Gruber P. Convex and Discrete Geometry

8 Convex Functions
Theorem 1.4. Let I be open and f : I → R convex. Then f ′ − and f ′ + exist, are nondecreasing
(not necessarily strictly), and f ′ − ≤ f ′ + on I . f is differentiable precisely
at those points x ∈ I where f ′ − is continuous. Hence f ′ (x) exists for all x ∈ I with a
set of exceptions which is at most countable and f ′ is non-decreasing on its domain
of definition.
Proof. Before embarking on the proof we state a simple proposition, the proof of
which is left to the interested reader.
(1) Consider a non-decreasing sequence of real continuous non-decreasing
functions on I with limit g, say. Then g is continuous on the left.
The proof of the theorem is split into several steps. First,
(2) f ′ − , f ′ +
exist on I and
Let x, y ∈ I, x < y. By Lemma 1.1,
f (w) − f (x)
w − x
≤
f ′ − (x) ≤ f ′ + (x) ≤ f ′ − (y) ≤ f ′ + (y) for x, y ∈ I, x < y.
f (z) − f (x)
z − x
≤
f (y) − f (x)
y − x
for w, z ∈ I,w < x < z < y,
and the first two expressions are non-decreasing in w, respectively, z. Thus their
limits as w → x − 0 and z → x + 0 exist and we deduce that
(3)
Similar arguments show that
(4)
f ′ − (x), f ′ + (x) exist and f ′ − (x) ≤ f ′ +
f (y) − f (x)
(x) ≤ .
y − x
f ′ − (y), f ′ f (y) − f (x)
+ (y) exist and ≤ f
y − x
′ − (y) ≤ f ′ + (y).
Propositions (3) and (4) yield (2). This settles the assertion about f ′ − and f ′ + .
Second,
(5) f ′ − is continuous on the left and f ′ +
The functions gn, n = 1, 2,...,defined by
on the right.
gn(x) = f (x − 1 n ) − f (x)
− 1 for x ∈ I such that x −
n
1
∈ I,
n
are continuous and non-decreasing by Lemma 1.1 and, again by Lemma 1.1, form a
non-decreasing sequence with limit f ′ − . It thus follows from (1) that f ′ − is continuous
on the left. This establishes (5) for f ′ − . The statement about f ′ + is shown similarly.
Third,
(6) f ′ (x) exists precisely for those x ∈ I for which f ′ − is continuous.