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. 16
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Theorem 15 Given a sectorial region G, suppose that f (z) ∼ f (z) in G.

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
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 ˜
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

˜ ˜
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.

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

˜
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.

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,

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,

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
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.

¯
(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.

˜
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

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( 61 .)



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