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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex function on $\mathbb{R}^n$. How to prove that $f$ is continuous?

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This seems to be special case of this question: Is a convex function defined on a convex open subset of $\mathbb R^n$ continuous? (Which has no satisfactory answer at the moment - the single answer which is there now consists of a dead link.) – Martin Sleziak Oct 18 '12 at 16:18
@MartinSleziak When I open the pdf given in that link, it occurs to be "404 Not Found" – Golbez Oct 18 '12 at 16:21
@Golbez: That’s why Martin said that it’s a dead link. – Brian M. Scott Oct 18 '12 at 16:26
For $n=1$ it is there:… I think, similar argument would work for $n$ dim. – Berci Oct 18 '12 at 17:02

1 Answer 1

up vote 1 down vote accepted

$\def\R{\mathbb R}\def\conv{\operatorname{conv}}$Note first, that a convex function is locally bounded: Let $x_0 \in \R^n$, and define $U := \conv\{x_0 \pm e_i \mid 1 \le i \le n\}$, then $U$ is a neighbourhood of $x_0$, given $x \in U$, there are $\lambda_{i,\pm}\ge 0$ with $\sum_{i=1}^n \lambda_{i,+} + \lambda_{i,-} = 1$ and $x = \sum_i \lambda_{i,+}(x_0+e_i) + \lambda_{i,-}(x_0 -e_i)$, convexity of $f$ gives \begin{align*} f(x) &\le \sum_i \lambda_{i,+}f(x_0 + e_i) + \lambda_{i,-}f(x_0 - e_i)\\ &\le \max\{f(x_0 \pm e_i)\mid 1 \le i \le n\}. \end{align*} Hence $f$ is locally bounded above. Now let $x \in U$, then $x_0 - (x-x_0)\in U$ and \[ f(x_0) \le \frac 12 f(x) + \frac 12 f(x- 2x_0) \iff f(x) \ge 2f(x_0) - f(x-2x_0) \] Hence, if $M$ denotes the upper bound of $f$ on $U$, $2f(x_0) - M$ is a lower bound.

So, $f$ is locally bounded, now we will show, that it is locally Lipschitz (and hence, continuous): Let $x_0 \in \mathbb R^n$. By the above, there is an $\epsilon > 0$, such that $|f| \le M$ on $U_{\epsilon}(x_0)$. We will show, that $f$ is Lipschitz on $U_{\epsilon/2}(x_0)$. Suppose not, then there were $x_1, x_2 \in U_{\epsilon/2}(x_0)$ with \[ \frac{\|f(x_1) - f(x_2)\|}{\|x_2 - x_1\|} > \frac {4M}\epsilon \|x_2 - x_1\| \] Let $x_3 := x_2 + \frac \epsilon2\frac{x_2 - x_1}{\|x_2 - x_1\|}$, then $x_3 \in U_\epsilon(x_0)$ and $x_1$, $x_2$, $x_3$ are on one line, and $\|x_2 -x_3\| = \frac \epsilon 2$. Hence \[ \frac{f(x_3) - f(x_2)}{\|x_3 - x_2\|} \ge \frac{f(x_2) - f(x_1)}{\|x_2 - x_1\|} > \frac {4M}{\epsilon} \] So $f(x_3) - f(x_2) > 2M$, contradicting $|f| \le M$.

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It is worth noting that by Rademacher theorem every convex function is almost everywhere differentiable. – Tomás Oct 18 '12 at 19:06

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