Gruber P. Convex and Discrete Geometry

Fifth, we show that
(13) G is differentiable almost everywhere on C.
2 Convex Functions of Several Variables 31
By (6), K = (I + G) −1 : D → E d is Lipschitz and thus can be extended to
a Lipschitz function of E d into E d by (1). Denote this extension also by K .Let
M be the set of points in E d where this K is not differentiable and by N the set
where it is differentiable but with singular derivative. Propositions (2)–(4) then show
that K (M ∪ N) has measure 0. The set L, from (7) and (8), is also of measure 0.
For the proof of (13), it is thus sufficient to show that G is differentiable for any
x ∈ C\ � K (M ∪ N) ∪ L � . For each such x Proposition (8) says that G is continuous
at x and the above definitions of M and N imply that K is differentiable at x + u
with non-singular derivative (note that x + u = (I + G)(x) = K −1 (x) �∈ M ∪ N).
Hence G is differentiable at x by (9). The proof of (13) is complete.
In the sixth, and final, step it will be shown that
(14) f (y) = f (x) + u · (y − x) + 1 2 (y − x)T H(y − x) + o(�y − x� 2 ) for almost
all x ∈ C and each y ∈ C. Hereu = G(x) and H is the derivative of G
at x.
The idea of the proof of (14) is to restrict f to a line segment on which it is differentiable
almost everywhere and then represent it as the integral of its derivative.
f and G are differentiable at almost every point x ∈ C, see (7) and (13). Let x
be such a point. Then, in particular,
(15) G(y) = u + H(y − x) + o(�y − x�) for y → x, y ∈ C,
where u = G(x) and H is the derivative of G at x.
In addition, (7) shows that, for almost every unit vector h and almost every t ≥ 0for
which x + th ∈ C, the function f is differentiable at x + th.Ifh and t are such a
unit vector and such a number, respectively, then
f (y) = f (x + th) + v · � y − (x + th) � + o � �y − (x + th)� � as y → x + th, y ∈ C,
where v = G(y + th). Thus, in particular, for y = x + sh,
f (x + sh) = f (x + th) + v · h (s − t) + o(|s − t|) as s → t, x + sh ∈ C.
Hence, for t ≥ 0 such that x +th ∈ C, the convex function f (x +th) of one variable
is differentiable at almost every t. Its derivative is v ·h, where v = G(x +th). Since a
convex function of one variable is absolutely continuous in the interior of its interval
of definition by Theorem 1.1, a theorem of Lebesgue shows that integration of its
derivative yields the original function, see the proof of Theorem 1.7. Hence
f (x + th) − f (x) =
=
�t
0
�t
0
G(x + sh) · hds
� u · h + h T Hhs + o(s) � ds = u · ht+ 1
2 hT Hht 2 + o(t 2 ) as t → 0