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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 —¦ g with a smooth curve c, but then g —¦ c is a Lipk -
curve, thus it is su¬cient to show that the composition of a Lipk curve c : R ’ U ⊆
E with a Lipk -mapping f : U ’ F is again Lipk , and that (f —¦c) (t) = df (c(t), c (t)).
This follows by induction on k for k ≥ 1 in the same way as we proved theorem
(12.8.2) ’ (12.8.1), using theorem (12.8) itself.

12.10. De¬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 ¤ j ¤
k + 1) c ’ δ j c from Lipk (R, F ) into the space of all F -valued maps in j + 1
pairwise di¬erent real variables (t0 , . . . , tj ) which are bounded on bounded
subsets, with the c∞ -complete locally convex topology of uniform convergence
on bounded subsets. In fact, the mappings δ 0 and δ k+1 are su¬cient.
(2) The initial structure with respect to the k + 2 linear mappings (for 0 ¤ j ¤
k +1) c ’ δeq c from Lipk (R, F ) into the space of all maps from R—(R\{0})
j

into F which are bounded on bounded subsets, with the c∞ -complete locally
convex topology of uniform convergence on bounded subsets. In fact, the
0 k+1
mappings δeq and δeq are su¬cient.
(3) The initial structure with respect to the derivatives of order j ¤ 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 di¬erence quotient on bounded subsets of {(t, s) ∈ R2 : t = s}.
The convenient vector space Lipk (R, F ) satis¬es the uniform boundedness principle
with respect to the point evaluations.

Proof. All three structures describe closed embeddings into ¬nite products of
spaces, which in (1) and (2) are obviously c∞ -complete. For (3) this follows, since
by (1) the structure on Lip0 (R, E) is convenient.

12.10
12.13 12. Di¬erentiability of ¬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 — for all ∈ 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. De¬nition and Proposition. Let E and F be convenient vector spaces
and U ⊆ E be c∞ -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 c— : Lipk (U, F ) ’
Lipk (R, F ) for all c ∈ C ∞ (R, F ).
(2) The initial structure with respect to the linear mappings c— : Lipk (U, F ) ’
Lipk (R, F ) for all c ∈ Lipk (R, F ).
This space satis¬es the uniform boundedness principle with respect to the evaluations
evx : Lipk (U, F ) ’ F for all x ∈ 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 ∈ Lipk (R, Lipk (R, F ))
then ( ‚t )p ( ‚s )q f § (t, s) exists if max(p, q) ¤ k. This describes a smaller space than
‚ ‚

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 c∞ -open sets U ⊆ E, V ⊆ F we have

Lipk (U, L(F, G)) ∼ L(F, Lipk (U, G)),
=
Lipk (U, Lipl (V, G)) ∼ Lipl (V, Lipk (U, G)),
=

see [Fr¨licher, Kriegl, 1988, 4.4.5, 4.5.1, 4.5.2]. For a mapping f : U — F ’ G which
o
is linear in F we have: f ∈ Lipk (U — F, G) if and only if f ∨ ∈ Lipk (U, L(E, F )),
see [Fr¨licher, Kriegl, 1988, 4.3.5]. The last property fails if we weaken Lipschitz to
o
continuous, see the following example.
1
12.13. Smolyanov™s Example. Let f : 2 ’ R be de¬ned by f := k≥1 k2 fk ,
where fk (x) := •(k(kxk ’ 1)) · j<k •(jxj ) and • : R ’ [0, 1] is smooth with
•(0) = 1 and •(t) = 0 for |t| ≥ 1 . We shall show that
4
2
’ R is Fr´chet di¬erentiable.
(1) f : e
2
’ ( 2 ) is not continuous.
(2) f :
2
— 2 ’ R is continuous.
(3) f :

12.13
126 Chapter III. Partitions of unity 12.13

1
2 2
Proof. Let A := {x ∈ : |kxk | ¤ for all k}. This is a closed subset of .
4
2
(1) Remark that for x ∈ at most one fk (x) can be unequal to 0. In fact fk (x) = 0
1 1 3
implies that |kxk ’ 1| ¤ 4k ¤ 4 , and hence kxk ≥ 4 and thus fj (x) = 0 for j > k.
For x ∈ A there exists a k > 0 with |kxk | > 1 and the set of points satisfying this
/ 4
condition is open. It follows that •(kxk ) = 0 and hence f = j<k j1 fj is smooth
2

on this open set.
On the other hand let x ∈ A. Then |kxk ’ 1| ≥ 4 > 1 and hence •(k(kxk ’ 1)) = 0
3
4
2
for all k and thus f (x) = 0. Let v ∈ 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 )’1| < 4k ¤ 4 . Since |kxk | ¤ 4 we conclude |kvk | ≥ 1’|k(xk +vk )’1|’
|kxk | ≥ 1 ’ 1 ’ 1 = 2 . Hence |f (x + v)| = k2 |fk (x + v)| ¤ k2 ¤ (2|vk |)2 ¤ 4 v 2 .
1 1 1
4 4
Thus f (x+v)’0’0 ¤ 4 v ’ 0 for v ’ 0, i.e. f is Fr´chet di¬erentiable at x
e
v
with derivative 0.
d1
(2) If fact take a ∈ R with • (a) = 0. Then f (t ek )(ek ) = dt k2 fk (t ek ) =
d1 1a
2
dt k2 •(k t ’ k) = • (k (k t ’ 1)) = • (a) if t = tk := k k + 1 , which goes to
0 for k ’ ∞. However f (0)(ek ) = 0 since 0 ∈ A.
(3) We have to show that f (xn )(v n ) ’ f (x)(v) for (xn , v n ) ’ (x, v). For x ∈ A
/
this is obviously satis¬ed, since then there exists a k with |kxk | > 1 and hence
4
1
f = j¤k j 2 fj locally around x.
If x ∈ A then f (x) = 0 and thus it remains to consider the case, where xn ∈ A. /
Let µ > 0 be given. Locally around xn at most one summand fk does not vanish:
If xn ∈ A then there is some k with |kxk | > 1/4 which we may choose minimal.
/
Thus |jxj | ¤ 1/4 for all j < k, so |j(jxj ’ 1)| ≥ 3j/4 and hence fj = 0 locally since
the ¬rst factor vanishes. For j > k we get fj = 0 locally since the second factor
vanishes. Thus we can evaluate the derivative:
1 •∞ 2n
|f (xn )(v n )| = 2 fk (xn )(v n ) ¤ n
k |vk | + j|vj | .
2
k k
j<k

Since v ∈ 2 we ¬nd a K1 such that ( j≥K1 |vj |2 )1/2 ¤ 2 •µ ∞ . Thus we conclude
from v n ’v 2 ’ 0 that |vj | ¤ • µ ∞ for j ≥ K1 and large n. For the ¬nitely many
n

small n we can increase K1 such that for these n and j ≥ K1 also |vj | ¤ • µ ∞ .
n

Furthermore there is a constant K2 ≥ 1 such that v n ∞ ¤ v n 2 ¤ K2 for all n.
Now choose N ≥ K1 so large that N 2 ≥ 1 • ∞ K2 K1 . Obviously n<N n2 fn is 1
2
µ
smooth. So it remains to consider those n for which the non-vanishing term has
index k ≥ N . For those terms we have
1 1
|f (xn )(v n )| = 2 fk (xn )(v n ) ¤ • ∞ |vk | + 2
n n
j|vj |
k k
j<k
1 1
n n n
¤ |vk | • j|vj | + j|vj | •
∞+ • ∞2 ∞
k2
k
K1 ¤j<k
j<K1
2
K1 n 1
¤µ+ • ∞ 2 v j µ ¤ µ + µ + µ = 3µ
∞+
k2
N
K1 ¤j<k
This shows the continuity.

12.13
127

13. Di¬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 ’ R on a
vector space E factors over Ep := E/ ker p and gives a norm on this space. Hence, it
˜˜ ˜
can be extended to a norm p : Ep ’ 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 ’ Ep is continuous. Thus, smoothness of
p o¬ 0 implies smoothness of p on its carrier, and so the case where E is a Banach
˜
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
di¬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 ’ R on a vector
space E the following statements are equivalent (see for example [Fr¨licher, Kriegl,
o
1988, p. 199]):
n n
(1) The function f is convex, i.e. f ( i=1 »i xi ) ¤ i=1 »i f (xi ) for »i ≥ 0
n
with i=1 »i = 1;
(2) The set Uf := {(x, t) ∈ E — R : f (x) ¤ t} is convex;
(3) The set Af := {(x, t) ∈ E — 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 c∞ -topology on a convenient vector space) and
any convex function f : E ’ R the following statements are equivalent:
(1) The function f is continuous;
(2) The set Af is open in E — R;
(3) The set f<t := {x ∈ E : f (x) < t} is open in E for all t ∈ R.

13.2. Result. Convex Lipschitz functions. Let f : E ’ 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 c∞ -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 [Fr¨licher, Kriegl,
o
1988, p. 200], for convenient vector spaces.

13.3. Basic de¬nitions. Let f : E ⊇ U ’ F be a mapping de¬ned on a c∞ -open
subset of a convenient vector space E with values in another one F . Let x ∈ U
and v ∈ E. Then the (one sided) directional derivative of f at x in direction v is
de¬ned as
f (x + t v) ’ 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 ∈ E the mapping v ’ 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
Gˆteaux-di¬erentiable at x, if the directional derivatives f (x)(v) exist for all v ∈ E
a
and v ’ f (x)(v) is a bounded linear mapping from E ’ F .
Even for Gˆteaux-di¬erentiable mappings the di¬erence quotient f (x+t v)’f (x) need
a t
not converge uniformly for v in bounded sets (or even in compact sets). Hence, one
de¬nes f to be Fr´chet-di¬erentiable at x if f is Gˆteaux-di¬erentiable at x and
e a
f (x+t v)’f (x)
’ f (x)(v) ’ 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 ’ F such that
f (x + v) ’ f (x) ’ f (x)(v)
lim = 0.
v
v’0

If f : E ⊇ U ’ F is Gˆteaux-di¬erentiable and the derivative f : E ⊇ U ’
a
L(E, F ) is continuous, then f is Fr´chet-di¬erentiable, and we will call such a
e
function C 1 . In fact, the fundamental theorem applied to t ’ f (x + t v) gives us
1
f (x + v) ’ f (x) = f (x + t v)(v) dt,
0
and hence
1
f (x + s v) ’ f (x)
’ f (x)(v) = f (x + t s v) ’ f (x) (v) dt ’ 0,
s 0
which converges to 0 for s ’ 0 uniformly for v in any bounded set, since f (x +
t s v) ’ f (x) uniformly on bounded sets for s ’ 0 and uniformly for t ∈ [0, 1] and
v in any bounded set, since f is assumed to be continuous.
Recall furthermore that a mapping f : E ⊇ U ’ F on a Banach space E is called
Lipschitz if
f (x1 ) ’ f (x2 )
: x1 , x2 ∈ U, x1 = x2 is bounded in F.
x1 ’ x2
It is called H¨lder of order 0 < p ¤ 1 if
o
f (x1 ) ’ f (x2 )
: x1 , x2 ∈ U, x1 = x2 is bounded in F.
x1 ’ x2 p


13.3
13.4 13. Di¬erentiability of seminorms 129

13.4. Lemma. Gˆteaux-di¬erentiability of convex functions. Every convex
a
function q : E ’ 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 Gˆteaux-di¬erentiable at x if and only if q (x) is
a
an odd function, i.e. q (x)(’v) = ’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 ’ )x + (x + t v) ¤ (1 ’ ) q(x) + q(x + t v).
t t t t

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

q(x + t v) ’ q(x)
lim
t
t 0

exists.
As a derivative q (x) automatically satis¬es q (x)(t v) = t q (x)(v) for all t ≥ 0. The
derivative q (x) is convex as limit of the convex functions v ’ q(x+tv)’q(x) . Hence
t
it is sublinear.
The convexity of q implies that

q(x) ’ q(x ’ v) ¤ q (x)(v) ¤ q(x + v) ’ 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) ¤ (a + b) + (’b) = (a + b) ’ (b)
and hence the converse triangle inequality.

Remark. If q is a seminorm, then q(x+tv)’q(x) ¤ q(x)+t q(v)’q(x)
= q(v), hence
t t
q(x+t x)’q(x)
q (x)(v) ¤ 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) ¤ 1} = 1.

Convention. Let q = 0 be a seminorm and let q(x) = 0. Then there exists a
v ∈ E with q(v) = 0, and we have q(x + tv) = |t| q(v), hence q (x)(±v) = q(v). So q
is not Gˆteaux di¬erentiable at x. Therefore, we call a seminorm smooth for some
a
di¬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. Di¬erentiability properties of convex functions f can be translated in geo-
metric properties of Af :

Lemma. Di¬erentiability of convex functions. Let f : E ’ R be a contin-
uous convex function on a Banach space E, and let x0 ∈ E. Then the following
statements are equivalent:
(1) The function f is Gˆteaux (Fr´chet) di¬erentiable at x0 ;
a e
(2) There exists a unique ∈ E with

(v) ¤ f (x0 + v) ’ f (x0 ) for all v ∈ E;

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

An element x— ∈ E — is said to expose a subset K ⊆ E if there exists a unique point
k0 ∈ K with x— (k0 ) = sup{x— (k) : k ∈ K}. It is said to strongly expose K, if
satis¬es in addition that x— (xn ) ’ x— (k0 ) implies xn ’ k0 .
By an a¬ne hyperplane H tangent to a convex set K at a point x ∈ K we mean
that x ∈ 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
Gˆteaux-di¬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) ’ (2) by the following arguments: We have f (x0 )(v) ¤ f (x0 +v)’f (x0 ),
and (v) ¤ f (x0 + v) ’ f (x0 ) implies (v) ¤ f (x0 +t v)’f (x0 ) , and hence (v) ¤
t
f (x)(v).
(2) ’ (1) The uniqueness of implies f (x0 ) = since otherwise we had a linear
functional µ = 0 with u ¤ f (x) ’ . Then µ + contradicts uniqueness.
(2) ” (3) Any hyperplane tangent to Af at (x0 , f (x0 )) is described by a functional
( , s) ∈ E — R such that (x) + s t ≥ (x0 ) + s f (x0 ) for all t ≥ f (x). Note that
the scalar s cannot be 0, since this would imply that (x) ≥ (x0 ) for all x. It has
to be positive, since otherwise the left side would go to ’∞ for f (x) ¤ t ’ +∞.
Without loss of generality we may thus assume that s = 1, so the linear functional
is uniquely determined by the hyperplane. Moreover, (x ’ x0 ) ≥ f (x0 ) ’ f (x) or,
by replacing by ’ , we have (x0 + v) ≥ f (x0 ) + (v) for all v ∈ E.

13.5
13.6 13. Di¬erentiability of seminorms 131

(3) ” (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) ¤ 1} has exactly one tangent hyperplane at x. Applying this to
the Minkowski-functional p of Af gives the desired result.
(5) ” (6) We show this ¬rst for Gˆteaux-di¬erentiability. We have to show that
a
there is a unique tangent hyperplane to x0 ∈ K := {x : f (x) ¤ 1} if and only if
x0 exposes K o := { ∈ E — : (x) ¤ 1 for all x ∈ K}. Let us assume 0 ∈ K and
0 = x0 ∈ ‚K. Then a tangent hyperplane to K at x0 is uniquely determined by
a linear functional ∈ E — with (x0 ) = 1 and (x) ¤ 1 for all x ∈ K. This is
equivalent to ∈ K o and (x0 ) = 1, since by Hahn-Banach there exists an ∈ K o
with (x0 ) = 1. From this the result follows.
This shows also (7) ” (8) ” (9) for Gˆteaux-di¬erentiability.
a
In order to show the statements for Fr´chet-di¬erentiability one has to show that
e
= f (x) is a Fr´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 — G ’ R be a dual pairing.
Let ,
(1) For f : F ’ R ∪ {+∞}, f = +∞ one de¬nes the dual function

f — : G ’ R ∪ {+∞}, f — (z) := sup{ z, y ’ f (y) : y ∈ F }.

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


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


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


(10) If E = F = R and f ≥ 0 with f (0) = 0, then f — (t) = sup{ts ’ f (s) : t ≥ 0}
for t ≥ 0.
(11) If γ ≥ 0 is convex and γ(t) ’ 0, then γ(t) > 0 for t > 0.
t
(12) Let (F, G) be a Banach space and its dual. If γ ≥ 0 is convex and γ(0) = 0,
and f (y) := γ( y ), then f — (z) = γ — ( z ).
(13) A convex function f on a Banach space is Fr´chet di¬erentiable at a with
e
derivative b := f (a) if and only if there exists a convex non-negative func-
tion γ, with γ(0) = 0 and limt’0 γ(t) = 0, such that
t

f (a + h) ¤ f (a) + f (a), h + γ( h ).


13.6
132 Chapter III. Partitions of unity 13.6

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

∀z : f — (z) ¤ g(z) ” ∀z, y : z, y ’ f (y) ¤ g(z)
” ∀z, y : z, y ’ g(z) ¤ f (y)
” ∀y : g — (y) ¤ f (y).

(5) Since (f — )— is convex and lower semi-continuous, this is true for f provided
f = (f — )— . Conversely, let g(b) = ’a and g(z) = +∞ otherwise. Then g — (y) =
¤ f ” f — (b) ¤ ’a. If f is
sup{ z, y ’ g(z) : z ∈ 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 — (b) ¤ ’a. Hence,
functionals a + b,
f —— (y) = sup{ z, y ’ f — (z) : z ∈ G} ≥ b, y + a and thus f = f —— .
(6) Let f (a+y) ≥ f (a)+ b, y . Then f — (b) = sup{ b, y ’f (y) : y ∈ F } = sup{ b, a+
v ’ f (a + v) : v ∈ F } ¤ sup{ b, a + b, v ’ f (a) ’ b, v : v ∈ F } = b, a ’ f (a).
(7) Let f1 (y) = f (y ’ a). Then

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

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

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




(9) Let f1 (y) = f (y) + b, y . Then

f1 (z) = sup{ z, y ’ f (y) ’ b, y : y ∈ F }
= sup{ z ’ b, y ’ f (y) : y ∈ F } = f — (z ’ b).

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

f — (s) = sup{s t ’ f (t) : t ∈ R} = sup{s t ’ f (t) : t ≥ 0}.

γ(t) γ(t)
(11) Let γ ≥ 0 with limt = 0, and let s > 0. Then there are t with s > t,
0 t
and hence
γ(t)
γ — (s) = sup{st ’ γ(t) : t ≥ 0} = sup{t(s ’ ) : t ≥ 0} > 0.
t
13.6
13.7 13. Di¬erentiability of seminorms 133

(12) Let f (y) = γ( y ). Then

f — (z) = sup{ z, y ’ γ( y ) : y ∈ F }
= sup{t z, y ’ γ(t) : y = 1, t ≥ 0}
= sup{sup{t z, y ’ γ(t) : y = 1}, t ≥ 0}
= sup{t z ’ γ(t) : y = 1, t ≥ 0}
= γ — ( z ).

(13) If f (a + h) ¤ f (a) + b, h + γ( h ), then we have

f (a + t h) ’ f (a) γ(t h )
¤ b, h + ,
t t
hence f (a)(h) ¤ b, h . Since h ’ 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) ’ f (a)
≥ b, h = f (a)(h).
t
So f is Fr´chet-di¬erentiable at a with derivative f (a)(h) = b, h , since the re-
e
mainder is bounded by γ( h ) which satis¬es γ( h ) ’ 0 for h ’ 0.
h


Conversely, assume that f is Fr´chet-di¬erentiable at a with derivative b. Then
e

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

and by convexity
g(h) := f (a + h) ’ f (a) ’ b, h ≥ 0.
Let γ(t) := sup{g(u) : u = |t|}. Since g is convex γ is convex, and obviously
γ(t) ∈ [0, +∞], γ(0) = 0 and γ(t) ’ 0 for t ’ 0. This is the required function.
t

13.7. Proposition. Continuity of the Fr´chet derivative. [Asplund, 1968].
e
The di¬erential f of any continuous convex function f on a Banach space is con-
tinuous on the set of all points where f is Fr´chet di¬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
≥ f (x) + f (x)(v). Suppose xn ’ x are points where f is Fr´chet
that f (x + v) e
Then we obtain f (xn )(v) ¤ f (xn + v) ’ f (xn ) which is bounded in
di¬erentiable.
n. Hence, the f (xn ) form a bounded sequence. We get

f (x) ≥ f (xn ), x ’ f — (f (xn )) since f (y) + f — (z) ≥ z, y
since f — (f (z)) + f (z) = f (z)(z)
= f (xn ), x + f (xn ) ’ f (xn ), xn
≥ f (xn ), x ’ xn + f (x) + f (x), xn ’ x since f (x + h) ≥ f (x) + f (x)(h)
= f (xn ) ’ f (x), x ’ xn + f (x).

13.7
134 Chapter III. Partitions of unity 13.8

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

lim f (xn ), x ’ f — (f (xn )) = f (x).
n’∞


Since f is convex and Fr´chet-di¬erentiable at a := x with derivative b := f (x),
e
there exists by (13.6.13) a γ with

f (h) ¤ f (a) + b, h ’ a + γ( h ’ a ).

By duality we obtain using (13.6.3)

f — (z) ≥ z, a ’ f (a) + γ — ( z ’ b ).

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

f — (f (xn )) ≥ f (xn ), x ’ f (x) + γ — ( f (xn ) ’ f (x) ).

Hence
γ — ( f (xn ) ’ f (x) ) ¤ f — (f (xn )) ’ f (xn ), x + f (x),

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

13.8. Asplund spaces and generic Fr´chet di¬erentiability. From (13.4)
e
it follows easily that a convex function f : R ’ R is di¬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 di¬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 ’ R is Fr´chet-di¬erentiable on a
e
dense Gδ -subset of E;
(2) Every continuous convex function f : E ’ R is Fr´chet-di¬erentiable on a
e
dense subset of E;
(3) Every locally Lipschitz function f : E ’ R is Fr´chet-di¬erentiable on a
e
dense subset of E;
(4) Every equivalent norm is Fr´chet-di¬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 — has the Radon-Nikodym property;
(8) Every linear mapping E ’ L1 (X, „¦, µ) which is integral is nuclear;
(9) Every closed convex bounded subset of E — is the closed convex hull of its
extremal points;
(10) Every bounded subset of E — is dentable.

13.8
13.8 13. Di¬erentiability of seminorms 135

A Banach space satisfying these equivalent conditions is called Asplund space.
Every Banach space with a Fr´chet di¬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 Fr´chet di¬eren-
e
tiable on a dense subset then it is so on a dense Gδ -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 ’ R with Lipschitz constant L
de¬ned on a convex open set U the function

˜
f (x) := inf{f (y) + L x ’ y : y ∈ U }

is a Lipschitz extension with constant L, and it is convex if f is.
(2) ’ (3) is due to [Preiss, 1990], Every locally Lipschitz function on an Asplund
space is Fr´chet di¬erentiable at points of a dense subset.
e
(3) ’ (2) follows from the fact that continuous convex functions are locally Lip-
schitz, see (13.2).
(2) ” (4) is mentioned in [Preiss, 1990] without any proof or reference.
(2) ” (10) is due to [Stegall, 1975]. A subset D of a Banach space is called dentable,
if and only if for every x ∈ D there exists an µ > 0 such that x is not in the closed
convex hull of {y ∈ D : y ’ x ≥ µ}.
(2) ” (5) is due to [John, Zizler, 1978]. A norm p is called rough, see also (13.23),
if and only if there exists an µ > 0 such that arbitrary close to each x ∈ X there
are points xi and u with u = 1 such that |p (x2 )(u) ’ p (x1 )(u)| ≥ µ. 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 w— -dentable. The unit ball is dentable if and only if
the dual norm is not rough.
(2) ” (6) is due to [Stegall, 1975].
(2) ” (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 ¬nite measure space
(„¦, Σ, µ) every µ-continuous countably additive function m : Σ ’ E of ¬nite vari-
ation with average range { m(A) : S ∈ Σ, µ(S) > 0} contained in K is representable
µ(S)
by a Bochner integrable function, i.e. there exists a Borel-measurable essentially
separably valued function f : „¦ ’ E with m(S) = S f dµ. 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) ” (8) can be found in [Stegall, 1975] and is due to [Grothendieck, 1955]. A
linear mapping E ’ F is called integral if and only if it has a factorization

wF w Fu ——
E

u
w L (K, µ)
1
C(K)

for some Radon-measure µ on a compact space K.
A linear mapping E ’ F is called nuclear if and only if there are x— ∈ E — and
n
— —
yn ∈ F such that n xn yn < ∞ and T = n xn — yn .
(2) ” (9) is due to [Stegall, 1981, p.516].

13.9. Results on generic Gˆteaux di¬erentiability of Lipschitz functions.
a
(1) [Mazur, 1933] & [Asplund, 1968] A Banach space E with the property that
every continuous convex function f : E ’ R is Gˆteaux-di¬erentiable on
a
a dense Gδ -subset is called weakly Asplund. Separable Banach spaces are
weakly Asplund.
ˇ
(2) In [Zivkov, 1983] it is mentioned that there are Lipschitz functions on R,
which fail to be di¬erentiable on a dense Gδ -subset.
(3) A Lipschitz function on a separable Banach space is “almost everywhere”
Gˆteaux-di¬erentiable, [Aronszajn, 1976].
a
(4) [Preiss, 1990] If the norm on a Banach space is B-di¬erentiable then every
Lipschitz function is B-di¬erentiable on a dense set. A function f : E ⊇
U ’ F is called B-di¬erentiable at x ∈ 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 ∈ B one has f (x+t v)’f (x) ’ f (x)(v) ’ 0 for t ’ 0
t
uniformly for v ∈ B.
ˇ
(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 Gˆteaux di¬erentiable on a non-meager
a
subset.
ˇ
(6) [Zivkov, 1983]. For every space with Fr´chet di¬erentiable norm any locally
e
Lipschitzian function which has directional derivatives for a dense set of
directions is generically Gˆteaux di¬erentiable.
a
(7) There exists a Lipschitz Gˆteaux di¬erentiable function f : L1 [0, 1] ’ R
a
which is nowhere Fr´chet di¬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 di¬erentiability are: [Phelps, 1989], [Preiss, 1984],
and [Zhivkov, 1987].

13.10. Lemma. Smoothness of 2n-norm. For n ∈ N the 2n-norm is smooth
on L2n \ {0}.

13.10
13.11 13. Di¬erentiability of seminorms 137

Proof. Since t ’ t1/2n is smooth on R+ it is enough to show that x ’ ( x 2n )2n
is smooth. Let p := 2n. Since (x1 , . . . , xn ) ’ x1 · . . . · xn is a n-linear contraction
p 1
from Lp — . . . — Lp ’ L1 by the H¨lder-inequality ( i=1 p = 1) and : L1 ’ R
o
is a linear contraction the mapping x ’ (x, . . . , x) ’ 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 ∈ 1 and j ∈ 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 ’ v 1 is given by p (x)(ei ) = 1 and p (x)(’ei ) = 1, and p is not di¬erentiable
at x. More generally we have:

Lemma. [Mazur, 1933, p.79]. Let “ be some set, and let p be the 1-norm given
1
γ∈“ |xγ | for x ∈ xγ =0 |hγ | +
by x 1 = p(x) := (“). Then p (x)(h) =
xγ =0 hγ sign xγ .

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 ≥ 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 |xγ + hγ | ’ |xγ | = ||xγ | + hγ sign xγ | ’ |xγ | ≥ |xγ | + hγ sign xγ ’ |xγ | =
hγ sign xγ , and is equal to |hγ | if xγ = 0. Summing up these (in)equalities we
obtain
p(x + h) ’ p(x) ’ |hγ | ’ hγ sign xγ ≥ 0.
xγ =0 xγ =0

µ
For µ > 0 choose a ¬nite set F ‚ “, such that γ ∈F |hγ | < 2 . Now choose t so
/
small that
|xγ | + t hγ sign xγ ≥ 0 for all γ ∈ F with xγ = 0.
We claim that

q(x + t h) ’ q(x)
’ |hγ | ’ hγ sign xγ ¤ µ.
t x γ =0 xγ =0


|xγ +t hγ |’|xγ |
= |hγ |, hence these terms cancel
Let ¬rst γ be such that xγ = 0. Then t
with ’ xγ =0 |hγ |.
Let now xγ = 0. For |xγ | + t hγ sign xγ ≥ 0 (hence in particular for γ ∈ F with
xγ = 0) we have

|xγ + t hγ | ’ |xγ | |xγ | + t hγ sign xγ ’ |xγ |
= = hγ sign xγ .
t t

Thus, these terms sum up to the corresponding sum hγ sign xγ .
γ


13.11
138 Chapter III. Partitions of unity 13.12

It remains to consider γ with xγ = 0 and |xγ | + t hγ sign xγ < 0. Then γ ∈ F and
/
|xγ + t hγ | ’ |xγ | ’|xγ | ’ t hγ sign xγ ’ |xγ | ’ t hγ sign xγ
’ hγ sign xγ =
t t
¤ ’2hγ sign xγ ,
µ
|hγ | <
and since these remaining terms sum up to something smaller than
γ ∈F
/ 2
µ.

Remark. The 1-norm is rough. This result shows that the 1-norm is Gˆteaux- a
di¬erentiable exactly at those points, where all coordinates are non-zero. Thus, if
“ is uncountable, the 1-norm is nowhere Gˆteaux-di¬erentiable.
a
In contrast to what is claimed in [Mazur, 1933, p.79], the 1-norm is nowhere Fr´chet
e
di¬erentiable. In fact, take 0 = x ∈ 1 (“). For γ with xγ = 0 and t > 0 we have
that
p(x + t (’ sign xγ eγ )) ’ p(x) ’ t p (x)(’ sign xγ eγ ) =
= |xγ ’ t sign xγ | ’ |xγ | + t = |xγ | ’ t ’ |xγ | + t ≥ t · 1,

provided t ≥ 2 |xγ |, since then |xγ | ’ t = t ’ |xγ | ≥ |xγ |. Obviously, for every t > 0
there are γ satisfying this required condition; either xγ = 0 then we have a corner,
or xγ = 0 then it gets arbitrarily small. Thus, the directional di¬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 ≥ 0
+1
p(x + t en ) = p(x) + |t|, hence p (x + t en )(en ) = .
’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 ∈ 1 with h = 1 and
|p (x1 )(h) ’ p (x2 )(h)| ≥ 2.

13.12. Derivative of the ∞-norm. On c0 the norm is not di¬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 ≥ 0 and
p(x + th) = |(x + th)(b)| = x for t ¤ 0. Thus, t ’ p(x + th) is not di¬erentiable
at 0 and thus p not at x.
If the norm of x is attained at a single coordinate a, then p is di¬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 ¤ x ’ sup{|x(t)| : t = a}.
Hence the directional di¬erence-quotient converges uniformly for h in the unit-ball.
Let x ∈ C[0, 1] be such that x ∞ = |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 ≥ 0 we have
|(x + t y)(s)| ¤ |x(a) + t y(a)| = (1 + t) x ∞ and hence x + t y ∞ = (1 + t) x ∞ .
For ’1 ¤ t ¤ 0 we have |(x + t y)(s)| ¤ |x(a)| and (x + t y)(b) = x(a) and hence
x + t y ∞ = x ∞ . Thus the directional derivative is given by p (x)(y) = x ∞
and p (x)(’y) = 0. More precisely we have the following results.

13.12
13.12 13. Di¬erentiability of seminorms 139

Lemma. [Banach, 1932, p. 168]. Let T be a compact metric space. Let x ∈
C(T, R) \ {0} and h ∈ C(T, R). By p we denote the ∞-norm x ∞ = p(x) :=
sup{|x(t)| : t ∈ 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 ≥ 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 ∈ T : |x(t)| = p(x)}. For given µ > 0 we ¬nd by the uniform
continuity of x and h a δ1 such that |x(t) ’ x(t )| < 1 and |h(t) ’ h(t )| < µ for
2
dist(t, t ) < δ1 . Then {t : dist(t, A) ≥ δ1 } is closed, hence compact. Therefore
δ := x ∞ ’ sup{|x(t)| : dist(t, A) ≥ δ1 } > 0.
Now we claim that for 0 < t < min{δ, 1} we have
p(x + t h) ’ p(x)
0¤ ’ sup{h(r) sign x(r) : r ∈ A} ¤ µ.
t
For all s ∈ A we have
p(x + t h) ≥ |(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 ¤ t ¤ 1, since |h(s)| ¤ p(h) = p(x). Hence
p(x + t h) ’ p(x)
≥ sup{h(t) sign x(t) : t ∈ 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) ≥ p(x) + t sup{h(r) sign x(r) : r ∈ A}, and hence
|x(s)| ≥ |(x + th)(s)| ’ t |h(s)| ≥ p(x + t h) ’ t p(h)
≥ p(x) ’ t p(h) ’ sup{h(r) sign x(r) : r ∈ A}

¤1
> p(x) ’ δ = sup{|x(r)| : dist(r, A) ≥ δ1 }.
Therefore dist(s, A) < δ1 , and thus there exists an a ∈ A with dist(s, a) < δ1 and
consequently |x(s) ’ x(a)| < 1 and |h(s) ’ h(a)| < µ. In particular, sign x(s) =
2
sign x(a) = 0. So we get
|x(s)| + t h(s) sign x(s) ’ p(x)
p(x + t h) ’ p(x) |(x + t h)(s)| ’ p(x)
= =
t t t
|x(s)| + t h(s) sign x(s) ’ p(x)
¤ h(s) sign x(a)
=
t
¤ |h(s) ’ h(a)| + h(a) sign x(a)
< µ + sup{h(r) sign x(r) : r ∈ A}.

13.12
140 Chapter III. Partitions of unity 13.13

This proves the claim which ¬nally implies

p(x + t h) ’ p(x)
= sup{h(r) sign x(r) : r ∈ A}.
p (x)(v) = lim
t
t 0



Remark. The ∞-norm is rough. This result shows that the points where the
∞-norm is Gˆteaux-di¬erentiable are exactly those x where the supremum p(x) is
a
attained in a single point a. The Gˆteaux-derivative is then given by p (x)(h) =
a
h(a) sign x(a). In general, this is however not the Fr´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 ’ a be
such that 0 < x(an ) < x(a) and let 0 < δn := x(a) ’ x(an ) < x(a). Now choose
sn := 2δn ’ 0 and hn ∈ C[0, 1] with p(hn ) ¤ 1, hn (a) = 0 and hn (an ) := 1 and
p(x + sn hn ) = (x + sn hn )(an ) = x(an ) + 2(x(a) ’ x(an )) = 2x(a) ’ x(an ). For this
choose (x + sn hn )(t) ¤ (x + sn hn )(an ) locally, i.e.. hn (t) ¤ 1 + (x(an ) ’ x(t))/sn
and 0 far away from x. Then p (x)(hn ) = 0 by (13.12) and

p(x + sn hn ) ’ p(x) 2x(a) ’ x(an ) ’ x(a)
’ p (x)(hn ) =
sn sn
δn 1
= ’0
=
2δn 2
Thus the limit is not uniform and p is not Fr´chet di¬erentiable at x.
e
The set of vectors x ∈ 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) = ’x(a), h(b) = x(b)
that
for t ≥ 0
+1
p(x + t h) = (1 + max{t, ’t}) p(x), hence p (x + t h)(h) = .
’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 ∈ C[0, 1] with
h = 1 and |p (x1 )(h) ’ p (x2 )(h)| ≥ 2.

13.13. Results on the di¬erentiability of p-norms. [Bonic, Frampton, 1966,
p.887].
For 1 < p < ∞ not an even integer the function t ’ |t|p is di¬erentiable of order
n if n < p, and the highest derivative (t ’ p (p ’ 1) . . . (p ’ n + 1) |t|p’n ) satis¬es
a H¨lder-condition with modulus p ’ n, one can show that the p-norm has exactly
o
these di¬erentiability properties, i.e.
(1) It is (p ’ 1)-times di¬erentiable with Lipschitzian highest derivative if p is
an integer.
(2) It is [p]-times di¬erentiable with highest derivative being H¨lderian of order
o
p ’ [p], otherwise.
(3) The norm has no higher H¨lder-di¬erentiability properties.
o
That the norm on Lp is C 1 for 1 < p < ∞ was already shown by [Mazur, 1933].


13.13
13.15 13. Di¬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 ’ 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 := p’1 (1) is a smooth submanifold (of codimension 1), see (27.11).
Explicitly, this can be shown as follows: For a ∈ S let ¦ : ker(p (a)) — R+ ’ E + :=
a+v
{x ∈ E : p (a)(x) > 0} be given by (v, t) ’ t p(a+v) . This is well-de¬ned, since
p(a + v) ≥ p(a) + p (a)(v) = p(a) = 0 for v ∈ ker(p (a)). Note that ¦(v, t) = y
implies that t = p(y) and v ∈ ker(p (a)) is such that a + v = µ y for some µ = 0,
1
i.e. µ p (a)(y) = p (a)(a + v) = p (a)(a) = p(a) = 1 and hence v = p (a)(y) y ’ a.
Thus ¦ is a di¬eomorphism that maps ker(p (a)) — {1} onto S © E + .
x0
Conversely, let x0 ∈ E \ {0} and a := p(x0 ) . Then a is in the unit sphere, hence
there exists locally around a a di¬eomorphism ¦ : E ⊇ U ’ ¦(U ) ⊆ E which maps
S © U ’ F © ¦(U ) for some closed linear subspace F ⊆ U . Let » : E ’ R be a
continuous linear functional with »(a) = 1 and » ¤ p. Note that b := ¦ (a)(a) = F ,
since otherwise t ’ ¦’1 (tb) is in S, but then »(¦’1 (tb)) ¤ 0 and hence 0 =
’1
(tb)) = »(¦ (a)’1 b) = »(a) = 1 gives a contradiction. Choose µ ∈ E
d
dt |t=0 »(¦
with µ|F = 0 and µ(b) = 1. We have to show that x ’ p(x) is C n locally around
1
x0 , or equivalently that this is true for g : x ’ p(x) . Then g(x) is solution of
the implicit equation •(x, g(x)) = 0, where • : E — R ’ F is given by (x, g) ’
f (g · x) with f := µ —¦ ¦. This solution is C n by the implicit function theorem,
since ‚2 •(x0 , g(x0 )) = f (g(x0 )x0 )(x0 ) = p(x0 ) f (a)(a) = p(x0 ) µ(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 ’ R be a convex function which is smooth on a neighborhood of
p’1 (1), and assume that U := {x ∈ E : p(x) < 1} is not empty. Then U is open,
and its boundary ‚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 ∈ 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 = ‚U . Let x0 ∈ U and x1 ∈ M . Since
t ’ p(x1 + t(x0 ’ 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) ≥ 1,
i.e. x belongs to ‚U . Conversely, let x ∈ ‚U . Since U is open we have p(x1 ) ≥ 1.
Suppose p(x1 ) > 1, then p(x) > 1 locally around x1 , a contradiction to x1 ∈ ‚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 di¬eomorphism a complemented subspace.
Let x1 ∈ M = ‚U . We consider the convex mapping t ’ p(x0 + t(x1 ’ x0 )). It
is locally around 1 di¬erentiable, and its value at 0 is strictly less than that at 1.
Thus, p (x1 )(x1 ’ x0 ) ≥ p(x1 ) ’ 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 ’ x0 ) > 0.
Without loss of generality we may assume that x0 = 0. Let U := {x ∈ E : p (0)x >
0 and p (x1 )x > 0} and V := (U ’ x1 ) © ker p (x1 ) ⊆ ker p (x1 ). A smooth mapping
from the open set U ⊆ E to the open set V — R ⊆ ker p (x1 ) — (p(0), +∞) is given
by x ’ (tx ’ x1 , p(x)), where t := p (x11)(x1 ) . This mapping is a di¬eomorphism,
p (x )(x)
since for (y, r) ∈ ker p (x1 ) — R the inverse image is given as t(y + x1 ) where t can be
calculated from r = p(t (y+x1 )). Since t ’ p(t (y+x1 )) is a di¬eomorphism between
the intervals (0, +∞) ’ (p(0), +∞) this t is uniquely determined. Furthermore, t
depends smoothly on (y, r): Let s ’ (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 ’ 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 di¬eomorphism v, locally
de¬ned on E near an intersection point a of the ray through c(0) with the boundary
‚U = {x : p(x) = 1}, such that ‚U corresponds to a closed linear subspace F ⊆ E.
Since U is convex there is a bounded linear functional » ∈ E with »(a) = 1 and
U ⊆ {x ∈ E : »(x) ¤ 1} by the theorem of Hahn-Banach. Then »(Ta (‚U )) = 0
since any smooth curve in ‚U through a stays inside {x : »(x) ¤ 1}. Furthermore,
b : ‚t |1 v(ta) ∈ F , since otherwise t ’ v ’1 (tb) ∈ ‚U but ‚t |1 »(v ’1 (tb)) = »(a) = 1.
‚ ‚
/
Put f := 1/pU —¦ c : R ’ R. Then f is a solution of the implicit equation (» —¦
dv ’1 (0) —¦ v)(f (t)c(t)) = 0 which has a unique smooth solution by the implicit
function theorem in dimension 2 since

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

‚s |s=f (t) (»

1
for t near 0, since for t = 0 we get »(c(0)) = f (0) . So pU is smooth on its carrier.

13.16. The space c0 (“). For an arbitrary set “ the space c0 (“) is the closure
of all functions on “ with ¬nite support in the Banach space ∞ (“) of globally
bounded functions on “ with the supremum norm. The supremum norm on c0 (“)
is not di¬erentiable on its carrier, see (13.12). Nevertheless, it was shown in [Bonic,
Frampton, 1965] that c0 is C ∞ -regular.

Proposition. Smooth norm on c0 . Due to Kuiper according to [Bonic, Framp-
ton, 1966]. There exists an equivalent norm on c0 (“) which is smooth o¬ 0.

Proof. To prove this let h : R ’ R be an unbounded symmetric smooth convex
function vanishing near 0. Let f : c0 (“) ’ R be given by f (x) := γ∈“ h(xγ ).
Locally on c0 (“) the function f is just a ¬nite sum, hence f is smooth. In fact let

13.16
13.17 13. Di¬erentiability of seminorms 143

h(t) = 0 for |t| ¤ δ. For x ∈ c0 (“) the set F := {γ : |xγ | ≥ δ/2} is ¬nite, and for
y ’ x < δ we have that f (y) = γ∈F h(yγ ).
The set U := {x : f (x) < 1} is open, and bounded: Let h(t) ≥ 1 for |t| ≥ ∆ and
f (x) < 1, then h(xγ ) < 1 and thus |xγ | ¤ ∆ for all γ. 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 ‚U = f ’1 (1) is a splitting submanifold of c0 (“) by the implicit
function theorem on Banach spaces, since df (x)x = 0 for x ∈ ‚U . In fact df (x)(x) =
γ h (xγ )xγ ≥ 0 and at least for one γ we have h(xγ ) > 0 and thus h (xγ ) = 0.
So by (13.14) the Minkowski functional pU is smooth o¬ 0. Obviously, it is an
equivalent norm.

13.17. Proposition. Inheritance properties for di¬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 (“) ’ E ’ 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 ’ K has the following properties:

A⊆B’A ⊆B
(a)
(A ∪ B) = A ∪ B
(b)
(A — B) = (A — B) ∪ (A — B )
(c)
1
{0} ∪ { n : n ∈ N} = {0}
(d)
K = … ” K discrete.
(e)

(5) If K is compact and K (ω) = … then C(K) has an equivalent C ∞ -norm, see
also (16.20).

Proof. (1) and (2) are obvious.
(4) (a) is obvious, since if {x} is open in B and x ∈ A, then it is also open in A
in the trace topology, hence A © (B \ B ) ⊆ A \ A and hence A = A \ (A \ A ) ⊆
(A \ A © (B \ B )) = A © B ⊆ B .
(b) By monotonicity we have ˜⊇™. Conversely let x ∈ A ∪ B , w.l.o.g. x ∈ A ,
suppose x ∈ (A ∪ B) , then {x} is open in A ∪ B and hence {x} = {x} © A would
/
be open in A, i.e. x ∈ A , a contradiction.
/
(c) is obvious, since {(x, y)} is open in A — B ” {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 Kuiper™s smooth norm for c0 . Let
π : E ’ F be the quotient mapping and the quotient norm on F . The dual
sequence 1 (A) ← E — ← F — splits (just de¬ne T : 1 (A) ’ E — by selection of
x— := T (ea ) ∈ E — with x— = 1 and x— |c0 (A) = eva using Hahn Banach). Note
a a a
that for every x ∈ E and µ > 0 the set {± : |x— (x)| ≥ π(x) + µ} is ¬nite. In fact,
±
by de¬nition of the quotient norm π(x) := sup{ x + y : y ∈ c0 (“)} there is a
y ∈ c0 (“) such that x + y ¤ π(x) + µ/2. The set “0 := {± : |y± | ≥ µ/2} is
¬nite. For all other ± we have

|x— (x)| ¤ |x— (x + y)| + |x— (y)| ¤ x— x + y + |y± | <
± ± ± ±
< 1 ( π(x) + µ/2) + µ/2 = π(x) + µ.

Furthermore, we have

x ¤ 2 π(x) + sup{|x— (x)| : ±}.
±


In fact,

x = sup{| x— , x | : x— ¤ 1}
¤ sup{| T (») + y — —¦ π, x | : » ¤ 1, y — ¤ 2}
1

= sup{|x— (x)| : ±} + 2 π(x) ,
±


since x— = T (») + x— ’ T (»), where » := x— |c0 (“) and hence » 1 ¤ x— ¤ 1,
and |T (»)(x)| ¤ » 1 sup{|x— (x)| : ±} ¤ x hence T (») ¤ » 1 , and y — —¦ π =
±

x ’ T (»). Let denote a norm on F which is smooth and is larger than the
quotient norm. Analogously to (13.16) we de¬ne

h(x— (x)),
f (x) := h(4 π(x) ) a
a∈A


where h : R ’ [0, 1] is smooth, even, 1 for |t| ¤ 1, 0 for |t| ≥ 2 and concave
on {t : h(t) ≥ 1/2}. Then f is smooth, since if π(x) > 1/2 then the ¬rst factor
vanishes locally, and if π(x) < 1 we have that “0 := {± : |x— (x)| ≥ 1 ’ µ}
±
is ¬nite, where µ := (1 ’ π(x) )/2, for y ’ x < µ also |x± (y) ’ x— (x)| < µ

±

and hence |x± (y)| < 1 ’ µ + µ = 1 for all ± ∈ “0 . So the product is locally
/
¬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 π(x) ) ≥ 1/2 and h(x— (x)) ≥ 1/2 for all
a

a. Thus 4 π(x) ¤ 2 and |xa (x)| ¤ 2 and thus x ¤ 2 · 1/2 + 2 = 3. For the
convexity note that xi ≥ 0, yi ≥ 0, 0 ¤ t ¤ 1, i xi ≥ 1/2, i yi ≥ 1/2 imply
i (txi +(1’t)yi ) ≥ 1/2, since log is concave. Since all factors of f have to be ≥ 1/2
and h is concave on this set, convexity follows. Since one factor of f (x) = ± f± (x)
has to be unequal to 1, the derivative f (x)(x) < 0, since f± (x)(x) ¤ 0 for all ± by
concavity and f± (x)(x) < 0 for all x with f± (x) < 1. So its Minkowski-functional
is an equivalent smooth norm on E.

13.17
13.18 13. Di¬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 ± = β + 1 one de¬nes
K (±) := (K (β) ) and for limit ordinals ± as β<± K (β) . For a compact space K the
equality K (ω) = … implies K (n) = … for some n ∈ ω, since K (n) is closed. Now one
shows this by induction. Let E := {f ∈ C(K) : f |K = 0}. By the Tietze-Urysohn
theorem one has a short exact sequence c0 (K \ K ) ∼ E ’ C(K) ’ C(K ). The
=
equality E = c0 (K \ K0 ) can be seen as follows:
Let f ∈ C(K) with f |K = 0. Suppose there is some µ > 0 such that {x : |f (x)| ≥ µ}
is not ¬nite. Then there is some accumulation point x∞ of this set and hence
|f (x∞ )| ≥ µ but x∞ ∈ K and so f (x∞ ) = 0. Conversely let f ∈ c0 (K \ K ) and
˜ ˜ ˜ ˜
de¬ne f by f |K := 0 and f |K\K = f . Then f is continuous on K \ K , since
˜
K \ K is discrete. For x ∈ K we have that f (x) = 0 and for each µ > 0 the set
˜ ˜
{y : |f (y)| ≥ µ} is ¬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 (ω) = {∞} but without a Gˆteaux-di¬erentiable
a
norm.
(3) [Talagrand, 1986] proved that for every ordinal number γ, the compact and
scattered space [0, γ] with the order topology is C 1 -normable.
(4) It was shown by [Toru´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 (di¬erentiable or linear) structures on Hilbert spaces.
If two Banach spaces are even C 1 -di¬eomorphic then the di¬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 [En¬‚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 (“1 ) ’ C(K) ’ c0 (“2 ) where
C(K) cannot be continuously injected into some c0 (“) but is Lipschitz equiv-
alent to c0 (“). 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) [En¬‚o, Lindenstrauss, Pisier, 1975] There exists a Banach space E not iso-
morphic to a Hilbert space and a short exact sequence 2 ’ E ’ 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 ∈ N by induction a space Kn with Kn = … and Kn = …. In fact
(A — B)(n) = i+j=n A(i) — B (j) . Next consider the 1-point compacti¬cation K∞
of the locally compact space n∈N Kn . Then K∞ = {∞} ∪ n∈N Kn . In fact
every neighborhood of {∞} contains all but ¬nitely many of the Kn , thus we
(n) (i)
have ⊇. The obvious relation is clear. Hence K∞ = {∞} ∪ i≥n Kn . And
(ω) (n)
K∞ = n<ω K∞ = {∞} = …. The space of [Haydon, 1990] is the one-point
±
compacti¬cation of a locally compact space L given as follows: L := ±<ω1 ω1 , i.e.
the space of functions ω1 ’ ω1 , which are de¬ned on some countable ordinal. It is
ordered by restriction, i.e. s t :” dom s ⊆ dom t and t|dom s = s.
(3) The order topology on X := [0, γ] 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 ∈ X : [inf X, x) is covered by ¬nitely many U ∈ U}.
Let s∞ := sup S. Note that x ∈ S implies that [inf X, x] is covered by ¬nitely many
sets in U. We have that s∞ ∈ S, since there is an U ∈ U with s∞ ∈ U . Then there
is an x with s∞ ∈ (x, s∞ ] ⊆ U , hence [inf X, x] is covered by ¬nitely many sets in
U since there is an s ∈ S with x < s, so [inf X, s∞ ] = [inf X, x] ∪ (x, s∞ ] is covered
by ¬nitely many sets, i.e. s∞ ∈ S.
The space X is scattered, i.e. X (±) = … for some ordinal ±. For this we have to show
that every closed non-empty subset K ⊆ X has open points. For every subset K
of X there is a minimum min K ∈ K, hence [inf X, min K + 1) © K = {min K} is
open in K.
For γ equal to the ¬rst in¬nite ordinal ω we have [0, γ] = N∞ , the one-point
compacti¬cation of the discrete space N. Thus C([0, γ]) ∼ c0 — 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 ’ F ’
E ’ E/F ’ 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 Fr´chet di¬erentiable at x;
e
(2) The following two equivalent conditions hold:

x+h + x’h ’2 x
lim = 0,
h
h’0

x + th + x ’ th ’ 2 x
= 0 uniformly in h ¤ 1;
lim
t
t’0

13.19
13.20 13. Di¬erentiability of seminorms 147

— — — — — —
yn = 1, zn = 1, yn (x) ’ 1, zn (x) ’ 1 ’ yn ’ zn ’ 0.
(3)
Proof. (1)’(2) This is obvious, since for the derivative of the norm at x we have
limh’0 x±h ’ h ’l(±h) = 0 and adding these equations gives (2).
x

x+th ’ x
(2) ’ (1) Since (h) := limt always exists, and since
0 t
x + th + x ’ th ’ 2 x x + th ’ x x + t(’h) ’ x
= +
t t t
≥ l(h) + l(’h) ≥ 0
x±th ’ x
’ (±h) ≥ 0, so the
we have (’h) = (h), thus is linear. Moreover t
limit is uniform for h ¤ 1.
(2) ’ (3) By (2) we have that for µ > 0 there exists a δ such that x+h + x’h ¤
— —
2 + µ h for all h < δ. For yn = 1 and zn = 1 we have
— —
yn (x + h) + zn (x ’ h) ¤ x + h + x ’ h .
— —
Since yn (x) ’ 1 and zn (x) ’ 1 we get for large n that
— — — —
(yn ’ zn )(h) ¤ 2 ’ yn (x) ’ zn (x) + µ h ¤ 2µδ,
— — — —
hence yn ’ zn ¤ 2µ, i.e. zn ’ yn ’ 0.
(3) ’ (2) Otherwise, there exists an µ > 0 and 0 = hn ’ 0, such that
x + hn + x ’ hn ≥ 2 + µ hn .
— —
Now choose yn = 1 and zn = 1 with
1 1
— —
yn (x + hn ) ≥ x + hn ’ hn and zn (x ’ hn ) ≥ x ’ hn ’ hn .
n n
— — — —
Then yn (x) = yn (x + hn ) ’ yn (hn ) ’ 1 and similarly zn (x) ’ 1. Furthermore
2
— —
yn (x + hn ) + zn (x ’ hn ) ≥ 2 + (µ ’ ) hn ,
n
hence
2 2
— — — —
(yn ’ zn )(hn ) ≥ 2 + (µ ’ ) hn ’ (yn + zn )(x) ≥ (µ ’ ) hn ,
n n
— — 2
thus yn ’ zn ≥ µ ’ n , a contradiction.

13.20. Proposition. Fr´chet di¬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 Fr´chet di¬erentiable on E.
e

A norm is called locally uniformly rotund if xn ’ x and x + xn ’ 2 x
implies xn ’ x. This is equivalent to 2( x 2 + xn 2 ) ’ x + xn 2 ’ 0 implies
xn ’ x, since
2
+ 2 xn 2 ) ’ x + xn 2 2 2
’ ( x + xn )2 = ( x ’ xn )2 .
≥2 x
2( x + 2 xn
— — — —
Proof. We use (13.19), so let x = 1, yn = 1, zn = 1, yn (x) ’ 1, zn (x) ’ 1.
Let x— = 1 with x— (x) = 1. Then 2 ≥ x— + yn ≥ (x— + yn )(x) ’ 2. Since
— —
— —
E is locally uniformly rotund we get yn ’ x and similarly zn ’ z, hence
— —
yn ’ zn ’ 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 ∞ (“) is not isomorphic to a locally
uniformly rotund Banach space. Every Banach space admitting a continuous linear
injection into some c0 (“) 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 re¬‚exive Banach space has an
equivalent Fr´chet di¬erentiable norm. In particular Lp has a Fr´chet di¬erentiable
e e
norm for 1 < p < ∞ and in fact the p-norm is itself Fr´chet di¬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 ’
such that T — (( 2 ) ) is dense in E (and obviously T — is weak— -continuous):
2

Take a dense subset {x— : i ∈ N} ⊆ E of {x— ∈ E : x— ¤ 1} with x— ¤ 1.
i i
2
De¬ne T : E ’ by

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