In particular, the c∞ -interior K o is convex, hence it is not only c∞ -open but open

in the locally convex topology (4.5).

Without loss of generality we now assume that 0 ∈ K o . We claim that the closure of

K is the set {x : tx ∈ K o for 0 < t < 1}. This implies the statement on closedness.

Let U := K o and consider the Minkowski-functional pU (x) := inf{t > 0 : x ∈ tU }.

Since U is convex, the function pU is convex, see (52.2). Using that U is c∞ -open

it can easily be shown that U = {x : pU (x) < 1}. From (13.2) we conclude that pU

is c∞ -continuous, and furthermore that it is even continuous for the locally convex

topology. Hence, the set {x : tx ∈ K o for 0 < t < 1} = {x : pU (x) ¤ 1} = {x :

pK (x) ¤ 1} is the closure of K in the locally convex topology by (52.3).

24.2. Theorem. Derivative of smooth maps.

Let K ⊆ E be a convex subset with non-void interior K o , and let f : K ’ R be a

smooth map. Then f |K o : K o ’ F is smooth, and its derivative (f |K o ) extends

(uniquely) to a smooth map K ’ L(E, F ).

Proof. Only the extension property is to be shown. Let us ¬rst try to ¬nd a

candidate for f (x)(v) for x ∈ K and v ∈ E with x + v ∈ K o . By convexity the

smooth curve cx,v : t ’ x + t2 v has for 0 < |t| < 1 values in K o and cx,v (0) =

x ∈ K, hence f —¦ cx,v is smooth. In the special case where x ∈ K o we have by

the chain rule that (f —¦ cx,v ) (t) = f (x)(cx,v (t))(cx,v (t)), hence (f —¦ cx,v ) (t) =

f (cx,v (t))(cx,v (t), cx,v (t)) + f (cx,v (t))(cx,v (t)), and for t = 0 in particular (f —¦

cx,v ) (0) = 2 f (x)(v). Thus we de¬ne

2 f (x)(v) := (f —¦ cx,v ) (0) for x ∈ K and v ∈ K o ’ x.

√

Note that for 0 < µ < 1 we have f (x)(µ v) = µ f (x)(v), since cx,µ v (t) = cx,v ( µ t).

Let us show next that f ( )(v) : {x ∈ K : x + v ∈ K o } ’ R is smooth. So let

s ’ x(s) be a smooth curve in K, and let v ∈ K 0 ’ x(0). Then x(s) + v ∈ K o for

all su¬ciently small s. And thus the map (s, t) ’ cx(s),v (t) is smooth from some

neighborhood of (0, 0) into K. Hence (s, t) ’ f (cx(s),v (t)) is smooth and also its

second derivative s ’ (f —¦ cx(s),v ) (0) = 2 f (x(s))(v).

In particular, let x0 ∈ K and v0 ∈ K o ’ x0 and x(s) := x0 + s2 v0 . Then

2f (x0 )(v) := (f —¦ cx0 ,v ) (0) = lim (f —¦ cx(s),v ) (0) = lim 2 f (x(s))(v),

s’0 s’0

with x(s) ∈ K o for 0 < |s| < 1. Obviously this shows that the given de¬nition of

f (x0 )(v) is the only possible smooth extension of f ( )(v) to {x0 } ∪ K o .

Now let v ∈ E be arbitrary. Choose a v0 ∈ K o ’ x0 . Since the set K o ’ x0 ’ v0 is

a c∞ -open neighborhood of 0, hence absorbing, there exists some µ > 0 such that

v0 + µv ∈ K o ’ x0 . Thus

f (x)(v) = 1 f (x)(µv) = 1

f (x)(v0 + µv) ’ f (x)(v0 )

µ µ

for all x ∈ K 0 . By what we have shown above the right side extends smoothly

to {x0 } ∪ K o , hence the same is true for the left side. I.e. we de¬ne f (x0 )(v) :=

24.2

24.3 24. Smooth mappings on non-open domains 249

lims’0 f (x(s))(v) for some smooth curve x : (’1, 1) ’ K with x(s) ∈ K o for

0 < |s| < 1. Then f (x) is linear as pointwise limit of f (x(s)) ∈ L(E, R) and is

bounded by the Banach-Steinhaus theorem (applied to EB ). This shows at the

same time, that the de¬nition does not depend on the smooth curve x, since for

v ∈ x0 + K o it is the unique extension.

In order to show that f : K ’ L(E, F ) is smooth it is by (5.18) enough to show

that

f evx

x

s ’ f (x(s))(v), R ’ K ’ L(E, F ) ’ F

is smooth for all v ∈ E and all smooth curves x : R ’ K. For v ∈ x0 + K o

this was shown above. For general v ∈ E, this follows since f (x(s))(v) is a linear

combination of f (x(s))(v0 ) for two v0 ∈ x0 + K o not depending on s locally.

By (24.2) the following lemma applies in particular to smooth maps.

24.3. Lemma. Chain rule. Let K ⊆ E be a convex subset with non-void interior

K o , let f : K ’ R be smooth on K o and let f : K ’ L(E, F ) be an extension of

(f |K o ) , which is continuous for the c∞ -topology of K, and let c : R ’ K ⊆ E be a

smooth curve. Then (f —¦ c) (t) = f (c(t))(c (t)).

Proof.

Claim Let g : K ’ L(E, F ) be continuous along smooth curves in K, then g : ˆ

K — E ’ F is also continuous along smooth curves in K — E.

In order to show this let t ’ (x(t), v(t)) be a smooth curve in K — E. Then

g —¦ x : R ’ L(E, F ) is by assumption continuous (for the bornological topology on

L(E, F )) and v — : L(E, F ) ’ C ∞ (R, F ) is bounded and linear (3.13) and (3.17).

Hence, the composite v — —¦ g —¦ x : R ’ C ∞ (R, F ) ’ C(R, F ) is continuous. Thus,

(v — —¦ g —¦ x)§ : R2 ’ F is continuous, and in particular when restricted to the

diagonal in R2 . But this restriction is just g —¦ (x, v).

Now choose a y ∈ K o . And let cs (t) := c(t) + s2 (y ’ c(t)). Then cs (t) ∈ K o for 0 <

|s| ¤ 1 and c0 = c. Furthermore, (s, t) ’ cs (t) is smooth and cs (t) = (1 ’ s2 )c (t).

And for s = 0

1 1

f (cs (t)) ’ f (cs (0)) 2

(f —¦ cs ) (t„ )d„ = (1 ’ s )

= f (cs (t„ ))(c (t„ ))d„ .

t 0 0

Now consider the speci¬c case where c(t) := x + tv with x, x + v ∈ K. Since

f is continuous along (t, s) ’ cs (t), the left side of the above equation converges

to f (c(t))’f (c(0)) for s ’ 0. And since f (·)(v) is continuous along (t, „, s) ’

t

cs (t„ ) we have that f (cs (t„ ))(v) converges to f (c(t„ ))(v) uniformly with respect

to 0 ¤ „ ¤ 1 for s ’ 0. Thus, the right side of the above equation converges to

1

f (c(t„ ))(v)d„ . Hence, we have

0

1 1

f (c(t)) ’ f (c(0))

f (c(t„ ))(v)d„ ’

= f (c(0))(v)d„ = f (c(0))(c (0))

t 0 0

24.3

250 Chapter V. Extensions and liftings of mappings 24.5

for t ’ 0.

Now let c : R ’ K be an arbitrary smooth curve. Then (s, t) ’ c(0)+s(c(t)’c(0))

is smooth and has values in K for 0 ¤ s ¤ 1. By the above consideration we have

for x = c(0) and v = (c(t) ’ c(0))/t that

1

f (c(t)) ’ f (c(0)) c(t) ’ c(0)

f c(0) + „ (c(t) ’ c(0))

=

t t

0

which converges to f (c(0))(c (0)) for t ’ 0, since f is continuous along smooth

curves in K and thus f (c(0) + „ (c(t) ’ c(0))) ’ f (c(0)) uniformly on the bounded

set { c(t)’c(0) : t near 0}. Thus, f —¦ c is di¬erentiable with derivative (f —¦ c) (t) =

t

f (c(t))(c (t)).

Since f can be considered as a map df : E — E ⊇ K — E ’ F it is important to

study sets A — B ⊆ E — F . Clearly, A — B is convex provided A ⊆ E and B ⊆ F

are. Remains to consider the openness condition. In the locally convex topology

(A — B)o = Ao — B o , which would be enough to know in our situation. However,

we are also interested in the corresponding statement for the c∞ -topology. This

topology on E — F is in general not the product topology c∞ E — c∞ F . Thus, we

cannot conclude that A — B has non-void interior with respect to the c∞ -topology

on E — F , even if A ⊆ E and B ⊆ F have it. However, in case where B = F

everything is ¬ne.

24.4. Lemma. Interior of a product.

Let X ⊆ E. Then the interior (X — F )o of X — F with respect to the c∞ -topology

on E — F is just X o — F , where X o denotes the interior of X with respect to the

c∞ -topology on E.

Proof. Let W be the saturated hull of (X — F )o with respect to the projection

pr1 : E — F ’ E, i.e. the c∞ -open set (X — F )o + {0} — F ⊆ X — F . Its projection

to E is c∞ -open, since it agrees with the intersection with E — {0}. Hence, it is

contained in X o , and (X — F )o ⊆ X o — F . The converse inclusion is obvious since

pr1 is continuous.

24.5. Theorem. Smooth maps on convex sets.

Let K ⊆ E be a convex subset with non-void interior K o , and let f : K ’ F be

a map. Then f is smooth if and only if f is smooth on K o and all derivatives

(f |K o )(n) extend continuously to K with respect to the c∞ -topology of K.

Proof. (’) It follows by induction using (24.2) that f (n) has a smooth extension

K ’ Ln (E; F ).

(⇐) By (24.3) we conclude that for every c : R ’ K the composite f —¦ c : R ’ F

is di¬erentiable with derivative (f —¦ c) (t) = f (c(t))(c (t)) =: df (c(t), c (t)).

The map df is smooth on the interior K o — E, linear in the second variable, and

its derivatives (df )(p) (x, w)(y1 , w1 ; . . . , yp , wp ) are universal linear combinations of

f (p+1) (x)(y1 , . . . , yp ; w) and of f (k+1) (x)(yi1 , . . . , yik ; wi0 ) for k ¤ p.

24.5

24.6 24. Smooth mappings on non-open domains 251

These summands have unique extensions to K — E. The ¬rst one is continuous

along smooth curves in K — E, because for such a curve (t ’ (x(t), w(t)) the

extension f (k+1) : K ’ L(E k , L(E, F )) is continuous along the smooth curve x,

and w— : L(E, F ) ’ C ∞ (R, F ) is continuous and linear, so the mapping t ’

(s ’ f (k+1) (x(t))(yi1 , . . . , yik ; w(s))) is continuous from R ’ C ∞ (R, F ) and thus

as map from R2 ’ F it is continuous, and in particular if restricted to the diagonal.

And the other summands only depend on x, hence have a continuous extension by

assumption.

So we can apply (24.3) inductively using (24.4), to conclude that f —¦ c : R ’ F is

smooth.

In view of the preceding theorem (24.5) it is important to know the c∞ -topology

c∞ X of X, i.e. the ¬nal topology generated by all the smooth curves c : R ’

X ⊆ E. So the ¬rst question is whether this is the trace topology c∞ E|X of the

c∞ -topology of E.

24.6. Lemma. The c∞ -topology is the trace topology.

In the following cases of subsets X ⊆ E the trace topology c∞ E|X equals the topol-

ogy c∞ X:

(1) X is c∞ E-open.

(2) X is convex and locally c∞ -closed.

(3) The topology c∞ E is sequential and X ⊆ E is convex and has non-void

interior.

(3) applies in particular to the case where E is metrizable, see (4.11). A topology

is called sequential if and only if the closure of any subset equals its adherence,

i.e. the set of all accumulation points of sequences in it. By (2.13) and (2.8) the

adherence of a set X with respect to the c∞ -topology, is formed by the limits of all

Mackey-converging sequences in X.

Proof. Note that the inclusion X ’ E is by de¬nition smooth, hence the identity

c∞ X ’ c∞ E|X is always continuous.

(1) Let U ⊆ X be c∞ X-open and let c : R ’ E be a smooth curve with c(0) ∈ U .

Since X is c∞ E-open, c(t) ∈ X for all small t. By composing with a smooth

map h : R ’ R which satis¬es h(t) = t for all small t, we obtain a smooth curve

c —¦ h : R ’ X, which coincides with c locally around 0. Since U is c∞ X-open we

conclude that c(t) = (c —¦ h)(t) ∈ U for small t. Thus, U is c∞ E-open.

¯

(2) Let A ⊆ X be c∞ X-closed. And let A be the c∞ E-closure of A. We have to

¯ ¯

show that A © X ⊆ A. So let x ∈ A © X. Since X is locally c∞ E-closed, there

exists a c∞ E-neighborhood U of x ∈ X with U © X c∞ -closed in U . For every

c∞ E-neighborhood U of x we have that x is in the closure of A © U in U with

respect to the c∞ E-topology (otherwise some open neighborhood of x in U does

not meet A © U , hence also not A). Let an ∈ A © U be Mackey converging to a ∈ U .

Then an ∈ X © U which is closed in U thus a ∈ X. Since X is convex the in¬nite

polygon through the an lies in X and can be smoothly parameterized by the special

24.6

252 Chapter V. Extensions and liftings of mappings 24.8

curve lemma (2.8). Using that A is c∞ X-closed, we conclude that a ∈ A. Thus,

A © U is c∞ U -closed and x ∈ A.

¯

(3) Let A ⊆ X be c∞ X-closed. And let A denote the closure of A in c∞ E. We

¯ ¯

have to show that A © X ⊆ A. So let x ∈ A © X. Since c∞ E is sequential there

is a Mackey converging sequence A an ’ x. By the special curve lemma (2.8)

the in¬nite polygon through the an can be smoothly parameterized. Since X is

convex this curve gives a smooth curve c : R ’ X and thus c(0) = x ∈ A, since A

is c∞ X-closed.

24.7. Example. The c∞ -topology is not trace topology.

Let A ⊆ E be such that the c∞ -adherence Adh(A) of A is not the whole c∞ -closure

¯ ¯

A of A. So let a ∈ A \ Adh(A). Then consider the convex subset K ⊆ E — R de¬ned

by K := {(x, t) ∈ E — R : t ≥ 0 and (t = 0 ’ x ∈ A ∪ {a})} which has non-empty

interior E — R+ . However, the topology c∞ K is not the trace topology of c∞ (E — R)

which equals c∞ (E) — R by (4.15).

Note that this situation occurs quite often, see (4.13) and (4.36) where A is even a

linear subspace.

Proof. Consider A = A — {0} ⊆ K. This set is closed in c∞ K, since E © K is

closed in c∞ K and the only point in (K © E) \ A is a, which cannot be reached by

a Mackey converging sequence in A, since a ∈ Adh(A).

/

It is however not the trace of a closed subset in c∞ (E) — R. Since such a set has to

¯

contain A and hence A a.

24.8. Theorem. Smooth maps on subsets with collar.

Let M ⊆ E have a smooth collar, i.e., the boundary ‚M of M is a smooth sub-

manifold of E and there exists a neighborhood U of ‚M and a di¬eomorphism

ψ : ‚M — R ’ U which is the identity on ‚M and such that ψ(M — {t ∈ R :

t ≥ 0}) = M © U . Then every smooth map f : M ’ F extends to a smooth map

˜

f : M ∪ U ’ F . Moreover, one can choose a bounded linear extension operator

˜

C ∞ (M, F ) ’ C ∞ (M ∪ U, F ), f ’ f .

Proof. By (16.8) there is a continuous linear right inverse S to the restriction

map C ∞ (R, R) ’ C ∞ (I, R), where I := {t ∈ R : t ≥ 0}. Now let x ∈ U and

(px , tx ) := ψ ’1 (x). Then f (ψ(px , ·)) : I ’ F is smooth, since ψ(px , t) ∈ M

for t ≥ 0. Thus, we have a smooth map S(f (ψ(px , ·))) : R ’ F and we de¬ne

˜ ˜

f (x) := S(f (ψ(px , ·)))(tx ). Then f (x) = f (x) for all x ∈ M © U , since for such

˜

an x we have tx ≥ 0. Now we extend the de¬nition by f (x) = f (x) for x ∈ M o .

˜

Remains to show that f is smooth (on U ). So let s ’ x(s) be a smooth curve

in U . Then s ’ (ps , ts ) := ψ ’1 (x(s)) is smooth. Hence, s ’ (t ’ f (ψ(ps , t))

is a smooth curve R ’ C ∞ (I, F ). Since S is continuous and linear the composite

s ’ (t ’ S(f ψ(ps , ·))(t)) is a smooth curve R ’ C ∞ (R, F ) and thus the associated

˜

map R2 ’ F is smooth, and also the composite f (xs ) of it with s ’ (s, ts ).

The existence of a bounded linear extension operator follows now from (21.2).

In particular, the previous theorem applies to the following convex sets:

24.8

24.10 24. Smooth mappings on non-open domains 253

24.9. Proposition. Convex sets with smooth boundary have a collar.

Let K ⊆ E be a closed convex subset with non-empty interior and smooth boundary

‚K. Then K has a smooth collar as de¬ned in (24.8).

Proof. Without loss of generality let 0 ∈ K o .

In order to show that the set U := {x ∈ E : tx ∈ K for some t > 0} is c∞ -open let

/

s ’ x(s) be a smooth curve R ’ E and assume that t0 x(0) ∈ K for some t0 > 0.

/

Since K is closed we have that t0 x(s) ∈ K for all small |s|.

/

1

For x ∈ U let r(x) := sup{t ≥ 0 : tx ∈ K o } > 0, i.e. r = pK o as de¬ned in the

proof of (24.1) and r(x)x is the unique intersection point of ‚K © (0, +∞)x. We

claim that r : U ’ R+ is smooth. So let s ’ x(s) be a smooth curve in U and

x0 := r(x(0))x(0) ∈ ‚K. Choose a local di¬eomorphism ψ : (E, x0 ) ’ (E, 0) which

maps ‚K locally to some closed hyperplane F ⊆ E. Any such hyperplane is the

kernel of a continuous linear functional : E ’ R, hence E ∼ F — R.

=

We claim that v := ψ (x0 )(x0 ) ∈ F . If this were not the case, then we consider the

/

smooth curve c : R ’ ‚K de¬ned by c(t) = ψ ’1 (’tv). Since ψ (x0 ) is injective its

derivative is c (0) = ’x0 and c(0) = x0 . Since 0 ∈ K o , we have that x0 + c(t)’c(0) ∈

t

c(t)’c(0)

o o

K for all small |t|. By convexity c(t) = x0 + t ∈ K for small t > 0, a

t

contradiction.

So we may assume that (ψ (x)(x)) = 0 for all x in a neighborhood of x0 .

For s small r(x(s)) is given by the implicit equation (ψ(r(x(s))x(s))) = 0. So let

g : R2 ’ R be the locally de¬ned smooth map g(t, s) := (ψ(tx(s))). For t = 0

its ¬rst partial derivative is ‚1 g(t, s) = (ψ (tx(s))(x(s))) = 0. So by the classical

implicit function theorem the solution s ’ r(x(s)) is smooth.

Now let Ψ : U —R ’ U be the smooth map de¬ned by (x, t) ’ e’t r(x)x. Restricted

to ‚K — R ’ U is injective, since tx = t x with x,x ∈ ‚K and t, t > 0 implies

x = x and hence t = t . Furthermore, it is surjective, since the inverse mapping is

1

given by x ’ (r(x)x, ln(r(x))). Use that r(»x) = » r(x). Since this inverse is also

smooth, we have the required di¬eomorphism Ψ. In fact, Ψ(x, t) ∈ K if and only if

e’t r(x) ¤ r(x), i.e. t ¤ 0.

That (24.8) is far from being best possible shows the

24.10. Proposition. Let K ⊆ Rn be the quadrant K := {x = (x1 , . . . , xn ) ∈

Rn : x1 ≥ 0, . . . , xn ≥ 0}. Then there exists a bounded linear extension operator

C ∞ (K, F ) ’ C ∞ (Rn , F ) for each convenient vector space F .

This can be used to obtain the same result for submanifolds with convex corners

sitting in smooth ¬nite dimensional manifolds.

Proof. Since K = (R+ )n ⊆ Rn and the inclusion is the product of inclusions

ι : R+ ’ R we can use the exponential law (23.2.3) to obtain C ∞ (K, F ) ∼ =

∞ ∞

n’1

C ((R+ ) , C (R+ , F )). By Seeley™s theorem (16.8) we have a bounded lin-

ear extension operator S : C ∞ (R+ , F ) ’ C ∞ (R, F ). We now proceed by induction

24.10

254 Chapter V. Extensions and liftings of mappings 25.1

on n. So we have an extension operator Sn’1 : C ∞ ((R+ )n’1 , G) ’ C ∞ (Rn’1 , G)

for the convenient vector space G := C ∞ (R, F ) by induction hypothesis. The

composite gives up to natural isomorphisms the required extension operator

S—

C ∞ (K, F ) ∼ C ∞ ((R+ )n’1 , C ∞ (R+ , F )) ’’ C ∞ ((R+ )n’1 , C ∞ (R, F )) ’

=

Sn’1

’ ’ C ∞ (Rn’1 , C ∞ (R, F )) ∼ C ∞ (Rn , F ).

’’ =

25. Real Analytic Mappings on Non-Open Domains

In this section we will consider real analytic mappings de¬ned on the same type of

convex subsets as in the previous section.

25.1. Theorem. Power series in Fr´chet spaces. Let E be a Fr´chet space and

e e

(F, F ) be a dual pair. Assume that a Baire vector space topology on E exists for

which the point evaluations are continuous. Let fk be k-linear symmetric bounded

functionals from E to F , for each k ∈ N. Assume that for every ∈ F and every x

∞

in some open subset W ⊆ E the power series k=0 (fk (xk ))tk has positive radius of

convergence. Then there exists a 0-neighborhood U in E, such that {fk (x1 , . . . , xk ) :

∞

k ∈ N, xj ∈ U } is bounded and thus the power series x ’ k=0 fk (xk ) converges

Mackey on some 0-neighborhood in E.

Proof. Choose a ¬xed but arbitrary ∈ F . Then —¦ fk satisfy the assumptions

of (7.14) for an absorbing subset in a closed cone C with non-empty interior. Since

this cone is also complete metrizable we can proceed with the proof as in (7.14)

to obtain a set AK,r ⊆ C whose interior in C is non-void. But this interior has

to contain a non-void open set of E and as in the proof of (7.14) there exists

some ρ > 0 such that for the ball Uρ in E with radius ρ and center 0 the set

{ (fk (x1 , . . . , xk )) : k ∈ N, xj ∈ Uρ } is bounded.

Now let similarly to (9.6)

{ ∈ F : | (fk (x1 , . . . , xk ))| ¤ Krk }

AK,r,ρ :=

k∈N x1 ,...xn ∈Uρ

for K, r, ρ > 0. These sets AK,r,ρ are closed in the Baire topology, since evaluation

at fk (x1 , . . . , xk ) is assumed to be continuous.

By the ¬rst part of the proof the union of these sets is F . So by the Baire property,

there exist K, r, ρ > 0 such that the interior U of AK,r,ρ is non-empty. As in the

proof of (9.6) we choose an 0 ∈ U . Then for every ∈ F there exists some µ > 0

such that µ := µ ∈ U ’ 0 . So | (y)| ¤ 1 (| µ (y) + 0 (y)| + | 0 (y)|) ¤ 2 Krn for

µ µ

every y = fk (x1 , . . . , xk ) with xi ∈ Uρ . Thus, {fk (x1 , . . . , xk ) : k ∈ N, xi ∈ U ρ } is

r

bounded.

On every smaller ball we have therefore that the power series with terms fk con-

verges Mackey.

Note that if the vector spaces are real and the assumption above hold, then the

conclusion is even true for the complexi¬ed terms by (7.14).

25.1

25.2 25. Real analytic mappings on non-open domains 255

25.2. Theorem. Real analytic maps I ’ R are germs.

Let f : I := {t ∈ R : t ≥ 0} ’ R be a map. Suppose t ’ f (t2 ) is real analytic

˜˜ ˜

R ’ R. Then f extends to a real analytic map f : I ’ R, where I is an open

neighborhood of I in R.

Proof. We show ¬rst that f is smooth. Consider g(t) := f (t2 ). Since g : R ’ R

is assumed to be real analytic it is smooth and clearly even. We claim that there

exists a smooth map h : R ’ R with g(t) = h(t2 ) (this is due to [Whitney, 1943]).