Prove that convex function is bounded from below

Question

Let $f: [a,b] \to \mathbb{R}$ be a convex function. Prove that $f$ is bounded from below, that $f$ has a maximum and that this maximum is $f(a)$ or $f(b)$.

I need some help. I think some information is missing, but probably not. Any ideas?

Answer 1

If $x \in [a,b]$ then we can write $x=ta+(1-t)b$ with $0 \leq t \leq 1$. Hence $f(x)\leq tf(a)+(1-t)f(b) \leq t\max \{f(a),f(b)\}+ (1-t) \max \{f(a),f(b)\}=\max \{f(a),f(b)\}$

Proof of the fact that $f$ is bounded below: On any closed sub interval of $(a,b)$ the function is continuous, Hence bounded. Hence, if $f$ is not bounded below then there exist a sequence $x_n$ converging to $a$ or $b$ such that $f(x_n) \to -\infty$. Suppose $x_n \to a$. Write $\frac {a+b} 2 $ as $\lambda_n x_n +(1-\lambda_n) b$. Compute $\lambda_n$ from this equation and observe that $\lambda_n \to \frac 1 2$. Now take limits in $f(\frac {a+b} 2 ) \leq \lambda_n f(x_n) +(1-\lambda_n) f(b)$ to get the contradiction $f(\frac {a+b} 2 )=-\infty$.