continuity of convex functions
We will prove below that every convex function on an open (http://planetmath.org/Open) convex subset of a finite-dimensional real vector space is continuous. This statement becomes false if we do not require to be open, since we can increase the value of at any point of which is not a convex combination of two other points without affecting the convexity of . An example of this is shown in Figure 1.
Let be an open convex set in a finite-dimensional vector space over , and let be a convex function. Let be arbitrary, and let be a parallelepiped centered at and lying completely inside . Here “a parallelepiped centered at ” means a subset of of the form
where is some basis of . Furthermore, let
denote the boundary of . We will show that is continuous at by showing that attains a maximum on and by estimating in of this maximum as .
The idea is to use the condition of convexity to ‘squeeze’ the graph of near , as is shown in Figure 2.
For and , the convexity of implies
On the other hand, for all we have
Dividing by and setting gives
Note that both and , and that is bounded on (hence in particular on ). Indeed, the convexity of implies that is bounded by its values at two faces of , and repeatedly applying this shows that attains a maximum at one of the corners of .
Write for the parallelepiped shrunk by a relative to :
Now the inequality (3) implies that for all and all , we have
Consequently, the same inequality holds for all and all in the open neighbourhood of . The right-hand of this inequality goes to zero as , from which we conclude that is continuous at .
|Title||continuity of convex functions|
|Date of creation||2013-03-22 15:28:00|
|Last modified on||2013-03-22 15:28:00|
|Last modified by||pbruin (1001)|