TITLE: convex function in open interval is continuous
QUESTION [11 upvotes]: How can I prove that a convex function ƒ defined on some open interval C is continuous on C?
Thank you.

REPLY [10 votes]: For each $x_0 \in C$, $f$ is both left- and right-differentiable at $x_0$, that is,
$$
\mathop {\lim }\limits_{x \uparrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R} \;\; {\rm and} \;\; \mathop {\lim }\limits_{x \downarrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R},
$$
respectively. For a proof, see Theorem 14.5(a) on p. 248 of this book or Theorem 1(2) here. It follows that $f$ is both left- and right-continuous at $x_0$, hence continuous there.
Remark: A convex function on a closed interval need not be continuous at the end points (for example, the function $f$ on $[0,1]$ defined by $f(x)=x^2$ for $|x|<1$, $f(x)=2$ for $x \in \lbrace -1,1 \rbrace)$.