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.