continuity of convex functions

We will prove below that every convex function on an open (http://planetmath.org/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

 $P=\left\{x+\sum_{i=1}^{n}\lambda_{i}b_{i}\colon-1\leq\lambda_{i}\leq 1\text{ % for }i=1,2,\ldots,n\right\},$

where $\{b_{1},\ldots,b_{n}\}$ is some basis of $V$. Furthermore, let

 $\partial P=\left\{x+\sum_{i=1}^{n}\lambda_{i}b_{i}\colon\max_{1\leq i\leq n}|% \lambda_{i}|=1\right\}$

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 $|f(y)-f(x)|$ in 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 ∂⁡P={y1,y2}, 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

 $\displaystyle f\big{(}(1-\lambda)x+\lambda y\big{)}$ $\displaystyle\leq$ $\displaystyle(1-\lambda)f(x)+\lambda f(y)$ (1) $\displaystyle=$ $\displaystyle f(x)+\lambda\big{(}f(y)-f(x)\big{)}.$

On the other hand, for all $\mu\in[0,1/2]$ we have

 $\displaystyle f(x)$ $\displaystyle=$ $\displaystyle f\left((1-\mu)\left[\frac{(1-2\mu)x}{1-\mu}+\frac{\mu y}{1-\mu}% \right]+\mu(2x-y)\right)$ $\displaystyle\leq$ $\displaystyle(1-\mu)f\left(\frac{(1-2\mu)x}{1-\mu}+\frac{\mu y}{1-\mu}\right)+% \mu f(2x-y).$

Dividing by $1-\mu$ and setting $\lambda=\frac{\mu}{1-\mu}\in[0,1]$ gives

 $(1+\lambda)f(x)\leq f\big{(}(1-\lambda)x+\lambda y\big{)}+\lambda f(2x-y).$ (2)
 $-\lambda\big{(}f(2x-y)-f(x)\big{)}\leq f\big{(}x+\lambda(y-x)\big{)}-f(x)\leq% \lambda\big{(}f(y)-f(x)\big{)}.$ (3)

Note that both $y$ and $2x-y$ $\partial P$, 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 faces of $P$, and repeatedly applying this shows that $f$ attains a maximum at one of the corners of $P$.

Write $P_{\lambda}$ for the parallelepiped $P$ shrunk by a $\lambda$ relative to $x$:

 $P_{\lambda}=\{x+\lambda(y-x)\colon y\in P\}.$

Now the inequality (3) implies that for all $\lambda\in[0,1]$ and all $z\in\partial P_{\lambda}$, we have

 $\left|f(z)-f(x)\right|\leq\lambda\left|\max_{y\in\partial P}f(y)-f(x)\right|.$

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 of this inequality goes to zero as $\lambda\to 0$, from which we conclude that $f$ is continuous at $x$.

Title continuity of convex functions ContinuityOfConvexFunctions 2013-03-22 15:28:00 2013-03-22 15:28:00 pbruin (1001) pbruin (1001) 5 pbruin (1001) Result msc 26A51 msc 26B25