continuity of convex functions

We will prove below that every convex function on an open ( convex subset A of a finite-dimensional real vector space is continuousMathworldPlanetmath. 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 combinationMathworldPlanetmath 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 spaceMathworldPlanetmath V over , and let f:A be a convex function. Let xA 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+i=1nλibi:-1λi1 for i=1,2,,n},

where {b1,,bn} is some basis of V. Furthermore, let


denote the boundary of P. We will show that f is continuous at x by showing that f attains a maximum on P and by estimating |f(y)-f(x)| in of this maximum as yx.

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 λ[0,1] and yP, the convexity of f implies

f((1-λ)x+λy) (1-λ)f(x)+λf(y) (1)
= f(x)+λ(f(y)-f(x)).

On the other hand, for all μ[0,1/2] we have

f(x) = f((1-μ)[(1-2μ)x1-μ+μy1-μ]+μ(2x-y))

Dividing by 1-μ and setting λ=μ1-μ[0,1] gives

(1+λ)f(x)f((1-λ)x+λy)+λf(2x-y). (2)

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

-λ(f(2x-y)-f(x))f(x+λ(y-x))-f(x)λ(f(y)-f(x)). (3)

Note that both y and 2x-y P, and that f is bounded on P (hence in particular on 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 Pλ for the parallelepiped P shrunk by a λ relative to x:


Now the inequality (3) implies that for all λ[0,1] and all zPλ, we have


Consequently, the same inequality holds for all λ(0,1] and all z in the open neighbourhood PλPλ of x. The right-hand of this inequality goes to zero as λ0, 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