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Assume $f:[a,b] \to \mathbb R$ is convex. We want to show $f(a+)=\displaystyle \lim_{x \to a^+} f(x)$ and $f(b-)=\displaystyle \lim_{x \to b^-} f(x)$ exist. I was thinking of first showing that $f$ has right and left derivatives everywhere in $(a,b)$ which are increasing, then use that is $x<y$ then slope of line joining $f(x), f(y)$ is greater than $f'_{+}(x)$ and lower than $f'_{+}(y)$ to establish that $f(x)$ is bounded and monotone in some $(a,d)$, $d$ small and use it to derive the existence of right limit to $a$ (similar for $b$).

It seems that this is a bit too complicated though... Is there a simpler way to do it?

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Well, it is not complicated. Convex functions are locally lipschitz whenever they are bounded, and therefore the have left and right limits. You are just using the definition of convexity, which is really powerful, if you think of it. – Siminore Sep 11 '12 at 7:45
Convex functions are Lipschitz on compact subsets if the INTERIOR of their domain. The crucial property here is that to prove $f$ is Lipschitz on $[c,d] \subset (a,b)$ you pick $a<c'<c$ and $d<d'<b$. Clearly you cannot do that at the endpoints. In fact at the endpoints f may not even be continuous as demonstrated by the well-known example $f(x)=0, x<1$ and $f(1)=1$. – Bernard Sep 11 '12 at 14:04
In the book by Niculescu, Convex functions and their applications, Proposition 1.3.4 solves your problem the same way as you did: by local monotonicity. I am unable to find a more direct approach... – Siminore Sep 11 '12 at 14:22

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