Let $f$ be a real valued function on the domain $[0,1]$, i.e., $$f:[0,1]\to\mathbb{R}$$ such that $$ f(\lambda x + (1 - \lambda) y) \geq \lambda f(x) + (1 - \lambda) f(y) \quad \text{for all } x, y, \lambda \in [0, 1]. $$ This means that $f$ is concave on the set $[0,1]$.
I need to show that the set $C_{\alpha}=\{u \in [0, 1] \mid f(u) \leq \alpha \}$ is closed for all $\alpha \in \mathbb{R}$.
As per my understanding, I need to show that for any sequence $\{u_n\}\subset C_{\alpha}$ such that $u_n\to u$, $u$ must be in $C_{\alpha}$.
This means it is given that $$f(u_n)\leq \alpha \ \text{for all}\ n,$$ and we need to show that $$f(u)\leq \alpha.$$ But I am not being able to use convexity of $f$ show this. I also tried to do it via contradiction but did not succeed.
Please help how to proceed.
