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1
f (x + v) ā’ f (x) = f (x + t v)(v) dt,
0
and hence
1
f (x + s v) ā’ f (x)
ā’ f (x)(v) = f (x + t s v) ā’ f (x) (v) dt ā’ 0,
s 0
which converges to 0 for s ā’ 0 uniformly for v in any bounded set, since f (x +
t s v) ā’ f (x) uniformly on bounded sets for s ā’ 0 and uniformly for t ā [0, 1] and
v in any bounded set, since f is assumed to be continuous.
Recall furthermore that a mapping f : E ā U ā’ F on a Banach space E is called
Lipschitz if
f (x1 ) ā’ f (x2 )
: x1 , x2 ā U, x1 = x2 is bounded in F.
x1 ā’ x2
It is called HĀØlder of order 0 < p ā¤ 1 if
o
f (x1 ) ā’ f (x2 )
: x1 , x2 ā U, x1 = x2 is bounded in F.
x1 ā’ x2 p

13.3
13.4 13. Diļ¬erentiability of seminorms 129

13.4. Lemma. GĖteaux-diļ¬erentiability of convex functions. Every convex
a
function q : E ā’ R has one sided directional derivatives. The derivative q (x) is
sublinear and locally bounded (or continuous) if q is locally bounded (or continuous).
In particular, such a function is GĖteaux-diļ¬erentiable at x if and only if q (x) is
a
an odd function, i.e. q (x)(ā’v) = ā’q (x)(v).

If E is not normed, then locally bounded-ness should mean bounded on bornologi-
cally compact sets.

Proof. For 0 < t < t we have by convexity that
t t t t
q(x + t v) = q (1 ā’ )x + (x + t v) ā¤ (1 ā’ ) q(x) + q(x + t v).
t t t t

Hence q(x+t v)ā’q(x) ā¤ q(x+t tv)ā’q(x) . Thus, the diļ¬erence quotient is monotone
t
falling for t ā’ 0. It is also bounded from below, since for t < 0 < t we have
t t
(x + t v) + (1 ā’
q(x) = q ) (x + t v)
tā’t tā’t
t t
ā¤ q(x + t v) + (1 ā’ ) q(x + t v),
tā’t tā’t
q(x+t v)ā’q(x) q(x+t v)ā’q(x)
ā¤
and hence . Thus, the one sided derivative
t t

q(x + t v) ā’ q(x)
lim
t
t 0

exists.
As a derivative q (x) automatically satisļ¬es q (x)(t v) = t q (x)(v) for all t ā„ 0. The
derivative q (x) is convex as limit of the convex functions v ā’ q(x+tv)ā’q(x) . Hence
t
it is sublinear.
The convexity of q implies that

q(x) ā’ q(x ā’ v) ā¤ q (x)(v) ā¤ q(x + v) ā’ q(x).

Therefore, the local boundedness of q at x implies that of q (x) at 0. Let := f (x),
then subadditivity and odd-ness implies (a) ā¤ (a + b) + (ā’b) = (a + b) ā’ (b)
and hence the converse triangle inequality.

Remark. If q is a seminorm, then q(x+tv)ā’q(x) ā¤ q(x)+t q(v)ā’q(x)
= q(v), hence
t t
q(x+t x)ā’q(x)
q (x)(v) ā¤ q(v), and furthermore q (x)(x) = limt 0 = limt 0 q(x) =
t
q(x). Hence we have

q (x) := sup{|q (x)(v)| : q(v) ā¤ 1} = 1.

Convention. Let q = 0 be a seminorm and let q(x) = 0. Then there exists a
v ā E with q(v) = 0, and we have q(x + tv) = |t| q(v), hence q (x)(Ā±v) = q(v). So q
is not GĖteaux diļ¬erentiable at x. Therefore, we call a seminorm smooth for some
a
diļ¬erentiability class, if and only if it is smooth on its carrier {x : q(x) > 0}.

13.4
130 Chapter III. Partitions of unity 13.5

13.5. Diļ¬erentiability properties of convex functions f can be translated in geo-
metric properties of Af :

Lemma. Diļ¬erentiability of convex functions. Let f : E ā’ R be a contin-
uous convex function on a Banach space E, and let x0 ā E. Then the following
statements are equivalent:
(1) The function f is GĖteaux (FrĀ“chet) diļ¬erentiable at x0 ;
a e
(2) There exists a unique ā E with

(v) ā¤ f (x0 + v) ā’ f (x0 ) for all v ā E;

(3) There exists a unique aļ¬ne hyperplane through (x0 , f (x0 )) which is tangent
to Af .
(4) The Minkowski functional of Af is GĖteaux diļ¬erentiable at (x0 , f (x0 )).
a
Moreover, for a sublinear function f the following statements are equivalent:
(5) The function f is GĖteaux (FrĀ“chet) diļ¬erentiable at x;
a e
(6) The point x0 (strongly) exposes the closed unit ball {x : f (x) ā¤ 1}.
In particular, the following statements are equivalent for a convex function f :
(7) The function f is GĖteaux (FrĀ“chet) diļ¬erentiable at x0 ;
a e
(8) The Minkowski functional of Af is GĖteaux (FrĀ“chet) diļ¬erentiable at the
a e
point (x0 , f (x0 ));
(9) The point (x0 , f (x0 )) (strongly) exposes the polar (Af )o .

An element xā— ā E ā— is said to expose a subset K ā E if there exists a unique point
k0 ā K with xā— (k0 ) = sup{xā— (k) : k ā K}. It is said to strongly expose K, if
satisļ¬es in addition that xā— (xn ) ā’ xā— (k0 ) implies xn ā’ k0 .
By an aļ¬ne hyperplane H tangent to a convex set K at a point x ā K we mean
that x ā H and K lies on one side of H.

Proof. Let f be a convex function. By (13.4) and continuity we know that f is
GĖteaux-diļ¬erentiable if and only if the sub-linear mapping f (x0 ) is linear. This
a
is exactly the case if f (x0 ) is minimal among all sub-linear mappings. From this
follows (1) ā’ (2) by the following arguments: We have f (x0 )(v) ā¤ f (x0 +v)ā’f (x0 ),
and (v) ā¤ f (x0 + v) ā’ f (x0 ) implies (v) ā¤ f (x0 +t v)ā’f (x0 ) , and hence (v) ā¤
t
f (x)(v).
(2) ā’ (1) The uniqueness of implies f (x0 ) = since otherwise we had a linear
functional Āµ = 0 with u ā¤ f (x) ā’ . Then Āµ + contradicts uniqueness.
(2) ā” (3) Any hyperplane tangent to Af at (x0 , f (x0 )) is described by a functional
( , s) ā E Ć— R such that (x) + s t ā„ (x0 ) + s f (x0 ) for all t ā„ f (x). Note that
the scalar s cannot be 0, since this would imply that (x) ā„ (x0 ) for all x. It has
to be positive, since otherwise the left side would go to ā’ā for f (x) ā¤ t ā’ +ā.
Without loss of generality we may thus assume that s = 1, so the linear functional
is uniquely determined by the hyperplane. Moreover, (x ā’ x0 ) ā„ f (x0 ) ā’ f (x) or,
by replacing by ā’ , we have (x0 + v) ā„ f (x0 ) + (v) for all v ā E.

13.5
13.6 13. Diļ¬erentiability of seminorms 131

(3) ā” (4) Since the graph of a sublinear functional p is just the cone of {(y, 1) :
p(y) = 1}, the set Ap has exactly one tangent hyperplane at (x, 1) if and only if
the set {y : p(y) ā¤ 1} has exactly one tangent hyperplane at x. Applying this to
the Minkowski-functional p of Af gives the desired result.
(5) ā” (6) We show this ļ¬rst for GĖteaux-diļ¬erentiability. We have to show that
a
there is a unique tangent hyperplane to x0 ā K := {x : f (x) ā¤ 1} if and only if
x0 exposes K o := { ā E ā— : (x) ā¤ 1 for all x ā K}. Let us assume 0 ā K and
0 = x0 ā ā‚K. Then a tangent hyperplane to K at x0 is uniquely determined by
a linear functional ā E ā— with (x0 ) = 1 and (x) ā¤ 1 for all x ā K. This is
equivalent to ā K o and (x0 ) = 1, since by Hahn-Banach there exists an ā K o
with (x0 ) = 1. From this the result follows.
This shows also (7) ā” (8) ā” (9) for GĖteaux-diļ¬erentiability.
a
In order to show the statements for FrĀ“chet-diļ¬erentiability one has to show that
e
= f (x) is a FrĀ“chet derivative if and only if x0 is a strongly exposing point. This
e

13.6. Lemma. Duality for convex functions. [Moreau, 1965].
: F Ć— G ā’ R be a dual pairing.
Let ,
(1) For f : F ā’ R āŖ {+ā}, f = +ā one deļ¬nes the dual function

f ā— : G ā’ R āŖ {+ā}, f ā— (z) := sup{ z, y ā’ f (y) : y ā F }.

(2) The dual function f ā— is convex and lower semi-continuous with respect to
the weak topology. Since a convex function g is lower semi-continuous if
and only if for all a ā R the set {x : g(x) > a} is open, equivalently the
convex set {x : g(x) ā¤ a} is closed, this is equivalent for every topology
which is compatible with the duality.
ā— ā—
(3) f1 ā¤ f2 ā’ f1 ā„ f2 .
(4) f ā— ā¤ g ā” g ā— ā¤ f .
(5) f ā—ā— = f if and only if f is lower semi-continuous and convex.
(6) Suppose z ā G satisļ¬es f (x + v) ā„ f (x) + z, v for all v (in particular, this
is true if z = f (x)). Then f (x) + f ā— (z) = z, x .
(7) If f1 (y) = f (y ā’ a), then f1 (z) = z, a + f ā— (z).
ā—

(8) If f1 (y) = f (y) + a, then f1 (z) = f ā— (z) ā’ a.
ā—

(9) If f1 (y) = f (y) + b, x , then f1 (z) = f ā— (z ā’ b).
ā—

(10) If E = F = R and f ā„ 0 with f (0) = 0, then f ā— (t) = sup{ts ā’ f (s) : t ā„ 0}
for t ā„ 0.
(11) If Ī³ ā„ 0 is convex and Ī³(t) ā’ 0, then Ī³(t) > 0 for t > 0.
t
(12) Let (F, G) be a Banach space and its dual. If Ī³ ā„ 0 is convex and Ī³(0) = 0,
and f (y) := Ī³( y ), then f ā— (z) = Ī³ ā— ( z ).
(13) A convex function f on a Banach space is FrĀ“chet diļ¬erentiable at a with
e
derivative b := f (a) if and only if there exists a convex non-negative func-
tion Ī³, with Ī³(0) = 0 and limtā’0 Ī³(t) = 0, such that
t

f (a + h) ā¤ f (a) + f (a), h + Ī³( h ).

13.6
132 Chapter III. Partitions of unity 13.6

Proof. (1) Since f = +ā, there is some y for which z, y ā’ f (y) is ļ¬nite, hence
f ā— (z) > ā’ā.
(2) The function z ā’ z, y ā’f (y) is continuous and linear, and hence the supremum
f ā— (z) is lower semi-continuous and convex. It remains to show that f ā— is not
constant +ā: This is not true. In fact, take f (t) = ā’t2 then f ā— (s) = sup{s tā’f (t) :
t ā R} = sup{s t + t2 : t ā R} = +ā. More generally, f ā— = +ā ā” f lies above
some aļ¬ne hyperplane, see (5).
ā— ā—
(3) If f1 ā¤ f2 then z, y ā’ f1 (z) ā„ z, y ā’ f2 (z), and hence f1 (z) ā„ f2 (z).
(4) One has

āz : f ā— (z) ā¤ g(z) ā” āz, y : z, y ā’ f (y) ā¤ g(z)
ā” āz, y : z, y ā’ g(z) ā¤ f (y)
ā” āy : g ā— (y) ā¤ f (y).

(5) Since (f ā— )ā— is convex and lower semi-continuous, this is true for f provided
f = (f ā— )ā— . Conversely, let g(b) = ā’a and g(z) = +ā otherwise. Then g ā— (y) =
ā¤ f ā” f ā— (b) ā¤ ā’a. If f is
sup{ z, y ā’ g(z) : z ā G} = b, y + a. Hence, a + b,
convex and lower semi-continuous, then it is the supremum of all continuous linear
below it, and this is exactly the case if f ā— (b) ā¤ ā’a. Hence,
functionals a + b,
f ā—ā— (y) = sup{ z, y ā’ f ā— (z) : z ā G} ā„ b, y + a and thus f = f ā—ā— .
(6) Let f (a+y) ā„ f (a)+ b, y . Then f ā— (b) = sup{ b, y ā’f (y) : y ā F } = sup{ b, a+
v ā’ f (a + v) : v ā F } ā¤ sup{ b, a + b, v ā’ f (a) ā’ b, v : v ā F } = b, a ā’ f (a).
(7) Let f1 (y) = f (y ā’ a). Then
ā—
f1 (z) = sup{ z, y ā’ f (y ā’ a) : y ā F }
= sup{ z, y + a ā’ f (y) : y ā F } = z, a + f ā— (z).

(8) Let f1 (y) = f (y) + a. Then

f1 (z) = sup{ z, y ā’ f (y) ā’ a : y ā F } = f ā— (z) ā’ a.
ā—

(9) Let f1 (y) = f (y) + b, y . Then
ā—
f1 (z) = sup{ z, y ā’ f (y) ā’ b, y : y ā F }
= sup{ z ā’ b, y ā’ f (y) : y ā F } = f ā— (z ā’ b).

(10) Let E = F = R and f ā„ 0 with f (0) = 0, and let s ā„ 0. Using that
s t ā’ f (t) ā¤ 0 for t ā¤ 0 and that s 0 ā’ f (0) = 0 we obtain

f ā— (s) = sup{s t ā’ f (t) : t ā R} = sup{s t ā’ f (t) : t ā„ 0}.

Ī³(t) Ī³(t)
(11) Let Ī³ ā„ 0 with limt = 0, and let s > 0. Then there are t with s > t,
0 t
and hence
Ī³(t)
Ī³ ā— (s) = sup{st ā’ Ī³(t) : t ā„ 0} = sup{t(s ā’ ) : t ā„ 0} > 0.
t
13.6
13.7 13. Diļ¬erentiability of seminorms 133

(12) Let f (y) = Ī³( y ). Then

f ā— (z) = sup{ z, y ā’ Ī³( y ) : y ā F }
= sup{t z, y ā’ Ī³(t) : y = 1, t ā„ 0}
= sup{sup{t z, y ā’ Ī³(t) : y = 1}, t ā„ 0}
= sup{t z ā’ Ī³(t) : y = 1, t ā„ 0}
= Ī³ ā— ( z ).

(13) If f (a + h) ā¤ f (a) + b, h + Ī³( h ), then we have

f (a + t h) ā’ f (a) Ī³(t h )
ā¤ b, h + ,
t t
hence f (a)(h) ā¤ b, h . Since h ā’ f (a)(h) is sub-linear and the linear functionals
are minimal among the sublinear ones, we have equality. By convexity we have

f (a + t h) ā’ f (a)
ā„ b, h = f (a)(h).
t
So f is FrĀ“chet-diļ¬erentiable at a with derivative f (a)(h) = b, h , since the re-
e
mainder is bounded by Ī³( h ) which satisļ¬es Ī³( h ) ā’ 0 for h ā’ 0.
h

Conversely, assume that f is FrĀ“chet-diļ¬erentiable at a with derivative b. Then
e

|f (a + h) ā’ f (a) ā’ b, h |
ā’ 0 for h ā’ 0,
h

and by convexity
g(h) := f (a + h) ā’ f (a) ā’ b, h ā„ 0.
Let Ī³(t) := sup{g(u) : u = |t|}. Since g is convex Ī³ is convex, and obviously
Ī³(t) ā [0, +ā], Ī³(0) = 0 and Ī³(t) ā’ 0 for t ā’ 0. This is the required function.
t

13.7. Proposition. Continuity of the FrĀ“chet derivative. [Asplund, 1968].
e
The diļ¬erential f of any continuous convex function f on a Banach space is con-
tinuous on the set of all points where f is FrĀ“chet diļ¬erentiable. In general, it is
e
however neither uniformly continuous nor bounded, see (15.8).

Proof. Let f (x)(h) denote the one sided derivative. From convexity we conclude
ā„ f (x) + f (x)(v). Suppose xn ā’ x are points where f is FrĀ“chet
that f (x + v) e
Then we obtain f (xn )(v) ā¤ f (xn + v) ā’ f (xn ) which is bounded in
diļ¬erentiable.
n. Hence, the f (xn ) form a bounded sequence. We get

f (x) ā„ f (xn ), x ā’ f ā— (f (xn )) since f (y) + f ā— (z) ā„ z, y
since f ā— (f (z)) + f (z) = f (z)(z)
= f (xn ), x + f (xn ) ā’ f (xn ), xn
ā„ f (xn ), x ā’ xn + f (x) + f (x), xn ā’ x since f (x + h) ā„ f (x) + f (x)(h)
= f (xn ) ā’ f (x), x ā’ xn + f (x).

13.7
134 Chapter III. Partitions of unity 13.8

Since xn ā’ x and f (xn ) is bounded, both sides converge to f (x), hence

lim f (xn ), x ā’ f ā— (f (xn )) = f (x).
nā’ā

Since f is convex and FrĀ“chet-diļ¬erentiable at a := x with derivative b := f (x),
e
there exists by (13.6.13) a Ī³ with

f (h) ā¤ f (a) + b, h ā’ a + Ī³( h ā’ a ).

By duality we obtain using (13.6.3)

f ā— (z) ā„ z, a ā’ f (a) + Ī³ ā— ( z ā’ b ).

If we apply this to z := f (xn ) we obtain

f ā— (f (xn )) ā„ f (xn ), x ā’ f (x) + Ī³ ā— ( f (xn ) ā’ f (x) ).

Hence
Ī³ ā— ( f (xn ) ā’ f (x) ) ā¤ f ā— (f (xn )) ā’ f (xn ), x + f (x),

and since the right side converges to 0, we have that Ī³ ā— ( f (xn ) ā’ f (x) ) ā’ 0.
Then f (xn ) ā’ f (x) ā’ 0 where we use that Ī³ is convex, Ī³(0) = 0, and Ī³(t) > 0
for t > 0, thus Ī³ is strictly monotone increasing.

13.8. Asplund spaces and generic FrĀ“chet diļ¬erentiability. From (13.4)
e
it follows easily that a convex function f : R ā’ R is diļ¬erentiable at all except
countably many points. This has been generalized by [Rademacher, 1919] to: Ev-
ery Lipschitz mapping from an open subset of Rn to R is diļ¬erentiable almost
everywhere. Recall that a locally bounded convex function is locally Lipschitz, see
(13.2).

Proposition. For a Banach space E the following statements are equivalent:
(1) Every continuous convex function f : E ā’ R is FrĀ“chet-diļ¬erentiable on a
e
dense GĪ“ -subset of E;
(2) Every continuous convex function f : E ā’ R is FrĀ“chet-diļ¬erentiable on a
e
dense subset of E;
(3) Every locally Lipschitz function f : E ā’ R is FrĀ“chet-diļ¬erentiable on a
e
dense subset of E;
(4) Every equivalent norm is FrĀ“chet-diļ¬erentiable at least at one point;
e
(5) E has no equivalent rough norm;
(6) Every (closed) separable subspace has a separable dual;
(7) The dual E ā— has the Radon-Nikodym property;
(8) Every linear mapping E ā’ L1 (X, ā„¦, Āµ) which is integral is nuclear;
(9) Every closed convex bounded subset of E ā— is the closed convex hull of its
extremal points;
(10) Every bounded subset of E ā— is dentable.

13.8
13.8 13. Diļ¬erentiability of seminorms 135

A Banach space satisfying these equivalent conditions is called Asplund space.
Every Banach space with a FrĀ“chet diļ¬erentiable bump function is Asplund, [Eke-
e
land, Lebourg, 1976, p. 203]. It is an open question whether the converse is true.
Every WCG-space (i.e. a Banach space for which a weakly compact subset K exists,
whose linear hull is the whole space) is Asplund, [John, Zizler, 1976].
The Asplund property is inherited by subspaces, quotients, and short exact se-
quences, [Stegall, 1981].

About the proof. (1) [Asplund, 1968]: If a convex function is FrĀ“chet diļ¬eren-
e
tiable on a dense subset then it is so on a dense GĪ“ -subset, i.e. a dense countable
intersection of open subsets.
(2) is in fact a local property, since in [Borwein, Fitzpatrick, Kenderov, 1991] it
is mentioned that for a Lipschitz function f : E ā’ R with Lipschitz constant L
deļ¬ned on a convex open set U the function

Ė
f (x) := inf{f (y) + L x ā’ y : y ā U }
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