‚ ‚

For any s ∈ R we have f (s v)(v) = ‚t |t=0 f (s v ‚t |t=s t f (v)

+ tv) = = f (v).

Using this homogeneity we show next, that it is also continuously di¬erentiable

along continuously di¬erentiable curves. So we have to show that (f —¦ c) (t) ’

(f —¦ c) (0) for t ’ 0. Again only the case c(0) = 0 is interesting. As before we

factor c as c(t) = t c1 (t). In the case, where c (0) = c1 (0) = 0 we have for t = 0

that

(f —¦ c) (t) ’ (f —¦ c) (0) = f (t c1 (t))(c (t)) ’ f (0)(c1 (0))

= f (c1 (t))(c (t)) ’ f (c1 (0))(c1 (0))

= f (c1 (t))(c (t)) ’ f (c1 (0))(c (0)),

which converges to 0 for t ’ 0, since (f )§ is continuous (and even smooth) on

(R2 \ {0}) — R2 .

In the other case, where c (0) = c1 (0) = 0 we consider ¬rst the values of t, for which

c(t) = 0. Then

(f —¦ c) (t) ’ (f —¦ c) (0) = f (0)(c (t)) ’ f (0)(c (0))

= f (c (t)) ’ f (c (0)) ’ 0,

since f is continuous. For the remaining values of t, where c(t) = 0, we factor

c(t) = |c(t)| e(t), with e(t) ∈ {x : x = 1}. Then

(f —¦ c) (t) ’ (f —¦ c) (0) = f (e(t))(c (t)) ’ 0 ’ 0,

since f (x)(c (t)) ’ 0 for t ’ 0 uniformly for x = 1, since c (t) ’ 0.

Furthermore, f —¦ c is smooth for all c which are smooth and nowhere in¬nitely

¬‚at. In fact, a smooth curve c with c(k) (0) = 0 for k < n can be factored as

c(t) = tn cn (t) with smooth cn , by Taylor™s formula with integral remainder. Since

c(n) (0) = n! cn (0), we may assume that n is chosen maximal and hence cn (0) = 0.

But then (f —¦ c)(t) = tn · (f —¦ cn )(t), and f —¦ cn is smooth.

A completely analogous argument shows also that f —¦ c is real analytic for all real

analytic curves c : R ’ R2 .

However, let us show that f —¦c is not Lipschitz di¬erentiable even for smooth curves

c. For x = 0 we have

‚2 ‚2

(‚2 )2 f (x, 0) = 1

|s=0 f (x, s) |s=0 f (1, x s)

=x =

‚s ‚s

‚2

1 a

|s=0 f (1, s)

= =: = 0.

x ‚s x

3.3

26 Chapter I. Calculus of smooth mappings 3.4

Now we choose a smooth curve c which passes for each n in ¬nite time tn through

1 1

( n2n+1 , 0) with locally constant velocity vector (0, nn ), by (2.10). Then for small t

we get

(f —¦ c) (tn + t) = ‚1 f (c(tn + t)) pr1 (c (tn + t)) +‚2 f (c(tn + t)) pr2 (c (tn + t))

=0

n2n+1

2 2

(f —¦ c) (tn ) = (‚2 ) f (c(tn )) (pr2 (c (tn ))) + 0 = a 2n = n a,

n

which is unbounded.

So although preservation of (continuous) di¬erentiability of curves is not enough to

ensure di¬erentiability of a function R2 ’ R, we now prove that smoothness can

be tested with smooth curves.

3.4. Boman™s theorem. [Boman, 1967] For a mapping f : R2 ’ R the following

assertions are equivalent:

(1) All iterated partial derivatives exist and are continuous.

(2) All iterated partial derivatives exist and are locally bounded.

(3) For v ∈ R2 the iterated directional derivatives

dn f (x) := ( ‚t )n |t=0 (f (x + tv))

‚

v

exist and are locally bounded with respect to x.

(4) For all smooth curves c : R ’ R2 the composite f —¦ c is smooth.

Proof. (1) ’ (2) is obvious.

(2) ’ (1) follows immediately, since the local boundedness of ‚1 f and ‚2 f imply

the continuity of f (see also the proof of (3.2)):

1 1

f (t, s) ’ f (0, 0) = t ‚1 f („ t, s)d„ + s ‚2 f (0, σs)dσ.

0 0

(1) ’ (4) is a direct consequence of the chain rule, namely that (f —¦ c) (t) =

‚1 f (c(t)) · x (t) + ‚2 f (c(t)) · y (t), where c = (x, y).

d

(4) ’ (3) Obviously, dp f (x) := ( dt )p |t=0 f (x + tv) exists, since t ’ x + tv is a

v

smooth curve. Suppose dp f is not locally bounded. So we may ¬nd a sequence

v

2

xn which converges fast to x, and such that |dp f (xn )| ≥ 2n . Let c be a smooth

v

t

curve with c(t + tn ) = xn + 2n v locally for some sequence tn ’ 0, by (2.8). Then

1

(f —¦ c)(p) (tn ) = dp f (xn ) 2np is unbounded, which is a contradiction.

v

(3) ’ (1) First we claim that dp f is continuous. We prove this by induction on

v

1 p+1

p p

p: dv f ( +tv) ’ dv f ( ) = t 0 dv f ( +t„ v)d„ ’ 0 for t ’ 0 uniformly on

bounded sets. Suppose now that |dp f (xn ) ’ dp f (x)| ≥ µ for some sequence xn ’ x.

v v

Without loss of generality we may assume that dp f (xn ) ’ dp f (x) ≥ µ. Then by the

v v

3.4

3.5 3. Smooth mappings and the exponential law 27

µ

uniform convergence there exists a δ > 0 such that dp f (xn + tv) ’ dp f (x + tv) ≥

v v 2

δ

for |t| ¤ δ. Integration 0 dt yields

dp’1 f (xn + δv) ’ dp’1 f (xn ) ’ dp’1 f (x + δv) ’ dp’1 f (x) ≥ µδ

2,

v v v v

but by induction hypothesis the left hand side converges towards

dp’1 f (x + δv) ’ dp’1 f (x) ’ dp’1 f (x + δv) ’ dp’1 f (x) = 0.

v v v v

To complete the proof we use convolution by an approximation of unity. So let

• ∈ C ∞ (R2 , R) have compact support, • = 1, and •(y) ≥ 0 for all y. De¬ne

•µ (x) := µ1 •( 1 x), and let

2 µ

f (x ’ y) •µ (y) dy = f (x ’ µy)•(y)dy.

fµ (x) := (f •µ )(x) =

R2 R2

Since the convolution fµ := f •µ of a continuous function f with a smooth function

•µ with compact support is di¬erentiable with directional derivative dv (f •µ )(x) =

(f dv •µ )(x), we obtain that fµ is smooth. And since f •µ ’ f in C(R2 , R) for

µ ’ 0 and any continuous function f , we conclude dp fµ = dp f •µ ’ dp f uniformly

v v v

on compact sets.

We remark now that for a smooth map g : R2 ’ R we have by the chain rule

d

g(x + tv) = ‚1 g(x + tv) · v1 + ‚2 g(x + tv) · v2

dv g(x + tv) =

dt

and by induction that

p

p i p’i i p’i

dp g(x) = v v ‚1 ‚2 g(x).

v

i 12

j=0

i p’i

Hence, we can calculate the iterated derivatives ‚1 ‚2 g(x) for 0 ¤ i ¤ p from

p + 1 many derivatives dpj g(x) provided the v j are chosen in such a way, that the

v

j j

Vandermonde™s determinant det((v1 )i (v2 )p’i )ij = 0. For this choose v2 = 1 and all

j j j

the v1 pairwise distinct, then det((v1 )i (v2 )p’i )ij = j>k (v1 ’ v1 ) = 0.

k

Hence, the iterated derivatives of fµ are linear combinations of the derivatives dp fµ

v

for p + 1 many vectors v, where the coe¬cients depend only on the v™s. So we

conclude that the iterated derivatives of fµ form a Cauchy sequence in C(R2 , R),

and hence converge to continuous functions f ± . Thus, all iterated derivatives ‚ ± f

of f exist and are equal to these continuous functions f ± , by the following lemma

(3.5).

3.5. Lemma. Let fµ ’ f in C(R2 , R) and dv fµ ’ fv in C(R2 , R). Then dv f

exists and equals fv .

Proof. We have to show that for ¬xed x, v ∈ R2 the curve

f (x+tv)’f (x)

for t = 0

t

c:t’

fv (x) otherwise

3.5

28 Chapter I. Calculus of smooth mappings 3.9

is continuous from R ’ R. The corresponding curve cµ for fµ can be rewritten

1

as cµ (t) = 0 dv fµ (x + „ t v) d„ , which converges by assumption uniformly for t in

1

compact sets to the continuous curve t ’ 0 fv (x + „ t v) d„ . Pointwise it converges

to c(t), hence c is continuous.

For the vector valued case of the exponential law we need a locally convex structure

on C ∞ (R, E).

3.6. De¬nition. Space of curves. Let C ∞ (R, E) be the locally convex vector

space of all smooth curves in E, with the pointwise vector operations, and with

the topology of uniform convergence on compact sets of each derivative separately.

dk

This is the initial topology with respect to the linear mappings C ∞ (R, E) ’ ’

C ∞ (R, E) ’ ∞ (K, E), where k runs through N, where K runs through all compact

subsets of R, and where ∞ (K, E) carries the topology of uniform convergence, see

also (2.15).

Note that the derivatives dk : C ∞ (R, E) ’ C ∞ (R, E), the point evaluations evt :

C ∞ (R, E) ’ E and the pull backs g — : C ∞ (R, E) ’ C ∞ (R, E) for all g ∈ C ∞ (R, R)

are continuous and linear.

3.7. Lemma. A space E is c∞ -complete if and only if C ∞ (R, E) is.

Proof. (’) The mapping c ’ (c(n) )n∈N is by de¬nition an embedding of C ∞ (R, E)

into the c∞ -complete product n∈N ∞ (R, E). Its image is a closed subspace, since

the previous lemma can be easily generalized to curves c : R ’ E.

(⇐) Consider the continuous linear mapping const : E ’ C ∞ (R, E) given by

x ’ (t ’ x). It has as continuous left inverse the evaluation at any point (e.g. ev0 :

C ∞ (R, E) ’ E, c ’ c(0)). Hence, E can be identi¬ed with the closed subspace of

C ∞ (R, E) given by the constant curves, and is thereby itself c∞ -complete.

3.8. Lemma. Curves into limits. A curve into a c∞ -closed subspace of a space

is smooth if and only if it is smooth into the total space. In particular, a curve is

smooth into a projective limit if and only if all its components are smooth.

Proof. Since the derivative of a smooth curve is the Mackey limit of the di¬erence

quotient, the c∞ -closedness implies that this limit belongs to the subspace. Thus,

we deduce inductively that all derivatives belong to the subspace, and hence the

curve is smooth into the subspace.

The result on projective limits now follows, since obviously a curve is smooth into

a product, if all its components are smooth.

We show now that the bornology, but obviously not the topology, on function spaces

can be tested with the linear functionals on the range space.

3.9. Lemma. Bornology of C ∞ (R, E). The family

: C ∞ (R, E) ’ C ∞ (R, R) : ∈ E — }

{ —

3.9

3.10 3. Smooth mappings and the exponential law 29

generates the bornology of C ∞ (R, E). This also holds for E — replaced by E .

A set in C ∞ (R, E) is bounded if and only if each derivative is uniformly bounded

on compact subsets.

Proof. A set B ⊆ C ∞ (R, E) is bounded if and only if the sets {dn c(t) : t ∈ I, c ∈ B}

are bounded in E for all n ∈ N and compact subsets I ‚ R.

This is furthermore equivalent to the condition that the set { (dn c(t)) = dn ( —¦c)(t) :

t ∈ I, c ∈ B} is bounded in R for all ∈ E — , n ∈ N, and compact subsets I ‚ R

and in turn equivalent to: { —¦ c : c ∈ B} is bounded in C ∞ (R, R).

For E — replaced by E ⊇ E — the statement holds, since ∈E

is bounded for all

—

by the explicit description of the bounded sets.

3.10. Proposition. Vector valued simplest exponential law. For a map-

ping f : R2 ’ E into a locally convex space (which need not be c∞ -complete) the

following assertions are equivalent:

(1) f is smooth along smooth curves.

All iterated directional derivatives dp f exist and are locally bounded.

(2) v

±

(3) All iterated partial derivatives ‚ f exist and are locally bounded.

f ∨ : R ’ C ∞ (R, E) exists as a smooth curve.

(4)

Proof. We prove this result ¬rst for c∞ -complete spaces E. Then each of the

statements (1-4) are valid if and only if the corresponding statement for —¦ f is

valid for all ∈ E — . Only (4) needs some arguments: In fact, f ∨ (t) ∈ C ∞ (R, E)

if and only if — (f ∨ (t)) = ( —¦ f )∨ (t) ∈ C ∞ (R, R) for all ∈ E — by (2.14). Since

C ∞ (R, E) is c∞ -complete, its bornologically isomorphic image in ∞

∈E — C (R, R)

is c∞ -closed. So f ∨ : R ’ C ∞ (R, E) is smooth if and only if — —¦ f ∨ = ( —¦ f )∨ :

R ’ C ∞ (R, R) is smooth for all ∈ E — . So the proof is reduced to the scalar valid

case, which was proved in (3.2) and (3.4).

Now the general case. For the existence of certain derivatives we know by (1.9) that

it is enough that we have some candidate in the space, which is the corresponding

derivative of the map considered as map into the c∞ -completion (or even some

larger space). Since the derivatives required in (1-4) depend linearly on each other,

this is true. In more detail this means:

(1) ’ (2) is obvious.

|±|

(2) ’ (3) is the fact that ‚ ± is a universal linear combination of dv f .

(3) ’ (1) follows from the chain rule which says that (f —¦c)(p) (t) is a universal linear

(p )

(p )

combination of ‚i1 . . . ‚iq f (c(t))ci1 1 (t) . . . ciq q (t) for ij ∈ {1, 2} and pj = p, see

also (10.4).

(3) ” (4) holds by (1.9) since (‚1 f )∨ = d(f ∨ ) and (‚2 f )∨ = d —¦ f ∨ = d— (f ∨ ).

For the general case of the exponential law we need a notion of smooth mappings

and a locally convex topology on the corresponding function spaces. Of course, it

3.10

30 Chapter I. Calculus of smooth mappings 3.12

would be also handy to have a notion of smoothness for locally de¬ned mappings.

Since the idea is to test smoothness with smooth curves, such curves should have

locally values in the domains of de¬nition, and hence these domains should be

c∞ -open.

3.11. De¬nition. Smooth mappings and spaces of them. A mapping f :

E ⊇ U ’ F de¬ned on a c∞ -open subset U is called smooth (or C ∞ ) if it maps

smooth curves in U to smooth curves in F .

Let C ∞ (U, F ) denote the locally convex space of all smooth mappings U ’ F with

pointwise linear structure and the initial topology with respect to all mappings

c— : C ∞ (U, F ) ’ C ∞ (R, F ) for c ∈ C ∞ (R, U ).

For U = E = R this coincides with our old de¬nition. Obviously, any composition

of smooth mappings is also smooth.

Lemma. The space C ∞ (U, F ) is the (inverse) limit of spaces C ∞ (R, F ), one for

each c ∈ C ∞ (R, U ), where the connecting mappings are pull backs g — along repa-

rameterizations g ∈ C ∞ (R, R).

Note that this limit is the closed linear subspace in the product

C ∞ (R, F )

c∈C ∞ (R,U )

consisting of all (fc ) with fc—¦g = fc —¦ g for all c and all g ∈ C ∞ (R, R).

Proof. The mappings c— : C ∞ (U, F ) ’ C ∞ (R, F ) de¬ne a continuous linear em-

g—

bedding C (U, F ) ’ limc {C (R, F ) ’ C ∞ (R, F )}, since c— (f ) —¦ g = f —¦ c —¦ g =

∞ ∞

’

(c —¦ g) (f ). It is surjective since for any (fc ) ∈ limc C ∞ (R, F ) one has fc = f —¦ c

—

where f is de¬ned as x ’ fconstx (0).

3.12. Theorem. Cartesian closedness. Let Ui ⊆ Ei be c∞ -open subsets in

locally convex spaces, which need not be c∞ -complete. Then a mapping f : U1 —

U2 ’ F is smooth if and only if the canonically associated mapping f ∨ : U1 ’

C ∞ (U2 , F ) exists and is smooth.

Proof. We have the following implications:

f ∨ : U1 ’ C ∞ (U2 , F ) is smooth.

f ∨ —¦c1 : R ’ C ∞ (U2 , F ) is smooth for all smooth curves c1 in U1 , by (3.11).

”

c— —¦ f ∨ —¦ c1 : R ’ C ∞ (R, F ) is smooth for all smooth curves ci in Ui , by

” 2

(3.11) and (3.8).

f —¦ (c1 — c2 ) = (c— —¦ f ∨ —¦ c1 )§ : R2 ’ F is smooth for all smooth curves ci

” 2

in Ui , by (3.10).

” f : U1 — U2 ’ F is smooth.

Here the last equivalence is seen as follows: Each curve into U1 — U2 is of the form

(c1 , c2 ) = (c1 — c2 ) —¦ ∆, where ∆ is the diagonal mapping. Conversely, f —¦ (c1 —

c2 ) : R2 ’ F is smooth for all smooth curves ci in Ui , since the product and the

composite of smooth mappings is smooth by (3.11) (and by (3.4)).

3.12

3.15 3. Smooth mappings and the exponential law 31

3.13. Corollary. Consequences of cartesian closedness. Let E, F , G, etc. be

locally convex spaces, and let U , V be c∞ -open subsets of such. Then the following

canonical mappings are smooth.

ev : C ∞ (U, F ) — U ’ F , (f, x) ’ f (x);

(1)

ins : E ’ C ∞ (F, E — F ), x ’ (y ’ (x, y));

(2)

( )§ : C ∞ (U, C ∞ (V, G)) ’ C ∞ (U — V, G);

(3)

( )∨ : C ∞ (U — V, G) ’ C ∞ (U, C ∞ (V, G));

(4)

comp : C ∞ (F, G) — C ∞ (U, F ) ’ C ∞ (U, G), (f, g) ’ f —¦ g;

(5)

C ∞ ( , ) : C ∞ (E2 , E1 ) — C ∞ (F1 , F2 ) ’

(6)

’ C ∞ (C ∞ (E1 , F1 ), C ∞ (E2 , F2 )), (f, g) ’ (h ’ g —¦ h —¦ f );

: C ∞ (Ei , Fi ) ’ C ∞ ( Ei , Fi ), for any index set.

(7)

Proof. (1) The mapping associated to ev via cartesian closedness is the identity

on C ∞ (U, F ), which is C ∞ , thus ev is also C ∞ .

(2) The mapping associated to ins via cartesian closedness is the identity on E — F ,

hence ins is C ∞ .

(3) The mapping associated to ( )§ via cartesian closedness is the smooth com-

position of evaluations ev —¦(ev — Id) : (f ; x, y) ’ f (x)(y).

(4) We apply cartesian closedness twice to get the associated mapping (f ; x; y) ’

f (x, y), which is just a smooth evaluation mapping.

(5) The mapping associated to comp via cartesian closedness is (f, g; x) ’ f (g(x)),

which is the smooth mapping ev —¦(Id — ev).

(6) The mapping associated to the one in question by applying cartesian closed twice

is (f, g; h, x) ’ g(h(f (x))), which is the C ∞ -mapping ev —¦(Id — ev) —¦ (Id — Id — ev).

(7) Up to a ¬‚ip of factors the mapping associated via cartesian closedness is the

product of the evaluation mappings C ∞ (Ei , Fi ) — Ei ’ Fi .

Next we generalize (3.4) to ¬nite dimensions.

3.14. Corollary. [Boman, 1967]. The smooth mappings on open subsets of Rn in

the sense of de¬nition (3.11) are exactly the usual smooth mappings.

Proof. Both conditions are of local nature, so we may assume that the open subset

of Rn is an open box and in turn even Rn itself.

(’) If f : Rn ’ F is smooth then by cartesian closedness (3.12), for each coordinate

the respective associated mapping f ∨i : Rn’1 ’ C ∞ (R, F ) is smooth, so again by

(3.12) we have ‚i f = (d— f ∨i )§ , so all ¬rst partial derivatives exist and are smooth.

Inductively, all iterated partial derivatives exist and are smooth, thus continuous,

so f is smooth in the usual sense.

(⇐) Obviously, f is smooth along smooth curves by the usual chain rule.

3.15. Di¬erentiation of an integral. We return to the question of di¬erentiat-

ing an integral. So let f : E — R ’ F be smooth, and let F be the completion of

the locally convex space F . Then we may form the function f0 : E ’ F de¬ned by

3.15

32 Chapter I. Calculus of smooth mappings 3.16

1

x ’ 0 f (x, t) dt. We claim that it is smooth, and that the directional derivative is

1

given by dv f0 (x) = 0 dv (f ( , t))(x) dt. By cartesian closedness (3.12) the associ-

1

ated mapping f ∨ : E ’ C ∞ (R, F ) is smooth, so the mapping 0 —¦f ∨ : E ’ F is

smooth since integration is a bounded linear operator, and

1

‚ ‚

dv f0 (x) = f0 (x + sv) = f (x + sv, t)dt

‚s s=0 ‚s s=0

0

1 1

‚

= f (x + sv, t)dt = dv (f ( , t))(x) dt.

‚s s=0

0 0

But we want to generalize this to functions f de¬ned only on some c∞ -open subset

U ⊆ E—R, so we have to show that the natural domain U0 := {x ∈ E : {x}—[0, 1] ⊆

U } of f0 is c∞ -open in E. We will do this in lemma (4.15). From then on the proof

runs exactly the same way as for globally de¬ned functions. So we obtain the

Proposition. Let f : E — R ⊇ U ’ F be smooth with c∞ -open U ⊆ E — R. Then

1

x ’ 0 f (x, t) dt is smooth on the c∞ -open set U0 := {x ∈ E : {x} — [0, 1] ⊆ U }

1

with values in the completion F and dv f0 (x) = 0 dv (f ( , t))(x) dt for all x ∈ U0

and v ∈ E.

Now we want to de¬ne the derivative of a general smooth map and prove the chain

rule for them.

3.16. Corollary. Smoothness of the di¬erence quotient. For a smooth curve

c : R ’ E the di¬erence quotient

c(t) ’ c(s)

±

for t = s

t’s

(t, s) ’

c (t) for t = s

is a smooth mapping R2 ’ E.

1

c(t)’c(s)

Proof. By (2.5) we have f : (t, s) ’ c (s + r(t ’ s))dr, and by (3.15)

=

t’s 0

it is smooth R2 ’ E. The left hand side has values in E, and for t = s this is also

true for all iterated directional derivatives. It remains to consider the derivatives

for t = s. The iterated directional derivatives are given by (3.15) as

1

dp f (t, s) dp c (s + r(t ’ s)) dr

=

v v

0

1

dp c (s + r(t ’ s)) dr,

= v

0

where dv acts on the (t, s)-variable. The later integrand is for t = s just a linear

combination of derivatives of c which are independent of r, hence dp f (t, s) ∈ E. By

v

(3.10) the mapping f is smooth into E.

3.16

3.18 3. Smooth mappings and the exponential law 33

3.17. De¬nition. Spaces of linear mappings. Let L(E, F ) denote the space

of all bounded (equivalently smooth by (2.11)) linear mappings from E to F . It is

a closed linear subspace of C ∞ (E, F ) since f is linear if and only if for all x, y ∈ E

and » ∈ R we have (evx +» evy ’ evx+»y )f = 0. We equip it with this topology

and linear structure.

So a mapping f : U ’ L(E, F ) is smooth if and only if the composite mapping

f

U ’ L(E, F ) ’ C ∞ (E, F ) is smooth.

’

3.18. Theorem. Chain rule. Let E and F be locally convex spaces, and let

U ⊆ E be c∞ -open. Then the di¬erentiation operator

d : C ∞ (U, F ) ’ C ∞ (U, L(E, F )),

f (x + tv) ’ f (x)

df (x)v := lim ,

t

t’0

exists, is linear and bounded (smooth). Also the chain rule holds:

d(f —¦ g)(x)v = df (g(x))dg(x)v.

Proof. Since t ’ x+tv is a smooth curve we know that d§§ : C ∞ (U, F )—U —E ’

F exists. We want to show that it is smooth, so let (f, x, v) : R ’ C ∞ (U, F )—U —E

be a smooth curve. Then

f (t)(x(t) + sv(t)) ’ f (t)(x(t))

d§§ (f (t), x(t), v(t)) = lim = ‚2 h(t, 0),

s

s’0

which is smooth in t, where the smooth mapping h : R2 ’ F is given by (t, s) ’

f § (t, x(t)+sv(t)). By cartesian closedness (3.12) the mapping d§ : C ∞ (U, F )—U ’

C ∞ (E, F ) is smooth.

Now we show that this mapping has values in the subspace L(E, F ): d§ (f, x)

is obviously homogeneous. It is additive, because we may consider the smooth

mapping (t, s) ’ f (x + tv + sw) and compute as follows, using (3.14).

‚

df (x)(v + w) = f (x + t(v + w))

‚t 0

‚ ‚

= f (x + tv + 0w) + f (x + 0v + tw) = df (x)v + df (x)w.

‚t 0 ‚t 0

So we see that d§ : C ∞ (U, F ) — U ’ L(E, F ) is smooth, and the mapping d :

C ∞ (U, F ) ’ C ∞ (U, L(E, F )) is smooth by (3.12) and obviously linear.

We ¬rst prove the chain rule for a smooth curve c instead of g. We have to show

that the curve

f (c(t))’f (c(0))

for t = 0

t

t’

df (c(0)).c (0) for t = 0

1

is continuous at 0. It can be rewritten as t ’ df (c(0) + s(c(t) ’ c(0))).c1 (t) ds,

0

where c1 is the smooth curve given by

c(t)’c(0)

for t = 0

t

t’ .

c (0) for t = 0

3.18

34 Chapter I. Calculus of smooth mappings 4.1

Since h : R2 ’ U — E given by

(t, s) ’ (c(0) + s(c(t) ’ c(0)), c1 (t))

is smooth, the map t ’ s ’ df (c(0) + s(c(t) ’ c(0))).c1 (t) is smooth R ’

1

C ∞ (R, F ), and hence t ’ df (c(0) + s(c(t) ’ c(0))).c1 (t) ds is smooth, and hence

0

continuous.

For general g we have

‚ ‚

d(f —¦ g)(x)(v) = (f —¦ g)(x + tv) = (df )(g(x + 0v))( ‚t (g(x + tv)))

‚t 0 0

= (df )(g(x))(dg(x)(v)).

3.19. Lemma. Two locally convex spaces are locally di¬eomorphic if and only if

they are linearly di¬eomorphic.

Any smooth and 1-homogeneous mapping is linear.

Proof. By the chain rule the derivatives at corresponding points give the linear

di¬eomorphisms.

‚

For a 1-homogeneous mapping f one has df (0)v = f (tv) = f (v), and this is

‚t 0

linear in v.

4. The c∞ -Topology

4.1. De¬nition. A locally convex vector space E is called bornological if and only

if the following equivalent conditions are satis¬ed:

(1) For any locally convex vector space F any bounded linear mapping T : E ’

F is continuous; it is su¬cient to know this for all Banach spaces F .

(2) Every bounded seminorm on E is continuous.

(3) Every absolutely convex bornivorous subset is a 0-neighborhood.

A radial subset U (i.e. [0, 1]U ⊆ U ) of a locally convex space E is called bornivorous

if it absorbs each bounded set, i.e. for every bounded B there exists r > 0 such that

[0, r]U ⊇ B.

Proof. (3 ’ 2), since for a > 0 the inverse images under bounded seminorms of

intervals (’∞, a) are absolutely convex and bornivorous. In fact, let B be bounded

and a > 0. Then by assumption p(B) is bounded, and so there exists a C > 0 with

p(B) ⊆ C · (’∞, a). Hence, B ⊆ C · p’1 (’∞, a).

(2 ’ 1), since p —¦ T is a bounded seminorm, for every continuous seminorm on F .

(2 ’ 3), since the Minkowski-functional p generated by an absolutely convex bor-

nivorous subset is a bounded seminorm.

(1 ’ 2) Since the canonical projection T : E ’ E/ker p is bounded, for any

bounded seminorm p, it is by assumption continuous. Hence, p = p—¦T is continuous,

˜

where p denotes the canonical norm on E/ker p induced from p.

˜

4.1

4. The c∞ -topology

4.3 35

4.2. Lemma. Bornologi¬cation. The bornologi¬cation Eborn of a locally convex

space can be described in the following equivalent ways:

(1) It is the ¬nest locally convex structure having the same bounded sets;

(2) It is the ¬nal locally convex structure with respect to the inclusions EB ’ E,

where B runs through all bounded (closed) absolutely convex subsets.

Moreover, Eborn is bornological. For any locally convex vector space F the contin-

uous linear mappings Eborn ’ F are exactly the bounded linear mappings E ’ F .

The continuous seminorms on Eborn are exactly the bounded seminorms of E. An

absolutely convex set is a 0-neighborhood in Eborn if and only if it is bornivorous,

i.e. absorbs bounded sets.

Proof. Let Eborn be the vector space E supplied with the ¬nest locally convex

structure having the same bounded sets as E.

(‘) Since all bounded absolutely convex sets B in E are bounded in Eborn , the

inclusions EB ’ Eborn are bounded and hence continuous. Thus, the ¬nal structure

on E induced by the inclusions EB ’ E is ¬ner than the structure of Eborn .

(“) Since every bounded subset of E is contained in some absolutely convex bounded

set B ⊆ E it has to be bounded in the ¬nal structure given by all inclusions

EB ’ E. Hence, this ¬nal structure has exactly the same bounded sets as E, and

we have equality between the ¬nal structure and that of Eborn .

A seminorm p on E is bounded, if and only if p(B) is bounded for all bounded B,

and this is exactly the case if p|EB is a bounded (=continuous) seminorm on EB

for all B, or equivalently that p is a continuous seminorm for the ¬nal structure

Eborn on E induced by the inclusions EB ’ E.

As a consequence, all bounded seminorms on Eborn are continuous, and hence Eborn

is bornological.

An absolutely convex subset U is a 0-neighborhood for the ¬nal structure induced

by EB ’ E if and only if U © EB is a 0-neighborhood, or equivalently if U absorbs

B, for all bounded absolutely convex B, i.e. U is bornivorous. All other assertions

follow from (4.1).

4.3. Corollary. Bounded seminorms. For a seminorm p and a sequence µn ’

∞ the following statements are equivalent:

(1) p is bounded;

(2) p is bounded on compact sets;

(3) p is bounded on M -converging sequences;

(4) p is bounded on µ-converging sequences;

p is bounded on images of bounded intervals under Lipk -curves (for ¬xed

(5)

0 ¤ k ¤ ∞).

The corresponding statement for subsets of E is the following. For a radial subset

U ⊆ E (i.e., [0, 1] · U ⊆ U ) the following properties are equivalent:

(1) U is bornivorous.

4.3

36 Chapter I. Calculus of smooth mappings 4.5

(2) For all absolutely convex bounded sets B, the trace U © EB is a 0-neighbor-

hood in EB .

(3) U absorbs all compact subsets in E.

(4) U absorbs all Mackey convergent sequences.

(4™) U absorbs all sequences converging Mackey to 0.

(5) U absorbs all µ-convergent sequences (for a ¬xed µ).

(5™) U absorbs all sequences which are µ-converging to 0.

(6) U absorbs the images of bounded sets under Lipk -curves (for a ¬xed 0 ¤

k ¤ ∞).

Proof. We prove the statement on radial subsets, for seminorms p it then follows

by considering the radial set U := {x ∈ E : p(x) ¤ 1} and using the equality

K · U = {x ∈ E : p(x) ¤ K}.

(1) ” (2) ’ (3) ’ (4) ’ (5) ’ (5™), (4) ’ (4™), (3) ’ (6), (4™) ’ (5™), are trivial.

(6) ’ (5™) Suppose that (xn ) is µ-converging to x but is not absorbed by U . Then

for each m ∈ N there is an nm with xnm ∈ mU and clearly we may suppose that

/

1/µnm is fast falling. The sequence (xnm )m is then fast falling and lies on some

compact part of a smooth curve by the special curve lemma (2.8). The set U

absorbs this by (6), a contradiction.

(5™) ’ (1) Suppose U does not absorb some bounded B. Hence, there are bn ∈ B

b

with bn ∈ µ2 U . However, µn is µ-convergent to 0, so it is contained in KU for

/n n

some K > 0. Equivalently, bn ∈ µn KU ⊆ µ2 U for all µn ≥ K, which gives a

n

contradiction.

4.4. Corollary. Bornologi¬cation as locally convex-i¬cation.

The bornologi¬cation of E is the ¬nest locally convex topology with one (hence all)

of the following properties:

(1) It has the same bounded sets as E.

(2) It has the same Mackey converging sequences as E.

(3) It has the same µ-converging sequences as E (for some ¬xed µ).

It has the same Lipk -curves as E (for some ¬xed 0 ¤ k ¤ ∞).

(4)

(5) It has the same bounded linear mappings from E into arbitrary locally convex

spaces.

(6) It has the same continuous linear mappings from normed spaces into E.

Proof. Since the bornologi¬cation has the same bounded sets as the original topol-

ogy, the other objects are also the same: they depend only on the bornology “ this

would not be true for compact sets. Conversely, we consider a topology „ which has

for one of the above mentioned types the same objects as the original one. Then

„ has by (4.3) the same bornivorous absolutely convex subsets as the original one.

Hence, any 0-neighborhood of „ has to be bornivorous for the original topology,

and hence is a 0-neighborhood of the bornologi¬cation of the original topology.

4.5. Lemma. Let E be a bornological locally convex vector space, U ⊆ E a convex

subset. Then U is open for the locally convex topology of E if and only if U is open

4.5

4. The c∞ -topology

4.8 37

for the c∞ -topology.

Furthermore, an absolutely convex subset U of E is a 0-neighborhood for the locally

convex topology if and only if it is so for the c∞ -topology.

Proof. (’) The c∞ -topology is ¬ner than the locally convex topology, cf. (4.2).

(⇐) Let ¬rst U be an absolutely convex 0-neighborhood for the c∞ -topology. Hence,

U absorbs Mackey-0-sequences. By (4.1.3) we have to show that U is bornivorous,

in order to obtain that U is a 0-neighborhood for the locally convex topology. But

this follows immediately from (4.3).

Let now U be convex and c∞ -open, let x ∈ U be arbitrary. We consider the c∞ -

open absolutely convex set W := (U ’ x) © (x ’ U ) which is a 0-neighborhood of

the locally convex topology by the argument above. Then x ∈ W + x ⊆ U . So U

is open in the locally convex topology.

4.6. Corollary. The bornologi¬cation of a locally convex space E is the ¬nest

locally convex topology coarser than the c∞ -topology on E.

4.7. In (2.12) we de¬ned the c∞ -topology on an arbitrary locally convex space E

as the ¬nal topology with respect to the smooth curves c : R ’ E. Now we will

compare the c∞ -topology with other re¬nements of a given locally convex topology.

We ¬rst specify those re¬nements.

De¬nition. Let E be a locally convex vector space.

(i) We denote by kE the Kelley-¬cation of the locally convex topology of E, i.e.

the vector space E together with the ¬nal topology induced by the inclusions of

the subsets being compact for the locally convex topology.

(ii) We denote by sE the vector space E with the ¬nal topology induced by the

curves being continuous for the locally convex topology, or equivalently the se-

quences N∞ ’ E converging in the locally convex topology. The equivalence holds

since the in¬nite polygon through a converging sequence can be continuously pa-

rameterized by a compact interval.

(iii) We recall that by c∞ E we denote the vector space E with its c∞ -topology, i.e.

the ¬nal topology induced by the smooth curves.

Using that smooth curves are continuous and that converging sequences N∞ ’ E

have compact images, the following identities are continuous: c∞ E ’ sE ’ kE ’

E.

If the locally convex topology of E coincides with the topology of c∞ E, resp. sE,

resp. kE then we call E smoothly generated, resp. sequentially generated, resp.

compactly generated.

4.8. Example. On E = RJ all the re¬nements of the locally convex topology

described in (4.7) above are di¬erent, i.e. c∞ E = sE = kE = E, provided the

cardinality of the index set J is at least that of the continuum.

Proof. It is enough to show this for J equipotent to the continuum, since RJ1 is

a direct summand in RJ2 for J1 ⊆ J2 .

4.8

38 Chapter I. Calculus of smooth mappings 4.9

(c∞ E = sE) We may take as index set J the set c0 of all real sequences converging

to 0. De¬ne a sequence (xn ) in E by (xn )j := jn . Since every j ∈ J is a 0-sequence

we conclude that the xn converge to 0 in the locally convex topology of the product,

hence also in sE. Assume now that the xn converge towards 0 in c∞ E. Then by

(1.8) some subsequence converges Mackey to 0. Thus, there exists an unbounded

sequence of reals »n with {»n xn : n ∈ N} bounded. Let j be a 0-sequence with

{jn »n : n ∈ N} unbounded (e.g. (jn )’2 := 1 + max{|»k | : k ¤ n}). Then the j-th

coordinate jn »n of »n xn is not bounded with respect to n, contradiction.

(sE = kE) Consider in E the subset

A := x ∈ {0, 1}J : xj = 1 for at most countably many j ∈ J .

It is clearly closed with respect to the converging sequences, hence closed in sE.

But it is not closed in kE since it is dense in the compact set {0, 1}J .

(kE = E) Consider in E the subsets

An := x ∈ E : |xj | < n for at most n many j ∈ J .

Each An is closed in E since its complement is the union of the open sets {x ∈ E :

|xj | < n for all j ∈ Jo } where Jo runs through all subsets of J with n + 1 elements.

We show that the union A := n∈N An is closed in kE. So let K be a compact

subset of E; then K ⊆ prj (K), and each prj (K) is compact, hence bounded in

R. Since the family ({j ∈ J : prj (K) ⊆ [’n, n]})n∈N covers J, there has to exist an

N ∈ N and in¬nitely many j ∈ J with prj (K) ⊆ [’N, N ]. Thus K © An = … for all

n > N , and hence, A © K = n∈N An © K is closed. Nevertheless, A is not closed

¯

in E, since 0 is in A but not in A.

4.9. c∞ -convergent sequences. By (2.13) every M -convergent sequence gives

a continuous mapping N∞ ’ c∞ E and hence converges in c∞ E. Conversely, a

sequence converging in c∞ E is not necessarily Mackey convergent, see [Fr¨licher,

o

Kriegl, 1985]. However, one has the following result.

Lemma. A sequence (xn ) is convergent to x in the c∞ -topology if and only if every

subsequence has a subsequence which is Mackey convergent to x.

Proof. (⇐) is true for any topological convergence. In fact if xn would not con-

verge to x, then there would be a neighborhood U of x and a subsequence of xn

which lies outside of U and hence cannot have a subsequence converging to x.

(’) It is enough to show that (xn ) has a subsequence which converges Mackey to x,

since every subsequence of a c∞ -convergent sequence is clearly c∞ -convergent to the

same limit. Without loss of generality we may assume that x ∈ A := {xn : n ∈ N}.

/

Hence, A cannot be c∞ -closed, and thus there is a sequence nk ∈ N such that

(xnk ) converges Mackey to some point x ∈ A. The set {nk : k ∈ N} cannot be

/

bounded, and hence we may assume that the nk are strictly increasing by passing

to a subsequence. But then (xnk ) is a subsequence of (xn ) which converges in c∞ E

to x and Mackey to x hence also in c∞ E. Thus x = x.

4.9

4. The c∞ -topology

4.11 39

Remark. A consequence of this lemma is, that there is no topology having as

convergent sequences exactly the M -convergent ones, since this topology obviously

would have to be coarser than the c∞ -topology.

One can use this lemma also to show that the c∞ -topology on a locally convex

vector space gives a so called arc-generated vector space. See [Fr¨licher, Kriegl,

o

1988, 2.3.9 and 2.3.13] for a discussion of this.

Let us now describe several important situations where at least some of these topolo-

gies coincide. For the proof we will need the following

4.10. Lemma. [Averbukh, Smolyanov, 1968] For any locally convex space E the

following statements are equivalent:

(1) The sequential closure of any subset is formed by all limits of sequences in

the subset.

(2) For any given double sequence (xn,k ) in E with xn,k convergent to some

xk for n ’ ∞ and k ¬xed and xk convergent to some x, there are strictly

increasing sequences i ’ n(i) and i ’ k(i) with xn(i),k(i) ’ x for i ’ ∞.

Proof. (1’2) Take an a0 ∈ E di¬erent from k · (xn+k,k ’ x) and from k · (xk ’ x)

1

for all k and n. De¬ne A := {an,k := xn+k,k ’ k · a0 : n, k ∈ N}. Then x is in the

1 1

sequential closure of A, since xn+k,k ’ k · a0 converges to xk ’ k · a0 as n ’ ∞,

1

and xk ’ k · a0 converges to x ’ 0 = x as k ’ ∞. Hence, by (1) there has to exist

a sequence i ’ (ni , ki ) with ani ,ki convergent to x. By passing to a subsequence

we may suppose that i ’ ki and i ’ ni are increasing. Assume that i ’ ki is

1

bounded, hence ¬nally constant. Then a subsequence xni +ki ,ki ’ ki ·a0 is converging

1 1

to xk ’ k · a0 = x if i ’ ni is unbounded, and to xn+k,k ’ k · a0 = x if i ’ ni

is bounded, which both yield a contradiction. Thus, i ’ ki can be chosen strictly

increasing. But then

1

xni +ki ,ki = ani ,ki + ki a0 ’ x.

(1) ⇐ (2) is obvious.

4.11. Theorem. For any bornological vector space E the following implications

hold:

(1) c∞ E = E provided the closure of subsets in E is formed by all limits of

sequences in the subset; hence in particular if E is metrizable.

(2) c∞ E = E provided E is the strong dual of a Fr´chet Schwartz space;

e

(3) c∞ E = kE provided E is the strict inductive limit of a sequence of Fr´chet

e

spaces.

(4) c∞ E = sE provided E satis¬es the M -convergence condition, i.e. every

sequence converging in the locally convex topology is M-convergent.

(5) sE = E provided E is the strong dual of a Fr´chet Montel space;

e

Proof. (1) Using the lemma (4.10) above one obtains that the closure and the

sequential closure coincide, hence sE = E. It remains to show that sE ’ c∞ E is

(sequentially) continuous. So suppose a sequence converging to x is given, and let

4.11

40 Chapter I. Calculus of smooth mappings 4.13

(xn ) be an arbitrary subsequence. Then xn,k := k(xn ’ x) ’ k · 0 = 0 for n ’ ∞,

and hence by lemma (4.10) there are subsequences ki , ni with ki · (xni ’ x) ’ 0,

i.e. i ’ xni is M-convergent to x. Thus, the original sequence converges in c∞ E

by (4.9).

(3) Let E be the strict inductive limit of the Fr´chet spaces En . By (52.8) every En

e

carries the trace topology of E, hence is closed in E, and every bounded subset of E

is contained in some En . Thus, every compact subset of E is contained as compact

subset in some En . Since En is a Fr´chet space such a subset is even compact in

e

c∞ En and hence compact in c∞ E. Thus, the identity kE ’ c∞ E is continuous.

(4) is valid, since the M-closure topology is the ¬nal one induced by the M-

converging sequences.

(5) Let E be the dual of any Fr´chet Montel space F . By (52.29) E is bornological.

e

First we show that kE = sE. Let K ⊆ E = F be compact for the locally convex

topology. Then K is bounded, hence equicontinuous since F is barrelled by (52.25).

Since F is separable by (52.27) the set K is metrizable in the weak topology σ(E, F )

by (52.21). By (52.20) this weak topology coincides with the topology of uniform

convergence on precompact subsets of F . Since F is a Montel space, this latter

topology is the strong one, and even the bornological one, as remarked at the

beginning. Thus, the (metrizable) topology on K is the initial one induced by the

converging sequences. Hence, the identity kE ’ sE is continuous, and therefore

sE = kE.

It remains to show kE = E. Since F is Montel the locally convex topology of

the strong dual coincides with the topology of uniform convergence on precom-

pact subsets of F . Since F is metrizable this topology coincides with the so-called

equicontinuous weak— -topology, cf. (52.22), which is the ¬nal topology induced by

the inclusions of the equicontinuous subsets. These subsets are by the Alao˜lu- g

Bourbaki theorem (52.20) relatively compact in the topology of uniform conver-

gence on precompact subsets. Thus, the locally convex topology of E is compactly

generated.

Proof of (2) By (5), and since Fr´chet Schwartz spaces are Montel by (52.24), we

e

have sE = E and it remains to show that c∞ E = sE. So let (xn ) be a sequence

converging to 0 in E. Then the set {xn : n ∈ N} is relatively compact, and by

[Fr¨licher, Kriegl, 1988, 4.4.39] it is relatively compact in some Banach space EB .

o

Hence, at least a subsequence has to be convergent in EB . Clearly its Mackey limit

has to be 0. This shows that (xn ) converges to 0 in c∞ E, and hence c∞ E = sE. One

can even show that E satis¬es the Mackey convergence condition, see (52.28).

4.12. Example. We give now a non-metrizable example to which (4.11.1) applies.

Let E denote the subspace of RJ of all sequences with countable support. Then

the closure of subsets of E is given by all limits of sequences in the subset, but

for non-countable J the space E is not metrizable. This was proved in [Balanzat,

1960].

4.13. Remark. The conditions (4.11.1) and (4.11.2) are rather disjoint since every

4.13

4. The c∞ -topology

4.16 41

locally convex space, that has a countable basis of its bornology and for which the

sequential adherence of subsets (the set of all limits of sequences in it) is sequentially

closed, is normable as the following proposition shows:

Proposition. Let E be a non-normable bornological locally convex space that has

a countable basis of its bornology. Then there exists a subset of E whose sequential

adherence is not sequentially closed.

Proof. Let {Bk : k ∈ N0 } be an increasing basis of the von Neumann bornology

with B0 = {0}. Since E is non-normable we may assume that Bk does not absorb

1

Bk+1 for all k. Now choose bn,k ∈ n Bk+1 with bn,k ∈ Bk . We consider the

/

double sequence {bk,0 ’ bn,k : n, k ≥ 1}. For ¬xed k the sequence bn,k converges

by construction (in EBk+1 ) to 0 for n ’ ∞. Thus, bk,0 ’ 0 is the limit of the

sequence bk,0 ’ bn,k for n ’ ∞, and bk,0 converges to 0 for k ’ ∞. Suppose

bk(i),0 ’ bn(i),k(i) converges to 0. So it has to be bounded, thus there must be

an N ∈ N with B1 ’ {bk(i),0 ’ bn(i),k(i) : i ∈ N} ⊆ BN . Hence, bn(i),k(i) =

bk(i),0 ’ (bk(i),0 ’ bn(i),k(i) ) ∈ BN , i.e. k(i) < N . This contradicts (4.10.2).

4.14. Lemma. Let U be a c∞ -open subset of a locally convex space, let µn ’ ∞

be a real sequence, and let f : U ’ F be a mapping which is bounded on each

µ-converging sequence in U . Then f is bounded on every bornologically compact

subset (i.e. compact in some EB ) of U .

Proof. By composing with linear functionals we may assume that F = R. Let

K ⊆ EB © U be compact in EB for some bounded absolutely convex set B. Assume

that f (K) is not bounded. So there is a sequence (xn ) in K with |f (xn )| ’ ∞.

Since K is compact in the normed space EB we may assume that (xn ) converges to

x ∈ K. By passing to a subsequence we may even assume that (xn ) is µ-converging.

Contradiction.

4.15. Lemma. Let U be c∞ -open in E — R and K ⊆ R be compact. Then

U0 := {x ∈ E : {x} — K ⊆ U } is c∞ -open in E.

Proof. Let x : R ’ E be a smooth curve in E with x(0) ∈ U0 , i.e. (x(0), t) ∈ U

for all t ∈ K. We have to show that x(s) ∈ U0 for all s near 0. So consider the

smooth map x — R : R — R ’ E — R. By assumption (x — R)’1 (U ) is open in

c∞ (R2 ) = R2 . It contains the compact set {0} — K and hence also a W — K for

some neighborhood W of 0 in R. But this amounts in saying that x(W ) ⊆ U0 .

4.16. The c∞ -topology of a product. Consider the product E — F of two

locally convex vector spaces. Since the projections onto the factors are linear and

continuous, and hence smooth, we always have that the identity mapping c∞ (E —

F ) ’ c∞ (E)—c∞ (F ) is continuous. It is not always a homeomorphism: Just take a

bounded separately continuous bilinear functional, which is not continuous (like the

evaluation map) on a product of spaces where the c∞ -topology is the bornological

topology.

However, if one of the factors is ¬nite dimensional the product is well behaved:

4.16

42 Chapter I. Calculus of smooth mappings 4.19

Corollary. For any locally convex space E the c∞ -topology of E —Rn is the product

topology of the c∞ -topologies of the two factors, so that we have c∞ (E — Rn ) =

c∞ (E) — Rn .

Proof. This follows recursively from the special case E — R, for which we can

proceed as follows. Take a c∞ -open neighborhood U of some point (x, t) ∈ E — R.

Since the inclusion map s ’ (x, s) from R into E — R is continuous and a¬ne, the

inverse image of U in R is an open neighborhood of t. Let™s take a smaller compact

neighborhood K of t. Then by the previous lemma U0 := {y ∈ E : {y} — K ⊆ U }

is a c∞ -open neighborhood of x, and hence U0 — K o is a neighborhood of (x, t) in

c∞ (E) — R, what was to be shown.

4.17. Lemma. Let U be c∞ -open in a locally convex space and x ∈ U . Then the

star stx (U ) := {x + v : x + »v ∈ U for all |»| ¤ 1} with center x in U is again

c∞ -open.

Proof. Let c : R ’ E be a smooth curve with c(0) ∈ stx (U ). The smooth mapping

f : (t, s) ’ (1 ’ s)x + sc(t) maps {0} — {s : |s| ¤ 1} into U . So there exists δ > 0

with f ({(t, s) : |t| < δ, |s| ¤ 1}) ⊆ U . Thus, c(t) ∈ stx (U ) for |t| < δ.

4.18. Lemma. The (absolutely) convex hull of a c∞ -open set is again c∞ -open.

Proof. Let U be c∞ -open in a locally convex vector space E.

For each x ∈ U the set

Ux := {x + t(y ’ x) : t ∈ [0, 1], y ∈ U } = U ∪ (x + t(U ’ x))

0<t¤1

is c∞ -open. The convex hull can be constructed by applying n times the operation

U ’ x∈U Ux and taking the union over all n ∈ N, which respects c∞ -openness.

The absolutely convex hull can be obtained by forming ¬rst {» : |»| = 1}.U =

∞

|»|=1 »U which is c -open, and then forming the convex hull.

4.19. Corollary. Let E be a bornological convenient vector space containing a

nonempty c∞ -open subset which is either locally compact or metrizable in the c∞ -

topology. Then the c∞ -topology on E is locally convex. In the ¬rst case E is ¬nite

dimensional, in the second case E is a Fr´chet space.

e

Proof. Let U ⊆ E be a c∞ -open metrizable subset. We may assume that 0 ∈ U .

Then there exists a countable neighborhood basis of 0 in U consisting of c∞ -open

sets. This is also a neighborhood basis of 0 for the c∞ -topology of E. We take

the absolutely convex hulls of these open sets, which are again c∞ -open by (4.18),

and obtain by (4.5) a countable neighborhood basis for the bornologi¬cation of the

locally convex topology, so the latter is metrizable and Fr´chet, and by (4.11) it

e

equals the c∞ -topology.

If U is locally compact in the c∞ -topology we may ¬nd a c∞ -open neighborhood V

of 0 with compact closure V in the c∞ -topology. By lemma (4.18) the absolutely

4.19

4. The c∞ -topology

4.23 43

convex hull of V is also c∞ -open, and by (4.5) it is also open in the bornologi¬cation

Eborn of E. The set V is then also compact in Eborn , hence precompact. So the

absolutely convex hull of V is also precompact by (52.6). Therefore, the absolutely

convex hull of V is a precompact neighborhood of 0 in Eborn , thus E is ¬nite

dimensional by (52.5). So Eborn = c∞ (E).

Now we describe classes of spaces where c∞ E = E or where c∞ E is not even a

topological vector space. Finally, we give an example where the c∞ -topology is not

completely regular. We begin with the relationship between the c∞ -topology and

the locally convex topology on locally convex vector spaces.

4.20. Proposition. Let E and F be bornological locally convex vector spaces. If

there exists a bilinear smooth mapping m : E — F ’ R that is not continuous with

respect to the locally convex topologies, then c∞ (E — F ) is not a topological vector

space.

We shall show in lemma (5.5) below that multilinear mappings are smooth if and

only if they are bounded.

Proof. Suppose that addition c∞ (E — F ) — c∞ (E — F ) ’ c∞ (E — F ) is continuous

with respect to the product topology. Using the continuous inclusions c∞ E ’

c∞ (E — F ) and c∞ F ’ c∞ (E — F ) we can write m as composite of continuous

+ m

maps as follows: c∞ E — c∞ F ’ c∞ (E — F ) — c∞ (E — F ) ’ c∞ (E — F ) ’ R.

’ ’

Thus, for every µ > 0 there are 0-neighborhoods U and V with respect to the

c∞ -topology such that m(U — V ) ⊆ (’µ, µ). Then also m( U — V ) ⊆ (’µ, µ)

where denotes the absolutely convex hull. By (4.5) one concludes that m is

continuous with respect to the locally convex topology, a contradiction.

4.21. Corollary. Let E be a non-normable bornological locally convex space. Then

c∞ (E — E ) is not a topological vector space.

Proof. By (4.20) it is enough to show that ev : E — E ’ R is not continuous for

the bornological topologies on E and E ; if it were so there was be a neighborhood

U of 0 in E and a neighborhood U of 0 in E such that ev(U — U ) ⊆ [’1, 1]. Since

U is absorbing, U is scalarwise bounded, hence a bounded neighborhood. Thus,

E is normable.

4.22. Remark. In particular, for a Fr´chet Schwartz space E and its dual E we

e

have c∞ (E —E ) = c∞ E —c∞ E , since by (4.11) we have c∞ E = E and c∞ E = E ,

so equality would contradict corollary (4.21).

In order to get a large variety of spaces where the c∞ -topology is not a topological

vector space topology the next three technical lemmas will be useful.

4.23. Lemma. Let E be a locally convex vector space. Suppose a double sequence

bn,k in E exists which satis¬es the following two conditions:

(b™) For every sequence k ’ nk the sequence k ’ bnk ,k has no accumulation

point in c∞ E.

(b”) For all k the sequence n ’ bn,k converges to 0 in c∞ E.

4.23

44 Chapter I. Calculus of smooth mappings 4.25

Suppose furthermore that a double sequence cn,k in E exists that satis¬es the fol-

lowing two conditions:

(c™) For every 0-neighborhood U in c∞ E there exists some k0 such that cn,k ∈ U

for all k ≥ k0 and all n.

(c”) For all k the sequence n ’ cn,k has no accumulation point in c∞ E.

Then c∞ E is not a topological vector space.

Proof. Assume that the addition c∞ E — c∞ E ’ c∞ E is continuous. In this

proof convergence is meant always with respect to c∞ E. We may without loss of

generality assume that cn,k = 0 for all n, k, since by (c”) we may delete all those

cn,k which are equal to 0. Then we consider A := {bn,k + µn,k cn,k : n, k ∈ N} where

the µn,k ∈ {’1, 1} are chosen in such a way that 0 ∈ A.

/

We ¬rst show that A is closed in the sequentially generated topology c∞ E: Let

bni ,ki +µni ,ki cni ,ki ’ x, and assume that (ki ) is unbounded. By passing if necessary

to a subsequence we may even assume that i ’ ki is strictly increasing. Then

cni ,ki ’ 0 by (c™), hence bni ,ki ’ x by the assumption that addition is continuous,

which is a contradiction to (b™). Thus, (ki ) is bounded, and we may assume it to be

constant. Now suppose that (ni ) is unbounded. Then bni ,k ’ 0 by (b”), and hence

µni ,k cni ,k ’ x, and for a subsequence where µ is constant one has cni ,k ’ ±x,

which is a contradiction to (c”). Thus, ni is bounded as well, and we may assume

it to be constant. Hence, x = bn,k + µn,k cn,k ∈ A.

By the assumed continuity of the addition there exists an open and symmetric

0-neighborhood U in c∞ E with U + U ⊆ E \ A. For K su¬ciently large and n

arbitrary one has cn,K ∈ U by (c™). For such a ¬xed K and N su¬ciently large

bN,K ∈ U by (b™). Thus, bN,K + µN,K cN,K ∈ A, which is a contradiction.

/

Let us now show that many spaces have a double sequence cn,k as in the above

lemma.

4.24. Lemma. Let E be an in¬nite dimensional metrizable locally convex space.

Then a double sequence cn,k subject to the conditions (c™) and (c”) of (4.23) exists.

Proof. If E is normable we choose a sequence (cn ) in the unit ball without accu-

1

mulation point and de¬ne cn,k := k cn . If E is not normable we take a countable

increasing family of non-equivalent seminorms pk generating the locally convex

1

topology, and we choose cn,k with pk (cn,k ) = k and pk+1 (cn,k ) > n.

Next we show that many spaces have a double sequence bn,k as in lemma (4.23).

4.25. Lemma. Let E be a non-normable bornological locally convex space hav-

ing a countable basis of its bornology. Then a double sequence bn,k subject to the

conditions (b™) and (b”) of (2.11) exists.

Proof. Let Bn (n ∈ N) be absolutely convex sets forming an increasing basis of

the bornology. Since E is not normable the sets Bn can be chosen such that Bn

1

does not absorb Bn+1 . Now choose bn,k ∈ n Bk+1 with bn,k ∈ Bk .

/

Using these lemmas one obtains the

4.25

4. The c∞ -topology

4.26 45

4.26. Proposition. For the following bornological locally convex spaces the c∞ -

topology is not a vector space topology:

(i) Every bornological locally convex space that contains as c∞ -closed subspaces

an in¬nite dimensional Fr´chet space and a space which is non-normable in

e

the bornological topology and having a countable basis of its bornology.

(ii) Every strict inductive limit of a strictly increasing sequence of in¬nite di-

mensional Fr´chet spaces.

e

(iii) Every product for which at least 2„µ0 many factors are non-zero.

(iv) Every coproduct for which at least 2„µ0 many summands are non-zero.

Proof. (i) follows directly from the last 3 lemmas.

(ii) Let E be the strict inductive limit of the spaces En (n ∈ N). Then E contains

the in¬nite dimensional Fr´chet space E1 as subspace. The subspace generated

e

by points xn ∈ En+1 \ En (n ∈ N) is bornologically isomorphic to R(N) , hence its

bornology has a countable basis. Thus, by (i) we are done.

(iii) Such a product E contains the Fr´chet space RN as complemented subspace.

e

We want to show that R(N) is also a subspace of E. For this we may assume that the

index set J is RN and all factors are equal to R. Now consider the linear subspace

E1 of the product generated by the elements xn ∈ E = RN , where (xn )j := j(n) for

every j ∈ J = RN . The linear map R(N) ’ E1 ⊆ E that maps the n-th unit vector

to xn is injective, since for a given ¬nite linear combination tn xn = 0 the j-th

|tn |. It is a morphism since R(N) carries

coordinate for j(n) := sign(tn ) equals

the ¬nest structure. So it remains to show that it is a bornological embedding.

We have to show that any bounded B ⊆ E1 is contained in a subspace generated

by ¬nitely many xn . Otherwise, there would exist a strictly increasing sequence

(nk ) and bk = n¤nk tk xn ∈ B with tk k = 0. De¬ne an index j recursively by

n n

k ’1 k

j(n) := n|tn | · sign m<n tm j(m) if n = nk and j(n) := 0 if n = nk for all k.

Then the absolute value of the j-th coordinate of bk evaluates as follows:

|(bk )j | = tk j(n) = tk j(n) + tk k j(nk )

n n n

n<nk

n¤nk

tk j(n) + |tk k j(nk )| ≥ |tk k j(nk )| ≥ nk .

= n n n

n<nk

Hence, the j-th coordinates of {bk : k ∈ N} are unbounded with respect to k ∈ N,

thus B is unbounded.

(iv) We can not apply lemma (4.23) since every double sequence has countable

support and hence is contained in the dual R(A) of a Fr´chet Schwartz space RA for

e

some countable subset A ‚ J. It is enough to show (iv) for R(J) where J = N ∪ c0 .

Let A := {jn (en + ej ) : n ∈ N, j ∈ c0 , jn = 0 for all n}, where en and ej denote

the unit vectors in the corresponding summand. The set A is M-closed, since its

intersection with ¬nite subsums is ¬nite. Suppose there exists a symmetric M-open

0-neighborhood U with U + U ⊆ E \ A. Then for each n there exists a jn = 0

with jn en ∈ U . We may assume that n ’ jn converges to 0 and hence de¬nes

4.26

46 Chapter I. Calculus of smooth mappings 4.27

an element j ∈ c0 . Furthermore, there has to be an N ∈ N with jN ej ∈ U , thus

jN (eN + ej ) ∈ (U + U ) © A, in contradiction to U + U ⊆ E \ A.

Remark. A nice and simple example where one either uses (i) or (ii) is RN • R(N) .

The locally convex topology on both factors coincides with their c∞ -topology (the

¬rst being a Fr´chet (Schwartz) space, cf. (i) of (4.11), the second as dual of the

e

¬rst, cf. (ii) of (4.11)); but the c∞ -topology on their product is not even a vector

space topology.

∞

From (ii) it follows also that each space Cc (M, R) of smooth functions with com-

pact support on a non-compact separable ¬nite dimensional manifold M has the

property, that the c∞ -topology is not a vector space topology.

4.27. Although the c∞ -topology on a convenient vector space is always functionally

separated, hence Hausdor¬, it is not always completely regular as the following

example shows.

Example. The c∞ -topology is not completely regular. The c∞ -topology of

RJ is not completely regular if the cardinality of J is at least 2„µ0 .

Proof. It is enough to show this for an index set J of cardinality 2„µ0 , since the

corresponding product is a complemented subspace in every product with larger

index set. We prove the theorem by showing that every function f : RJ ’ R

which is continuous for the c∞ -topology is also continuous with respect to the

locally convex topology. Hence, the completely regular topology associated to the

c∞ -topology is the locally convex topology of E. That these two topologies are

di¬erent was shown in (4.8). We use the following theorem of [Mazur, 1952]: Let

E0 := {x ∈ RJ : supp(x) is countable}, and let f : E0 ’ R be sequentially

continuous. Then there is some countable subset A ‚ J such that f (x) = f (xA ),

where in this proof xA is de¬ned as xA (j) := x(j) for j ∈ A and xA (j) = 0 for

j ∈ A. Every sequence which is converging in the locally convex topology of E0

/

is contained in a metrizable complemented subspace RA for some countable A and

therefore is even M-convergent. Thus, this theorem of Mazur remains true if f is

assumed to be continuous for the M-closure topology. This generalization follows

also from the fact that c∞ E0 = E0 , cf. (4.12). Now let f : RJ ’ R be continuous

for the c∞ -topology. Then f |E0 : E0 ’ R is continuous for the c∞ -topology, and

hence there exists a countable set A0 ‚ J such that f (x) = f (xA0 ) for any x ∈ E0 .

We want to show that the same is true for arbitrary x ∈ RJ . In order to show this

we consider for x ∈ RJ the map •x : 2J ’ R de¬ned by •x (A) := f (xA )’f (xA©A0 )

for any A ⊆ J, i.e. A ∈ 2J . For countable A one has xA ∈ E0 , hence •x (A) = 0.

Furthermore, •x is sequentially continuous where one considers on 2J the product

topology of the discrete factors 2 = {0, 1}. In order to see this consider a converging

sequence of subsets An ’ A, i.e. for every j ∈ J one has for the characteristic

functions χAn (j) = χA (j) for n su¬ciently large. Then {n(xAn ’ xA ) : n ∈ N} is

bounded in RJ since for ¬xed j ∈ J the j-th coordinate equals 0 for n su¬ciently

large. Thus, xAn converges Mackey to xA , and since f is continuous for the c∞ -

topology •x (An ) ’ •x (A). Now we can apply another theorem of [Mazur, 1952]:

4.27

4. The c∞ -topology

4.30 47

Any function f : 2J ’ R that is sequentially continuous and is zero on all countable

subsets of J is identically 0, provided the cardinality of J is smaller than the ¬rst

inaccessible cardinal. Thus, we conclude that 0 = •x (J) = f (x) ’ f (xAn ) for all

x ∈ RJ . Hence, f factors over the metrizable space RA0 and is therefore continuous

for the locally convex topology.

In general, the trace of the c∞ -topology on a linear subspace is not its c∞ -topology.

However, for c∞ -closed subspaces this is true:

4.28. Lemma. Closed embedding lemma. Let E be a linear c∞ -closed sub-

space of F . Then the trace of the c∞ -topology of F on E is the c∞ -topology on

E

Proof. Since the inclusion is continuous and hence bounded it is c∞ -continuous.

Therefore, it is enough to show that it is closed for the c∞ -topologies. So let A ⊆ E

be c∞ E-closed. And let xn ∈ A converge Mackey towards x in F . Then x ∈ E,

since E is assumed to be c∞ -closed, and hence xn converges Mackey to x in E.

Since A is c∞ -closed in E, we have that x ∈ A.

We will give an example in (4.33) below which shows that c∞ -closedness of the

subspace is essential for this result. Another example will be given in (4.36).

4.29. Theorem. The c∞ -completion. For any locally convex space E there

˜

exists a unique (up to a bounded isomorphism) convenient vector space E and a

˜

bounded linear injection i : E ’ E with the following universal property:

Each bounded linear mapping : E ’ F into a convenient vector space F

has a unique bounded extension ˜ : E ’ F such that ˜ —¦ i = .

˜

˜

Furthermore, i(E) is dense for the c∞ -topology in E.

˜

Proof. Let E be the c∞ -closure of E in the locally convex completion Eborn of the

˜

bornologi¬cation Eborn of E. The inclusion i : E ’ E is bounded (not continuous

˜

in general). By (4.28) the c∞ -topology on E is the trace of the c∞ -topology on

˜

Eborn . Hence, i(E) is dense also for the c∞ -topology in E.

Using the universal property of the locally convex completion the mapping has

a unique extension ˆ : Eborn ’ F into the locally convex completion of F , whose

restriction to E has values in F , since F is c∞ -closed in F , so it is the desired ˜.

˜

˜

Uniqueness follows, since i(E) is dense for the c∞ -topology in E.

4.30. Proposition. c∞ -completion via c∞ -dense embeddings. Let E be

c∞ -dense and bornologically embedded into a c∞ -complete locally convex space F .

If E ’ F has the extension property for bounded linear functionals, then F is

bornologically isomorphic to the c∞ -completion of E.

Proof. We have to show that E ’ F has the universal property for extending

bounded linear maps T into c∞ -complete locally convex spaces G. Since we are

4.30

48 Chapter I. Calculus of smooth mappings 4.31

only interested in bounded mappings, we may take the bornologi¬cation of G and

hence may assume that G is bornological. Consider the following diagram

y w F@

E

’@

’u @

’ R@

’ ‘ @@—¦ T

’ »

’δ pr‘‘@

T

’

G

“@

‘A

’ »

u”!

»

wR

G

The arrow δ, given by δ(x)» := »(x), is a bornological embedding, i.e. the image of

a set is bounded if and only if the set is bounded, since B ⊆ G is bounded if and

only if »(B) ⊆ R is bounded for all » ∈ G , i.e. δ(B) ⊆ G R is bounded.

By assumption, the dashed arrow on the right hand side exists, hence by the uni-

˜

versal property of the product the dashed vertical arrow (denoted T ) exists. It

˜

remains to show that it has values in the image of δ. Since T is bounded we have

c∞ c∞

c∞

˜ ˜ ˜

) ⊆ T (E) ⊆ δ(G)

T (F ) = T (E = δ(G),

since G is c∞ -complete and hence also δ(G), which is thus c∞ -closed.

˜

The uniqueness follows, since as a bounded linear map T has to be continuous

for the c∞ -topology (since it preserves the smooth curves by (2.11) which in turn

generate the c∞ -topology), and E lies dense in F with respect to this topology.

4.31. Proposition. Inductive representation of bornological locally con-

vex spaces. For a locally convex space E the bornologi¬cation Eborn is the colimit

of all the normed spaces EB for the absolutely convex bounded sets B. The col-

˜

imit of the respective completions EB is the linear subspace of the c∞ -completion

˜ ˜

E consisting of all limits in E of Mackey Cauchy sequences in E.

˜

Proof. Let E (1) be the Mackey adherence of E in the c∞ -completion E, by which

˜ ˜

we mean the limits in E of all sequences in E which converge Mackey in E. Then

E (1) is a subspace of the locally convex completion Eborn . For every absolutely

convex bounded set B we have the continuous inclusion EB ’ Eborn , and by

˜

passing to the c∞ -completion we get mappings EB = EB ’ E. These mappings

commute with the inclusions EB ’ EB for B ⊆ B and have values in the Mackey

adherence of E, since every point in EB is the limit of a sequence in EB , and hence

its image is the limit of this Mackey Cauchy sequence in E.

We claim that the Mackey adherence E (1) together with these mappings has the

universal property of the colimit ’ B EB . In fact, let T : E (1) ’ F be a linear

lim

’

(1)

mapping, such that EB ’ E ’ F is continuous for all B. In particular T |E :

E ’ F has to be bounded, and hence T |Eborn : Eborn ’ F is continuous. Thus,

it has a unique continuous extension T : E (1) ’ F , and it remains to show that

4.31

4. The c∞ -topology

4.33 49

this extension is T . So take a point x ∈ E (1) . Then there exists a sequence

(xn ) in E, which converges Mackey to x. Thus, the xn form a Cauchy-sequence

in some EB and hence converge to some y in EB . Then ιB (y) = x, since the

mapping ιB : EB ’ E (1) is continuous. Since the trace of T to EB is continuous

T (xn ) converges to T (ιB (y)) = T (x) and T (xn ) = T (xn ) converges to T (x), i.e.

T (x) = T (x).

In spite of (1) in (4.36) we can use the Mackey adherence to describe the c∞ -closure

in the following inductive way:

4.32. Proposition. Mackey adherences. For ordinal numbers ± the Mackey

adherence A(±) of order ± is de¬ned recursively by:

M-Adh(A(β) ) if ± = β + 1

(±)

A :=

A(β) if ± is a limit ordinal number.

β<±

Then the closure A of A in the c∞ -topology coincides with A(ω1 ) , where ω1 denotes

the ¬rst uncountable ordinal number, i.e. the set of all countable ordinal numbers.

Proof. Let us ¬rst show that A(ω1 ) is c∞ -closed. So take a sequence xn ∈ A(ω1 ) =

(±)

±<ω1 A , which converges Mackey to some x. Then there are ±n < ω1 with

xn ∈ A(±n ) . Let ± := supn ±n . Then ± is a again countable and hence less than

ω1 . Thus, xn ∈ A(±n ) ⊆ A(±) , and therefore x ∈ M-Adh(A(±) ) = A(±+1) ⊆ A(ω1 )

since ± + 1 ¤ ω1 .

It remains to show that A(±) is contained in A for all ±. We prove this by trans¬nite

induction. So assume that for all β < ± we have A(β) ⊆ A. If ± is a limit

ordinal number then A(±) = β<± A(β) ⊆ A. If ± = β + 1 then every point in

A(±) = M-Adh(A(β) ) is the Mackey-limit of some sequence in A(β) ⊆ A, and since

A is c∞ -closed, this limit has to belong to it. So A(±) ⊆ A in all cases.

4.33. Example. The trace of the c∞ -topology is not the c∞ -topology, in general.

1 1

Proof. Consider E = RN —R(N) , A := {an,k := ( n χ{1,..,k} , k χ{n} ) : n, k ∈ N} ⊆ E.

Let F be the linear subspace of E generated by A. We show that the closure of A

with respect to the c∞ -topology of F is strictly smaller than that with respect to

the trace topology of the c∞ -topology of E.

A is closed in the c∞ -topology of F : Assume that a sequence (anj ,kj ) is M-

converging to (x, y). Then the second component of anj ,kj has to be bounded.

Thus, j ’ nj has to be bounded and may be assumed to have constant value n∞ .

If j ’ kj were unbounded, then (x, y) = ( n1 χN , 0), which is not an element of F .

∞

Thus, j ’ kj has to be bounded too and may be assumed to have constant value

k∞ . Thus, (x, y) = an∞ ,k∞ ∈ A.

A is not closed in the trace topology since (0,0) is contained in the closure of A

with respect to the c∞ -topology of E: For k ’ ∞ and ¬xed n the sequence an,k is

1 1

M-converging to ( n χN , 0), and n χN is M-converging to 0 for n ’ ∞.

4.33

50 Chapter I. Calculus of smooth mappings 4.35

4.34. Example. We consider the space ∞ (X) := ∞ (X, R) as de¬ned in (2.15)

for a set X together with a family B of subsets called bounded. We have the

subspace Cc (X) := {f ∈ ∞ (X) : supp f is ¬nite}. And we want to calculate its

c∞ -closure in ∞ (X).

Claim: The c∞ -closure of Cc (X) equals

∞

c0 (X) := {f ∈ (X) : f |B ∈ c0 (B) for all B ∈ B},

provided that X is countable.

Proof. The right hand side is just the intersection c0 (X) := B∈B ι’1 (c0 (B)),

B

∞ ∞

(X) ’

where ιB : (B) denotes the restriction map. We use the notation

c0 (X), since in the case where X is bounded this is exactly the space {f ∈ ∞ (X) :

{x : |f (x)| ≥ µ} is ¬nite for all µ > 0}. In particular, this applies to the bounded

space N, where c0 (N) = c0 . Since ∞ (X) carries the initial structure with respect

to these maps c0 (X) is closed. It remains to show that Cc (X) is c∞ -dense in c0 (X).

So take f ∈ c0 (X). Let {x1 , x2 , . . . } := {x : f (x) = 0}.

We consider ¬rst the case, where there exists some δ > 0 such that |f (xn )| ≥ δ for

all n. Then we consider the functions fn := f · χx1 ,...,xn ∈ Cc (X). We claim that

n(f ’ fn ) is bounded in ∞ (X, R). In fact, let B ∈ B. Then {n : xn ∈ B} = {n :

xn ∈ B and |f (xn )| ≥ δ} is ¬nite. Hence, {n(f ’ fn )(x) : x ∈ B} is ¬nite and thus

bounded, i.e. fn converges Mackey to f .

1

Now the general case. We set Xn := {x ∈ X : |f (x)| ≥ n } and de¬ne fn := f · χXn .

1

Then each fn satis¬es the assumption of the particular case with δ = n and hence

is a Mackey limit of a sequence in Cc (X). Furthermore, n(f ’ fn ) is uniformly

bounded by 1, since for x ∈ Xn it is 0 and otherwise |n(f ’fn )(x)| = n|f (x)| < 1. So

after forming the Mackey adherence (i.e. adding the limits of all Mackey-convergent

sequences contained in the set, see (4.32) for a formal de¬nition) twice, we obtain

c0 (X).

Now we want to show that c0 (X) is in fact the c∞ -completion of Cc (X).

4.35. Example. c0 (X). We claim that c0 (X) is the c∞ -completion of the sub-