We will prove below that every convex function on an open convex subset $A$ of a finite-dimensional real vector space is continuous. This statement becomes false if we do not require $A$ to be open, since we can increase the value of $f$ at any point of $A$ which is not a convex combination of two other points without affecting the convexity of $f$ . An example of this is shown in Figure 1.

Figure 1: A convex function on a non-open set need not be continuous.

Let $A$ be an open convex set in a finite-dimensional vector space $V$ over $\mathbb{R}$ , and let $f\colon A\to\mathbb{R}$ be a convex function. Let $x\in A$ be arbitrary, and let $P$ be a parallelepiped centered at $x$ and lying completely inside $A$ . Here ``a parallelepiped centered at $x$ '' means a subset of $V$ of the form $$ P=\left\{x+\sum_{i=1}^n \lambda_ib_i\colon -1\le\lambda_i\le 1 \text{ for }i=1,2,\ldots,n\right\}, $$ where $\{b_1,\ldots,b_n\}$ is some basis of $V$ . Furthermore, let $$ \partial P=\left\{x+\sum_{i=1}^n \lambda_ib_i\colon\max_{1\le i\le n} \vert\lambda_i\vert=1\right\} $$ denote the boundary of $P$ . We will show that $f$ is continuous at $x$ by showing that $f$ attains a maximum on $\partial P$ and by estimating $\vert f(y)-f(x)\vert$ in terms of this maximum as $y\to x$ .

The idea is to use the condition of convexity to `squeeze' the graph of $f$ near $x$ , as is shown in Figure 2.

Figure 2: Given the values of $f$ in $x$ and on $\partial P=\{y_1,y_2\}$ , the convexity condition restricts the graph of $f$ to the grey area.

For $\lambda\in[0,1]$ and $y\in\partial P$ , the convexity of $f$ implies \begin{eqnarray} \label{ineq1} \nonumber f\big((1-\lambda)x+\lambda y\big)&\le&(1-\lambda) f(x)+\lambda f(y) \\ &=&f(x)+\lambda\big(f(y)-f(x)\big). \end{eqnarray}On the other hand, for all $\mu\in[0,1/2]$ we have \begin{eqnarray*} f(x)&=&f\left((1-\mu)\left[\frac{(1-2\mu)x}{1-\mu} +\frac{\mu y}{1-\mu}\right]+\mu(2x-y)\right) \\ &\le&(1-\mu)f\left(\frac{(1-2\mu)x}{1-\mu} +\frac{\mu y}{1-\mu}\right)+\mu f(2x-y). \end{eqnarray*}Dividing by $1-\mu$ and setting $\lambda=\frac{\mu}{1-\mu}\in[0,1]$ gives \begin{equation} \label{ineq2} (1+\lambda)f(x)\le f\big((1-\lambda)x+\lambda y\big)+\lambda f(2x-y). \end{equation}From the two inequalities () and () we obtain \begin{equation} \label{ineq3} -\lambda\big(f(2x-y)-f(x)\big)\le f\big(x+\lambda(y-x)\big)-f(x) \le\lambda\big(f(y)-f(x)\big). \end{equation}Note that both $y$ and $2x-y$ lie on $\partial P$ , and that $f$ is bounded on $P$ (hence in particular on $\partial P$ ). Indeed, the convexity of $f$ implies that $f$ is bounded by its values at two opposite faces of $P$ , and repeatedly applying this property shows that $f$ attains a maximum at one of the corners of $P$ .

Write $P_\lambda$ for the parallelepiped $P$ shrunk by a factor $\lambda$ relative to $x$ : $$ P_\lambda=\{x+\lambda(y-x)\colon y\in P\}. $$ Now the inequality () implies that for all $\lambda\in[0,1]$ and all $z\in\partial P_\lambda$ , we have $$ \left\vert f(z)-f(x)\right\vert\le\lambda \max_{y\in\partial P}\left\vert f(y)-f(x)\right\vert. $$ Consequently, the same inequality holds for all $\lambda\in(0,1]$ and all $z$ in the open neighbourhood $P_\lambda\setminus\partial P_\lambda$ of $x$ . The right-hand side of this inequality goes to zero as $\lambda\to 0$ , from which we conclude that $f$ is continuous at $x$ .