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 ℝ, and let f:A→ℝ be a convex
function. Let x∈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+n∑i=1λibi:-1≤λi≤1 for i=1,2,…,n}, |
where {b1,…,bn} is some basis of V. Furthermore, let
∂P={x+n∑i=1λibi:max1≤i≤n|λi|=1} |
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 y→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 λ∈[0,1] and y∈∂P, 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)) | ||
≤ | (1-μ)f((1-2μ)x1-μ+μy1-μ)+μf(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 inequalities (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:
Pλ={x+λ(y-x):y∈P}. |
Now the inequality (3) implies that for all λ∈[0,1] and all z∈∂Pλ, we have
|f(z)-f(x)|≤λ|maxy∈∂Pf(y)-f(x)|. |
Consequently, the same inequality holds for all λ∈(0,1] and all z in the open neighbourhood Pλ∖∂Pλ of x. The right-hand of this inequality goes to zero as λ→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 |