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Let $f$ be a real-valued function on the interval $[0, 1]$ such that: $ f(\lambda x + (1 - \lambda) y) \leq \lambda f(x) + (1 - \lambda) f(y) \quad \text{for all } x, y, \lambda \in [0, 1]. $

(a) The set $\{u \in [0, 1] \mid f(u) \leq \alpha \}$ is closed for all real $\alpha$

(b) The set $\{u \in [0, 1] \mid f(u) \geq \alpha \}$ is closed for all real $\alpha$ .
We have to mark true false .

I am stuck in this problem. Please suggest some ways for proceeding.

My approach: (b) part I have constructed an example $f(x)=-x^2$ when $x>0$ and $f(x)=-1$ when $x=0$ Now if we take $\alpha=-\frac{1}{2}$ then clearly the set is open . Hence part (b) is false.

I am stuck in part(a) let $A$$=$ $\{u \in [0, 1] \mid f(u) \leq \alpha \}$ let's assume $x_n \in A $ and $x_n \to x$
My claim is $x \in A$ Please help me to proceed further

RRR
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  • Does this help? https://math.stackexchange.com/questions/258511/prove-that-every-convex-function-is-continuous – Koro Dec 13 '23 at 16:01
  • careful, though, as continuity can fail at the endpoints, as gets pointed out in comments at the link Koro posted. – M W Dec 13 '23 at 17:43
  • given function is concave I think and how to use continuity please provide outline – RRR Dec 13 '23 at 18:13

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