ńņš. 6 |

12.9. Corollary. Chain rule. The composition of Lipk -mappings is again Lipk ,

and the usual formula for the derivative of the composite holds.

Proof. We have to compose f ā—¦ g with a smooth curve c, but then g ā—¦ c is a Lipk -

curve, thus it is suļ¬cient to show that the composition of a Lipk curve c : R ā’ U ā

E with a Lipk -mapping f : U ā’ F is again Lipk , and that (f ā—¦c) (t) = df (c(t), c (t)).

This follows by induction on k for k ā„ 1 in the same way as we proved theorem

(12.8.2) ā’ (12.8.1), using theorem (12.8) itself.

12.10. Deļ¬nition and Proposition. Let F be a convenient vector space. The

space Lipk (R, F ) of all Lipk -curves in F is again a convenient vector space with

the following equivalent structures:

(1) The initial structure with respect to the k + 2 linear mappings (for 0 ā¤ j ā¤

k + 1) c ā’ Ī“ j c from Lipk (R, F ) into the space of all F -valued maps in j + 1

pairwise diļ¬erent real variables (t0 , . . . , tj ) which are bounded on bounded

subsets, with the cā -complete locally convex topology of uniform convergence

on bounded subsets. In fact, the mappings Ī“ 0 and Ī“ k+1 are suļ¬cient.

(2) The initial structure with respect to the k + 2 linear mappings (for 0 ā¤ j ā¤

k +1) c ā’ Ī“eq c from Lipk (R, F ) into the space of all maps from RĆ—(R\{0})

j

into F which are bounded on bounded subsets, with the cā -complete locally

convex topology of uniform convergence on bounded subsets. In fact, the

0 k+1

mappings Ī“eq and Ī“eq are suļ¬cient.

(3) The initial structure with respect to the derivatives of order j ā¤ k considered

as linear mappings into the space of Lip0 -curves, with the locally convex

topology of uniform convergence of the curve on bounded subsets of R and

of the diļ¬erence quotient on bounded subsets of {(t, s) ā R2 : t = s}.

The convenient vector space Lipk (R, F ) satisļ¬es the uniform boundedness principle

with respect to the point evaluations.

Proof. All three structures describe closed embeddings into ļ¬nite products of

spaces, which in (1) and (2) are obviously cā -complete. For (3) this follows, since

by (1) the structure on Lip0 (R, E) is convenient.

12.10

12.13 12. Diļ¬erentiability of ļ¬nite order 125

All structures satisfy the uniform boundedness principle for the point evaluations

by (5.25), and since spaces of all bounded mappings on some (bounded) set satisfy

this principle. This can be seen by composing with ā— for all ā E , since Banach

spaces do this by (5.24).

By applying this uniform boundedness principle one sees that all these structures

are indeed equivalent.

12.11. Deļ¬nition and Proposition. Let E and F be convenient vector spaces

and U ā E be cā -open. The space Lipk (U, F ) of all Lipk -mappings from U to F

is again a convenient vector space with the following equivalent structures:

(1) The initial structure with respect to the linear mappings cā— : Lipk (U, F ) ā’

Lipk (R, F ) for all c ā C ā (R, F ).

(2) The initial structure with respect to the linear mappings cā— : Lipk (U, F ) ā’

Lipk (R, F ) for all c ā Lipk (R, F ).

This space satisļ¬es the uniform boundedness principle with respect to the evaluations

evx : Lipk (U, F ) ā’ F for all x ā U .

Proof. The structure (1) is convenient since by (12.1) it is a closed subspace of the

product space which is convenient by (12.10). The structure in (2) is convenient

since it is closed by (12.9). The uniform boundedness principle for the point evalu-

ations now follows from (5.25) and (12.10), and this in turn gives us the equivalence

of the two structures.

12.12. Remark. We want to call the attention of the reader to the fact that there

is no general exponential law for Lipk -mappings. In fact, if f ā Lipk (R, Lipk (R, F ))

then ( ā‚t )p ( ā‚s )q f ā§ (t, s) exists if max(p, q) ā¤ k. This describes a smaller space than

ā‚ ā‚

Lipk (R2 , F ), which is not invariantly describable.

However, some partial results still hold, namely for convenient vector spaces E, F ,

and G, and for cā -open sets U ā E, V ā F we have

Lipk (U, L(F, G)) ā¼ L(F, Lipk (U, G)),

=

Lipk (U, Lipl (V, G)) ā¼ Lipl (V, Lipk (U, G)),

=

see [FrĀØlicher, Kriegl, 1988, 4.4.5, 4.5.1, 4.5.2]. For a mapping f : U Ć— F ā’ G which

o

is linear in F we have: f ā Lipk (U Ć— F, G) if and only if f āØ ā Lipk (U, L(E, F )),

see [FrĀØlicher, Kriegl, 1988, 4.3.5]. The last property fails if we weaken Lipschitz to

o

continuous, see the following example.

1

12.13. Smolyanovā™s Example. Let f : 2 ā’ R be deļ¬ned by f := kā„1 k2 fk ,

where fk (x) := Ļ•(k(kxk ā’ 1)) Ā· j<k Ļ•(jxj ) and Ļ• : R ā’ [0, 1] is smooth with

Ļ•(0) = 1 and Ļ•(t) = 0 for |t| ā„ 1 . We shall show that

4

2

ā’ R is FrĀ“chet diļ¬erentiable.

(1) f : e

2

ā’ ( 2 ) is not continuous.

(2) f :

2

Ć— 2 ā’ R is continuous.

(3) f :

12.13

126 Chapter III. Partitions of unity 12.13

1

2 2

Proof. Let A := {x ā : |kxk | ā¤ for all k}. This is a closed subset of .

4

2

(1) Remark that for x ā at most one fk (x) can be unequal to 0. In fact fk (x) = 0

1 1 3

implies that |kxk ā’ 1| ā¤ 4k ā¤ 4 , and hence kxk ā„ 4 and thus fj (x) = 0 for j > k.

For x ā A there exists a k > 0 with |kxk | > 1 and the set of points satisfying this

/ 4

condition is open. It follows that Ļ•(kxk ) = 0 and hence f = j<k j1 fj is smooth

2

on this open set.

On the other hand let x ā A. Then |kxk ā’ 1| ā„ 4 > 1 and hence Ļ•(k(kxk ā’ 1)) = 0

3

4

2

for all k and thus f (x) = 0. Let v ā be such that f (x + v) = 0. Then there exists

a unique k such that fk (x + v) = 0 and therefore |j(xj + vj )| < 1 for j < k and

4

1 1 1

|k(xk +vk )ā’1| < 4k ā¤ 4 . Since |kxk | ā¤ 4 we conclude |kvk | ā„ 1ā’|k(xk +vk )ā’1|ā’

|kxk | ā„ 1 ā’ 1 ā’ 1 = 2 . Hence |f (x + v)| = k2 |fk (x + v)| ā¤ k2 ā¤ (2|vk |)2 ā¤ 4 v 2 .

1 1 1

4 4

Thus f (x+v)ā’0ā’0 ā¤ 4 v ā’ 0 for v ā’ 0, i.e. f is FrĀ“chet diļ¬erentiable at x

e

v

with derivative 0.

d1

(2) If fact take a ā R with Ļ• (a) = 0. Then f (t ek )(ek ) = dt k2 fk (t ek ) =

d1 1a

2

dt k2 Ļ•(k t ā’ k) = Ļ• (k (k t ā’ 1)) = Ļ• (a) if t = tk := k k + 1 , which goes to

0 for k ā’ ā. However f (0)(ek ) = 0 since 0 ā A.

(3) We have to show that f (xn )(v n ) ā’ f (x)(v) for (xn , v n ) ā’ (x, v). For x ā A

/

this is obviously satisļ¬ed, since then there exists a k with |kxk | > 1 and hence

4

1

f = jā¤k j 2 fj locally around x.

If x ā A then f (x) = 0 and thus it remains to consider the case, where xn ā A. /

Let Īµ > 0 be given. Locally around xn at most one summand fk does not vanish:

If xn ā A then there is some k with |kxk | > 1/4 which we may choose minimal.

/

Thus |jxj | ā¤ 1/4 for all j < k, so |j(jxj ā’ 1)| ā„ 3j/4 and hence fj = 0 locally since

the ļ¬rst factor vanishes. For j > k we get fj = 0 locally since the second factor

vanishes. Thus we can evaluate the derivative:

1 Ļ•ā 2n

|f (xn )(v n )| = 2 fk (xn )(v n ) ā¤ n

k |vk | + j|vj | .

2

k k

j<k

Since v ā 2 we ļ¬nd a K1 such that ( jā„K1 |vj |2 )1/2 ā¤ 2 Ļ•Īµ ā . Thus we conclude

from v n ā’v 2 ā’ 0 that |vj | ā¤ Ļ• Īµ ā for j ā„ K1 and large n. For the ļ¬nitely many

n

small n we can increase K1 such that for these n and j ā„ K1 also |vj | ā¤ Ļ• Īµ ā .

n

Furthermore there is a constant K2 ā„ 1 such that v n ā ā¤ v n 2 ā¤ K2 for all n.

Now choose N ā„ K1 so large that N 2 ā„ 1 Ļ• ā K2 K1 . Obviously n<N n2 fn is 1

2

Īµ

smooth. So it remains to consider those n for which the non-vanishing term has

index k ā„ N . For those terms we have

1 1

|f (xn )(v n )| = 2 fk (xn )(v n ) ā¤ Ļ• ā |vk | + 2

n n

j|vj |

k k

j<k

1 1

n n n

ā¤ |vk | Ļ• j|vj | + j|vj | Ļ•

ā+ Ļ• ā2 ā

k2

k

K1 ā¤j<k

j<K1

2

K1 n 1

ā¤Īµ+ Ļ• ā 2 v j Īµ ā¤ Īµ + Īµ + Īµ = 3Īµ

ā+

k2

N

K1 ā¤j<k

This shows the continuity.

12.13

127

13. Diļ¬erentiability of Seminorms

A desired separation property is that the smooth functions generate the topology.

Since a locally convex topology is generated by the continuous seminorms it is

natural to look for smooth seminorms. Note that every seminorm p : E ā’ R on a

vector space E factors over Ep := E/ ker p and gives a norm on this space. Hence, it

ĖĖ Ė

can be extended to a norm p : Ep ā’ R on the completion Ep of the space Ep which

is normed by this factorization. If E is a locally convex space and p is continuous,

then the canonical quotient mapping E ā’ Ep is continuous. Thus, smoothness of

p oļ¬ 0 implies smoothness of p on its carrier, and so the case where E is a Banach

Ė

space is of central importance.

Obviously, every seminorm is a convex function, and hence we can generalize our

treatment slightly by considering convex functions instead. The question of their

diļ¬erentiability properties is exactly the topic of this section.

Note that since the smooth functions depend only on the bornology and not on

the locally convex topology the same is true for the initial topology induced by all

smooth functions. Hence, it is appropriate to make the following

Convention. In this chapter the locally convex topology on all convenient vector

spaces is assumed to be the bornological one.

13.1. Remark. It can be easily seen that for a function f : E ā’ R on a vector

space E the following statements are equivalent (see for example [FrĀØlicher, Kriegl,

o

1988, p. 199]):

n n

(1) The function f is convex, i.e. f ( i=1 Ī»i xi ) ā¤ i=1 Ī»i f (xi ) for Ī»i ā„ 0

n

with i=1 Ī»i = 1;

(2) The set Uf := {(x, t) ā E Ć— R : f (x) ā¤ t} is convex;

(3) The set Af := {(x, t) ā E Ć— R : f (x) < t} is convex.

Moreover, for any translation invariant topology on E (and hence in particular for

the locally convex topology or the cā -topology on a convenient vector space) and

any convex function f : E ā’ R the following statements are equivalent:

(1) The function f is continuous;

(2) The set Af is open in E Ć— R;

(3) The set f<t := {x ā E : f (x) < t} is open in E for all t ā R.

13.2. Result. Convex Lipschitz functions. Let f : E ā’ R be a convex func-

tion on a convenient vector space E. Then the following statements are equivalent:

(1) It is locally Lipschitzian;

(2) It is continuous for the locally convex topology;

continuous for the cā -topology;

(3) It is

(4) It is bounded on Mackey converging sequences;

If f is a seminorm, then these further are equivalent to

(5) It is bounded on bounded sets.

13.2

128 Chapter III. Partitions of unity 13.3

If E is normed this further is equivalent to

(6) It is locally bounded.

The proof is due to [Aronszajn, 1976] for Banach spaces and [FrĀØlicher, Kriegl,

o

1988, p. 200], for convenient vector spaces.

13.3. Basic deļ¬nitions. Let f : E ā U ā’ F be a mapping deļ¬ned on a cā -open

subset of a convenient vector space E with values in another one F . Let x ā U

and v ā E. Then the (one sided) directional derivative of f at x in direction v is

deļ¬ned as

f (x + t v) ā’ f (x)

f (x)(v) = dv f (x) := lim .

t

t0

Obviously, if f (x)(v) exists, then so does f (x)(s v) for s > 0 and equals s f (x)(v).

Even if f (x)(v) exists for all v ā E the mapping v ā’ f (x)(v) may not be linear

in general, and if it is linear it will not be bounded in general. Hence, f is called

GĖteaux-diļ¬erentiable at x, if the directional derivatives f (x)(v) exist for all v ā E

a

and v ā’ f (x)(v) is a bounded linear mapping from E ā’ F .

Even for GĖteaux-diļ¬erentiable mappings the diļ¬erence quotient f (x+t v)ā’f (x) need

a t

not converge uniformly for v in bounded sets (or even in compact sets). Hence, one

deļ¬nes f to be FrĀ“chet-diļ¬erentiable at x if f is GĖteaux-diļ¬erentiable at x and

e a

f (x+t v)ā’f (x)

ā’ f (x)(v) ā’ 0 uniformly for v in any bounded set. For a Banach

t

space E this is equivalent to the existence of a bounded linear mapping denoted

f (x) : E ā’ F such that

f (x + v) ā’ f (x) ā’ f (x)(v)

lim = 0.

v

vā’0

If f : E ā U ā’ F is GĖteaux-diļ¬erentiable and the derivative f : E ā U ā’

a

L(E, F ) is continuous, then f is FrĀ“chet-diļ¬erentiable, and we will call such a

e

function C 1 . In fact, the fundamental theorem applied to t ā’ f (x + t v) gives us

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. 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 }

is a Lipschitz extension with constant L, and it is convex if f is.

(2) ā’ (3) is due to [Preiss, 1990], Every locally Lipschitz function on an Asplund

space is FrĀ“chet diļ¬erentiable at points of a dense subset.

e

(3) ā’ (2) follows from the fact that continuous convex functions are locally Lip-

schitz, see (13.2).

(2) ā” (4) is mentioned in [Preiss, 1990] without any proof or reference.

(2) ā” (10) is due to [Stegall, 1975]. A subset D of a Banach space is called dentable,

if and only if for every x ā D there exists an Īµ > 0 such that x is not in the closed

convex hull of {y ā D : y ā’ x ā„ Īµ}.

(2) ā” (5) is due to [John, Zizler, 1978]. A norm p is called rough, see also (13.23),

if and only if there exists an Īµ > 0 such that arbitrary close to each x ā X there

are points xi and u with u = 1 such that |p (x2 )(u) ā’ p (x1 )(u)| ā„ Īµ. The usual

norms on C[0, 1] and on 1 are rough by (13.12) and (13.13). A norm is not rough

if and only if the dual ball is wā— -dentable. The unit ball is dentable if and only if

the dual norm is not rough.

(2) ā” (6) is due to [Stegall, 1975].

(2) ā” (7) is due to [Stegall, 1978]. A closed bounded convex subset K of a Banach

space E is said to have the Radon-Nikodym property if for any ļ¬nite measure space

(ā„¦, Ī£, Āµ) every Āµ-continuous countably additive function m : Ī£ ā’ E of ļ¬nite vari-

ation with average range { m(A) : S ā Ī£, Āµ(S) > 0} contained in K is representable

Āµ(S)

by a Bochner integrable function, i.e. there exists a Borel-measurable essentially

separably valued function f : ā„¦ ā’ E with m(S) = S f dĀµ. This function f is

then called the Radon-Nikodym derivative of m. A Banach space is said to have

the Radon-Nikodym property if every closed bounded convex subset has it. See

also [Diestel, 1975]. A subset K is a Radon-Nikodym set if and only if every closed

convex subset of K is the closed convex hull of its strongly exposed points.

13.8

136 Chapter III. Partitions of unity 13.10

(7) ā” (8) can be found in [Stegall, 1975] and is due to [Grothendieck, 1955]. A

linear mapping E ā’ F is called integral if and only if it has a factorization

wF w Fu ā—ā—

E

u

w L (K, Āµ)

1

C(K)

for some Radon-measure Āµ on a compact space K.

A linear mapping E ā’ F is called nuclear if and only if there are xā— ā E ā— and

n

ā— ā—

yn ā F such that n xn yn < ā and T = n xn ā— yn .

(2) ā” (9) is due to [Stegall, 1981, p.516].

13.9. Results on generic GĖteaux diļ¬erentiability of Lipschitz functions.

a

(1) [Mazur, 1933] & [Asplund, 1968] A Banach space E with the property that

every continuous convex function f : E ā’ R is GĖteaux-diļ¬erentiable on

a

a dense GĪ“ -subset is called weakly Asplund. Separable Banach spaces are

weakly Asplund.

Ė

(2) In [Zivkov, 1983] it is mentioned that there are Lipschitz functions on R,

which fail to be diļ¬erentiable on a dense GĪ“ -subset.

(3) A Lipschitz function on a separable Banach space is āalmost everywhereā

GĖteaux-diļ¬erentiable, [Aronszajn, 1976].

a

(4) [Preiss, 1990] If the norm on a Banach space is B-diļ¬erentiable then every

Lipschitz function is B-diļ¬erentiable on a dense set. A function f : E ā

U ā’ F is called B-diļ¬erentiable at x ā U for some family B of bounded sub-

sets, if there exists a continuous linear mapping (denoted f (x)) in L(E, F )

such that for every B ā B one has f (x+t v)ā’f (x) ā’ f (x)(v) ā’ 0 for t ā’ 0

t

uniformly for v ā B.

Ė

(5) [Kenderov, 1974], see [Zivkov, 1983]. Every locally Lipschitzian function

on a separable Banach space which has one sided directional derivatives for

each direction in a dense subset is GĖteaux diļ¬erentiable on a non-meager

a

subset.

Ė

(6) [Zivkov, 1983]. For every space with FrĀ“chet diļ¬erentiable norm any locally

e

Lipschitzian function which has directional derivatives for a dense set of

directions is generically GĖteaux diļ¬erentiable.

a

(7) There exists a Lipschitz GĖteaux diļ¬erentiable function f : L1 [0, 1] ā’ R

a

which is nowhere FrĀ“chet diļ¬erentiable, [Sova, 1966a], see also [Gieraltow-

e

ska-Kedzierska, Van Vleck, 1991]. Hence, this is an example of a weakly

Asplund but not Asplund space.

Further references on generic diļ¬erentiability are: [Phelps, 1989], [Preiss, 1984],

and [Zhivkov, 1987].

13.10. Lemma. Smoothness of 2n-norm. For n ā N the 2n-norm is smooth

on L2n \ {0}.

13.10

13.11 13. Diļ¬erentiability of seminorms 137

Proof. Since t ā’ t1/2n is smooth on R+ it is enough to show that x ā’ ( x 2n )2n

is smooth. Let p := 2n. Since (x1 , . . . , xn ) ā’ x1 Ā· . . . Ā· xn is a n-linear contraction

p 1

from Lp Ć— . . . Ć— Lp ā’ L1 by the HĀØlder-inequality ( i=1 p = 1) and : L1 ā’ R

o

is a linear contraction the mapping x ā’ (x, . . . , x) ā’ x2n is smooth. Note that

since we have a real Banach space and p = 2n is even we can drop the absolute

value in the formula of the norm.

13.11. Derivative of the 1-norm. Let x ā 1 and j ā N be such that xj = 0.

Let ej be the characteristic function of {j}. Then x + t ej 1 = x 1 + |t| since

the supports of x and ej are disjoint. Hence, the directional derivative of the norm

p : v ā’ v 1 is given by p (x)(ei ) = 1 and p (x)(ā’ei ) = 1, and p is not diļ¬erentiable

at x. More generally we have:

Lemma. [Mazur, 1933, p.79]. Let Ī“ be some set, and let p be the 1-norm given

1

Ī³āĪ“ |xĪ³ | for x ā xĪ³ =0 |hĪ³ | +

by x 1 = p(x) := (Ī“). Then p (x)(h) =

xĪ³ =0 hĪ³ sign xĪ³ .

The basic idea behind this result is, that the unit sphere of the 1-norm is a hyper-

octahedra, and the points on the faces are those, for which no coordinate vanishes.

Proof. Without loss of generality we may assume that p(x) = 1 = p(h), since for

d d s

r > 0 and s ā„ 0 we have p (r x)(s h) = dt |t=0 p(r x + t s h) = dt |t=0 r p(x + t ( r h)) =

s

r p (x)( r h) = s p (x)(h).

We have |xĪ³ + hĪ³ | ā’ |xĪ³ | = ||xĪ³ | + hĪ³ sign xĪ³ | ā’ |xĪ³ | ā„ |xĪ³ | + hĪ³ sign xĪ³ ā’ |xĪ³ | =

hĪ³ sign xĪ³ , and is equal to |hĪ³ | if xĪ³ = 0. Summing up these (in)equalities we

obtain

p(x + h) ā’ p(x) ā’ |hĪ³ | ā’ hĪ³ sign xĪ³ ā„ 0.

xĪ³ =0 xĪ³ =0

Īµ

For Īµ > 0 choose a ļ¬nite set F ā‚ Ī“, such that Ī³ āF |hĪ³ | < 2 . Now choose t so

/

small that

|xĪ³ | + t hĪ³ sign xĪ³ ā„ 0 for all Ī³ ā F with xĪ³ = 0.

We claim that

q(x + t h) ā’ q(x)

ā’ |hĪ³ | ā’ hĪ³ sign xĪ³ ā¤ Īµ.

t x Ī³ =0 xĪ³ =0

|xĪ³ +t hĪ³ |ā’|xĪ³ |

= |hĪ³ |, hence these terms cancel

Let ļ¬rst Ī³ be such that xĪ³ = 0. Then t

with ā’ xĪ³ =0 |hĪ³ |.

Let now xĪ³ = 0. For |xĪ³ | + t hĪ³ sign xĪ³ ā„ 0 (hence in particular for Ī³ ā F with

xĪ³ = 0) we have

|xĪ³ + t hĪ³ | ā’ |xĪ³ | |xĪ³ | + t hĪ³ sign xĪ³ ā’ |xĪ³ |

= = hĪ³ sign xĪ³ .

t t

Thus, these terms sum up to the corresponding sum hĪ³ sign xĪ³ .

Ī³

13.11

138 Chapter III. Partitions of unity 13.12

It remains to consider Ī³ with xĪ³ = 0 and |xĪ³ | + t hĪ³ sign xĪ³ < 0. Then Ī³ ā F and

/

|xĪ³ + t hĪ³ | ā’ |xĪ³ | ā’|xĪ³ | ā’ t hĪ³ sign xĪ³ ā’ |xĪ³ | ā’ t hĪ³ sign xĪ³

ā’ hĪ³ sign xĪ³ =

t t

ā¤ ā’2hĪ³ sign xĪ³ ,

Īµ

|hĪ³ | <

and since these remaining terms sum up to something smaller than

Ī³ āF

/ 2

Īµ.

Remark. The 1-norm is rough. This result shows that the 1-norm is GĖteaux- a

diļ¬erentiable exactly at those points, where all coordinates are non-zero. Thus, if

Ī“ is uncountable, the 1-norm is nowhere GĖteaux-diļ¬erentiable.

a

In contrast to what is claimed in [Mazur, 1933, p.79], the 1-norm is nowhere FrĀ“chet

e

diļ¬erentiable. In fact, take 0 = x ā 1 (Ī“). For Ī³ with xĪ³ = 0 and t > 0 we have

that

p(x + t (ā’ sign xĪ³ eĪ³ )) ā’ p(x) ā’ t p (x)(ā’ sign xĪ³ eĪ³ ) =

= |xĪ³ ā’ t sign xĪ³ | ā’ |xĪ³ | + t = |xĪ³ | ā’ t ā’ |xĪ³ | + t ā„ t Ā· 1,

provided t ā„ 2 |xĪ³ |, since then |xĪ³ | ā’ t = t ā’ |xĪ³ | ā„ |xĪ³ |. Obviously, for every t > 0

there are Ī³ satisfying this required condition; either xĪ³ = 0 then we have a corner,

or xĪ³ = 0 then it gets arbitrarily small. Thus, the directional diļ¬erence quotient

does not converge uniformly on the unit-sphere.

1

The set of points x in where at least for one n the coordinate xn vanishes is

dense, and one has

for t ā„ 0

+1

p(x + t en ) = p(x) + |t|, hence p (x + t en )(en ) = .

ā’1 for t < 0

Hence the derivative of p is uniformly discontinuous, i.e., in every non-empty open

set there are points x1 , x2 for which there exists an h ā 1 with h = 1 and

|p (x1 )(h) ā’ p (x2 )(h)| ā„ 2.

13.12. Derivative of the ā-norm. On c0 the norm is not diļ¬erentiable at points

x, where the norm is attained in at least two points. In fact let |x(a)| = x = |x(b)|

and let h := sign x(a) ea . Then p(x + th) = |(x + th)(a)| = x + t for t ā„ 0 and

p(x + th) = |(x + th)(b)| = x for t ā¤ 0. Thus, t ā’ p(x + th) is not diļ¬erentiable

at 0 and thus p not at x.

If the norm of x is attained at a single coordinate a, then p is diļ¬erentiable at x.

In fact p(x + th) = |(x + th)(a)| = | sign(x(a)) x + th(a) sign2 (x(a))| = | x +

th(a) sign(x(a))| = x + th(a) sign(x(a)) for |t| h ā¤ x ā’ sup{|x(t)| : t = a}.

Hence the directional diļ¬erence-quotient converges uniformly for h in the unit-ball.

Let x ā C[0, 1] be such that x ā = |x(a)| = |x(b)| for a = b. Choose a y with

y(s) between 0 and x(s) for all s and y(a) = x(a) but y(b) = 0. For t ā„ 0 we have

|(x + t y)(s)| ā¤ |x(a) + t y(a)| = (1 + t) x ā and hence x + t y ā = (1 + t) x ā .

For ā’1 ā¤ t ā¤ 0 we have |(x + t y)(s)| ā¤ |x(a)| and (x + t y)(b) = x(a) and hence

x + t y ā = x ā . Thus the directional derivative is given by p (x)(y) = x ā

and p (x)(ā’y) = 0. More precisely we have the following results.

13.12

13.12 13. Diļ¬erentiability of seminorms 139

Lemma. [Banach, 1932, p. 168]. Let T be a compact metric space. Let x ā

C(T, R) \ {0} and h ā C(T, R). By p we denote the ā-norm x ā = p(x) :=

sup{|x(t)| : t ā T }. Then p (x)(h) = sup{h(t) sign x(t) : |x(t)| = p(x).

The idea here is, that the unit-ball is a hyper-cube, and the points on the faces are

exactly those for which the supremum is attained only in one point.

Proof. Without loss of generality we may assume that p(x) = 1 = p(h), since for

d d s

r > 0 and s ā„ 0 we have p (r x)(s h) = dt |t=0 p(r x + t s h) = dt |t=0 r p(x + t ( r h)) =

s

r p (x)( r h) = s p (x)(h).

Let A := {t ā T : |x(t)| = p(x)}. For given Īµ > 0 we ļ¬nd by the uniform

continuity of x and h a Ī“1 such that |x(t) ā’ x(t )| < 1 and |h(t) ā’ h(t )| < Īµ for

2

dist(t, t ) < Ī“1 . Then {t : dist(t, A) ā„ Ī“1 } is closed, hence compact. Therefore

Ī“ := x ā ā’ sup{|x(t)| : dist(t, A) ā„ Ī“1 } > 0.

Now we claim that for 0 < t < min{Ī“, 1} we have

p(x + t h) ā’ p(x)

0ā¤ ā’ sup{h(r) sign x(r) : r ā A} ā¤ Īµ.

t

For all s ā A we have

p(x + t h) ā„ |(x + t h)(s)| = |x(s)| sign x(s) + t h(s) sign x(s)2

= |x(s)| + t h(s) sign x(s) = p(x) + t h(s) sign x(s)

for 0 ā¤ t ā¤ 1, since |h(s)| ā¤ p(h) = p(x). Hence

p(x + t h) ā’ p(x)

ā„ sup{h(t) sign x(t) : t ā A}.

t

This shows the left inequality.

Let s be a point where the supremum p(x+t h) is attained. From the left inequality

it follows that

p(x + t h) ā„ p(x) + t sup{h(r) sign x(r) : r ā A}, and hence

|x(s)| ā„ |(x + th)(s)| ā’ t |h(s)| ā„ p(x + t h) ā’ t p(h)

ā„ p(x) ā’ t p(h) ā’ sup{h(r) sign x(r) : r ā A}

ā¤1

> p(x) ā’ Ī“ = sup{|x(r)| : dist(r, A) ā„ Ī“1 }.

Therefore dist(s, A) < Ī“1 , and thus there exists an a ā A with dist(s, a) < Ī“1 and

consequently |x(s) ā’ x(a)| < 1 and |h(s) ā’ h(a)| < Īµ. In particular, sign x(s) =

2

sign x(a) = 0. So we get

|x(s)| + t h(s) sign x(s) ā’ p(x)

p(x + t h) ā’ p(x) |(x + t h)(s)| ā’ p(x)

= =

t t t

|x(s)| + t h(s) sign x(s) ā’ p(x)

ā¤ h(s) sign x(a)

=

t

ā¤ |h(s) ā’ h(a)| + h(a) sign x(a)

< Īµ + sup{h(r) sign x(r) : r ā A}.

13.12

140 Chapter III. Partitions of unity 13.13

This proves the claim which ļ¬nally implies

p(x + t h) ā’ p(x)

= sup{h(r) sign x(r) : r ā A}.

p (x)(v) = lim

t

t 0

Remark. The ā-norm is rough. This result shows that the points where the

ā-norm is GĖteaux-diļ¬erentiable are exactly those x where the supremum p(x) is

a

attained in a single point a. The GĖteaux-derivative is then given by p (x)(h) =

a

h(a) sign x(a). In general, this is however not the FrĀ“chet derivative:

e

Let x = 0. Without loss we may assume (that p(x) = 1 and) that there is a unique

point a, where |x(a)| = p(x). Moreover, we may assume x(a) > 0. Let an ā’ a be

such that 0 < x(an ) < x(a) and let 0 < Ī“n := x(a) ā’ x(an ) < x(a). Now choose

sn := 2Ī“n ā’ 0 and hn ā C[0, 1] with p(hn ) ā¤ 1, hn (a) = 0 and hn (an ) := 1 and

p(x + sn hn ) = (x + sn hn )(an ) = x(an ) + 2(x(a) ā’ x(an )) = 2x(a) ā’ x(an ). For this

choose (x + sn hn )(t) ā¤ (x + sn hn )(an ) locally, i.e.. hn (t) ā¤ 1 + (x(an ) ā’ x(t))/sn

and 0 far away from x. Then p (x)(hn ) = 0 by (13.12) and

p(x + sn hn ) ā’ p(x) 2x(a) ā’ x(an ) ā’ x(a)

ā’ p (x)(hn ) =

sn sn

Ī“n 1

= ā’0

=

2Ī“n 2

Thus the limit is not uniform and p is not FrĀ“chet diļ¬erentiable at x.

e

The set of vectors x ā C[0, 1] which attain their norm at least at two points a and

b is dense, and one has for appropriately chosen h with h(a) = ā’x(a), h(b) = x(b)

that

for t ā„ 0

+1

p(x + t h) = (1 + max{t, ā’t}) p(x), hence p (x + t h)(h) = .

ā’1 for t < 0

Therefore, the derivative of the norm is uniformly discontinuous, i.e., in every non-

empty open set there are points x1 , x2 for which there exists an h ā C[0, 1] with

h = 1 and |p (x1 )(h) ā’ p (x2 )(h)| ā„ 2.

13.13. Results on the diļ¬erentiability of p-norms. [Bonic, Frampton, 1966,

p.887].

For 1 < p < ā not an even integer the function t ā’ |t|p is diļ¬erentiable of order

n if n < p, and the highest derivative (t ā’ p (p ā’ 1) . . . (p ā’ n + 1) |t|pā’n ) satisļ¬es

a HĀØlder-condition with modulus p ā’ n, one can show that the p-norm has exactly

o

these diļ¬erentiability properties, i.e.

(1) It is (p ā’ 1)-times diļ¬erentiable with Lipschitzian highest derivative if p is

an integer.

(2) It is [p]-times diļ¬erentiable with highest derivative being HĀØlderian of order

o

p ā’ [p], otherwise.

(3) The norm has no higher HĀØlder-diļ¬erentiability properties.

o

That the norm on Lp is C 1 for 1 < p < ā was already shown by [Mazur, 1933].

13.13

13.15 13. Diļ¬erentiability of seminorms 141

13.14. Proposition. Smooth norms on a Banach space. A norm on a

Banach space is of class C n on E \ {0} if and only if its unit sphere is a C n -

submanifold of E.

d

Proof. Let p : E ā’ R be a smooth norm. Since p (x)(x) = dt |t=0 p(x + tx) =

d

dt |t=0 (1 + t)p(x) = p(x), we see that p(x) = 1 is a regular equation and hence the

unit sphere S := pā’1 (1) is a smooth submanifold (of codimension 1), see (27.11).

Explicitly, this can be shown as follows: For a ā S let Ī¦ : ker(p (a)) Ć— R+ ā’ E + :=

a+v

{x ā E : p (a)(x) > 0} be given by (v, t) ā’ t p(a+v) . This is well-deļ¬ned, since

p(a + v) ā„ p(a) + p (a)(v) = p(a) = 0 for v ā ker(p (a)). Note that Ī¦(v, t) = y

implies that t = p(y) and v ā ker(p (a)) is such that a + v = Āµ y for some Āµ = 0,

1

i.e. Āµ p (a)(y) = p (a)(a + v) = p (a)(a) = p(a) = 1 and hence v = p (a)(y) y ā’ a.

Thus Ī¦ is a diļ¬eomorphism that maps ker(p (a)) Ć— {1} onto S ā© E + .

x0

Conversely, let x0 ā E \ {0} and a := p(x0 ) . Then a is in the unit sphere, hence

there exists locally around a a diļ¬eomorphism Ī¦ : E ā U ā’ Ī¦(U ) ā E which maps

S ā© U ā’ F ā© Ī¦(U ) for some closed linear subspace F ā U . Let Ī» : E ā’ R be a

continuous linear functional with Ī»(a) = 1 and Ī» ā¤ p. Note that b := Ī¦ (a)(a) = F ,

since otherwise t ā’ Ī¦ā’1 (tb) is in S, but then Ī»(Ī¦ā’1 (tb)) ā¤ 0 and hence 0 =

ā’1

(tb)) = Ī»(Ī¦ (a)ā’1 b) = Ī»(a) = 1 gives a contradiction. Choose Āµ ā E

d

dt |t=0 Ī»(Ī¦

with Āµ|F = 0 and Āµ(b) = 1. We have to show that x ā’ p(x) is C n locally around

1

x0 , or equivalently that this is true for g : x ā’ p(x) . Then g(x) is solution of

the implicit equation Ļ•(x, g(x)) = 0, where Ļ• : E Ć— R ā’ F is given by (x, g) ā’

f (g Ā· x) with f := Āµ ā—¦ Ī¦. This solution is C n by the implicit function theorem,

since ā‚2 Ļ•(x0 , g(x0 )) = f (g(x0 )x0 )(x0 ) = p(x0 ) f (a)(a) = p(x0 ) Āµ(b) = p(x) = 0,

because f is a regular equation at a.

Although this proof uses the implicit function theorem on Banach spaces we can

do without as the following theorem shows:

13.15. Theorem. Characterization of smooth seminorms. Let E be a con-

venient vector space.

(1) Let p : E ā’ R be a convex function which is smooth on a neighborhood of

pā’1 (1), and assume that U := {x ā E : p(x) < 1} is not empty. Then U is open,

and its boundary ā‚U equals {x : p(x) = 1}, a smooth splitting submanifold of E.

(2) If U is a convex absorbing open subset of E whose boundary is a smooth sub-

manifold of E then the Minkowski functional pU is a smooth sublinear mapping,

and U = {x ā E : pU (x) < 1}.

Proof. (1) The set U is obviously convex and open by (4.5) and (13.1). Let

M := {x : p(x) = 1}. We claim that M = ā‚U . Let x0 ā U and x1 ā M . Since

t ā’ p(x1 + t(x0 ā’ x1 )) is convex it is strictly decreasing in a neighborhood of 0.

Hence, there are points x close to x1 with p(x) < p(x1 ) and such with p(x) ā„ 1,

i.e. x belongs to ā‚U . Conversely, let x ā ā‚U . Since U is open we have p(x1 ) ā„ 1.

Suppose p(x1 ) > 1, then p(x) > 1 locally around x1 , a contradiction to x1 ā ā‚U .

13.15

142 Chapter III. Partitions of unity 13.16

Now we show that M is a smooth splitting submanifold of E, i.e. every point has

a neighborhood, in which M is up to a diļ¬eomorphism a complemented subspace.

Let x1 ā M = ā‚U . We consider the convex mapping t ā’ p(x0 + t(x1 ā’ x0 )). It

is locally around 1 diļ¬erentiable, and its value at 0 is strictly less than that at 1.

Thus, p (x1 )(x1 ā’ x0 ) ā„ p(x1 ) ā’ p(x0 ) > 0, and hence we may replace x0 by some

point on the segment from x0 to x1 closer to x1 , such that p (x0 )(x1 ā’ x0 ) > 0.

Without loss of generality we may assume that x0 = 0. Let U := {x ā E : p (0)x >

0 and p (x1 )x > 0} and V := (U ā’ x1 ) ā© ker p (x1 ) ā ker p (x1 ). A smooth mapping

from the open set U ā E to the open set V Ć— R ā ker p (x1 ) Ć— (p(0), +ā) is given

by x ā’ (tx ā’ x1 , p(x)), where t := p (x11)(x1 ) . This mapping is a diļ¬eomorphism,

p (x )(x)

since for (y, r) ā ker p (x1 ) Ć— R the inverse image is given as t(y + x1 ) where t can be

calculated from r = p(t (y+x1 )). Since t ā’ p(t (y+x1 )) is a diļ¬eomorphism between

the intervals (0, +ā) ā’ (p(0), +ā) this t is uniquely determined. Furthermore, t

depends smoothly on (y, r): Let s ā’ (y(s), r(s)) be a smooth curve, then t(s) is

given by the implicit equation p(t (y(s) + x1 )) = r(s), and by the 2-dimensional

implicit function theorem the solution s ā’ t(s) is smooth.

(2) By general principles pU is a sublinear mapping, and U = {x : pU (x) < 1} since

U is open. Thus it remains to show that pU is smooth on its open carrier. So let c be

a smooth curve in the carrier. By assumption, there is a diļ¬eomorphism v, locally

deļ¬ned on E near an intersection point a of the ray through c(0) with the boundary

ā‚U = {x : p(x) = 1}, such that ā‚U corresponds to a closed linear subspace F ā E.

Since U is convex there is a bounded linear functional Ī» ā E with Ī»(a) = 1 and

U ā {x ā E : Ī»(x) ā¤ 1} by the theorem of Hahn-Banach. Then Ī»(Ta (ā‚U )) = 0

since any smooth curve in ā‚U through a stays inside {x : Ī»(x) ā¤ 1}. Furthermore,

b : ā‚t |1 v(ta) ā F , since otherwise t ā’ v ā’1 (tb) ā ā‚U but ā‚t |1 Ī»(v ā’1 (tb)) = Ī»(a) = 1.

ā‚ ā‚

/

Put f := 1/pU ā—¦ c : R ā’ R. Then f is a solution of the implicit equation (Ī» ā—¦

dv ā’1 (0) ā—¦ v)(f (t)c(t)) = 0 which has a unique smooth solution by the implicit

function theorem in dimension 2 since

ā—¦ dv ā’1 (0) ā—¦ v)(sc(t)) = Ī»dv ā’1 (0)dv(f (t)c(t))c(t) = 0

ā‚

ā‚s |s=f (t) (Ī»

1

for t near 0, since for t = 0 we get Ī»(c(0)) = f (0) . So pU is smooth on its carrier.

13.16. The space c0 (Ī“). For an arbitrary set Ī“ the space c0 (Ī“) is the closure

of all functions on Ī“ with ļ¬nite support in the Banach space ā (Ī“) of globally

bounded functions on Ī“ with the supremum norm. The supremum norm on c0 (Ī“)

is not diļ¬erentiable on its carrier, see (13.12). Nevertheless, it was shown in [Bonic,

Frampton, 1965] that c0 is C ā -regular.

Proposition. Smooth norm on c0 . Due to Kuiper according to [Bonic, Framp-

ton, 1966]. There exists an equivalent norm on c0 (Ī“) which is smooth oļ¬ 0.

Proof. To prove this let h : R ā’ R be an unbounded symmetric smooth convex

function vanishing near 0. Let f : c0 (Ī“) ā’ R be given by f (x) := Ī³āĪ“ h(xĪ³ ).

Locally on c0 (Ī“) the function f is just a ļ¬nite sum, hence f is smooth. In fact let

13.16

13.17 13. Diļ¬erentiability of seminorms 143

h(t) = 0 for |t| ā¤ Ī“. For x ā c0 (Ī“) the set F := {Ī³ : |xĪ³ | ā„ Ī“/2} is ļ¬nite, and for

y ā’ x < Ī“ we have that f (y) = Ī³āF h(yĪ³ ).

The set U := {x : f (x) < 1} is open, and bounded: Let h(t) ā„ 1 for |t| ā„ ā and

f (x) < 1, then h(xĪ³ ) < 1 and thus |xĪ³ | ā¤ ā for all Ī³. The set U is also absolutely

convex: Since h is convex, so is f and hence U . Since h is symmetric, so is f and

hence U .

The boundary ā‚U = f ā’1 (1) is a splitting submanifold of c0 (Ī“) by the implicit

function theorem on Banach spaces, since df (x)x = 0 for x ā ā‚U . In fact df (x)(x) =

Ī³ h (xĪ³ )xĪ³ ā„ 0 and at least for one Ī³ we have h(xĪ³ ) > 0 and thus h (xĪ³ ) = 0.

So by (13.14) the Minkowski functional pU is smooth oļ¬ 0. Obviously, it is an

equivalent norm.

13.17. Proposition. Inheritance properties for diļ¬erentiable norms.

(1) The product of two spaces with C n -norm has again a C n -norm given by

x1 2 + x2 2 . More generally, the 2 -sum of C n -normable

(x1 , x2 ) :=

Banach spaces is C n -normable.

(2) A subspace of a space with a C n -norm has a C n -norm.

(3) [Godefroy, Pelant, et. al., 1988]. If c0 (Ī“) ā’ E ā’ F is a short exact sequence

of Banach spaces, and F has a C k -norm, then E has a C k -norm. See also

(14.12.1) and (16.19).

(4) For a compact space K let K be the set of all accumulation points of K.

The operation K ā’ K has the following properties:

AāBā’A āB

(a)

(A āŖ B) = A āŖ B

(b)

(A Ć— B) = (A Ć— B) āŖ (A Ć— B )

(c)

1

{0} āŖ { n : n ā N} = {0}

(d)

K = ā… ā” K discrete.

(e)

(5) If K is compact and K (Ļ) = ā… then C(K) has an equivalent C ā -norm, see

also (16.20).

Proof. (1) and (2) are obvious.

(4) (a) is obvious, since if {x} is open in B and x ā A, then it is also open in A

in the trace topology, hence A ā© (B \ B ) ā A \ A and hence A = A \ (A \ A ) ā

(A \ A ā© (B \ B )) = A ā© B ā B .

(b) By monotonicity we have ā˜āā™. Conversely let x ā A āŖ B , w.l.o.g. x ā A ,

suppose x ā (A āŖ B) , then {x} is open in A āŖ B and hence {x} = {x} ā© A would

/

be open in A, i.e. x ā A , a contradiction.

/

(c) is obvious, since {(x, y)} is open in A Ć— B ā” {x} is open in A and {y} is open

in B.

(d) and (e) are trivial.

13.17

144 Chapter III. Partitions of unity 13.17

For (3) a construction is used similar to that of Kuiperā™s smooth norm for c0 . Let

Ļ : E ā’ F be the quotient mapping and the quotient norm on F . The dual

sequence 1 (A) ā E ā— ā F ā— splits (just deļ¬ne T : 1 (A) ā’ E ā— by selection of

xā— := T (ea ) ā E ā— with xā— = 1 and xā— |c0 (A) = eva using Hahn Banach). Note

a a a

that for every x ā E and Īµ > 0 the set {Ī± : |xā— (x)| ā„ Ļ(x) + Īµ} is ļ¬nite. In fact,

Ī±

by deļ¬nition of the quotient norm Ļ(x) := sup{ x + y : y ā c0 (Ī“)} there is a

y ā c0 (Ī“) such that x + y ā¤ Ļ(x) + Īµ/2. The set Ī“0 := {Ī± : |yĪ± | ā„ Īµ/2} is

ļ¬nite. For all other Ī± we have

|xā— (x)| ā¤ |xā— (x + y)| + |xā— (y)| ā¤ xā— x + y + |yĪ± | <

Ī± Ī± Ī± Ī±

< 1 ( Ļ(x) + Īµ/2) + Īµ/2 = Ļ(x) + Īµ.

Furthermore, we have

x ā¤ 2 Ļ(x) + sup{|xā— (x)| : Ī±}.

Ī±

In fact,

x = sup{| xā— , x | : xā— ā¤ 1}

ā¤ sup{| T (Ī») + y ā— ā—¦ Ļ, x | : Ī» ā¤ 1, y ā— ā¤ 2}

1

= sup{|xā— (x)| : Ī±} + 2 Ļ(x) ,

Ī±

since xā— = T (Ī») + xā— ā’ T (Ī»), where Ī» := xā— |c0 (Ī“) and hence Ī» 1 ā¤ xā— ā¤ 1,

and |T (Ī»)(x)| ā¤ Ī» 1 sup{|xā— (x)| : Ī±} ā¤ x hence T (Ī») ā¤ Ī» 1 , and y ā— ā—¦ Ļ =

Ī±

ā—

x ā’ T (Ī»). Let denote a norm on F which is smooth and is larger than the

quotient norm. Analogously to (13.16) we deļ¬ne

h(xā— (x)),

f (x) := h(4 Ļ(x) ) a

aāA

where h : R ā’ [0, 1] is smooth, even, 1 for |t| ā¤ 1, 0 for |t| ā„ 2 and concave

on {t : h(t) ā„ 1/2}. Then f is smooth, since if Ļ(x) > 1/2 then the ļ¬rst factor

vanishes locally, and if Ļ(x) < 1 we have that Ī“0 := {Ī± : |xā— (x)| ā„ 1 ā’ Īµ}

Ī±

is ļ¬nite, where Īµ := (1 ā’ Ļ(x) )/2, for y ā’ x < Īµ also |xĪ± (y) ā’ xā— (x)| < Īµ

ā—

Ī±

ā—

and hence |xĪ± (y)| < 1 ā’ Īµ + Īµ = 1 for all Ī± ā Ī“0 . So the product is locally

/

ļ¬nite. The set {x : f (x) > 1 } is open, bounded and absolutely convex and has

2

a smooth boundary {x : f (x) = 1 }. It is symmetric since f is symmetric. It is

2

bounded, since f (x) > 1/2 implies h(4 Ļ(x) ) ā„ 1/2 and h(xā— (x)) ā„ 1/2 for all

a

ā—

a. Thus 4 Ļ(x) ā¤ 2 and |xa (x)| ā¤ 2 and thus x ā¤ 2 Ā· 1/2 + 2 = 3. For the

convexity note that xi ā„ 0, yi ā„ 0, 0 ā¤ t ā¤ 1, i xi ā„ 1/2, i yi ā„ 1/2 imply

i (txi +(1ā’t)yi ) ā„ 1/2, since log is concave. Since all factors of f have to be ā„ 1/2

and h is concave on this set, convexity follows. Since one factor of f (x) = Ī± fĪ± (x)

has to be unequal to 1, the derivative f (x)(x) < 0, since fĪ± (x)(x) ā¤ 0 for all Ī± by

concavity and fĪ± (x)(x) < 0 for all x with fĪ± (x) < 1. So its Minkowski-functional

is an equivalent smooth norm on E.

13.17

13.18 13. Diļ¬erentiability of seminorms 145

Statement (5) follows from (3). First recall that K is the set of accumulation

points of K, i.e. those points x for which every neighborhood meets K \ {x}, i.e. x

is not open. Thus K \ K is discrete. For successor ordinals Ī± = Ī² + 1 one deļ¬nes

K (Ī±) := (K (Ī²) ) and for limit ordinals Ī± as Ī²<Ī± K (Ī²) . For a compact space K the

equality K (Ļ) = ā… implies K (n) = ā… for some n ā Ļ, since K (n) is closed. Now one

shows this by induction. Let E := {f ā C(K) : f |K = 0}. By the Tietze-Urysohn

theorem one has a short exact sequence c0 (K \ K ) ā¼ E ā’ C(K) ā’ C(K ). The

=

equality E = c0 (K \ K0 ) can be seen as follows:

Let f ā C(K) with f |K = 0. Suppose there is some Īµ > 0 such that {x : |f (x)| ā„ Īµ}

is not ļ¬nite. Then there is some accumulation point xā of this set and hence

|f (xā )| ā„ Īµ but xā ā K and so f (xā ) = 0. Conversely let f ā c0 (K \ K ) and

Ė Ė Ė Ė

deļ¬ne f by f |K := 0 and f |K\K = f . Then f is continuous on K \ K , since

Ė

K \ K is discrete. For x ā K we have that f (x) = 0 and for each Īµ > 0 the set

Ė Ė

{y : |f (y)| ā„ Īµ} is ļ¬nite, hence its complement is a neighborhood of x, and f is

continuous at x. So the result follows by induction.

13.18. Results.

(1) We do not know whether the quotient of a C n -normable space is again C n -

normable. Compare however with [Fitzpatrick, 1980].

(2) The statement (13.17.5) is quite sharp, since by [Haydon, 1990] there is

a compact space K with K (Ļ) = {ā} but without a GĖteaux-diļ¬erentiable

a

norm.

(3) [Talagrand, 1986] proved that for every ordinal number Ī³, the compact and

scattered space [0, Ī³] with the order topology is C 1 -normable.

(4) It was shown by [ToruĀ“czyk, 1981] that two Banach spaces are homeomor-

n

phic if and only if their density number is the same. Hence, one can view Ba-

nach spaces as exotic (diļ¬erentiable or linear) structures on Hilbert spaces.

If two Banach spaces are even C 1 -diļ¬eomorphic then the diļ¬erential (at 0)

gives a continuous linear homeomorphism. It was for some time unknown

if also uniformly homeomorphic (or at least Lipschitz homeomorphic) Ba-

nach spaces are already linearly homeomorphic. By [Enļ¬‚o, 1970] a Banach

space which is uniformly homeomorphic to a Hilbert space is linearly home-

omorphic to it. A counter-example to the general statement was given by

[Aharoni, Lindenstrauss, 1978], and another one is due to [Ciesielski, Pol,

1984]: There exists a short exact sequence c0 (Ī“1 ) ā’ C(K) ā’ c0 (Ī“2 ) where

C(K) cannot be continuously injected into some c0 (Ī“) but is Lipschitz equiv-

alent to c0 (Ī“). For these and similar questions see [Tzafriri, 1980].

(5) A space all of whose closed subspaces are complemented is a Hilbert space,

[Lindenstrauss, Tzafriri, 1971].

(6) [Enļ¬‚o, Lindenstrauss, Pisier, 1975] There exists a Banach space E not iso-

morphic to a Hilbert space and a short exact sequence 2 ā’ E ā’ 2 .

(7) [Bonic, Reis, 1966]. If the norm of a Banach space and its dual norm are

C 2 then the space is a Hilbert space.

(8) [Deville, Godefroy, Zizler, 1990]. This yields also an example that existence

13.18

146 Chapter III. Partitions of unity 13.19

of smooth norms is not a three-space property, cf. (14.12).

Notes. (2) Note that K \ K is discrete, open and dense in K. So we get for

(n) (n+1)

every n ā N by induction a space Kn with Kn = ā… and Kn = ā…. In fact

(A Ć— B)(n) = i+j=n A(i) Ć— B (j) . Next consider the 1-point compactiļ¬cation Kā

of the locally compact space nāN Kn . Then Kā = {ā} āŖ nāN Kn . In fact

every neighborhood of {ā} contains all but ļ¬nitely many of the Kn , thus we

(n) (i)

have ā. The obvious relation is clear. Hence Kā = {ā} āŖ iā„n Kn . And

(Ļ) (n)

Kā = n<Ļ Kā = {ā} = ā…. The space of [Haydon, 1990] is the one-point

Ī±

compactiļ¬cation of a locally compact space L given as follows: L := Ī±<Ļ1 Ļ1 , i.e.

the space of functions Ļ1 ā’ Ļ1 , which are deļ¬ned on some countable ordinal. It is

ordered by restriction, i.e. s t :ā” dom s ā dom t and t|dom s = s.

(3) The order topology on X := [0, Ī³] has the sets {x : x < a} and {x : x > a}

as basis. In particular open intervals (a, b) := {x : a < x < b} are open. It is

compact, since every subset has a greatest lower bound. In fact let U on X be a

covering. Consider S := {x ā X : [inf X, x) is covered by ļ¬nitely many U ā U}.

Let sā := sup S. Note that x ā S implies that [inf X, x] is covered by ļ¬nitely many

sets in U. We have that sā ā S, since there is an U ā U with sā ā U . Then there

is an x with sā ā (x, sā ] ā U , hence [inf X, x] is covered by ļ¬nitely many sets in

U since there is an s ā S with x < s, so [inf X, sā ] = [inf X, x] āŖ (x, sā ] is covered

by ļ¬nitely many sets, i.e. sā ā S.

The space X is scattered, i.e. X (Ī±) = ā… for some ordinal Ī±. For this we have to show

that every closed non-empty subset K ā X has open points. For every subset K

of X there is a minimum min K ā K, hence [inf X, min K + 1) ā© K = {min K} is

open in K.

For Ī³ equal to the ļ¬rst inļ¬nite ordinal Ļ we have [0, Ī³] = Nā , the one-point

compactiļ¬cation of the discrete space N. Thus C([0, Ī³]) ā¼ c0 Ć— R and the result

=

follows in this case from (13.16).

(5) For splitting short exact sequences the result analogous to (13.17.3) is by

(13.17.1) obviously true. By (5) there are non-splitting exact sequences 0 ā’ F ā’

E ā’ E/F ā’ 0 for every Banach space which is not Hilbertizable.

(8) By (6) there is a sort exact sequence with hilbertizable ends, but with middle

term E not hilbertizable. So neither the sequence nor the dualized sequence splits.

If E and E would have a C 2 -norm then E would be hilbertizable by (7).

13.19. Proposition. Let E be a Banach space, x = 1. Then the following

statements are equivalent:

(1) The norm is FrĀ“chet diļ¬erentiable at x;

e

(2) The following two equivalent conditions hold:

x+h + xā’h ā’2 x

lim = 0,

h

hā’0

x + th + x ā’ th ā’ 2 x

= 0 uniformly in h ā¤ 1;

lim

t

tā’0

13.19

13.20 13. Diļ¬erentiability of seminorms 147

ā— ā— ā— ā— ā— ā—

yn = 1, zn = 1, yn (x) ā’ 1, zn (x) ā’ 1 ā’ yn ā’ zn ā’ 0.

(3)

Proof. (1)ā’(2) This is obvious, since for the derivative of the norm at x we have

limhā’0 xĀ±h ā’ h ā’l(Ā±h) = 0 and adding these equations gives (2).

x

x+th ā’ x

(2) ā’ (1) Since (h) := limt always exists, and since

0 t

x + th + x ā’ th ā’ 2 x x + th ā’ x x + t(ā’h) ā’ x

= +

t t t

ā„ l(h) + l(ā’h) ā„ 0

xĀ±th ā’ x

ā’ (Ā±h) ā„ 0, so the

we have (ā’h) = (h), thus is linear. Moreover t

limit is uniform for h ā¤ 1.

(2) ā’ (3) By (2) we have that for Īµ > 0 there exists a Ī“ such that x+h + xā’h ā¤

ā— ā—

2 + Īµ h for all h < Ī“. For yn = 1 and zn = 1 we have

ā— ā—

yn (x + h) + zn (x ā’ h) ā¤ x + h + x ā’ h .

ā— ā—

Since yn (x) ā’ 1 and zn (x) ā’ 1 we get for large n that

ā— ā— ā— ā—

(yn ā’ zn )(h) ā¤ 2 ā’ yn (x) ā’ zn (x) + Īµ h ā¤ 2ĪµĪ“,

ā— ā— ā— ā—

hence yn ā’ zn ā¤ 2Īµ, i.e. zn ā’ yn ā’ 0.

(3) ā’ (2) Otherwise, there exists an Īµ > 0 and 0 = hn ā’ 0, such that

x + hn + x ā’ hn ā„ 2 + Īµ hn .

ā— ā—

Now choose yn = 1 and zn = 1 with

1 1

ā— ā—

yn (x + hn ) ā„ x + hn ā’ hn and zn (x ā’ hn ) ā„ x ā’ hn ā’ hn .

n n

ā— ā— ā— ā—

Then yn (x) = yn (x + hn ) ā’ yn (hn ) ā’ 1 and similarly zn (x) ā’ 1. Furthermore

2

ā— ā—

yn (x + hn ) + zn (x ā’ hn ) ā„ 2 + (Īµ ā’ ) hn ,

n

hence

2 2

ā— ā— ā— ā—

(yn ā’ zn )(hn ) ā„ 2 + (Īµ ā’ ) hn ā’ (yn + zn )(x) ā„ (Īµ ā’ ) hn ,

n n

ā— ā— 2

thus yn ā’ zn ā„ Īµ ā’ n , a contradiction.

13.20. Proposition. FrĀ“chet diļ¬erentiable norms via locally uniformly

e

rotund duals. [Lovaglia, 1955] If the dual norm of a Banach space E is locally

uniformly rotund on E then the norm is FrĀ“chet diļ¬erentiable on E.

e

A norm is called locally uniformly rotund if xn ā’ x and x + xn ā’ 2 x

implies xn ā’ x. This is equivalent to 2( x 2 + xn 2 ) ā’ x + xn 2 ā’ 0 implies

xn ā’ x, since

2

+ 2 xn 2 ) ā’ x + xn 2 2 2

ā’ ( x + xn )2 = ( x ā’ xn )2 .

ā„2 x

2( x + 2 xn

ā— ā— ā— ā—

Proof. We use (13.19), so let x = 1, yn = 1, zn = 1, yn (x) ā’ 1, zn (x) ā’ 1.

Let xā— = 1 with xā— (x) = 1. Then 2 ā„ xā— + yn ā„ (xā— + yn )(x) ā’ 2. Since

ā— ā—

ā— ā—

E is locally uniformly rotund we get yn ā’ x and similarly zn ā’ z, hence

ā— ā—

yn ā’ zn ā’ 0.

13.20

148 Chapter III. Partitions of unity 13.22

13.21. Remarks on locally uniformly rotund spaces. By [Kadec, 1959] and

[Kadec, 1961] every separable Banach space is isomorphic to a locally uniformly

rotund Banach space. By [Day, 1955] the space ā (Ī“) is not isomorphic to a locally

uniformly rotund Banach space. Every Banach space admitting a continuous linear

injection into some c0 (Ī“) is locally uniformly rotund renormable, see [Troyanski,

1971]. By (53.21) every WCG-Banach space has such an injection, which is due

to [Amir, Lindenstrauss, 1968]. By [Troyanski, 1968] every Banach space with

unconditional basis (see [Jarchow, 1981, 14.7]) is isomorphic to a locally uniformly

rotund Banach space.

In particular, it follows from these results that every reļ¬‚exive Banach space has an

equivalent FrĀ“chet diļ¬erentiable norm. In particular Lp has a FrĀ“chet diļ¬erentiable

e e

norm for 1 < p < ā and in fact the p-norm is itself FrĀ“chet diļ¬erentiable, see

e

(13.13).

13.22. Proposition. If E is separable then E admits an equivalent norm, whose

dual norm is locally uniform rotund.

Proof. Let E be separable. Then there exists a bounded linear operator T : E ā’

such that T ā— (( 2 ) ) is dense in E (and obviously T ā— is weakā— -continuous):

2

Take a dense subset {xā— : i ā N} ā E of {xā— ā E : xā— ā¤ 1} with xā— ā¤ 1.

i i

2

Deļ¬ne T : E ā’ by

ńņš. 6 |