Download PDF

proof of criterion for convexity∗
rspuzio†
2013-03-21 22:41:49
Theorem 1. Suppose f : (a, b) → R is continuous and that, for all x, y ∈ (a, b),
x+y
f (x) + f (y)
f
≤
.
2
2
Then f is convex.
Proof. We begin by showing that, for any natural numbers n and m ≤ 2n ,
mx + (2n − m)y
mf (x) + (2n − m)f (y)
f
≤
n
2
2n
by induction. When n = 1, there are three possibilities: m = 1, m = 0, and
m = 2. The first possibility is a hypothesis of the theorem being proven and
the other two possibilities are trivial.
Assume that
0
m x + (2n − m0 )y
m0 f (x) + (2n − m0 )f (y)
f
≤
n
2
2n
for some n and all m0 ≤ 2n . Let m be a number less than or equal to 2n+1 .
Then either m ≤ 2n or 2n+1 − m ≤ 2n . In the former case we have
1 mx + (2n − m)y y
1
mx + (2n − m)y
f
+
≤
f
+
f
(y)
2
2n
2
2
2n
n
1 mf (x) + (2 − m)f (y) f (y)
+
≤
2
2n
2
mf (x) + (2n+1 − m)f (y)
=
.
2n+1
In the other case, we can reverse the roles of x and y.
∗ hProofOfCriterionForConvexityi created: h2013-03-21i by: hrspuzioi version: h39288i
Privacy setting: h1i hProofi h52A41i h26A51i h26B25i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
Now, every real number s has a binary expansion; in other words, there
exists a sequence {mn } of integers such that limn→∞ mn /2n = s. If 0 ≤ s ≤ 1,
then we also have 0 ≤ mn ≤ 2n so, by what we proved above,
mn f (x) + (2n − mn )f (y)
mn x + (2n − mn )y
≤
.
f
2n
2n
Since f is assumed to be continuous, we may take the limit of both sides and
conclude
f (sx + (1 − s)y) ≤ sf (x) + (1 − s)f (y),
which implies that f is convex.
2