absolutely convex set B.

First Proof. By the obvious extension of lemma (1.8) for smooth mappings R2 ⊃

D ’ E the curve c factors locally to a Lip1 -curve into some complete EB . Since it

has complex linear derivative, by theorem (7.4) it is holomorphic.

Second direct proof. Let W be a relatively compact neighborhood of some point

in D. Then c(W ) is bounded in E. It su¬ces to show that for the absolutely convex

closed hull B of c(W ) the Taylor series of c at each z ∈ W converges in EB , i.e.

that c|W : W ’ EB is holomorphic. This follows from the

Vector valued Cauchy inequalities. If r > 0 is smaller than the radius of

convergence at z of c then

r k (k)

(z) ∈ B

k! c

where B is the closed absolutely convex hull of { c(w) : |w ’ z| = r}. (By the

Hahn-Banach theorem this follows directly from the scalar valued case.)

Thus, we get

r k (k)

m m

w’z k w’z k

· ∈ ·B

k=n ( r ) k! c (z) k=n ( r )

c(k) (z)

’ z)k is convergent in EB for |w ’ z| < r.

and so k! (w

k

7.6

7.10 7. Calculus of holomorphic mappings 83

This proof also shows that holomorphic curves with values in complex convenient

vector spaces are topologically and bornologically holomorphic in the sense analo-

gous to (9.4).

7.7. Lemma. Let E be a regular (i.e. every bounded set is contained and bounded

in some step E± ) inductive limit of complex locally convex spaces E± ⊆ E, let

c : C ⊇ U ’ E be a holomorphic mapping, and let W ⊆ C be open and such that

the closure W is compact and contained in U . Then there exists some ±, such that

c|W : W ’ E± is well de¬ned and holomorphic.

Proof. By lemma (7.6) the restriction of c to W factors to a holomorphic curve

c|W : W ’ EB for a suitable bounded absolutely convex set B ⊆ E. Since B is

contained and bounded in some E± one has c|W : W ’ EB = (E± )B ’ E± is

holomorphic.

7.8. De¬nition. Let E and F be convenient vector spaces and let U ⊆ E be

c∞ -open. A mapping f : U ’ F is called holomorphic, if it maps holomorphic

curves in U to holomorphic curves in F .

It is remarkable that [Fantappi´, 1930] already gave this de¬nition. Connections to

e

other concepts of holomorphy are discussed in [Kriegl, Nel, 1985, 2.19].

So by (7.4) f is holomorphic if and only if —¦f —¦c : D ’ C is a holomorphic function

for all ∈ F — and holomorphic curve c.

Clearly, any composition of holomorphic mappings is again holomorphic.

For ¬nite dimensions this coincides with the usual notion of holomorphic mappings,

by the ¬nite dimensional Hartogs™ theorem.

7.9. Hartogs™ Theorem. Let E1 , E2 , and F be convenient vector spaces with U

c∞ -open in E1 — E2 . Then a mapping f : U ’ F is holomorphic if and only if it

is separately holomorphic, i.e. f ( , y) and f (x, ) are holomorphic.

Proof. If f is holomorphic then f ( , y) is holomorphic on the c∞ -open set E1 —

{y} © U = incl’1 (U ), likewise for f (x, ).

y

If f is separately holomorphic, for any holomorphic curve (c1 , c2 ) : D ’ U ⊆ E1 —E2

we consider the holomorphic mapping c1 — c2 : D2 ’ E1 — E2 . Since the ck are

smooth by (7.4.8) also c1 — c2 is smooth and thus (c1 — c2 )’1 (U ) is open in C2 .

For each » ∈ F — the mapping » —¦ f —¦ (c1 — c2 ) : (c1 — c2 )’1 (U ) ’ C is separately

holomorphic and so holomorphic by the usual Hartogs™ theorem. By composing

with the diagonal mapping we see that » —¦ f —¦ (c1 , c2 ) is holomorphic, thus f is

holomorphic.

7.10. Lemma. Let f : E ⊇ U ’ F be holomorphic from a c∞ -open subset in

a convenient vector space to another convenient vector space. Then the derivative

(df )§ : U — E ’ F is again holomorphic and complex linear in the second variable.

Proof. (z, v, w) ’ f (v + zw) is holomorphic. We test with all holomorphic curves

‚

and linear functionals and see that (v, w) ’ ‚z |z=0 f (v + zw) =: df (v)w is again

holomorphic, C-homogeneous in w by (7.4).

7.10

84 Chapter II. Calculus of holomorphic and real analytic mappings 7.13

Now w ’ df (v)w is a holomorphic and C-homogeneous mapping E ’ F . But

any such mapping is automatically C-linear: Composed with a bounded linear

functional on F and restricted to any two dimensional subspace of E this is a ¬nite

dimensional assertion.

7.11. Remark. In the de¬nition of holomorphy (7.8) one could also have admitted

subsets U which are only open in the ¬nal topology with respect to holomorphic

curves. But then there is a counterexample to (7.10), see [Kriegl, Nel, 1985, 2.5].

7.12. Theorem. A multilinear mapping between convenient vector spaces is holo-

morphic if and only if it is bounded.

This result is false for not c∞ -complete vector spaces, see [Kriegl, Nel, 1985, 1.4].

Proof. Since both conditions can be tested in each factor separately by Hartogs™

theorem (7.9) and by (5.19), and by testing with linear functionals, we may restrict

our attention to linear mappings f : E ’ C only.

By theorem (7.4.2) a bounded linear mapping is holomorphic. Conversely, suppose

that f : E ’ C is a holomorphic but unbounded linear functional. So there exists

a sequence (an ) in E with |f (an )| > 1 and {2n an } bounded. Consider the power

∞

series n=0 (an ’ an’1 )(2z)n . This describes a holomorphic curve c in E, by (7.3)

and (7.4.2). Then f —¦ c is holomorphic and thus has a power series expansion

∞

f (c(z)) = n=0 bn z n . On the other hand

N

(f (an ) ’ f (an’1 ))(2z)n + (2z)N f (an ’ an’1 )(2z)n’N

f (c(z)) = .

n=0 n>N

So bn = 2n (f (an ) ’ f (an’1 )) and we get the contradiction

∞

(f (an ) ’ f (an’1 )) = lim f (an ).

0 = f (0) = f (c(1/2)) =

n’∞

n=0

Parts of the following results (7.13) to (10.2) can be found in [Bochnak, Siciak,

1971]. For x in any vector space E let xk denote the element (x, . . . , x) ∈ E k .

7.13. Lemma. Polarization formulas. Let f : E — · · · — E ’ F be an k-linear

symmetric mapping between vector spaces. Then we have:

1

k

(’1)k’Σµj f (x0 +

1

(1) f (x1 , . . . , xk ) = µj xj ) .

k!

µ1 ,...,µk =0

k

k

k

(’1)k’j

1

f ((a + jx)k ).

(2) f (x ) = k! j

j=0

k

kk j

k

k

(’1)k’j f ((a + k x)k ).

(3) f (x ) = k! j

j=0

1

»Σµj f (xµ1 , . . . , xµk ).

f (x0 »x1 , . . . , x0 »x1 )

(4) + + =

1 1 k k 1 k

µ1 ,...,µk =0

7.13

7.14 7. Calculus of holomorphic mappings 85

√

’1 in the passage to the complexi¬cation.

Formula (4) will mainly be used for » =

Proof. (1). (see [Mazur, Orlicz, 1935]). By multilinearity and symmetry the right

hand side expands to

Aj0 ,...,jk

f (x0 , . . . , x0 , . . . , xk , . . . , xk ),

j0 ! · · · jk !

j0 +···+jk =k j0 jk

where the coe¬cients are given by

1

(’1)k’Σµj µj1 · · · µjk .

Aj0 ,...,jk = 1 k

µ1 ,...,µk =0

The only nonzero coe¬cient is A0,1,...,1 = 1.

(2). In formula (1) we put x0 = a and all xj = x.

(3). In formula (2) we replace a by ka and pull k out of the k-linear expression

f ((ka + jx)k ).

(4) is obvious.

7.14. Lemma. Power series. Let E be a real or complex Fr´chet space and let

e

fk be a k-linear symmetric scalar valued bounded functional on E, for each k ∈ N.

Then the following statements are equivalent:

k

(1) k fk (x ) converges pointwise on an absorbing subset of E.

k

(2) k fk (x ) converges uniformly and absolutely on some neighborhood of 0.

(3) {fk (xk ) : k ∈ N, x ∈ U } is bounded for some neighborhood U of 0.

(4) {fk (x1 , . . . , xk ) : k ∈ N, xj ∈ U } is bounded for some neighborhood U of 0.

If any of these statements are satis¬ed over the reals, then also for the complexi¬-

cation of the functionals fk .

Proof. (1) ’ (3) The set AK,r := {x ∈ E : |fk (xk )| ¤ Krk for all k} is closed

in E since every bounded multilinear mapping is continuous. The countable union

K,r AK,r is E, since the series converges pointwise on an absorbing subset. Since

E is Baire there are K > 0 and r > 0 such that the interior U of AK,r is non

void. Let x0 ∈ U and let V be an absolutely convex neighborhood of 0 contained

in U ’ x0

From (7.13) (3) we get for all x ∈ V the following estimate:

k

k j

k

k k

|f ((x0 + k x)k )|

|f (x )| ¤ k! j

j=0

kk k k

¤ K(2re)k .

¤ k! 2 Kr

1

Now we replace V by V and get the result.

2re

7.14

86 Chapter II. Calculus of holomorphic and real analytic mappings 7.17

(3) ’ (4) From (7.13) (1) we get for all xj ∈ U the estimate:

1

k

1

|f (x1 , . . . , xk )| ¤ |f ( |

µj xj )

k!

µ1 ,...,µk =0

1 k

µj xj

k

1

= ( µj ) f

k! µj

µ1 ,...,µk =0

1

k

1

¤ ( µj ) C

k!

µ1 ,...,µk =0

k

k

1

j k C ¤ C(2e)k .

¤ k! j

j=0

1

Now we replace U by U and get (4).

2e

Proof of (4) ’ (2) The series converges on rU uniformly and absolutely for any

0 < r < 1.

(2) ’ (1) is clear.

√

(4), real case, ’ (4), complex case, by (7.13.4) for » = ’1.

7.15. Lemma. Let E be a complex convenient vector space and let fk be a k-linear

symmetric scalar valued bounded functional on E, for each k ∈ N. If k fk (xk )

converges pointwise on E and x ’ f (x) := k fk (xk ) is bounded on bounded sets,

then the power series converges uniformly on bounded sets.

Proof. Let B be an absolutely convex bounded set in E. For x ∈ 2B we apply the

vector valued Cauchy inequalities from (7.6) to the holomorphic curve z ’ f (zx)

at z = 0 for r = 1 and get that fk (xk ) is contained in the closed absolutely convex

hull of {f (zx) : |z| = 1}. So {fk (xk ) : x ∈ 2B, k ∈ N} is bounded and the series

converges uniformly on B.

7.16. Example. We consider the power series k k(xk )k on the Hilbert space

2

= {x = (xk ) : k |xk |2 < ∞}. This series converges pointwise everywhere, it

yields a holomorphic function f on 2 by (7.19.5) which however is unbounded on

the unit sphere, so convergence cannot be uniform on the unit sphere.

The function g : C(N) — 2 ’ C given by g(x, y) := k xk f (kx1 y) is holomor-

phic since it is a ¬nite sum locally along each holomorphic curve by (7.7), but its

Taylor series at 0 does not converge uniformly on any neighborhood of 0 in the

locally convex topology: A typical neighborhood is of the form {(x, y) : |xk | ¤

µk for all k, y 2 ¤ µ} and so it contains points (x, y) with |xk f (kx1 y)| ≥ 1, for all

large k. This shows that lemma (7.14) is not true for arbitrary convenient vector

spaces.

7.17. Corollary. Let E be a real or complex Fr´chet space and let fk be a k-

e

linear symmetric scalar valued bounded functional on E, for each k ∈ N such that

7.17

7.18 7. Calculus of holomorphic mappings 87

fk (xk ) converges to f (x) for x near 0 in E. Let ak z k be

the power series k≥1

a power series in E which converges to a(z) ∈ E for z near 0 in C.

Then the composite

fn (ak1 , . . . , akn ) z k

k≥0 n≥0 k1 ,...,kn ∈N

k1 +···+kn =k

of the power series converges to f —¦ a near 0.

Proof. By (7.14) there exists a 0-neighborhood U in E such that {fk (x1 , . . . , xk ) :

k ∈ N, xj ∈ U } is bounded. Since the series for a converges there is r > 0 such that

r

ak rk ∈ U for all k. For |z| < 2 we have

ak1 z k1 , . . . , akn z kn

f (a(z)) = fn

k1 ≥1 kn ≥1

n≥0

fn ak1 , . . . , akn z k1 +···+kn

···

=

n≥0 k1 ≥1 kn ≥1

fn (ak1 , . . . , akn ) z k ,

=

k≥0 n≥0 k1 ,...,kn ∈N

k1 +···+kn =k

since the last complex series converges absolutely: the coe¬cient of z k is a sum of

2k ’ 1 terms which are bounded when multiplied by rk . The second equality follows

from boundedness of all fk .

7.18. Almost continuous functions. In the proof of the next theorem we will

need the following notion: A (real valued) function on a topological space is called

almost continuous if removal of a meager set yields a continuous function on the

remainder.

Lemma. [Hahn, 1932, p. 221] A pointwise limit of a sequence of almost continuous

functions on a Baire space is almost continuous.

Proof. Let (fk ) be a sequence of almost continuous real valued functions on a Baire

space X which converges pointwise to f . Since the complement of a meager set in

a Baire space is again Baire we may assume that each function fk is continuous

on X. We denote by Xn the set of all x ∈ X such that there exists N ∈ N and a

1

neighborhood U of x with |fk (y) ’ f (y)| < n for all k ≥ N and all y ∈ U . The set

Xn is clearly open.

We claim that each Xn is dense: Let V be a nonempty open subset of X. For

1

N ∈ N the set VN := {x ∈ V : |fk (x) ’ f (x)| ¤ 2n for all k, ≥ N } is closed

in V and V = N VN since the sequence (fk ) converges pointwise. Since V is a

Baire space, some VN contains a nonempty open set W . For each y ∈ W we have

1

|fk (y) ’ f (y)| ¤ 2n for all k, ≥ N . We take the pointwise limit for ’ ∞ and

see that W ⊆ V © Xn .

Since X is Baire, the set n Xn has a meager complement and obviously the re-

striction of f on this set is continuous.

7.18

88 Chapter II. Calculus of holomorphic and real analytic mappings 7.19

7.19. Theorem. Let f : E ⊇ U ’ F be a mapping from a c∞ -open subset in

a convenient vector space to another convenient vector space. Then the following

assertions are equivalent:

(1) f is holomorphic.

(2) For all ∈ F — and absolutely convex closed bounded sets B the mapping

—¦ f : EB ’ C is holomorphic.

(3) f is holomorphic along all a¬ne (complex) lines and is c∞ -continuous.

(4) f is holomorphic along all a¬ne (complex) lines and is bounded on bornolog-

ically compact sets (i.e. those compact in some EB ).

(5) f is holomorphic along all a¬ne (complex) lines and at each point the ¬rst

derivative is a bounded linear mapping.

(6) f is c∞ -locally a convergent series of bounded homogeneous complex poly-

nomials.

(7) f is holomorphic along all a¬ne (complex) lines and in every connected

component for the c∞ -topology there is at least one point where all deriva-

tives are bounded multilinear mappings.

(8) f is smooth and the derivative is complex linear at every point.

(9) f is Lip1 in the sense of (12.1) and the derivative is complex linear at every

point.

Proof. (1) ” (2) By (7.6) every holomorphic curve factors locally over some EB

and we test with linear functionals on F .

So for the rest of the proof we may assume that F = C. We prove the rest of the

theorem ¬rst for the case where E is a Banach space.

(1) ’ (5) By lemma (7.10) the derivative of f is holomorphic and C-linear in the

second variable. By (7.12) f (z) is bounded.

(5) ’ (6) Choose a ¬xed point z ∈ U . Since f is holomorphic along each complex

line through z it is given there by a pointwise convergent power series. By the

classical Hartogs™ theorem f is holomorphic along each ¬nite dimensional linear

subspace. The mapping f : E ⊇ U ’ E is well de¬ned by assumption and is also

holomorphic along each a¬ne line since we may test this by all point evaluations:

using (5.18) we see that it is smooth and by (7.4.8) it is a holomorphic curve. So

the mapping

v ’ f (n+1) (z)(v, v1 , . . . , vn ) = (f ( )(v))(n) (z)(v1 , . . . , vn )

= (f )(n) (z)(v1 , . . . , vn )(v).

is bounded, and by symmetry of higher derivatives at z they are thus separately

bounded in all variables. By (5.19) f is given by a power series of bounded homo-

geneous polynomials which converges pointwise on the open set {z + v : z + »v ∈

U for all |»| ¤ 1}. Now (6) follows from lemma (7.14).

(6) ’ (3) By lemma (7.14) the series converges uniformly and hence f is continuous.

(3) ’ (4) is obvious.

7.19