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n
n

1—
x— := x n.

2n
n=1


We claim that is the required norm.


So we show ¬rst, that it is an equivalent norm. For x— = 1 we have x— n ≥

min{1/(2 n T — ), 1/2}. In fact if y — ≥ 1/(2 T — ) then x— ’T — y — 2 + n y — 2 ≥
1

1/(2n2 T — 2 ) and if y — ¤ 1/(2 T — ) then x— ’ T — y — ≥ x ’ T — y — ≥ 1 ’ 1 =
2

1
2 . Furthermore if we take y := 0 then we see that x n ¤ x . Thus n and
are equivalent norms, and hence also ∞.

Note ¬rst, that a dual norm is the supremum of the weak— (lower semi-)continuous
functions x— ’ |x— (x)| for x ¤ 1. Conversely the unit ball B has to be weak—

13.22
13.22 13. Di¬erentiability of seminorms 149

closed in E since the norm is assumed to be weak— lower semi-continuous and B
is convex. Let Bo be its polar in E. By the bipolar-theorem (Bo )o = B, and thus
the dual of the Minkowski functional of Bo is the given norm.
Next we show that the in¬mum de¬ning n is in fact a minimum, i.e. for each n
and x— there exists a y — with x— n = x— ’T — y — 2 + n y — 2 . Since fx : y — ’ x— ’
1
2
T — y — 2 + n y — 2 is weak— lower semi-continuous and satis¬es limy— ’∞ fx (y — ) =
1

+∞, hence it attains its minimum on some large (weak— -compact) ball.
We have that x n ’ 0 for n ’ ∞.
In fact since the image of T — is dense in E , there is for every µ > 0 a y — with
x— ’ T — y — < µ, and so for large n we have x— 2 ¤ x— ’ T — y — 2 + n y 2 < µ2 .
1
n
Let us next show that ∞ is a dual norm. For this it is enough to show that n
is a dual norm, i.e. is weak— lower semi-continuous. So let x— be a net converging
i
— — — —2 — ——2 —
1
weak to x . Then we may choose yi with xi n = xi ’ T yi + n yi 2 . Then
{x— : i} is bounded, and hence also yi 2 . Let thus y — be a weak— cluster point

i
of the (yi ). Without loss of generality we may assume that yi ’ y — . Since the
— —

original norms are weak— lower semicontinuous we have
1— 1 —2
x— ¤ x— ’ T — y — ¤ lim inf ( x— ’ T — yi

yi ) = lim inf x—
2 2 2 2 n
+ y + 2.
n i i
n n
i i


So n is weak lower semicontinuous.
Here we use that a function f : E ’ R is lower semicontinuous if and only if
x∞ = limi xi ’ f (x∞ ) ¤ lim inf i f (xi ).
(’) otherwise for some subnet (which we again denote by xi ) we have f (x∞ ) >
limi f (xi ) and this contradicts the fact that f ’1 ((a, ∞)) has to be a neighborhood
of x∞ for 2a := f (x∞ ) + limi f (xi ).
(’) otherwise there exists some x∞ and an a < f (x∞ ) such that in every neigh-
borhood U of x∞ there is some xU with f (xU ) ¤ a. Hence limU xU = x∞ and
lim inf U f (xU ) ¤ lim supU f (xU ) ¤ a < f (x∞ ).
Let us ¬nally show that is locally uniform rotund.

So let x— , x— ∈ E with
j

2( x— + x— ’ x— + x—
2 2 2
’ 0,
∞)
∞ ∞
j j

or equivalently

x— ’ x— and x— + x— ’ 2 x— ∞.
∞ ∞ ∞
j j

Thus also
x— ’ x— and x— + x— ’ 2 x—
n n n n
j j

and equivalently
2( x— + x— ’ x— + x—
2 2 2
’ 0.
n)
n j j n

Now we may choose y — and yj such that



1— 1 —2
x— = x— ’ T — y — and x— = x— ’ T — yj

n 2 2 n 2
+ y + y.
2 j 2 j
nj
n
13.22
150 Chapter III. Partitions of unity 13.23

We calculate as follows:

2( x— + x— ’ x— + x—
2 2 2

n)
n j j
1— 1—
≥ 2( x— ’ T — y — + x— ’ T — yj

2 2 2 2
+ y + y
j
nj
n
1—
’ x— + x— ’ T — (y — + yj )
— —
2 2
’ y + yj
j
n
1— 1—
≥ 2( x— ’ T — y — + x— ’ T — yj

2 2 2 2
+ y + y
j
nj
n
1—
’ ( x— ’ T — (y — ) + x— ’ T — (yj ) )2 ’
— — 2
y + yj
j
n
≥ ( x— ’ T — y — ’ x— ’ T — yj )2 +

j
1
+ (2 y — 2 + 2 yj 2 ’ y — + yj 2 ) ≥ 0,
— —
n

hence

x— ’ T — yj ’ x— ’ T — y — and 2( y —

+ yj 2 ) ’ y — + yj
— —
2 2
’ 0.
j


is locally uniformly rotund on ( 2 )— we get that yj ’ y — . Hence

Since

lim sup x— ’ x— ¤ lim sup( x— ’ T — y — + T — (y — ’ yj ) + x— ’ T — yj )
— —
j j
j j
= 2 x— ’ T — y — ¤ 2 x— n.


Since x— ’ 0 for n ’ ∞ we get x— ’ x— .
n j

13.23. Proposition. [Leach, Whit¬eld, 1972]. For the norm = p on a
Banach space E the following statements are equivalent:
(1) The norm is rough, i.e. p is uniformly discontinuous, see (13.8.5).

(2) There exists an µ > 0 such that for all x ∈ E with x = 1 and all yn ,
— — — — —
zn ∈ E with yn = 1 = zn and limn yn (x) = 1 = limn zn (x) we have:

— —
lim sup yn ’ zn ≥ µ;
n


(3) There exists an µ > 0 such that for all x ∈ E with x = 1 we have that

x+h + x’h ’2
≥ µ;
lim sup
h
h’0


(4) There exists an µ > 0 such that for every x ∈ E with x = 1 and δ > 0
there is an h ∈ E with h ¤ 1 and x + th ≥ x + µ|t| ’ δ for all |t| ¤ 1.

Note that we always have

x+h + x’h ’2 x
0¤ ¤ 2,
x

13.23
13.23 13. Di¬erentiability of seminorms 151

1
hence µ in (3) satis¬es µ ¤ 2. For and C[0, 1] the best choice is µ = 2, see (13.11)
and (13.12).

Proof. (3)’(2) is due to [Cudia, 1964]. Let µ > 0 such that for all x = 1 there
— —
are 0 = hn ’ 0 with x + hn + x ’ hn ’ 2 ≥ µ hn . Now choose yn , zn ∈ E
with yn = 1 = zn — , yn (x + hn ) = x + hn and zn (x ’ hn ) = x ’ hn . Then
— — —
— —
limn yn (x) = x = 1 and also limn zn (x) = 1. Moreover,

— —
yn (x + hn ) + zn (x ’ hn ) ≥ 2 + µ hn

and hence
— — — —
(yn ’ zn )(hn ) ≥ 2 ’ yn (x) ’ zn (x) + µ hn ≥ µ hn ,

thus (2) is satis¬ed.
— —
(2)’(1) By (2) we have an µ > 0 such that for all x = 1 there are yn and zn
— — — —
with yn = 1 = zn , limn yn (x) = 1 = limn zn (x) and hn with hn = 1 and
— —
(yn ’ zn )(hn ) ≥ µ. Let 0 < δ < µ/2 and t > 0. Then

δ2 δ2
— —
>1’ >1’
yn (x) and zn (x) for large n.
4 4

Thus
δ2
— —
x + thn ≥ + thn ) ≥ 1 ’
yn (x + tyn (hn )
4
and hence

δ2

t p (x + thn )(hn ) ≥ x + thn ’ x ≥ ’ ’
tyn (hn )
4
δ2

p (x + thn )(hn ) ≥ ’
yn (hn )
4t
δ2

and similarly ’ p (x ’ thn )(hn ) ≥ ’zn (hn ) ’
4t

If we choose 0 < t < δ such that δ 2 /(2t) < δ we get

δ2 µ
— —
p (x + thn )(hn ) ’ p (x ’ thn )(hn ) ≥ ’ ’ >µ’δ > .
(yn zn )(hn )
2t 2

(1)’(4) Using the uniform discontinuity assumption of p we get xj ∈ E with
p(xj ’ x) ¤ ·/4 and u ∈ E with p(u) = 1 such that (p (x2 ) ’ p (x1 ))(u) ≥ µ. Let
µ := (p (x1 ) + p (x2 ))(u)/(2p(x)) and v := u ’ µ x.
Since p (x1 )(u) ¤ p (x2 )(u) ’ µ we get (p (x1 ) + p (x2 ))(u))/2 ¤ p (x2 )(u) ’ µ/2 ¤
p(u) ’ µ/2 < 1 and (p (x1 ) + p (x2 ))(u)/2 ≥ p (x1 )(u) + µ ≥ ’p(u) + µ/2 > 1, i.e.
|(p (x1 ) + p (x2 ))(u)/2| < 1, so 0 < p(v) < 2. For 0 ¤ t ¤ p(x) and s := 1 ’ t µ we
get
t t
x + tv = sx + tu = s(x + u) = s (x2 + u) + (x ’ x2 ) .
s s
13.23
152 Chapter III. Partitions of unity 14

Thus 0 < s < 2 and
t
p(x + tv) ≥ s(p(x2 + u) ’ p(x ’ x2 ))
s
t
> s p(x2 ) + p (x2 )u ’ ·/4 since p(y + w) ≥ p(y) + p (y)(w)
s
> s p(x) + t p (x2 )(u) ’ s ·/2 since p(x) ¤ p(x2 ) + p(x ’ x2 )
= p(x) + (t/2) (p (x2 ) ’ p (x1 ))(u) ’ s ·/2
> p(x) + tµ/2 ’ ·.
If ’p(x) ¤ t < 0 we proceed with the role of x1 and x2 exchanged and obtain
p(x + tv) > s p(x) + t p (x1 )(u) ’ s ·/2
= p(x) + (’t/2) (p (x2 ) ’ p (x1 ))(u) ’ s ·/2
> p(x) + |t| µ/2 ’ ·.
Thus
p(x + tv) ≥ p(x) + |t| µ/2 ’ ·.
(4)’(3) By (4) there exists an µ > 0 such that for every x ∈ E with x = 1 and
δ > 0 there is an h ∈ E with h ¤ 1 and x + th ≥ x + µ|t| ’ δ for all |t| ¤ 1.
If we put t := 1/n we have
n( x + hn /n + x ’ hn /n ’ 2) ≥ µ ’ 1/n > µ/2 for large n.

13.24. Results on the non-existence of C 1 -norms on certain spaces.
(1) [Restrepo, 1964] and [Restrepo, 1965]. A separable Banach space has an
equivalent C 1 -norm if and only if E — is separable. This will be proved in
(16.11).
(2) [Kadec, 1965]. More generally, if for a Banach space dens E < dens E — then
no C 1 -norm exists. This will be proved by showing the existence of a rough
norm in (14.10) and then using (14.9). The density number dens X of a
topological space X is the minimum of the cardinalities of all dense subsets
of X.
(3) [Haydon, 1990]. There exists a compact space K, such that K (ω1 ) = {—}, in
particular K (ω1 +1) = …, but C(K) has no equivalent Gˆteaux di¬erentiable
a
norm, see also (13.18.2).
One can interpret these results by saying that in these spaces every convex body
necessarily has corners.


14. Smooth Bump Functions

In this section we return to the original question whether the smooth functions
generate the topology. Since we will use the results given here also for manifolds,
and since the existence of charts is of no help here, we consider fairly general non-
linear spaces. This allows us at the same time to treat all considered di¬erentiability
classes in a uni¬ed way.

14
14.4 14. Smooth bump functions 153

14.1. Convention. We consider a Hausdor¬ topological space X with a subalge-
bra S ⊆ C(X, R), whose elements will be called the smooth or S-functions on X.
We assume that for functions h ∈ C ∞ (R, R) (at least for those being constant o¬
some compact set, in some cases) one has h— (S) ⊆ S, and that f ∈ S provided it is
locally in S, i.e., there exists an open covering U such that for every U ∈ U there
exists an fU ∈ S with f = fU on U . In particular, we will use for S the classes of
C ∞ - and of Lipk -mappings on c∞ -open subsets X of convenient vector spaces with
the c∞ -topology and the class of C n -mappings on open subsets of Banach spaces,
as well as subclasses formed by boundedness conditions on the derivatives or their
di¬erence quotients.
1
Under these assumptions on S one has that f ∈ S provided f ∈ S with f (x) > 0
for all x ∈ X: Just choose everywhere positive hn ∈ C ∞ (R, R) with hn (t) = 1 for
t
1 1 1
t ≥ n . Then hn —¦ f ∈ S and f = hn —¦ f on the open set {x : f (x) > n }. Hence,
1
f ∈ S.
For a (convenient) vector space F the carrier carr(f ) of a mapping f : X ’ F
is the set {x ∈ X : f (x) = 0}. The zero set of f is the set where f vanishes,
{x ∈ X : f (x) = 0}. The support of f support(f ) is the closure of carr(f ) in X.
We say that X is smoothly regular (with respect to S) or S-regular if for any
neighborhood U of a point x there exists a smooth function f ∈ S such that
f (x) = 1 and carr(f ) ⊆ U . Such a function f is called a bump function.

14.2. Proposition. Bump functions and regularity. [Bonic, Frampton,
1966]. A Hausdor¬ space is S-regular if and only if its topology is initial with
respect to S.

Proof. The initial topology with respect to S has as a subbasis the sets f ’1 (I),
where f ∈ S and I is an open interval in R. Let x ∈ U , with U open for the initial
topology. Then there exist ¬nitely many open intervals I1 , . . . , In and f1 , . . . , fn ∈ S
n
with x ∈ i=1 fi’1 (Ii ). Without loss of generality we may assume that Ii = {t :
|fi (x) ’ t| < µi } for certain µi > 0. Let h ∈ C ∞ (R, R) be chosen such that h(0) = 1
n
and h(t) = 0 for |t| ≥ 1. Set f (x) := i=1 h( fiµi ). Then f is the required bump
(x)

function.

14.3. Corollary. Smooth regularity is inherited by products and sub-
spaces. Let Xi be topological spaces and Si ⊆ C(Xi , R). On a space X we con-
sider the initial topology with respect to mappings fi : X ’ Xi , and we assume that
S ⊆ C(X, R) is given such that fi— (Si ) ⊆ S for all i. If each Xi is Si -regular, then
X is S-regular.

Note however that the c∞ -topology on a locally convex subspace is not the trace
of the c∞ -topology in general, see (4.33) and (4.36.5). However, for c∞ -closed
subspaces this is true, see (4.28).

14.4. Proposition. [Bonic, Frampton, 1966]. Every Banach space with S-norm
is S-regular.

14.4
154 Chapter III. Partitions of unity 14.6

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