<s>[INST] Prove if $f$ is convex on $(a,b)$, then $f$ is bounded on every closed subinterval of $(a,b)$ Prove :

$f : (a,b) \to \mathbb{R} $ is convex, then $f$ is bounded on every closed subinterval of $(a,b)$

where $f$ is convex if $f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda)f(y), \forall x,y \in (a,b), \forall \lambda \in [0,1]$

Try
$f$ is bounded above
Let $J = [\alpha, \beta] \subset (a,b)$. 
$\forall c \in [\alpha, \beta]$, $\exists \lambda_0 \in [0,1]$ s.t. $c = \lambda_0 \alpha + (1-\lambda_0) \beta$, and
$$
f(c) = f(\lambda_0 \alpha + (1-\lambda_0) \beta) \le \lambda_0 f(\alpha) + (1-\lambda_0) f(\beta)
$$
thus, $f$ is bounded above on $[\alpha, \beta]$
But I'm stuck at how I should proceed to prove that $f$ is bounded below.
 [/INST] Here is an analytic proof that a convex function $f:[\alpha,\beta] \to \mathbb{R}$ is bounded on the closed interval.
Take $M = \max (f(\alpha),f(\beta))$. Note that any $x \in [\alpha,\beta]$ is of the form $x = \lambda\alpha + (1-\lambda)\beta$ where $0 \leqslant \lambda \leqslant 1$.
Hence, we have for all $x \in [\alpha, \beta]$, the upper bound
$$f(x) \leqslant \lambda f(\alpha) + (1-\lambda)f(\beta) \leqslant \lambda M + (1-\lambda)M = M$$
To find a lower bound, write $x = \frac{a+b}{2} + \theta$.  Since $\frac{a+b}{2} = \frac{1}{2} \left(\frac{a+b}{2} + \theta  \right) +  \frac{1}{2} \left(\frac{a+b}{2} - \theta  \right) $, we have by convexity
$$f\left(\frac{a+b}{2}\right) \leqslant \frac{1}{2}f \left(\frac{a+b}{2} + \theta  \right) +   \frac{1}{2}f \left(\frac{a+b}{2} - \theta  \right),$$
and
$$f(x) = f\left(\frac{a+b}{2} + \theta  \right) \geqslant 2f\left(\frac{a+b}{2}\right)- f \left(\frac{a+b}{2} - \theta  \right)$$
But from the upper bound we have $-f \left(\frac{a+b}{2} - \theta  \right) \geqslant -M$, and so we have the lower bound
$$f(x) \geqslant 2f\left(\frac{a+b}{2}\right)-M = m$$
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