Gruber P. Convex and Discrete Geometry

2 Convex Functions of Several Variables 29
(3) Let K : E d → E d be a Lipschitz mapping and let N ⊆ E d be the set of
all points of E d where K is differentiable but with singular derivative, i.e.
det A = 0. Then the image K (N) of N has measure 0.
The final tool is well known.
(4) Let K : E d → E d be Lipschitz and let N ⊆ E d have measure 0. Then
K (N) also has measure 0.
Second, the notion of subgradient and some of its properties will be considered.
Let x ∈ C. Then the set G(x) of all vectors u ∈ E d for which the affine function
a : E d → R defined by a(y) = f (x) + u · (y − x) for y ∈ E d is an affine support
of f at x is the subgradient of f at x. The mapping G : x → G(x) for x ∈ C is
set-valued. If G(x) is a singleton, say G(x) ={u}, then u is the gradient of f at x,
see Theorem 2.7. Now we show that
(5) G is monotone, i.e.
(x − y) · (u − v) ≥ 0forx, y ∈ C, u ∈ G(x), v ∈ G(y).
The definitions of G(x) and G(y) imply that f (y) ≥ f (x) + u · (y − x) and f (x) ≥
f (y) + v · (x − y), respectively. Adding these inequalities yields (5). Let I denote
the identity mapping, respectively, the d × d unit matrix.
(6) Let D = � {x + G(x) : x ∈ C}. Then the set-valued mapping
K = (I + G) −1 : D → E d is Lipschitz and thus single-valued.
(Here x + G(x) ={x + u : u ∈ G(x)}, forx + u ∈ D the set K (x + u) is the
set of all t ∈ C with x + u ∈ t + G(t), and when we say that K is Lipschitz this
means that, if w ∈ K (x + u) and z ∈ K (y + v) then the following inequality holds:
�z − w� ≤�y + v − x − u�.) To prove (6), let x + u, y + v ∈ D and choose
w ∈ K (x + u), z ∈ K (y + v). Then x + u ∈ w + G(w), y + v ∈ z + G(z). The
monotonicity of G then shows that
(z − w) · (y + v − z − x − u + w) ≥ 0, or
�z − w� 2 ≤ (y + v − x − u) · (z − w) ≤�y + v − x − u� �z − w�
by the Cauchy–Schwarz inequality, and thus �z − w� ≤�y + v − x − u�.
Third, G is said to be continuous at x ∈ C if G is single-valued at x,sayG(x) =
u, and for any neighborhood of u the set G(y) is contained in this neighborhood if
y ∈ C and �y − x� is sufficiently small. By Reidemeister’s theorem 2.6,
Then
(7) f is differentiable on C\L, where the set L ⊆ C has measure 0.
(8) G is continuous on C\L.
To see this, the following must be shown: let x ∈ C\L, x1, x2, ··· ∈ C such that
xn → x, and choose u1 ∈ G(x1), u2 ∈ G(x2),..., arbitrarily. Then un → u =