continuity of convex functions

We will prove below that every convex function on anopen (http://planetmath.org/Open) convex subset $A$ of a finite-dimensionalreal vector space is continuous. This statement becomes false if wedo not require $A$ to be open,since we can increase the value of $f$ at any point of $A$ which isnot a convex combination of two other points without affecting theconvexity 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 $ℝ$, and let $f:A\to ℝ$ be a convexfunction. Let $x\in A$ be arbitrary, and let $P$ be a parallelepipedcentered at $x$ and lying completely inside $A$. Here “aparallelepiped centered at $x$” means a subset of $V$ of the form

$$P=\{x+\sum _{i=1}^n{\lambda }_i{b}_i:-1\le {\lambda }_i\le 1\text{ for }i=1,2,\mathrm{\dots },n\},$$

where $\{b_1,\mathrm{\dots },b_n\}$ is some basis of $V$. Furthermore, let

$$\partial P=\{x+\sum _{i=1}^n{\lambda }_i{b}_i:\underset{1\le i\le n}{\max }|{\lambda }_i|=1\}$$

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 byestimating $|f(y)-f(x)|$ in of this maximum as $y\to x$.

The idea is to use the condition of convexity to ‘squeeze’ the graphof $f$ 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 thegraph of $f$ to the grey area.

For $\lambda \in [0,1]$ and $y\in \partial P$, the convexity of $f$implies

$f\left((1-\lambda )x+\lambda y\right)$$\le$$(1-\lambda )f(x)+\lambda f(y)$(1)

$=$$f(x)+\lambda \left(f(y)-f(x)\right).$

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

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

$\le$$(1-\mu )f\left({\displaystyle \frac{(1-2\mu )x}{1-\mu }}+{\displaystyle \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

$$(1+\lambda )f(x)\le f\left((1-\lambda )x+\lambda y\right)+\lambda f(2x-y).$$(2)

From the two inequalities (1) and (2) we obtain

$$-\lambda \left(f(2x-y)-f(x)\right)\le f\left(x+\lambda (y-x)\right)-f(x)\le \lambda \left(f(y)-f(x)\right).$$(3)

Note that both $y$ and $2x-y$ $\partial P$, and that $f$ isbounded on $P$ (hence in particular on $\partial P$). Indeed, theconvexity of $f$ implies that $f$ is bounded by its values at two faces of $P$, and repeatedly applyingthis showsthat $f$ attains a maximum at one of the corners of $P$.

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

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

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

$$\left|f(z)-f(x)\right|\le \lambda \left|\underset{y\in \partial P}{\max }f(y)-f(x)\right|.$$

Consequently, the same inequality holds for all $\lambda \in (0,1]$ andall $z$ in the open neighbourhood $P_{\lambda }\setminus \partial P_{\lambda }$ of $x$. The right-hand of this inequality goes tozero as $\lambda \to 0$, from which we conclude that $f$ is continuousat $x$.