closed we use (2.13). Let xn ’ x∞ be fast falling, xn ∈ E but x∞ ∈ F . By

(2.8) the polygon c through (xn ) can be smoothly parameterized. Hence c is

smooth and has values in the vector space generated by {xn : n = ∞}, which is

contained in E. Its anti-derivative c2 is up to a constant equal to c, and by (2)

x1 ’ x∞ = c(1) ’ c(0) = c2 (1) ’ c2 (0) lies in E. So x∞ ∈ E.

(3) ’ (5) Let F be the completion E of E. Any Mackey Cauchy sequence in E

has a limit in F , and since E is by assumption c∞ -closed in F the limit lies in E.

Hence, the sequence converges in E.

(6) ’ (7) Let f : F ’ E be a continuous mapping on a normed space F . Since the

image of the unit ball is bounded, it is a bounded mapping into EB for some closed

absolutely convex B. But into EB it can be extended to the completion, since EB

is complete.

2.14

2.15 2. Completeness 21

(7) ’ (1) Let c : R ’ E be a Lipschitz curve. Then c is locally a continuous curve

into EB for some absolutely convex bounded set B. The inclusion of EB into E

has a continuous extension to the completion of EB , and c is Riemann integrable

in this Banach space, so also in E.

(4) ’ (3) Let E be embedded in some space F . We use again (2.13) in order to

show that E is c∞ -closed in F . So let xn ’ x∞ fast falling, xn ∈ E for n = 0,

but x∞ ∈ F . By (2.8) the polygon c through (xn ) can be smoothly symmetrically

parameterized in F , and c(t) ∈ E for t = 0. We consider c(t) := tc(t). This is a

˜

curve in E which is smooth in F , so it is scalarwise smooth in E, thus smooth in

E by (4). Then x∞ = c (0) ∈ E.

˜

2.15. Theorem. Inheritance of c∞ -completeness. The following construc-

tions preserve c∞ -completeness: limits, direct sums, strict inductive limits of se-

quences of closed embeddings, as well as formation of ∞ (X, ), where X is a set

together with a family B of subsets of X containing the ¬nite ones, which are called

bounded and ∞ (X, F ) denotes the space of all functions f : X ’ F , which are

bounded on all B ∈ B, supplied with the topology of uniform convergence on the

sets in B.

Note that the de¬nition of the topology of uniform convergence as initial topology

shows, that adding all subsets of ¬nite unions of elements in B to B does not change

this topology. Hence, we may always assume that B has this stability property; this

is the concept of a bornology on a set.

Proof. The projective limit (52.8) of F is the c∞ -closed linear subspace

(x± ) ∈ F(±) : F(f )x± = xβ for all f : ± ’ β ,

hence is c∞ -complete, since the product of c∞ -complete factors is obviously c∞ -

complete.

Since the coproduct (52.7) of spaces X± is the topological direct sum, and has as

bounded sets those which are contained and bounded in some ¬nite subproduct, it

is c∞ -complete if all factors are.

For colimits this is in general not true. For strict inductive limits of sequences of

closed embeddings it is true, since bounded sets are contained and bounded in some

step, see (52.8).

For the result on ∞ (X, F ) we consider ¬rst the case, where X itself is bounded.

Then c∞ -completeness can be proved as in (52.4) or reduced to this result. In fact

let B be bounded in ∞ (X, F ). Then B(X) is bounded in F and hence contained

in some absolutely convex bounded set B, for which FB is a Banach space. So we

may assume that B := {f ∈ ∞ (X, F ) : f (X) ⊆ B}. The space ∞ (X, F )B is just

the space ∞ (X, FB ) with the supremum norm, which is a Banach space by (52.4).

∞

(X, F ) ’ ∞ (B, F )

Let now X and B be arbitrary. Then the restriction maps

give an embedding ι of ∞ (X, F ) into the product B∈B ∞

(B, F ). Since this

2.15

22 Chapter I. Calculus of smooth mappings 3.2

product is complete, by what we have shown above, it is enough to show that this

embedding has a closed image. So let f± |B converge to some fB in ∞ (B, F ).

De¬ne f (x) := f{x} (x). For any B ∈ B containing x we have that fB (x) =

(lim± f± |B )(x) = lim± (f± (x)) = lim± f± |{x} = f{x} (x) = f (x), and f (B) is boun-

ded for all B ∈ B, since f |B = fB ∈ ∞ (B, F ).

Example. In general, a quotient and an inductive limit of c∞ -complete spaces

need not be c∞ -complete. In fact, let ED := {x ∈ RN : supp x ⊆ D} for any

subset D ⊆ N of density dens D := lim sup{ |D©[1,n]| } = 0. It can be shown that

n

E := dens D=0 ED ‚ R is the inductive limit of the Fr´chet subspaces ED ∼ RD .

N

e =

∞

It cannot be c -complete, since ¬nite sequences are contained in E and are dense

in RN ⊃ E.

3. Smooth Mappings and the Exponential Law

Now let us start proving the exponential law C ∞ (U — V, F ) ∼ C ∞ (U, C ∞ (V, F ))

=

¬rst for U = V = F = R.

3.1. Proposition. For a continuous map f : R — [0, 1] ’ R the partial derivative

‚1 f exists and is continuous if and only if f ∨ : R ’ C([0, 1], R) is continuously

1 1‚

di¬erentiable. And in this situation I((f ∨ ) (t)) = dt 0 f (t, s) ds = 0 ‚t f (t, s) ds,

d

where I : C([0, 1], R) ’ R is integration.

Proof. We assume that ‚1 f exists and is continuous. Hence, (‚1 f )∨ : R ’

C([0, 1], R) is continuous. We want to show that f ∨ : R ’ C([0, 1], R) is dif-

ferentiable (say at 0) with this function (at 0) as derivative. So we have to show

∨ ∨

that the mapping t ’ f (t)’f (0) is continuously extendable to R by de¬ning its

t

∨

value at 0 as (‚1 f ) (0). Or equivalently, by what is obvious for continuous maps,

that the map

f (t,s)’f (0,s)

for t = 0

t

(t, s) ’

‚1 f (0, s) otherwise

is continuous. This follows immediately from the continuity of ‚1 f and of integra-

1

tion since it can be written as 0 ‚1 f (r t, s) dr by the fundamental theorem.

1

So we arrive under this assumption at the conclusion, that f (t, s) ds is di¬eren-

0

tiable with derivative

1 1

d ‚

∨

f (t, s) ds = I((f ) (t)) = f (t, s) ds.

dt ‚t

0 0

The converse implication is obvious.

3.2. Theorem. Simplest case of exponential law. Let f : R2 ’ R be an

arbitrary mapping. Then all iterated partial derivatives exist and are locally bounded

if and only if the associated mapping f ∨ : R ’ C ∞ (R, R) exists as a smooth curve,

3.2

3.2 3. Smooth mappings and the exponential law 23

where C ∞ (R, R) is considered as the Fr´chet space with the topology of uniform

e

convergence of each derivative on compact sets. Furthermore, we have (‚1 f )∨ =

d(f ∨ ) and (‚2 f )∨ = d —¦ f ∨ = d— (f ∨ ).

Proof. We have several possibilities to prove this result. Either we show Mackey

convergence of the di¬erence quotients, via the boundedness of 1 c(t)’c(0) ’ c (0) ,

t t

∼ ∞ (X, ∞ (Y, R)); or we use

∞

(X —Y, R) =

and then use the trivial exponential law

the induction step proved in (3.1), namely that f ∨ : R ’ C(R, R) is di¬erentiable

if and only if ‚1 f exists and is continuous R2 ’ R, together with the exponential

law C(R2 , R) ∼ C(R, C(R, R)). We choose the latter method.

=

For this we have to note ¬rst that if for a function g the partial derivatives ‚1 g and

‚2 g exist and are locally bounded then g is continuous:

g(x, y) ’ g(0, 0) = g(x, y) ’ g(x, 0) + g(x, 0) ’ g(0, 0)

= y‚2 g(x, r2 y) + x‚1 g(r1 x, 0)

for suitable r1 , r2 ∈ [0, 1], which goes to 0 with (x, y).

Proof of (’) By what we just said, all iterated partial derivatives of f are contin-

uous. First observe that f ∨ : R ’ C ∞ (R, R) makes sense and that for all t ∈ R we

have

q

dq (f ∨ (t)) = (‚2 f )∨ (t).

(1)

Next we claim that f ∨ : R ’ C ∞ (R, R) is di¬erentiable, with derivative d(f ∨ ) =

(‚1 f )∨ , or equivalently that for all a the curve

f ∨ (t+a)’f ∨ (a)

for t = 0

t

c:t’ ∨

(‚1 f ) (a) otherwise

is continuous as curve R ’ C ∞ (R, R). Without loss of generality we may assume

that a = 0. Since C ∞ (R, R) carries the initial structure with respect to the linear

mappings dp : C ∞ (R, R) ’ C(R, R) we have to show that dp —¦ c : R ’ C(R, R)

is continuous, or equivalently by the exponential law for continuous maps, that

(dp —¦ c)§ : R2 ’ R is continuous. For t = 0 and s ∈ R we have

f ∨ (t) ’ f ∨ (0)

§

p p p

(d —¦ c) (t, s) = d (c(t))(s) = d (s)

t

p p

‚2 f (t, s) ’ ‚2 f (0, s)

= by (1)

t

1

p

= ‚1 ‚2 f (t „, s) d„ by the fundamental theorem.

0

For t = 0 we have

(dp —¦ c)§ (0, s) = dp (c(0))(s) = dp ((‚1 f )∨ (0))(s)

p

= (‚2 (‚1 f ))∨ (0)(s) by (1)

p

= ‚2 ‚1 f (0, s)

p

= ‚1 ‚2 f (0, s) by the theorem of Schwarz.

3.2

24 Chapter I. Calculus of smooth mappings 3.3

1 p

So we see that (dp —¦ c)§ (t, s) = 0 ‚1 ‚2 f (t „, s) d„ for all (t, s). This function is con-

p p

tinuous in (t, s), since ‚1 ‚2 f : R2 ’ R is continuous, hence (t, s, „ ) ’ ‚1 ‚2 f (t „, s)

p

is continuous, and therefore also (t, s) ’ („ ’ ‚1 ‚2 f (t „, s)) from R2 ’ C([0, 1], R).

1

Composition with the continuous linear mapping 0 : C([0, 1], R) ’ R gives the

continuity of (dp —¦ c)§ .

Now we proceed by induction. By the induction hypothesis applied to ‚1 f , we

obtain that d(f ∨ ) = (‚1 f )∨ and (‚1 f )∨ : R ’ C ∞ (R, R) is n times di¬erentiable,

so f ∨ is (n + 1)-times di¬erentiable.

Proof of (⇐) First remark that for a smooth map f : R ’ C ∞ (R, R) the asso-

ciated map f § : R2 ’ R is locally bounded: Since f is smooth f (I1 ) is com-

pact, hence bounded in C ∞ (R, R) for all compact intervals I1 . In particular,

f (I1 )(I2 ) = f § (I1 — I2 ) has to be bounded in R for all compact intervals I1 and I2 .

Since f is smooth both curves df and d —¦ f = d— f are smooth (use (1.3) and that

d is continuous and linear). An easy calculation shows that the partial derivatives

of f § exist and are given by ‚1 f § = (df )§ and ‚2 f § = (d —¦ f )§ . So one obtains

inductively that all iterated derivatives of f § exist and are locally bounded, since

they are associated to smooth curves R ’ C ∞ (R, R).

In order to proceed to more general cases of the exponential law we need a de¬nition

of C ∞ -maps de¬ned on in¬nite dimensional spaces. This de¬nition should at least

guarantee the chain rule, and so one could take the weakest notion that satis¬es

the chain rule. However, consider the following

3.3. Example. We consider the following 3-fold “singular covering” f : R2 ’ R2

given in polar coordinates by (r, •) ’ (r, 3•). In cartesian coordinates we obtain

the following formula for the values of f :

(r cos(3•), r sin(3•)) = r (cos •)3 ’ 3 cos •(sin •)2 , 3 sin •(cos •)2 ’ (sin •)3

x3 ’ 3xy 2 3x2 y ’ y 3

= ,2 .

x2 + y 2 x + y2

Note that the composite from the left with any orthonormal projection is just the

composite of the ¬rst component of f with a rotation from the right (Use that f

intertwines the rotation with angle δ and the rotation with angle 3δ).

Obviously, the map f is smooth on R2 \ {0}. It is homogeneous of degree 1, and

‚

hence the directional derivative is f (0)(v) = ‚t |t=0 f (tv) = f (v). However, both

components are nonlinear with respect to v and thus are not di¬erentiable at (0, 0).

Obviously, f : R2 ’ R2 is continuous.

We claim that f is di¬erentiable along di¬erentiable curves, i.e. (f —¦ c) (0) exists,

provided c (0) exists.

Only the case c(0) = 0 is not trivial. Since c is di¬erentiable at 0 the curve c1

de¬ned by c1 (t) := c(t) for t = 0 and c (0) for t = 0 is continuous at 0. Hence

t

f (c(t))’f (c(0)) f (t c1 (t))’0

= = f (c1 (t)). This converges to f (c1 (0)), since f is con-

t t

tinuous.

3.3

3.3 3. Smooth mappings and the exponential law 25

Furthermore, if f (x)(v) denotes the directional derivative, which exists everywhere,

then (f —¦ c) (t) = f (c(t))(c (t)). Indeed for c(t) = 0 this is clear and for c(t) = 0 it

follows from f (0)(v) = f (v).

The directional derivative of the 1-homogeneous mapping f is 0-homogeneous: In

fact, for s = 0 we have

‚ ‚ t 1

f (sx)(v) = f (s x + tv) = s f (x + v) = s f (x)( v) = f (x)(v).

‚t ‚t s s

t=0 t=0

‚ ‚

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