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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-
space Cc (X) in ∞ (X) formed by the ¬nite sequences.
We may assume that the bounded sets of X are formed by those subsets B, for
which f (B) is bounded for all f ∈ ∞ (X). Obviously, any bounded set has this
property, and the space ∞ (X) is not changed by adding these sets. Furthermore,
the restriction map ιB : ∞ (X) ’ ∞ (B) is also bounded for such a B, since us-
ing the closed graph theorem (52.10) we only have to show that evb —¦ιB = ι{b} is
bounded for every b ∈ B, which is obviously the case.
By proposition (4.30) it is enough to show the universal property for bounded
linear functionals. In analogy to Banach-theory, we only have to show that the
dual Cc (X) is just

1
(X) := {g : X ’ R : supp g is bounded and g is absolutely summable}.

4.35
4. The c∞ -topology
4.36 51

In fact, any such g acts even as bounded linear functional on ∞ (X, R) by g, f :=

x g(x) f (x), since a subset is bounded in (X) if and only if it is uniformly
bounded on all bounded sets B ⊆ X. Conversely, let : Cc (X) ’ R be bounded
and linear and de¬ne g : X ’ R, by g(x) := (ex ), where ex denotes the function
given by ex (y) := 1 for x = y and 0 otherwise. Obviously (f ) = g, f for all
f ∈ Cc (X). Suppose indirectly supp g = {x : (ex ) = 0} is not bounded. Then
there exists a sequence xn ∈ supp g and a function f ∈ ∞ (X) such that |f (xn )| ≥ n.
In particular, the only bounded subsets of {xn : n ∈ N} are the ¬nite ones. Hence
n
{ |g(xn )| exn } is bounded in Cc (X), but the image under is not. Furthermore, g
has to be absolutely summable since the set of ¬nite subsums of x sign g(x) ex is
bounded in Cc (X) and its image under are the subsums of x |g(x)|.

4.36. Corollary. Counter-examples on c∞ -topology. The following state-
ments are false:
(1) The c∞ -closure of a subset (or of a linear subspace) is given by the Mackey
adherence, i.e. the set formed by all limits of sequences in this subset which
are Mackey convergent in the total space.
(2) A subset U of E that contains a point x and has the property, that every
sequence which M -converges to x belongs to it ¬nally, is a c∞ -neighborhood
of x.
(3) A c∞ -dense subspace of a c∞ -complete space has this space as c∞ -comple-
tion.
(4) If a subspace E is c∞ -dense in the total space, then it is also c∞ -dense in
each linear subspace lying in between.
(5) The c∞ -topology of a linear subspace is the trace of the c∞ -topology of the
whole space.
(6) Every bounded linear functional on a linear subspace can be extended to such
a functional on the whole space.
(7) A linear subspace of a bornological locally convex space is bornological.
(8) The c∞ -completion preserves embeddings.

Proof. (1) For this we give an example, where the Mackey adherence of Cc (X) is
not all of c0 (X).
Let X = N — N, and take as bounded sets all sets of the form Bµ := {(n, k) : n ¤
µ(k)}, where µ runs through all functions N ’ N. Let f : X ’ R be de¬ned by
1
f (n, k) := k . Obviously, f ∈ c0 (X), since for given j ∈ N and function µ the set of
points (n, k) ∈ Bµ for which f (n, k) = k ≥ 1 is the ¬nite set {(n, k) : k ¤ j, n ¤
1
j
µ(k)}.
Assume there is a sequence fn ∈ Cc (X) Mackey convergent to f . By passing to a
subsequence we may assume that n2 (f ’ fn ) is bounded. Now choose µ(k) to be
larger than all of the ¬nitely many n, with fk (n, k) = 0. If k 2 (f ’ fk ) is bounded
on Bµ , then in particular {k 2 (f ’ fk )(µ(k), k) : k ∈ N} has to be bounded, but
1
k 2 (f ’ fk )(µ(k), k) = k 2 k ’ 0 = k.
(2) Let A be a set for which (1) fails, and choose x in the c∞ -closure of A but not
in the M -adherence of A. Then U := E \ A satis¬es the assumptions of (2). In

4.36
52 Chapter I. Calculus of smooth mappings 5.1

fact, let xn be a sequence which converges Mackey to x, and assume that it is not
¬nally in U . So we may assume without loss of generality that xn ∈ U for all n, but
/
then A xn ’ x would imply that x is in the Mackey adherence of A. However,
U cannot be a c∞ -neighborhood of x. In fact, such a neighborhood must meet A
since x is assumed to be in the c∞ -closure of A.
(3) Let F be a locally convex vector space whose Mackey adherence in its c∞ -com-
pletion E is not all of E, e.g. Cc (X) ⊆ c0 (X) as in the previous counter-example.
Choose a y ∈ E that is not contained in the Mackey adherence of F , and let F1
be the subspace of E generated by F ∪ {y}. We claim that F1 ⊆ E cannot be the
c∞ -completion although F1 is obviously c∞ -dense in the convenient vector space
E. In order to see this we consider the linear map : F1 ’ R characterized by
(F ) = 0 and (y) = 1. Clearly is well de¬ned.
: F1 ’ R is bornological: For any bounded B ⊆ F1 there exists an N such that
B ⊆ F + [’N, N ]y. Otherwise, bn = xn + tn y ∈ B would exist with tn ’ ∞ and
xn ∈ F . This would imply that bn = tn ( xn + y), and thus ’ xn would converge
n n
t t
Mackey to y; a contradiction.
Now assume that a bornological extension ¯ to E exists. Then F ⊆ ker( ¯) and
ker( ¯) is c∞ -closed, which is a contradiction to the c∞ -denseness of F in E. So
F1 ⊆ E does not have the universal property of a c∞ -completion.
This shows also that (6) fails.
(4) Furthermore, it follows that F is c∞ F1 -closed in F1 , although F and hence F1
are c∞ -dense in E.
(5) The trace of the c∞ -topology of E to F1 cannot be the c∞ -topology of F1 , since
for the ¬rst one F is obviously dense.
(7) Obviously, the trace topology of the bornological topology on E cannot be
bornological on F1 , since otherwise the bounded linear functionals on F1 would be
continuous and hence extendable to E.
(8) Furthermore, the extension of the inclusion ι : F • R ∼ F1 ’ E to the com-
=
pletion is given by (x, t) ∈ E • R ∼ F • R = F1 ’ x + ty ∈ E and has as kernel
=˜ ˜
the linear subspace generated by (y, ’1). Hence, the extension of an embedding to
the c∞ -completions need not be an embedding anymore, in particular the inclusion
functor does not preserve injectivity of morphisms.



5. Uniform Boundedness Principles and Multilinearity

5.1. The category of locally convex spaces and smooth mappings. The
category of all smooth mappings between bornological vector spaces is a subcate-
gory of the category of all smooth mappings between locally convex spaces which
is equivalent to it, since a locally convex space and its bornologi¬cation (4.4) have
the same bounded sets and smoothness depends only on the bornology by (1.8).
So it is also cartesian closed, but the topology on C ∞ (E, F ) from (3.11) has to be

5.1
5.3 5. Uniform boundedness principles and multilinearity 53

bornologized. For an example showing the necessity see [Kriegl, 1983, p. 297] or
[Fr¨licher, Kriegl, 1988, 5.4.19]: The topology on C ∞ (R, R(N) ) is not bornological.
o
We will in general, however, work in the category of locally convex spaces and
smooth mappings, so function spaces carry the topology of (3.11).
The category of bounded (equivalently continuous) linear mappings between bor-
nological vector spaces is in the same way equivalent to the category of all bounded
linear mappings between all locally convex spaces, since a linear mapping is smooth
if and only if it is bounded, by (2.11). It is closed under formation of colimits and
under quotients (this is an easy consequence of (4.1.1)). The Mackey-Ulam theo-
rem [Jarchow, 1981, 13.5.4] tells us that a product of non trivial bornological vector
spaces is bornological if and only if the index set does not admit a Ulam measure,
i.e. a non trivial {0, 1}-valued measure on the whole power set. A cardinal admit-
ting a Ulam measure has to be strongly inaccessible, so we can restrict set theory
to exclude measurable cardinals.
Let L(E1 , . . . , En ; F ) denote the space of all bounded n-linear mappings from E1 —
. . . — En ’ F with the topology of uniform convergence on bounded sets in E1 —
. . . — En .

5.2. Proposition. Exponential law for L. There are natural bornological
isomorphisms

L(E1 , . . . , En+k ; F ) ∼ L(E1 , . . . , En ; L(En+1 , . . . , En+k ; F )).
=


Proof. We proof this for bilinear maps, the general case is completely analogous.
We already know that bilinearity translates into linearity into the space of linear
functions. Remains to prove boundedness. So let B ⊆ L(E1 , E2 ; F ) be given. Then
B is bounded if and only if B(B1 — B2 ) ⊆ F is bounded for all bounded Bi ⊆ Ei .
This however is equivalent to B ∨ (B1 ) is contained and bounded in L(E2 , F ) for all
bounded B ⊆ E1 , i.e. B ∨ is contained and bounded in L(E1 , L(E2 , F )).

Recall that we have already put a structure on L(E, F ) in (3.17), namely the initial
one with respect to the inclusion in C ∞ (E, F ). Let us now show that bornologically
these de¬nitions agree:

5.3. Lemma. Structure on L. A subset is bounded in L(E, F ) ⊆ C ∞ (E, F )
if and only if it is uniformly bounded on bounded subsets of E, i.e. L(E, F ) ’
C ∞ (E, F ) is initial.

Proof. Let B ⊆ L(E, F ) be bounded in C ∞ (E, F ), and assume that it is not
uniformly bounded on some bounded set B ⊆ E. So there are fn ∈ B, bn ∈ B, and
∈ F — with | (fn (bn ))| ≥ nn . Then the sequence n1’n bn converges fast to 0, and
hence lies on some compact part of a smooth curve c by the special curve lemma
(2.8). So B cannot be bounded, since otherwise C ∞ ( , c) = — —¦ c— : C ∞ (E, F ) ’
C ∞ (R, R) ’ ∞ (R, R) would have bounded image, i.e. { —¦ fn —¦ c : n ∈ N} would
be uniformly bounded on any compact interval.

5.3
54 Chapter I. Calculus of smooth mappings 5.6

Conversely, let B ⊆ L(E, F ) be uniformly bounded on bounded sets and hence
in particular on compact parts of smooth curves. We have to show that dn —¦ c— :
L(E, F ) ’ C ∞ (R, F ) ’ ∞ (R, F ) has bounded image. But for linear smooth maps
we have by the chain rule (3.18), recursively applied, that dn (f —¦ c)(t) = f (c(n) (t)),
and since c(n) is still a smooth curve we are done.

Let us now generalize this result to multilinear mappings. For this we ¬rst charac-
terize bounded multilinear mappings in the following two ways:

5.4. Lemma. A multilinear mapping is bounded if and only if it is bounded on
each sequence which converges Mackey to 0.

Proof. Suppose that f : E1 — . . . — Ek ’ F is not bounded on some bounded set
B ⊆ E1 — . . . — Ek . By composing with a linear functional we may assume that
F = R. So there are bn ∈ B with »k+1 := |f (bn )| ’ ∞. Then |f ( »1 bn )| = »n ’ ∞,
n n
1
but ( »n bn ) is Mackey convergent to 0.

5.5. Lemma. Bounded multilinear mappings are smooth. Let f : E1 —
. . . — En ’ F be a multilinear mapping. Then f is bounded if and only if it is
smooth. For the derivative we have the product rule:
n
df (x1 , . . . , xn )(v1 , . . . , vn ) = f (x1 , . . . , xi’1 , vi , xi+1 , . . . , xn ).
i=1

In particular, we get for f : E ⊇ U ’ R, g : E ⊇ U ’ F and x ∈ U , v ∈ E the
Leibniz formula

(f · g) (x)(v) = f (x)(v) · g(x) + f (x) · g (x)(v).


Proof. We use induction on n. The case n = 1 is corollary (2.11). The induction
goes as follows:
f is bounded
f (B1 — . . . — Bn ) = f ∨ (B1 — . . . — Bn’1 )(Bn ) is bounded for all bounded
⇐’
sets Bi in Ei ;
f ∨ (B1 — . . . — Bn’1 ) ⊆ L(En , F ) ⊆ C ∞ (En , F ) is bounded, by (5.3);
⇐’
f ∨ : E1 — . . . — En’1 ’ C ∞ (En , F ) is bounded;
⇐’
f ∨ : E1 — . . . — En’1 ’ C ∞ (En , F ) is smooth by the inductive assumption;
⇐’
f ∨ : E1 — . . . — En ’ F is smooth by cartesian closedness (3.13).
⇐’
The particular case follows by application to the scalar multiplication R — F ’ F .

Now let us show that also the structures coincide:

5.6. Proposition. Structure on space of multilinear maps. The injection
of L(E1 , . . . , En ; F ) ’ C ∞ (E1 — . . . — En , F ) is a bornological embedding.

Proof. We can show this by induction. In fact, let B ⊆ L(E1 , . . . , En ; F ). Then
B is bounded

5.6
5.7 5. Uniform boundedness principles and multilinearity 55

⇐’ B(B1 — . . . — Bn ) = B ∨ (B1 — . . . — Bn’1 )(Bn ) is bounded for all bounded
Bi in Ei ;
⇐’ B ∨ (B1 — . . . — Bn’1 ) ⊆ L(En , F ) ⊆ C ∞ (En , F ) is bounded, by (5.3);
⇐’ B ∨ ⊆ C ∞ (E1 — . . . — En’1 , C ∞ (En , F )) is bounded by the inductive as-
sumption;
⇐’ B ⊆ C ∞ (E1 — . . . — En , F ) is bounded by cartesian closedness (3.13).

5.7. Bornological tensor product. It is natural to consider the universal prob-
lem of linearizing bounded bilinear mappings. The solution is given by the bornolo-
gical tensor product E —β F , i.e. the algebraic tensor product with the ¬nest locally
convex topology such that E — F ’ E — F is bounded. A 0-neighborhood basis of
this topology is given by those absolutely convex sets, which absorb B1 — B2 for all
bounded B1 ⊆ E1 and B2 ⊆ E2 . Note that this topology is bornological since it is
the ¬nest locally convex topology with given bounded linear mappings on it.

Theorem. The bornological tensor product is the left adjoint functor to the Hom-
functor L(E, ) on the category of bounded linear mappings between locally convex
spaces, and one has the following bornological isomorphisms:

L(E —β F, G) ∼ L(E, F ; G) ∼ L(E, L(F, G))
= =
E —β R ∼ E
=
E —β F ∼ F —β E
=
(E —β F ) —β G ∼ E —β (F —β G)
=

Furthermore, the bornological tensor product preserves colimits. It neither preserves
embeddings nor countable products.

Proof. We show ¬rst that this topology has the universal property for bounded
bilinear mappings f : E1 — E2 ’ F . Let U be an absolutely convex zero neighbor-
hood in F , and let B1 , B2 be bounded sets. Then f (B1 — B2 ) is bounded, hence
˜ ˜
it is absorbed by U . Then f ’1 (U ) absorbs —(B1 — B2 ), where f : E1 — E2 ’ F is
˜
the canonically associated linear mapping. So f ’1 (U ) is in the zero neighborhood
˜
basis of E1 —β E2 described above. Therefore, f is continuous.
A similar argument for sets of mappings shows that the ¬rst isomorphism L(E —β
F, G) ∼ L(E, F ; G) is bornological.
=
The topology on E1 —β E2 is ¬ner than the projective tensor product topology, and

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