T (x)i := i i .

2

Then for the basic unit vector ei ∈ ( 2 ) we have

x— (x)

—

T (ei )(x) = ei (T (x)) = T (x)i = i i ,

2

i.e. T — (ei ) = 2’i x— .

i

Note that the canonical norm on 2 is locally uniformly rotund. We now claim that

E has a dual locally uniform rotund norm. For x— ∈ E and n ∈ N we de¬ne

1—

x— := inf{ x— ’ T — y — : y — ∈ ( 2 ) } and

2 2 2

+ y

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

More general, a convenient vector space is smoothly regular if its c∞ -topology is

generated by seminorms which are smooth on their respective carriers. For example,

nuclear Fr´chet spaces have this property.

e

Proof. Namely, g —¦ p is a smooth bump function with carrier contained in {x :

p(x) < 1} if g is a suitably chosen real function, i.e., g(t) = 1 for t ¤ 0 and g(t) = 0

for t ≥ 1.

Nuclear spaces have a basis of Hilbert-seminorms (52.34), and on Fr´chet spaces the

e

∞

c -topology coincides with the locally convex one (4.11.1), hence nuclear Fr´chet

e

spaces are c∞ -regular.

14.5. Open problem. Has every non-separable S-regular Banach space an equiv-

alent S-norm? Compare with (16.11).

A partial answer is given in:

14.6. Proposition. Let E be a C ∞ -regular Banach space. Then there exists a

smooth function h : E ’ R+ , which is positively homogeneous and smooth on

E \ {0}.

Proof. Let f : E \{0} ’ {t ∈ R : t ≥ 0} be a smooth function, such that carr(f ) is

bounded in E and f (x) ≥ 1 for x near 0. Let U := {x : f (tx) = 0 for some t ≥ 1}.

Then there exists a smooth function M f : E \ {0} ’ R with (M f ) (x)(x) < 0 for

x ∈ U , limx’0 f (x) = +∞ and carr M f ⊆ U .

The idea is to construct out of the smooth function f ≥ 0 another smooth function

M f with (M f ) (x)(x) = ’f (x) ¤ 0, i.e. (M f ) (tx)(tx) = ’f (tx) and hence

d f (tx)

M f (tx) = (M f ) (tx)(x) = ’ for t = 0.

dt t

Since we want bounded support for M f , we get

∞ ∞

∞ d f (tx)

M f (x) = ’ M f (tx) =’ M f (tx) dt = dt,

dt t

t=1 1 1

and we take this as a de¬nition of M f . Since the support of f is bounded, we may

N

replace the integral locally by 1 for some large N , hence M f is smooth on E \ {0}

and (M f ) (x)(x) = ’f (x).

Since f (x) > µ for all x < δ, we have that

N

1

M f (x) ≥ f (tx) dt ≥ log(N )µ

t

1

δ

for all x < N , i.e. limx’0 M f (x) = +∞.

Furthermore carr(M f ) ⊆ U , since f (tx) = 0 for all t ≥ 1 and x ∈ U .

/

Now consider M 2 f := M (M f ) : E \ {0} ’ R. Since (M f ) (x)(x) ¤ 0, we have

∞

(M 2 f ) (x)(x) = 1 (M f ) (tx)(x) dt ¤ 0 and it is < 0 if for some t ≥ 1 we have

(M f ) (tx)(x) < 0, in particular this is the case if M 2 f (x) > 0.

Thus Uµ := {x : M 2 f (x) ≥ µ} is radial set with smooth boundary, and the

Minkowski-functional is smooth on E \ {0}. Moreover Uµ ∼ E via x ’ M 2x (x) .

= f

14.6

14.9 14. Smooth bump functions 155

14.7. Lemma. Existence of smooth bump functions.

For a class S on a Banach space E in the sense of (14.1) the following statements

are equivalent:

(1) E is not S-regular;

(2) For every f ∈ S, every 0 < r1 < r2 and µ > 0 there exists an x with

r1 ¤ x ¤ r2 and |f (x) ’ f (0)| < µ;

(3) For every f ∈ S with f (0) = 0 there exists an x with 1 ¤ x ¤ 2 and

|f (x)| ¤ x

Proof. (1) ’ (2) Assume that there exists an f and 0 < r1 < r2 and µ > 0 such

that |f (x) ’ f (0)| ≥ µ for all r1 ¤ x ¤ r2 . Let h : R ’ R be a smooth bump

function on R. Let g(x) := h( 1 f (r1 x)’f (0)). Then g is of the corresponding class,

µ

g(0) = h(0) = 1, and for all x with 1 ¤ x ¤ r2 we have |f (r1 x) ’ f (0)| ≥ µ, and

r1

hence g(x) = 0. By rede¬ning g on {x : x ≥ r2 } as 0, we obtain the required

r1

bump function.

(2) ’ (3) Take r1 = 1 and r2 = 2 and µ = 1.

(3) ’ (1) Assume a bump function g exists, i.e., g(0) = 1 and g(x) = 0 for all

x ≥ 1. Take f := 2 ’ g. Then f (0) = 0 and f (x) = 2 for x ≥ 1, a contradiction

to (3).

14.8. Proposition. Boundary values for smooth mappings. [Bonic, Framp-

ton, 1966] Let E and F be convenient vector spaces, let F be S-regular but E not

S-regular. Let U ⊆ E be c∞ -open and f ∈ C(U , F ) with f — (S) ⊆ S. Then

f (‚U ) ⊇ f (U ). Hence, f = 0 on ‚U implies f = 0 on U .

Proof. Since f (U ) ⊆ f (U ) it is enough to show that f (U ) ⊆ f (‚U ). Suppose

f (x) ∈ f (‚U ) for some x ∈ U . Choose a smooth h on F such that h(f (x)) = 1 and

/

h = 0 on a neighborhood of f (‚U ). Let g = h —¦ f on U and 0 outside. Then g is a

smooth bump function on E, a contradiction.

14.9. Theorem. C 1 -regular spaces admit no rough norm. [Leach, Whit-

¬eld, 1972]. Let E be a Banach space whose norm p = has uniformly dis-

continuous directional derivative. If f is Fr´chet di¬erentiable with f (0) = 0 then

e

there exists an x ∈ E with 1 ¤ x < 2 and f (x) ¤ x .

By (14.7) this result implies that on a Banach space with rough norm there exists

no Fr´chet di¬erentiable bump function. In particular, C([0, 1]) and 1 are not

e

C 1 -regular by (13.11) and (13.12), which is due to [Kurzweil, 1954].

Proof. We try to reach the exterior of the unit ball by a recursively de¬ned se-

quence xn in {x : f (x) ¤ p(x)} starting at 0 with large step-length ¤ 1 in directions,

where p is large. Given xn we consider the set

(1) f (y) ¤ p(y),

±

Mn := y ∈ E : (2) p(y ’ xn ) ¤ 1 and .

(3) p(y) ’ p(xn ) ≥ (µ/8) p(y ’ xn )

14.9

156 Chapter III. Partitions of unity 14.10

Since xn ∈ Mn , this set is not empty and hence Mn := sup{p(y’xn ) : y ∈ Mn } ¤ 1

is well-de¬ned and it is possible to choose xn+1 ∈ Mn with

(4) p(xn+1 ’ xn ) ≥ Mn /2.

We claim that p(xn ) ≥ 1 for some n, since then x := xn for the minimal n satis¬es

the conclusion of the theorem:

Otherwise p(xn ) is bounded by 1 and increasing by (3), hence a Cauchy-sequence.

By (3) we then get that (xn ) is a Cauchy-sequence. So let z be its limit. If z = 0

then Mn = {0} and hence f (y) > p(y) for all |y| ¤ 1. Thus f is not di¬erentiable.

Then p(z) ¤ 1 and f (z) ¤ p(z). Since f is Fr´chet-di¬erentiable at z there exists a

e

δ > 0 such that

f (z + u) ’ f (z) ’ f (z)(u) ¤ µp(u)/8 for all p(u) < δ.

Without loss of generality let δ ¤ 1 and δ ¤ 2 p(z). By (13.23.4) there exists a v

such that p(v) < 2 and p(z + tv) > p(z) + µ|t|/2 ’ µδ/8 for all |t| ¤ p(z). Now let

t := ’ sign(f (z)(v)) δ/2. Then

(1) p(z + tv) > p(z) + µδ/8 ≥ f (z) + µp(tv)/8 ≥ f (z + tv),

(2) p(z + tv ’ z) = |t|p(v) < δ ¤ 1,

(3) p(z + tv) ’ p(z) > µδ/8 > µp(tv)/8.

Since f and p are continuous the z + tv satisfy (1)-(3) for large n and hence Mn ≥

p(z + t v ’ xn ). From p(z + tv ’ z) > µδ/8 we get Mn > µδ/8 and so p(xn+1 ’ xn ) >

µδ/16 by (4) contradicts the convergence of xn .

14.10. Proposition. Let E be a Banach-space with dens E < dens E . Then there

is an equivalent rough norm on E.

Proof. The idea is to describe the unit ball of a rough norm as intersection of hyper

planes {x ∈ E : x— (x) ¤ 1} for certain functionals x— ∈ E . The fewer functionals

we use the more ˜corners™ the unit ball will have, but we have to use su¬ciently

many in order that this ball is bounded and hence that its Minkowski-functional is

an equivalent norm. We call a set X large, if and only if |X| > dens(E) and small

otherwise. For x ∈ E and µ > 0 let Bµ (x) := {y ∈ E : x ’ y ¤ µ}. Now we

choose using Zorn™s lemma a subset D ⊆ E maximal with respect to the following

conditions:

(1) 0 ∈ D;

(2) x— ∈ D ’ ’x— ∈ D;

(3) x— , y — ∈ D, x— = y — ’ x— ’ y — > 1.

Note that D is then also maximal with respect to (3) alone, since otherwise, we

could add a point x— with x— ’ y — > 1 for all y — ∈ D and also add the point ’x— ,

and obtain a larger set satisfying all three conditions.

1

Claim. D∞ := n∈N n D is dense in E , and hence |D∞ | ≥ dens(E ):

Assume indirectly, that there is some x— ∈ E and n ∈ N with B1/n (x— ) © D∞ =

14.10

14.10 14. Smooth bump functions 157

…. Then B1 (n x— ) © D = … and hence we may add x— to D, contradicting the

maximality.

Without loss of generality we may assume that D is at least countable. Then |D| =

1

| n∈N n D| ≥ dens(E ) > dens(E), i.e. D is large. Since D = n∈N D © Bn (0), we

¬nd some n such that D©Bn (0) is large. Let y — ∈ E be arbitrary and w— := 4n+2 y — .

1

For every x— ∈ D there is a z — ∈ 1 D such that x— + w— ’ z — ¤ 1 (otherwise

2 2

we could add 2(x— + w— ) to D). Thus we may de¬ne a mapping D ’ 2 D by 1

x— ’ z — . This mapping is injective, since x— + w— ’ z — ¤ 2 for j ∈ {1, 2} implies

1

j

x— ’ x— ¤ 1 and hence x— = x— . If we restrict it to the large set D © Bn (0) it

1 2 1 2

has image in 2 D © Bn+1/2 (w ), since z — ’ w— ¤ z — + x— ’ w— + x— ¤ 2 + n.

—

1 1

Hence also 2(4n+2) D © B1/4 (y — ) = 4n+2 1 D © Bn+1/2 (w— ) is large.

1 1

2

In particular for y — := 0 and 1/4 replaced by 1 we get that A := 1

© B1 (0)

4(2n+1) D

is large. Now let

U := x ∈ E : ∃A0 ⊆ A small, ∀x— ∈ A \ A0 : x— (x) ¤ 1 .

Since A is symmetric, the set U is absolutely convex (use that the union of two

small exception sets is small). It is a 0-neighborhood, since {x : x ¤ 1} ⊆ U

(x— (x) ¤ x— · x = x ¤ 1 for x— ∈ A). It is bounded, since for x ∈ E we may

¬nd by Hahn-Banach an x— ∈ E with x— (x) = x and x— = 1. For all y — in the

large set A © B1/4 ( 3 x— ) we have y — (x) = (y — ’ 3 x— )(x) + 3 x— (x) ≥ 3 x ’ 1 x ≥

4 4 4 4 4

1

2 x . For x > 2 we thus get x ∈ U . Now let σ be the Minkowski-functional

/

generated by U and σ — the dual norm on E . Let ∆ ⊆ E be a small dense subset.

Then {x— ∈ A : σ — (x— ) > 1} is small, since σ — (x— ) > 1 for x— ∈ A implies that there

exists an x ∈ ∆ with x— (x) > σ(x), but this is n∈N {x— ∈ A : x— (x) > σ(x) + n }, 1

and each of these sets is small by construction of σ(x). Since ∆ is small so is the

union over all x ∈ ∆. Thus A1 := {x— ∈ A : σ(x— ) ¤ 1} is large.

1

Now let µ := 8(2n+1) , let x ∈ E, and let 0 < · < µ. We may choose two di¬erent

x— ∈ A1 for i ∈ {1, 2} with x— (x) > σ(x) ’ · 2 /2. This is possible, since this is true

i i

for all but a small set of x ∈ A. Thus σ — (x— ’ x— ) ≥ x— ’ x— > 2µ, and hence

—

1 2 1 2

— —

there is an h ∈ E with σ(h) = 1 and (x1 ’ x2 )(h) > 2µ. Let now t > 0. Then

·2

x— (x x— (x) tx— (h) + tx— (h),

σ(x + th) ≥ > σ(x) ’

+ th) = +

1 1 1 1

2

·2

x— (x ’ tx— (h).

σ(x ’ th) ≥ ’ th) > σ(x) ’

2 2

2

Furthermore σ(x) ≥ σ(x + th) ’ tσ (x + th)(h) implies

·2

σ(x + th) ’ σ(x) —

σ (x + th)(h) ≥ > x1 (h) ’ ,

t 2t

·2

—

’σ (x ’ th)(h) ≥ ’x2 (h) ’ .

2t

Adding the last two inequalities gives

·2

— —

σ (x + th)(h) ’ σ (x ’ th)(h) ≥ (x2 ’ x1 )(h) ’ > µ,

t

·2

since (x— ’ x— )(h) > 2µ and we choose t < · such that < µ.

2 1 t

14.10

158 Chapter III. Partitions of unity 14.12

14.11. Results. Spaces which are not smoothly regular. For Banach spaces

one has the following results:

(1) [Bonic, Frampton, 1965]. By (14.9) no Fr´chet-di¬erentiable bump function

e

exists on C[0, 1] and on 1 . Hence, most in¬nite dimensional C — -algebras

are not regular for 1-times Fr´chet-di¬erentiable functions, in particular

e

those for which a normal operator exists whose spectrum contains an open

interval.

(2) [Leduc, 1970]. If dens E < dens E — then no C 1 -bump function exists. This

follows from (14.10), (14.9), and (14.7). See also (13.24.2).

(3) [John, Zizler, 1978]. A norm is called strongly rough if and only if there

exists an µ > 0 such that for every x with x = 1 there exists a unit

vector y with lim supt 0 x+ty + t x’ty ’2 ≥ µ. The usual norm on 1 (“) is

strongly rough, if “ is uncountable. There is however an equivalent non-

rough norm on 1 (“) with no point of Gˆteaux-di¬erentiability. If a Banach

a

space has Gˆteaux di¬erentiable bump functions then it does not admit a

a

strongly rough norm.

(4) [Day, 1955]. On 1 (“) with uncountable “ there is no Gˆteaux di¬erentiable

a

continuous bump function.

(5) [Bonic, Frampton, 1965]. E < p , dim E = ∞: If p = 2n + 1 then E is

not Dp -regular. If p ∈ N then E is not S-regular, where S denotes the

/

C [p] -functions whose highest derivative satis¬es a H¨lder like condition of

o

order p ’ [p] but with o( ) instead of O( ).

14.12. Results.

(1) [Deville, Godefroy, Zizler, 1990]. If c0 (“) ’ E ’ F is a short exact se-

quence of Banach spaces and F has C k -bump functions then also E has

them. Compare with (16.19).

(2) [Meshkov, 1978] If a Banach space E and its dual E — admit C 2 -bump func-

tions, then E is isomorphic to a Hilbert space. Compare with (13.18.7).

(3) Smooth bump functions are not inherited by short exact sequences.

Notes. (1) As in (13.17.3) one chooses x— ∈ E — with x— |c0 (“) = eva . Let g be a

a a

∞

smooth bump function on E/F and h ∈ C (R, [0, 1]) with compact support and

equal to 1 near 0. Then f (x) := g(x + F ) a∈“ h(x— (x)) is the required bump

a

function.

(3) Use the example mentioned in (13.18.6), and apply (2).

Open problems. Is the product of C ∞ -regular convenient vector spaces again

C ∞ -regular? Beware of the topology on the product!

Is every quotient of any S-regular space again S-regular?

14.12

159

15. Functions with Globally Bounded Derivatives

In many problems (like Borel™s theorem (15.4), or the existence of smooth functions

with given carrier (15.3)) one uses in ¬nite dimensions the existence of smooth

functions with bounded derivatives. In in¬nite dimensions C k -functions have lo-

cally bounded k-th derivatives, but even for bump functions this need not be true

globally.

k

15.1. De¬nitions. For normed spaces we use the following notation: CB := {f ∈

C k : f (k) (x) ¤ B for all x ∈ E} and Cb := B>0 CB . For general convenient

k k

∞

vector spaces we may still de¬ne Cb as those smooth functions f : U ’ F for

which the image dk f (U ) of each derivative is bounded in the space Lk (E, F ) of

sym

bounded symmetric multilinear mappings.

Let Lipk denote the space of C k -functions with global Lipschitz-constant K for

K

the k-th derivatives and Lipk k k’1

k k

K>0 LipK . Note that CK = C © LipK .

global :=

15.2. Lemma. Completeness of C n . Let fj be C n -functions on some Banach

(k)

space such that fj converges uniformly on bounded sets to some function f k for

each k ¤ n. Then f := f 0 is C n , and f (k) = f k for all k ¤ n.

Proof. It is enough to show this for n = 1. Since fn ’ f 1 uniformly, we have that

1

f 1 is continuous, and hence 0 f 1 (x + t h)(h) dt makes sense and

1 1

f 1 (x + t h)(h) dt

fn (x + h) ’ fn (x) = fn (x + t h)(h) dt ’

0 0

for x and h ¬xed. Since fn ’ f pointwise, this limit has to be f (x + h) ’ f (x).

Thus we have

1

f (x + h) ’ f (x) ’ f 1 (x)(h) 1

(f 1 (x + t h) ’ f 1 (x))(h) dt

=

h h 0

1

f 1 (x + t h) ’ f 1 (x)) dt

¤

0

which goes to 0 for h ’ 0 and ¬xed x, since f 1 is continuous. Thus, f is di¬eren-

tiable and f = f 1 .

15.3. Proposition. When are closed sets zero-sets of smooth functions.

[Wells, 1973]. Let E be a separable Banach space and n ∈ N. Then E has a

Cb -bump function if and only if every closed subset of E is the zero-set of a C n -

n

function.

For n = ∞ and E a convenient vector space we still have (’), provided all Lk (E; R)

satisfy the second countability condition of Mackey, i.e. for every countable family

of bounded sets Bk there exist tk > 0 such that k tk Bk is bounded.

n

Proof. (’) Suppose ¬rst that E has a Cb -bump function. Let A ⊆ E be closed

and U := E \ A be the open complement. For every x ∈ U there exists an fx ∈

15.3

160 Chapter III. Partitions of unity 15.4

n

Cb (E) with fx (x) = 1 and carr(fx ) ⊆ U . The family of carriers of the fx is

an open covering of U . Since E is separable, those points in a countable dense

subset that lie in U are dense in the metrizable space U . Thus, U is Lindel¨f, and

o

consequently we can ¬nd a sequence of points xn such that for the corresponding

functions fn := fxn the carriers still cover U . Now choose constants tn > 0 such

(j) 1

that tn · sup{ fn (x) : x ∈ E} ¤ 2n’j for all j < n. Then f := n tn fn

converges uniformly in all derivatives, hence represents by (15.2) a C n -function on

E that vanishes on A. Since the carriers of the fn cover U , it is strictly positive on

U , and hence the required function has as 0-set exactly A.

(⇐) Consider a vector a = 0, and let A := E \ n∈N {x : x ’ 21 a < 2n+1 }. Since

1

n

A is closed there exists by assumption a C n -function f : E ’ R with f ’1 (0) = A

(without loss of generality we may assume f (E) ⊆ [0, 1]). By continuity of the

derivatives we may assume that f (n) is bounded on some neighborhood U of 0.

Choose n so large that D := {x : x ’ 21 a < 21 } ⊆ U , and let g := f on A ∪ D

n n

and 0 on E \ D. Then f ∈ C n and f (n) is bounded. Up to a¬ne transformations

this is the required bump function.

∞

15.4. Borel™s theorem. [Wells, 1973]. Suppose a Banach space E has Cb -

bump functions. Then every formal power series with coe¬cients in Ln (E; F )

sym

for another Banach space F is the Taylor-series of a smooth mapping E ’ F .

Moreover, if G is a second Banach space, and if for some open set U ⊆ G we are

given bk ∈ Cb (U, Lk (E, F )), then there is a smooth f ∈ C ∞ (E — U, F ) with

∞

sym

k

d (f ( , y))(0) = bk (y) for all y ∈ U and k ∈ N. In particular, smooth curves can

be lifted along the mapping C ∞ (E, F ) ’ k Lk (E; F ).

sym

∞ ∞

Proof. Let ρ ∈ Cb (E, R) be a Cb -bump function, which equals 1 locally at 0.

We shall use the notation bk (x, y) := bk (y)(xk ). De¬ne

1

fk (x, y) := bk (x, y) ρ(x)

k!

and

1

fk (tk · x, y)

f (x, y) :=

tk

k≥0 k

with appropriately chosen tk > 0. Then fk ∈ C ∞ (E — U, F ) and fk has carrier

inside of carr(ρ) — U , i.e. inside {x : x < 1} — U . For the derivatives of bk we

have

ji

‚1 ‚2 bk (x, y)(ξ, ·) = k (k ’ 1) . . . (k ’ j) (di bk (y)(·))(xk’j , ξ j ).

Hence, for x ¤ 1 this derivative is bounded by

(k)j sup di bk (y) L(F,Lk (E;G)) ,

sym

y∈U

where (k)j := k(k ’ 1) . . . (k ’ j). Using the product rule we see that for j ≥ k the

ji

derivative ‚1 ‚2 fk of fk is globally bounded by

j

sup{ ρ(j’l) (x) : x ∈ E} (k)l sup di bk (y) < ∞.

l y∈U

l¤k

15.4

15.4 15. Functions with globally bounded derivatives 161

The partial derivatives of f would be

tj j i

ji k

‚1 ‚2 fk (x, y) = ‚ ‚ fk (tk x, y).

tk 1 2

k

k

We now choose the tk ≥ 1 such that these series converge uniformly. This is the

case if,

1 ji

sup{ ‚1 ‚2 fk (x, y) : x ∈ E, y ∈ U } ¤

tk’j

k

1 1

ji

¤ sup{ ‚1 ‚2 fk (x, y) : x ∈ E, y ∈ U } ¤ ,

k’(j+i) 2k’(j+i)

tk

and thus if

1

ji

tk ≥ 2. sup{ ‚1 ‚2 fk (x, y) : x ∈ E, y ∈ U, j + i < k}.

k’(j+i)

j j

1

Since we have ‚1 fk (0, y)(ξ) = k! (k)j bk (y)(0k’j , ξ j ) ρ(0) = δk bk (y), we conclude

j

the desired result ‚1 f (0, y) = bk (y).

Remarks on Borel™s theorem.

(1) [Colombeau, 1979]. Let E be a strict inductive limit of a non-trivial se-

quence of Fr´chet spaces En . Then Borel™s theorem is wrong for f : R ’ E.

e

The idea is to choose bn = f (n) (0) ∈ En+1 \ En and to use that locally every

smooth curve has to have values in some En .

(2) [Colombeau, 1979]. Let E = RN . Then Borel™s theorem is wrong for f :

E ’ R. In fact, let bn : E — . . . — E ’ R be given by bn := prn — · · · — prn .

Assume f ∈ C ∞ (E, R) exists with f (n) (0) = bn . Let fn be the restriction of

(n)

f to the n-th factor R in E. Then fn ∈ C ∞ (R, R) and fn (0) = 1. Since f :

Rn ’ (Rn ) = R(N) is continuous, the image of B := {x : |xn | ¤ 1 for all n}

in R(N) is bounded, hence contained in some RN ’1 . Since fN is not constant

on the interval (’1, 1) there exists some |tN | < 1 with fN (tN ) = 0. For

xN := (0, . . . , 0, tN , 0, . . . ) we obtain

f (xN )(y) = fN (tN )(yN ) + ai yi ,

i=N

a contradiction to f (xn ) ∈ RN ’1 .

(3) [Colombeau, 1979] showed that Borel™s theorem is true for mappings f :

E ’ F , where E has a basis of Hilbert-seminorms and for any countable

∞

family of 0-neighborhoods Un there exist tn > 0 such that n=1 tn Un is a

0-neighborhood.

(4) If theorem (15.4) would be true for G = k Lk (E; F ) and bk = prk , then

sym

∞

the quotient mapping C (E, F ) ’ G = k Lk (E; F ) would admit a

sym

smooth and hence a linear section. This is well know to be wrong even for

E = F = R, see (21.5).

15.4

162 Chapter III. Partitions of unity 15.6

∞

15.5. Proposition. Hilbert spaces have Cb -bump functions. [Wells, 1973]

If the norm is given by the n-th root of a homogeneous polynomial b of even degree

∞

n, then x ’ ρ(b(xn )) is a Cb -bump function, where ρ : R ’ R is smooth with

ρ(t) = 1 for t ¤ 0 and ρ(t) = 0 for t ≥ 1.

Proof. As before in the proof of (15.4) we see that the j-th derivative of x ’ b(xn )

is bounded by (n)j on the closed unit ball. Hence, by the chain-rule and the

global boundedness of all derivatives of ρ separately, the composite has bounded

derivatives on the unit ball, and since it is zero outside, even everywhere. Obviously,

ρ(b(0)) = ρ(0) = 1.

In [Bonic, Frampton, 1966] it is shown that Lp is Lipn

global -smooth for all n if p is

[p’1]

an even integer and is Lipglobal -smooth otherwise. This follows from the fact (see

loc. cit., p. 140) that d(p+1) x p = 0 for even integers p and

p!

dk x + h p

’ dk x p

h||p’k

¤

k!

otherwise, cf. (13.13).

15.6. Estimates for the remainder in the Taylor-expansion. The Taylor

formula of order k of a C k+1 -function is given by

k 1

(1 ’ t)k (k+1)

1 (j)

f (x)(hj ) + (x + th)(hk+1 ) dt,

f (x + h) = f

j! k!

0

j=0

1

which can easily be seen by repeated partial integration of f (x + th)(h) dt =

0

f (x + h) ’ f (x).

2

For a CB function we have

1

1

(1 ’ t) f (2) (x + th) 2

h 2.

|f (x + h) ’ f (x) ’ f (x)(h)| ¤ dt ¤ B

h

2!

0

If we take the Taylor formula of f up to order 0 instead, we obtain

1

f (x + h) = f (x) + f (x + th)(h) dt

0

1

and usage of f (x)(h) = f (x)(h) dt gives

0

1

f (x + th) ’ f (x) 1

2

h 2,

|f (x + h) ’ f (x) ’ f (x)(h)| ¤ dt ¤ B

h

th 2!

0

so it is in fact enough to assume f ∈ C 1 with f satisfying a Lipschitz-condition

with constant B.

15.6

15.7 15. Functions with globally bounded derivatives 163

3

For a CB function we have

1

|f (x + h) ’ f (x) ’ f (x)(h) ’ f (x)(h2 )| ¤

2

1

(1 ’ t)2 (3) 1

3

h 3.

¤ dt ¤ B

f (x + th) h

2! 3!

0

If we take the Taylor formula of f up to order 1 instead, we obtain

1

(1 ’ t) f (x + th)(h2 ) dt,

f (x + h) = f (x) + f (x)(h) +

0

1

and using 1 f (x)(h2 ) = ’ t) f (x)(h2 ) dt we get

(1

2 0

1

|f (x + h) ’ f (x) ’ f (x)(h) ’ f (x)(h2 )| ¤

2

1

f (x + th) ’ f (x) 1

3

h 3.

¤ (1 ’ t)t dt ¤ B

h

th 3!

0

Hence, it is in fact enough to assume f ∈ C 2 with f satisfying a Lipschitz-condition

with constant B.

1

k

Let f ∈ CB be ¬‚at of order k at 0. Applying f (h) ’ f (0) = 0 f (th)(h) dt ¤

sup{ f (th) : t ∈ [0, 1]} h to f (j) ( )(h1 , . . . , hj ) gives using f (k) (x) ¤ B

inductively

f (k’1) (x) ¤ B · x

1 1

B

(k’2) (k’1) 2 2

(x) ¤ (tx)(x, . . . ) dt ¤ B

f f t dt x = x

2

0 0

.

.

.

B

f (j) (x) ¤ k’j

x .

(k ’ j)!

15.7. Lemma. Lip1 n N

global -functions on R . [Wells, 1973]. Let n := 2 and

E = Rn with the ∞-norm. Suppose f ∈ Lip1 (E, R) with f (0) = 0 and f (x) ≥ 1

M

for x ≥ 1. Then M ≥ 2N .

The idea behind the proof is to construct recursively a sequence of points xk :=

k’1

j<k σj hj of norm N starting at x0 = 0, such that the increment along the

segment is as small as possible. In order to evaluate this increment one uses the

Taylor-formula and chooses the direction hk such that the derivative at xk vanishes.

Proof. Let A be the set of all edges of a hyper-cube, i.e.

A := {x : xi = ±1 for all i except one i0 and |xi0 | ¤ 1}.

15.7

164 Chapter III. Partitions of unity 15.9

Then A is symmetric. Let x ∈ E be arbitrary. We want to ¬nd h ∈ A with

f (x)(h) = 0. By permuting the coordinates we may assume that i ’ |f (x)(ei )|

is monotone decreasing. For 2 ¤ i ¤ n we choose recursively hi ∈ {±1} such that

i i j 1

j=2 hj f (x)(ej ) is an alternating sum. Then | j=2 f (x)(e )hj | ¤ |f (x)(e )|.

Finally, we choose h1 ¤ 1 such that f (x)(h) = 0.

1

Now we choose inductively hi ∈ N A and σi ∈ {±1} such that f (xi )(hi ) = 0 for

N ’i i

x := j<i σj hj and xi has at least 2 coordinates equal to N . For the last

statement we have that xi+1 = xi + σi hi and at least 2N ’i coordinates of xi are

i 1

N . Among those coordinates all but at most 1 of the hi are ± N . Now let σi be

the sign which occurs more often and hence at least 2N ’i /2 times. Then those

2N ’(i+1) many coordinates of xi+1 are i+1 .

N

i

for i ¤ N , since at least one coordinate has this value. Furthermore

Thus xi = N

we have

N ’1

1 = |f (xN ) ’ f (x0 )| ¤ |f (xk+1 ) ’ f (xk ) ’ f (xk )(hk )|

k=0

N

M M1

2

¤ ¤N

hk ,

2 N2

2

k=1

hence M ≥ 2N .

15.8. Corollary. c0 is not Lip1

global -regular. [Wells, 1973]. The space c0 is not

Lip1

global -smooth.

Proof. Suppose there exists an f ∈ Lip1 global with f (0) = 1 and f (x) = 0 for all

x ≥ 1. Then the previous lemma applied to 1 ’ f restricted to ¬nite dimensional

subspaces shows that the Lipschitz constant M of the derivative has to be greater

or equal to N for all N , a contradiction.

This shows even that there exist no di¬erentiable bump functions on c0 (A) which

have uniformly continuous derivative. Since otherwise there would exist an N ∈ N

such that

1

1

f (x + h) ’ f (x) ’ f (x)h ¤ f (x + t h) ’ f (x) h dt ¤ h,

2

0

for h ¤ N . Hence, the estimation in the proof of (15.7) would give 1 ¤ N 1 N = 1 ,

1 1

2 2

a contradiction.

15.9. Positive results on Lip1

global -functions. [Wells, 1973].

(1) Every closed subset of a Hilbert space is the zero-set of a Lip1 global -function.

(2) For every two closed subsets of a Hilbert space which have distance d > 0

there exists a Lip1 2 -function which has value 0 on one set and 1 on the

4/d

other.

(3) Whitney™s extension theorem is true for Lip1 global -functions on closed subsets

of Hilbert spaces.

15.9

165

16. Smooth Partitions of Unity and Smooth Normality

16.1. De¬nitions. We say that a Hausdor¬ space X is smoothly normal with

respect to a subalgebra S ⊆ C(X, R) or S-normal, if for two disjoint closed subsets

A0 and A1 of X there exists a function f : X ’ R in S with f |Ai = i for i = 0, 1.

If an algebra S is speci¬ed, then by a smooth function we will mean an element of

S. Otherwise it is a C ∞ -function.

A S-partition of unity on a space X is a set F of smooth functions f : X ’ R

which satisfy the following conditions:

(1) For all f ∈ F and x ∈ X one has f (x) ≥ 0.

(2) The set {carr(f ) : f ∈ F} of all carriers is a locally ¬nite covering of X.

(3) The sum f ∈F f (x) equals 1 for all x ∈ X.

Since a family of open sets is locally ¬nite if and only if the family of the closures

is locally ¬nite, the foregoing condition (2) is equivalent to:

(2™) The set {supp(f ) : f ∈ F} of all supports is a locally ¬nite covering of X.

The partition of unity is called subordinated to an open covering U of X, if for

every f ∈ F there exists an U ∈ U with carr(f ) ⊆ U .

We say that X is smoothly paracompact with respect to S or S-paracompact if every

open cover U admits a S-partition F of unity subordinated to it. This implies that

X is S-normal.

The partition of unity can then even be chosen in such a way that for every f ∈ F

there exists a U ∈ U with supp(f ) ⊆ U . This is seen as follows. Since the family of

carriers is a locally ¬nite open re¬nement of U, the topology of X is paracompact.

˜ ˜

So we may ¬nd a ¬ner open cover {U : U ∈ U} such that the closure of U is

contained in U for all U ∈ U, see [Bourbaki, 1966, IX.4.3]. The partition of unity

subordinated to this ¬ner cover has the support property for the original one.

Lemma. Let S be an algebra which is closed under sums of locally ¬nite families

of functions. If F is an S-partition of unity subordinated to an open covering U,

then we may ¬nd an S-partition of unity (fU )U ∈U with carr(fU ) ⊆ U .

Proof. For every f ∈ F we choose a Uf ∈ U with carr(f ) ∈ Uf . For U ∈ U put

FU := {f : Uf = U } and let fU := f ∈FU f ∈ S.

16.2. Proposition. Characterization of smooth normality. Let X be a

Hausdor¬ space with S ⊆ C(X, R) as in (14.1) Consider the following statements:

(1) X is S-normal;

(2) For any two closed disjoint subsets Ai ⊆ X there is a function f ∈ S with

f |A0 = 0 and 0 ∈ f (A1 );

/

(3) Every locally ¬nite open covering admits S-partitions of unity subordinated

to it.

(4) For any two disjoint zero-sets A0 and A1 of continuous functions there exists

a function g ∈ S with g|Aj = j for j = 0, 1 and g(X) ⊆ [0, 1];

16.2

166 Chapter III. Partitions of unity 16.2

(5) For any continuous function f : X ’ R there exists a function g ∈ S with

f ’1 (0) ⊆ g ’1 (0) ⊆ f ’1 (R \ {1}).

(6) The set S is dense in the algebra of continuous functions with respect to the

topology of uniform convergence;

(7) The set of all bounded functions in S is dense in the algebra of continuous

bounded functions on X with respect to the supremum norm;

ˇ

(8) The bounded functions in S separate points in the Stone-Cech-compacti¬ca-

tion βX of X.

The statements (1)-(3) are equivalent, and (4)-(8) are equivalent as well. If X is

metrizable all statements are equivalent.

If every open set is the carrier set of a smooth function then X is S-normal. If X

is S-normal, then it is S-regular.

A space is S-paracompact if and only if it is paracompact and S-normal.

Proof. (2) ’ (1). By assumption, there is a smooth function f0 with f0 |A1 = 0

and 0 ∈ f0 (A0 ), and again by assumption, there is a smooth function f1 with

/

f1 |A0 = 0 and 0 ∈ f1 ({x : f0 (x) = 0}). The function f = f0f1 1 has the required

/ +f

properties.

(1) ’ (2) is obvious.

(3) ’ (1) Let A0 and A1 be two disjoint closed subset. Then U := {X \ A1 , X \ A0 }

admits a S-partition of unity F subordinated to it, and

{f ∈ F : carr f ⊆ X \ A0 }

is the required bump function.

(1) ’ (3) Let U be a locally ¬nite covering of X. The space X is S-normal, so

its topology is also normal, and therefore for every U ∈ U there exists an open set

VU such that VU ⊆ U and {VU : U ∈ U} is still an open cover. By assumption,

there exist smooth functions gU ∈ S such that VU ⊆ carr(gU ) ⊆ U , cf. (16.1). The

function g := U gU is well de¬ned, positive, and smooth since U is locally ¬nite,

and {fU := gU /g : U ∈ U} is the required partition of unity.

’1

(5) ’ (4) Let Aj := fj (aj ) for j = 0, 1. By replacing fj by (fj ’ aj )2 we may

’1

assume that fj ≥ 0 and Aj = fj (0). Then (f1 + f2 )(x) > 0 for all x ∈ X, since

A1 ©A2 = …. Thus, f := f0f0 1 is a continuous function in C(X, [0, 1]) with f |Aj = j

+f

for j = 0, 1.

Now we reason as in ((2) ’ (1)): By (4) there exists a g0 ∈ S with A0 ⊆ f ’1 (0) ⊆

’1

g0 (0) ⊆ f ’1 (R \ {1}) = X \ f ’1 (1) ⊆ X \ A1 . By replacing g0 by g0 we may

2

assume that g0 ≥ 0.

’1

Applying the same argument to the zero-sets A1 and g0 (0) we obtain a g1 ∈ S

’1 ’1

with A1 ⊆ g1 (0) ⊆ X \ g0 (0). Thus, (g0 + g1 )(x) > 0, and hence g := g0g0 1 ∈ S

+g

satis¬es g|Aj = j for j = 0, 1 and g(X) ⊆ [0, 1].

(4) ’ (6) Let f be continuous. Without loss of generality we may assume f ≥ 0

(decompose f = f+ ’ f’ ). Let µ > 0. Then choose gk ∈ S with image in [0, 1], and

gk (x) = 0 for all x with f (x) ¤ k µ, and gk (x) = 1 for all x with f (x) ≥ (k + 1) µ.

16.2

16.2 16. Smooth partitions of unity and smooth normality 167

Let k be the largest integer less or equal to f (x) . Then gj (x) = 1 for all j < k, and

µ

gj (x) = 0 for all j > k. Hence, the sum g := µ k∈N gk ∈ S is locally ¬nite, and

|f (x) ’ g(x)| < 2 µ.

(6) ’ (7) This is obvious, since for any given bounded continuous f and for any

µ > 0, by (6) there exists g ∈ S with |f (x) ’ g(x)| < µ for all x ∈ X, hence

f ’ g ∞ ¤ µ and g ∞ ¤ f ∞ + f ’ g ∞ < ∞.

(7) ” (8) This follows from the Stone-Weierstraß theorem, since obviously the

bounded functions in S form a subalgebra in Cb (X) = C(βX). Hence, it is dense

if and only if it separates points in the compact space βX.

(7) ’ (4) By cutting o¬ f at 0 and at 1, we may assume that f is bounded. By

(7) there exists a bounded g0 ∈ S with f ’ g0 ∞ < 1 . Let h ∈ C ∞ (R, R) be

2

1

such that h(t) = 0 ” t ¤ 2 . Then g := h —¦ g0 ∈ S, and f (x) = 0 ’ g0 (x) ¤

|g0 (x)| ¤ |f (x)| + f ’ g0 ∞ ¤ 1 ’ g(x) = h(g0 (x)) = 0 and also f (x) = 1 ’

2

g0 (x) ≥ f (x) ’ f ’ g0 ∞ > 1 ’ 2 = 1 ’ g(x) = 0.

1

2

If X is metrizable and A ⊆ X is closed, then dist( , A) : x ’ sup{dist(x, a) : a ∈

A} is a continuous function with f ’1 (0) = A. Thus, (1) and (4) are equivalent.

Let every open subset be the carrier of a smooth mapping, and let A0 and A1 be

closed disjoint subsets of X. By assumption, there is a smooth function f with

carr(f ) = X \ A0 .

Obviously, every S-normal space is S-regular. Take as second closed set in (2) a

single point. If we take instead the other closed set as single point, then we have

what has been called small zero-sets in (19.8).

That a space is S-paracompact if and only if it is paracompact and S-normal can

be shown as in the proof that a paracompact space admits continuous partitions of

unity, see [Engelking, 1989, 5.1.9].

In [Kriegl, Michor, Schachermayer, 1989] it is remarked that in an uncountable

product of real lines there are open subsets, which are not carrier sets of continuous

functions.

Corollary. Denseness of smooth functions. Let X be S-paracompact, let F

be a convenient vector space, and let U ⊆ X — F be open such that for all x ∈ X

the set ι’1 (U ) ⊆ F is convex and non-empty, where ιx : F ’ X — F is given by

x

y ’ (x, y). Then there exists an f ∈ S whose graph is contained in U .

Under the following assumption this result is due to [Bonic, Frampton, 1966]: For

U := {(x, y) : p(y ’ g(x)) < µ(x)}, where g : X ’ F , µ : X ’ R+ are continuous

and p is a continuous seminorm on F .

Proof. For every x ∈ X let yx be chosen such that (x, yx ) ∈ U . Next choose open

neighborhoods Ux of x such that Ux — {yx } ⊆ U . Since X is S-paracompact there

exists a S-partition of unity F subordinated to the covering {Ux : x ∈ X}. In

particular, for every • ∈ F there exists an x• ∈ X with carr • ⊆ Ux• . Now de¬ne

16.2

168 Chapter III. Partitions of unity 16.4

yx• •. Then f ∈ S and for every x ∈ X we have

f := •∈F

yx• •(x) ∈ ι’1 (U ),

f (x) = yx• •(x) = x

x∈carr •

•∈F

since ι’1 (U ) is convex, contains yx• for x ∈ carr(•) ⊆ Ux• , and •(x) ≥ 0 with

x

1 = • •(x) = x∈carr • •(x).

16.3. Lemma. Lip2 -functions on Rn . [Wells, 1973]. Let B ∈ N and A := {x ∈

RN : xi ¤ 0 for all i and x ¤ 1}. Suppose that f ∈ CB (RN , R) with f |A = 0 and

3

f (x) ≥ 1 for all x with dist(x, A) ≥ 1. Then N < B 2 + 36 B 4 .

Proof. Suppose N ≥ B 2 +36 B 4 . We may assume that f is symmetric by replacing

f with x ’ N ! σ f (σ — x), where σ runs through all permutations, and σ — just

1

permutes the coordinates. Consider points xj ∈ RN for j = 0, . . . , B 2 of the form

xj = 1 1 1 1

B , . . . , B , ’ B , . . . , ’ B , 0, . . . , 0 .

>36 B 4

B 2 ’j

j

2

Then xj = 1, x0 ∈ A and d(xB , A) ≥ 1. Since f is symmetric and y j :=

1 j j+1

) has vanishing j, B 2 + 1, . . . , N coordinates, we have for the partial

2 (x + x

derivatives ‚j f (y j ) = ‚k f (y j ) for k = B 2 + 1, . . . , N . Thus

N

f (y j ) 2

f (y j ) 2

1 1

2

|‚j f (y j )|2 = j 2

|‚k f (y )| ¤ ¤

= ,

N ’ B2 36 B 4 36 B 4 36 B 2

k=B 2 +1

since from f |A = 0 we conclude that f (0) = f (0) = f (0) = f (0) and hence

f (j) (h) ¤ B h 3’j for j ¤ 3, see (15.6).

From |f (x + h) ’ f (x) ’ f (x)(h) ’ 1 f (x)(h2 )| ¤ B 3! h

1 3

we conclude that

2

|f (x + h) ’ f (x ’ h)| ¤ |f (x + h) ’ f (x) ’ f (x)(h) ’ 1 f (x)(h2 )|

2

+ |f (x ’ h) ’ f (x) + f (x)(h) ’ 1 f (x)(h2 )|

2

+ 2|f (x)(h)|

3

2

¤ Bh + 2|f (x)(h)|.

3!

1

If we apply this to x = y j and h = B ej , where ej denotes the j-th unit vector, then

we obtain

2 1 1 2

|f (xj+1 ) ’ f (xj )| ¤ B 3 + 2|‚j f (y j )| ¤ .

3B 2

3! B B

2 2

2

Summing up yields 1 ¤ |f (xB )| = |f (xB ) ’ f (x0 )| ¤ < 1, a contradiction.

3

16.4. Corollary. 2 is not Lip2 -normal. [Wells, 1973]. Let A0 := {x ∈ 2 :

glob

xj ¤ 0 for all j and x ¤ 1} and A1 := {x ∈ 2 : d(x, A) ≥ 1} and f ∈ C 3 ( 2 , R)

with f |Aj = j for j = 0, 1. Then f (3) is not bounded.

Proof. By the preceding lemma a bound B of f (3) must satisfy for f restricted to

RN , that N < B 2 + 36B 4 . This is not for all N possible.

16.4

16.6 16. Smooth partitions of unity and smooth normality 169

16.5. Corollary. Whitney™s extension theorem is false on 2 . [Wells, 1973].

Let E := R — 2 ∼ 2 and π : E ’ R be the projection onto the ¬rst factor.

=

For subsets A ⊆ 2 consider the cone CA := {(t, ta) : t ≥ 0, a ∈ A} ⊆ E. Let

A := C(A0 ∪ A1 ) with A0 and A1 as in (16.4). Let a jet (f j ) on A be de¬ned by

f j = 0 on the cone CA1 and f j (x)(v 1 , . . . , v j ) = h(j) (π(x))(π(v 1 ), . . . , π(v j )) for

all x in the cone of CA0 , where h ∈ C ∞ (R, R) is in¬nite ¬‚at at 0 but with h(t) = 0

for all t = 0. This jet has no C 3 -prolongation to E.

Proof. Suppose that such a prolongation f exists. Then f (3) would be bounded

1

locally around 0, hence fa (x) := 1 ’ h(a) f (a, ax) would be a CB function on 2 for

3

small a, which is 1 on A1 and vanishes on A0 . This is a contradiction to (16.4).

So it remains to show that the following condition of Whitney (22.2) is satis¬ed:

k’j

1 j+i

f j (y) ’ f (x)(y ’ x)j = o( x ’ y k’j

x, y ’ a.

) for A

i!

i=0

j j

Let f1 := 0 and f0 (x) := h(j) (π(x)) —¦ (π — . . . — π). Then both are smooth on R • 2 ,

and thus Whitney™s condition is satis¬ed on each cone separately. It remains to

show this when x is in one cone and y in the other and both tend to 0. Thus,

we have to replace f at some places by f1 and at others by f0 . Since h is in¬nite

j

¬‚at at 0 we have f0 (z) = o( z n ) for every n. Furthermore for xi ∈ CAi for

i = 0, 1 we have that x1 ’ x0 ≥ sin(arctan 2 ’ arctan 1) max{ x0 , x1 }. Thus,

j j

we may replace f0 (y) by f1 (y) and vice versa. So the condition is reduced to the

case, where y and z are in the same cone CAi .

16.6. Lemma. Smoothly regular strict inductive limits. Let E be the strict

inductive limit of a sequence of C ∞ -normal convenient vector spaces En such that

En ’ En+1 is closed and has the extension property for smooth functions. Then

E is C ∞ -regular.

Proof. Let U be open in E and 0 ∈ U . Then Un := U © En is open in En . We