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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,

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