Gruber P. Convex and Discrete Geometry

2 Convex Functions of Several Variables 21
Theorem 2.2. Let f : C → R be convex. Then f is Lipschitz on each compact
subset of int C. Thus, in particular, f is continuous on int C.
Proof. It is sufficient to show the following.
(1) Let x ∈ int C. Then f is locally Lipschitz at x.
The proof of (1) is divided into several steps. For ε > 0letN(ε) denote the
ε-neighborhood of x.
First,
(2) There are α, ε > 0 such that N(2ε) ⊆ int C and f is bounded above by
α on N(2ε).
For the proof of (2) it is sufficient to show that f is bounded above on some simplex
in C which contains x in its interior. Let S be a simplex in C with vertices
x1,...,xd+1, say, and x ∈ int S. Ify ∈ S, then y = λ1x1 +···+λd+1xd+1, where
λ1,...,λd+1 ≥ 0,λ1 +···+λd+1 = 1. Jensen’s inequality then yields the desired
upper bound:
f (y) = f (λ1x1 +···+λd+1xd+1) ≤ λ1 f (x1) +···+λd+1 f (xd+1)
Second,
≤|f (x1)|+···+|f (xd+1)| =α, say.
(3) There is a γ>0 such that | f | is bounded above by γ on N(2ε).
Let y ∈ N(2ε). Then 2x − y = x − (y − x) ∈ N(2ε) ⊆ C and thus
f (x) = f
�
1 1
�
y + (2x − y) ≤
2 2 1 1
f (y) + f (2x − y)
2 2
by the convexity of f . This, together with (2), then shows that
Hence
Third,
α ≥ f (y) ≥ 2 f (x) − f (2x − y) ≥ 2 f (x) − α.
| f (y)| ≤max{α, |2 f (x) − α|} = γ, say.
(4) f is Lipschitz with Lipschitz constant L = 2γ
on N(ε).
ε
Let y, z ∈ N(ε), y �= z. Choose w ∈ N(2ε) such that z ∈[y,w] and �w − z� =ε.
Since the restriction of f to the line-segment [y,w] is convex, Lemma 1.1, together
with (3), yields the following:
f (z) − f (y)
�z − y�
≤ f (w) − f (z)
�w − z�
≤ 2γ
ε
or f (z) − f (y) ≤ 2γ
ε
Clearly, the same conclusion holds with y and z exchanged. Thus
�z − y�.