ńņš. 20 |

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. GĖteaux-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 GĖteaux-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 GĖteaux 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 GĖteaux (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 GĖteaux diļ¬erentiable at (x0 , f (x0 )).

a

Moreover, for a sublinear function f the following statements are equivalent:

(5) The function f is GĖteaux (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 GĖteaux (FrĀ“chet) diļ¬erentiable at x0 ;

a e

(8) The Minkowski functional of Af is GĖteaux (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

GĖteaux-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 GĖteaux-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 GĖteaux-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

is left to the reader, see also (13.19) for a more general result.

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 āŖ {+ā}, f = +ā one deļ¬nes the dual function

f ā— : G ā’ R āŖ {+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

āz : f ā— (z) ā¤ g(z) ā” āz, y : z, y ā’ f (y) ā¤ g(z)

ā” āz, y : z, y ā’ g(z) ā¤ f (y)

ā” āy : 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 }

ńņš. 20 |