=ˆ

Then z z

∼ f (z), ∼

ˆ ˆ

f (z) = f (w) dw = f (w) dw in G.

0 0

2

Proof: Follows easily from Proposition 8 (p. 66).

The next result implies surjectivity of the mapping J de¬ned in (4.5)

and was shown for E = C in [233], but the proof easily generalizes:

Theorem 16 (Ritt™s Theorem) Given any sectorial region G and any

f (z) ∈ E [[z]], there exists f (z) ∈ A(G, E ) such that f (z) ∼ f (z) in G.

ˆ =ˆ

Proof: Without loss of generality, let G be a sector S(d, ±) of in¬nite

fn z n be given. For β = π/±, and cn = ( fn n!)’1

ˆ

radius, and let f (z) =

4.4 Asymptotic Expansions 69

in case fn = 0, resp. cn = 0 otherwise, let wn (z) = 1 ’ exp[’cn /(ze’id )β ].

Using Exercise 1, we ¬nd fn |z|n |wn (z)| ¤ |z|n’β /n! for every n ≥ 0

and every z ∈ G. This proves absolute and locally uniform convergence of

∞

f (z) = n=0 fn z n wn (z) in G, so that f ∈ H(G, E ). Moreover,

N ’1

fn z n’N exp[’cn /(ze’id )β ],

rf (z, N ) = fN (z) ’

n=0

∞

where fN (z) = n=N fn z n’N wn (z) is bounded in G, and the other terms

¯

tend to zero as z ’ 0 in every closed subsector S of G, and hence are

2

bounded at the origin.

The following result characterizes the invertible elements in A(G, E ) in

case E is a Banach algebra with unit element:

Theorem 17 Let E be a Banach algebra with unit element e. Given a

sectorial region G, suppose that f (z) ∼ f (z) in G. Moreover, assume that

=ˆ

ˆ

the constant term f0 of f (z) and all values f (z), z ∈ G, are invertible

elements of E . Then

f ’1 (z) ∼ f ’1 (z)

=ˆ in G,

if f ’1 (z) denotes the unique formal power series g (z) with f (z) g (z) = e,

ˆ ˆˆ

ˆ ˆ

the unit element of E [[z]].

Proof: Compare Exercise 5 on p. 65 for existence of f ’1 (z) =

ˆ ˜

fn z n .

¯

Given a closed subsector S of G, there exist cN > 0 such that rf (z, N ) ¤

˜

¯

cN for N ≥ 0 and z ∈ S and, using (4.4), fN ¤ cN for every N ≥ 0. The

N ’1 ˜

identity rf ’1 (z, N ) f (z) = ’ m=0 fm rf (z, N ’ m), and the fact that for

¯

z ∈ S we have f (z) ≥ δ > 0, then imply

N ’1

’1

rf ’1 (z, N ) ¤ δ cm cN ’m ,

m=0

2

completing the proof.

Exercises:

1. Show |1 ’ e’z | < |z| for z in the right half-plane.

2. Suppose f (z) is holomorphic in a sectorial region G, and f (z) ∼ f (z)

=ˆ

ˆ ˜

in G for some f ∈ E [[z]]. If p is a natural number and G is such that

˜

˜ ˜

z ∈ G ⇐’ z p ∈ G, show that f (z) = f (z p ) is holomorphic in G, and

f (z) ∼ f (z p ) in G.

˜ =ˆ ˜

70 4. Asymptotic Power Series

3. For arbitrary b ∈ E , show e’z b ∼ ˆ in the right half-plane, where ˆ

=0 0

stands for the power series with all coe¬cients equal to zero.

4. Suppose f (z) is holomorphic in a sectorial region G, and f (z) ∼ 0

=ˆ

˜ ˜

in G. If k > 0 and G is such that z ∈ G ⇐’ z ∈ G, show that

k

f (z) = f (z k ) is holomorphic in G, and f (z) ∼ ˆ in G.

˜ ˜ =0

˜ ˜

5. Show that the mapping J de¬ned in (4.5) is never injective.

6. Let fn ∈ A(G, E ), n ≥ 0. Assume that for every m ≥ 0 the sequence

(m)

(fn )n converges uniformly on every closed subsector of G. Show

f = lim fn ∈ A(G, E ).

4.5 Gevrey Asymptotics

Given s > 0, we say that a function f , holomorphic in a sectorial region

ˆ ˆ

fn z n ∈ E [[z]] of Gevrey order s, or f

G, asymptotically equals f (z) =

¯

is the asymptotic expansion of order s of f , if to every closed subsector S

of G there exist c, K > 0 such that for every non-negative integer N and

¯

every z ∈ S

rf (z, N ) ¤ c K N “(1 + sN ). (4.7)

If this is so, we write for short f (z) ∼s f (z) in G.

=ˆ

This terminology di¬ers slightly from the one in [21] but agrees with

the classical one used in most papers on Gevrey expansions. While for a

general asymptotic, the bounds cN for the remainders rf (z, N ) may be

completely arbitrary, we note that for a Gevrey asymptotic of order s their

growth with respect to N is restricted. As we shall see, this has important

consequences!

Observe that f (z) ∼s f (z) in G implies f (z) ∼ f (z) in G in the previous

=ˆ =ˆ

sense; hence from Proposition 7, part (a) (p. 65) we conclude that f (z) ∼s=

ˆ(z) in G implies f (z) ∈ E [[z]]s . For s > s, Stirling™s formula implies

ˆ

f ˜

“(1 + sN )/“(1 + sN ) ’ 0 as N ’ ∞; hence f (z) ∼s f (z) in G implies

=ˆ

˜

f (z) ∼s f (z) in G. Also note that (4.7) remains meaningful for s = 0 and

=˜ ˆ

ˆ

is equivalent to f (z) being holomorphic at the origin and f (z) being its

power series expansion.

Proposition 9 Let f be holomorphic in a sectorial region G, and let s ≥ 0.

Then the following statements are equivalent:

(a) f (z) ∼s f (z) in G.

=ˆ

4.5 Gevrey Asymptotics 71

(b) All derivatives f (n) (z) are continuous at the origin, and for every

¯

closed subsector S of G there exist constants c, K such that

1

sup f (n) (z) ¤ c K n “(1 + sn)

n! z∈S

¯

for every n ≥ 0.

2

Proof: Proceed analogously to the proof of Proposition 8 (p. 66).

Let As (G, E ) denote the set of all f ∈ H(G, E ), having an asymptotic

expansion of order s. The following results are the direct analogues to the

ones in the previous section, proving that As (G, E ) is again a di¬erential

algebra provided that E is a Banach algebra, and J : As (G, E ) ’ E [[z]]s

is a homomorphism.

Theorem 18 Given a sectorial region G and s ≥ 0, suppose that f1 , f2 ∈

As (G, E ). Then f1 + f2 ∈ As (G, E ) and J(f1 + f2 ) = J f1 + J f2 . In other

words,

fj (z) ∼s fj (z) in G, 1 ¤ j ¤ 2,

=ˆ

implies

f1 (z) + f2 (z) ∼s f1 (z) + f2 (z) in G.

=ˆ ˆ

2

Proof: Follows directly from the de¬nition.

Theorem 19 Suppose that E , F are both Banach spaces, G is a secto-

rial region, and s ≥ 0. Let f ∈ As (G, E ), ± ∈ As (G, C ), and T ∈

As (G, L(E , F)). Then

T f ∈ As (G, F), J(T f ) = (J T )(J f ),

± f ∈ As (G, E ), J(± f ) = (J ±)(J f ).

Proof: Set cN = c K N “(1 + sN ) in the proof of Theorem 14, and further

estimate (4.6) by c2 K N “(1 + sN ) (N + 1). Observing 1 + N ¤ 2N then

2

completes the proof.

Corollary to Theorem 19 In case E is a Banach algebra, let f1 , f2 ∈

As (G, E ). Then f1 f2 ∈ As (G, E ) and J(f1 f2 ) = (J f1 )(J f2 ). In other

words,

fj (z) ∼s fj (z) in G, 1 ¤ j ¤ 2,

=ˆ

implies

f1 (z) f2 (z) ∼s f1 (z) f2 (z) in G.

=ˆ ˆ

72 4. Asymptotic Power Series

Theorem 20 Given s > 0 and a sectorial region G, suppose that f (z) ∼s

=

ˆ

f (z) in G. Then

z z

f (z) ∼s f (z), f (w) dw ∼s

=ˆ ˆ

f (w) dw in G.

=

0 0

2

Proof: Follows directly from Proposition 9.

Theorem 21 Let E be a Banach algebra with unit element e. Given s > 0

and a sectorial region G, suppose that f (z) ∼s f (z) in G. Moreover, assume

=ˆ

ˆ

that the constant term f0 of f (z) and all values f (z), z ∈ G, are invertible

elements of E . Then

f ’1 (z) ∼s f ’1 (z)

=ˆ in G,

if f ’1 (z) denotes the unique formal power series g (z) with f (z) g (z) = e,

ˆ ˆˆ

ˆ ˆ

the unit element of E [[z]].

∞˜

Proof: From Exercise 7 on p. 65 we conclude that f ’1 (z) = 0 fn z n ∈

ˆ

E [[z]]s . Using the same arguments as in the proof of Theorem 17, one can

2

obtain the necessary estimates.

ˆ

Exercises: Let s > 0, let f ∈ E [[z]], and let f (z) be holomorphic in a

sectorial region G with values in E .

1. (a) Given integers p ≥ 1 and q ≥ 0, assume that to every closed

¯ ¯

subsector S of G there exist c, K > 0 such that for every z ∈ S

and every N of the form N = pM + q, M ∈ N, we have (4.7).

Show that then f (z) ∼s f (z) in G.

=ˆ

¯

(b) Given µ > 0, assume that to every closed subsector S of G with

¯

radius of S smaller than µ, there exist c, K > 0 such that for

¯

every z ∈ S and every non-negative integer N we have (4.7).

Show that then f (z) ∼s f (z) in G.

=ˆ

˜

2. Let p be a natural number. For G = {z| z p ∈ G}, de¬ne g(z) = f (z p ),

z ∈ G, and g (z) = f (z p ). Show that then f (z) ∼s f (z) in G if and