Question

Let $X$ be a convex set and $A: X \to [0,1]$ be the membership degree of $x$ to the set $X$. Let also $$A_\alpha =\{x\in X \mid A(x) \ge \alpha\},\quad \alpha \in[0,1]$$ We want to prove the following proposition:

$A_\alpha$ is convex for all $\alpha\in\mathbb{R}$ iff for each $x_1 , x_2 \in X , \lambda \in[0,1]$ $$A\left(\lambda x_1 + (1-\lambda)x_2\right) \ge \min {\{A(x_1),A(x_2)\}}$$

Proof:

$(\Leftarrow)$: suppose that $x_1 , x_2 \in A_\alpha$. Since $A(x_1),A(x_2)\ge\alpha$, $$A(\lambda x_1 + (1-\lambda)x_2) \ge \min{\{A(x_1),A(x_2)\}} \ge \alpha$$ and we have $(\lambda x_1 + (1-\lambda)x_2) \in A_\alpha$.

Can you help me to prove the other side of this proposition?

Answer 1

First, you need to assume that $X$ is a convex set (even for the part you have proven), otherwise you do not known whether $\lambda x_1 + (1-\lambda) x_2 \in X$.

Let us now prove the converse. Assume that the sets $A_\alpha = \{x\in X: A(x) \geq \alpha\}$ are convex for all $\alpha\in\mathbb{R}$. We need to prove that $A(\lambda x_1 + (1-\lambda)x_2) \geq \min \{A(x_1), A(x_2)\}$ for all $\lambda \in (0,1)$ and all $x_1, x_2\in X$. Assume to the contrary that

$$ A(\lambda x_1 + (1-\lambda)x_2) < \beta < \min \{A(x_1), A(x_2)\}. $$

Then $A_\beta$ is not convex because $A_\beta\ni x_1$ and $A_\beta \ni x_2$, but $\lambda x_1 + (1-\lambda)x_2 \notin A_\beta$ - a contradiction.