Gruber P. Convex and Discrete Geometry

24 Convex Functions
First-Order Differentiability and Partial Derivatives
A function f : C → R is differentiable at a point x ∈ C in the sense of Stolz or
Fréchet if there is a (necessarily unique) vector u ∈ E d , such that
f (y) = f (x) + u · (y − x) + o(�y − x�) as y → x, y ∈ C.
A weaker notion of differentiability is differentiability in the sense of Gâteaux. Since
both notions of differentiability coincide for convex functions in finite dimensions,
only the former will be considered.
Theorem 2.5. Let f : C → R be convex and x ∈ int C. Then the following statements
are equivalent:
(i) f is differentiable at x.
(ii) The partial derivatives fxi (x), i = 1,...,d, exist.
Proof. (i)⇒(ii) This is well known from calculus and easy to prove.
(ii)⇒(i) Let ui = fxi (x), i = 1,...,d, let u = (u1,...,ud) and let {b1,...,bd}
be the standard basis of E d . Then
f (x + tbi) = f (x) + uit + o(|t|) as t → 0fori = 1,...,d.
Combined with Jensen’s inequality, this shows that
(7) f (y) = f � x + (y − x) �
= f
�
1 � � 1 � �
x + d(y1 − x1)b1 +···+ x + d(yd − xd)bd
d
d
�
≤ 1
d f � � 1
x + d(y1 − x1)b1 +···+
d f � �
x + d(yd − xd)bd
= f (x) + u1(y1 − x1) +···+ud(yd − xd)
+ o(|y1 − xy|) +···+o(|yd − xd|)
= f (x) + u · (y − x) + o(�y − x�) as y → x, y ∈ C.
A similar argument yields the following:
(8) f (2x − y) ≤ f (x) + u · (x − y) + o(�y − x�) as y → x, 2x − y ∈ C.
The final inequality we are seeking follows from the convexity of f :
�
1 1
�
(9) f (x) = f y + (2x − y) ≤
2 2 1 1
f (y) + f (2x − y) for y, 2x − y ∈ C.
2 2
The inequalities (9) and (8) now imply that
f (x) + u · (y − x) ≤ 1 1
f (y) + f (2x − y) + u · (y − x)
2 2
≤ 1 1
f (y) +
2 2
1
f (x) + u · (y − x) + o(�y − x�) as y → x, y ∈ C,
2