# 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.

Let $A$ be an open convex set in a finite-dimensional vector space^{} $V$
over $\mathbb{R}$, and let $f:A\to \mathbb{R}$ 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

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

where $\{{b}_{1},\mathrm{\dots},{b}_{n}\}$ is some basis of $V$. Furthermore, let

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

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 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.

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 $f$ 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 $x$:

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

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

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

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 |
---|---|

Canonical name | ContinuityOfConvexFunctions |

Date of creation | 2013-03-22 15:28:00 |

Last modified on | 2013-03-22 15:28:00 |

Owner | pbruin (1001) |

Last modified by | pbruin (1001) |

Numerical id | 5 |

Author | pbruin (1001) |

Entry type | Result |

Classification | msc 26A51 |

Classification | msc 26B25 |