direction d with 2 |d ’ d ’ π| < ± + π(2 ’ s). Since ± may be arbitrarily

ˆ

small, it may happen for 0 < k < 1/2 that there is no such d. This is

another reason why values of k below, or even equal to, 1/2 are special in

our theory. Moreover, it is worthwhile to observe for later applications that

for ψ ∈ As,0 (G, E ) and a, b ∈ G, the function CHa ψ’CHb ψ is holomorphic

3

at the origin.

ˆ

The formal series f in Proposition 17 will sometimes be denoted as CHa ψ

and named formal Cauchy-Heine transform of ψ.

Exercises: In the following exercises we shall compute the Cauchy-Heine

transform of functions ψ that are analytic near the origin; this will not be

used later on, but is of interest in its own right.

1. For ¬xed a = 0 and n ≥ 0, show

n

a

wn dw am z n’m

= z n log(1 ’ a/z) + ,

w’z m

0 m=1

for |z| > |a|, and then conclude by holomorphy that the formula stays

valid for z in the complex plane with a “cut” from 0 to a.

2. For ψ continuous along the line segment from 0 to a and holomorphic

in D(0, µ), for some µ > 0, show for |z| < µ, arg z = arg a:

a

ψ(w) dw

= ψ(z) log(1 ’ a/z) + φ(z),

w’z

0

with a function φ(z) which is analytic near the origin.

118 7. Cauchy-Heine Transform

7.2 Normal Coverings

We frequently consider sectors with openings larger than 2π and conse-

quently have to work on the Riemann surface of the logarithm. Nonethe-

less, we have seen in Chapter 6 that for k-summability we should identify

directions d that di¬er by integer multiples of 2π, hence should think of

rays arg z = d in the complex plane. Quite similarly, the notion of a normal

covering to be de¬ned is best understood if pictured in the complex plane:

For a natural number m, let d0 < d1 < . . . < dm’1 < d0 + 2π = dm be

given directions. Moreover, let ±j > 0, j = 0, . . . , m ’ 1, be so that with

±m = ±0 we have dj ’ ±j /2 < dj’1 + ±j’1 /2, 1 ¤ j ¤ m. If this is so, and

ρ > 0 is arbitrarily given, then we say that the sectors Sj = S(dj , ±j , ρ),

0 ¤ j ¤ m, form a normal covering. One should observe that on the

Riemann surface Sm is directly above S0 . One can also de¬ne dj and ±j ,

for arbitrary integers j, so that dj+m = dj + 2π, ±j+m = ±j for every j;

then the sectors Sj = S(dj , ±j , ρ) cover the whole Riemann surface, and

m + 1 consecutive ones always are a normal covering.

Using this notion of normal covering, we now prove the key result for

what is to follow:

Theorem 39 Let Sj = S(dj , ±j , ρ), 0 ¤ j ¤ m, be a normal covering. For

k > 0, s = 1/k, and 1 ¤ j ¤ m, let ψj ∈ As,0 (Sj’1 © Sj , E ) and choose

˜

aj ∈ Sj’1 © Sj . For 0 < ρ ¤ |aj | (1 ¤ j ¤ m), let Sj = S(dj , ±j , ρ),

˜ ˜

0 ¤ j ¤ m, and de¬ne

j m

(CHaµ ψµ )(ze2πi )

fj (z) = (CHaµ ψµ )(z) +

µ=1 µ=j+1

˜

for z ∈ Sj , 0 ¤ j ¤ m, interpreting empty sums as zero. Then we have

m

˜

fj ∈ As (Sj , E ), with J(fj ) = µ=1 CHaµ ψµ . Hence in particular, J(fj )

is independent of j. Moreover, fj (z) ’ fj’1 (z) = ψj (z), 1 ¤ j ¤ m,

fm (ze2πi ) = f0 (z).

Proof: For each µ, CHaµ ψµ is holomorphic for dµ ’ ±µ /2 < arg z <

2π + dµ’1 + ±µ’1 /2, |z| < ρ. So one easily checks for 1 ¤ µ ¤ j, resp.

˜

j +1 ¤ µ ¤ m, that (CHaµ ψµ )(z), resp. (CHaµ ψµ )(ze2πi ), are holomorphic

˜

in the sectors Sj , 0 ¤ j ¤ m. The proof then is easily completed, using

2

Proposition 17.

Exercises:

1. Check that the above theorem for m = 1 coincides with Proposi-

tion 17.

7.3 Decomposition Theorems 119

2. Given a normal covering, try to de¬ne what one might call a ¬ner

normal covering, and check that for any two normal coverings one

can always consider a common re¬nement.

7.3 Decomposition Theorems

ˆ

For ψ ∈ As,0 (G, E ), the coe¬cients of f = CHa ψ, for arbitrary a ∈ S,

ˆ

have the form of a moment sequence, so that f appears, at ¬rst glance,

to be rather special. The following theorem shows, however, that arbitrary

ˆ

f ∈ C [[z]]s admit a decomposition into a convergent series plus ¬nitely

many formal Cauchy-Heine transforms:

ˆ ˆ

Theorem 40 For s > 0 and f ∈ E [[z]], we have f ∈ E [[z]]s if and only if

for every normal covering Sj = S(dj , ±j , ρ) with 0 < ±j ¤ sπ, 0 ¤ j ¤ m,

there exist ψj ∈ As,0 (Sj’1 © Sj , E ), 1 ¤ j ¤ m, so that for arbitrarily

chosen aj ∈ Sj’1 © Sj we have

m

ˆˆ ˆ

CHaj ψj , with f0 ∈ E {z}.

f = f0 + (7.3)

j=1

ˆ

Proof: Let f ∈ C [[z]]s , and let Sj be as stated, 0 ¤ j ¤ m. Then

˜ ˜ ˆ

Proposition 10 (p. 73) implies existence of fj ∈ As (Sj , E ) with J(fj ) = f ,

for j = 0, . . . , m ’ 1. De¬ning fm (z) = f0 (ze’2πi ), z ∈ Sm , the same

˜ ˜

˜ ˜

holds for j = m. Let ψj (z) = fj (z) ’ fj’1 (z), z ∈ Sj’1 © Sj ; then

ψj ∈ As,0 (Sj’1 © Sj , E ), 1 ¤ j ¤ m. With fj as in Theorem 39, we then

˜

see that f (z) = fj (z) ’ fj (z) is independent of j, and holomorphic and

single-valued in a punctured neighborhood of the origin. Moreover, f has

an asymptotic expansion there; hence Proposition 7 (p. 65) implies holo-

˜

morphy of f at the origin. Applying J to fj (z) = f (z) + fj (z), we conclude

ˆ

(7.3) with convergent f0 = J(f ). The converse statement follows directly

2

from Proposition 17.

The above theorem and its proof remain correct if we allow normal

coverings of possibly larger openings provided that fj ∈ As (Sj , E ) with

ˆ

J(fj ) = f , 0 ¤ j ¤ m ’ 1, still exist. This will be used in the proof of the

following characterisation of k-summable power series:

ˆ

Corollary to Theorem 40 Let s > 0, d ∈ R and f ∈ E [[z]] be given.

Then the following holds:

ˆ

(a) For k = 1/s ≥ 1/2, k-summability of f in direction d is equivalent to

ˆ

the existence of a decomposition of f as in Theorem 40, with

|d + π ’ arg aj | ¤ π(1 ’ s/2), 1 ¤ j ¤ m.

120 7. Cauchy-Heine Transform

ˆ

(b) For k = 1/s ¤ 1/2, k-summability of f in direction d is equivalent to

ˆ

the existence of a decomposition of f as in Theorem 40, with m = 1

and

S0 © S1 = S(d + π, ±, ρ), ± > (s ’ 2)π.

ˆ

Proof: Let f ∈ E {z}k,d and k ≥ 1/2, and take a normal covering Sj ,

j = 0, . . . , m as follows: The sector S0 has bisecting direction d and opening

ˆ

larger than sπ and is so that f0 = Sk,d f is holomorphic in S0 . The openings

of the other sectors are at most sπ. Since we are free to choose any aj ∈

Sj’1 © Sj , it is easy to see that we can make the inequality in statement

(a) hold. Thus, we obtain a decomposition as in Theorem 40. In case 0 <

ˆ ˆ

k ¤ 1/2, we have that f ∈ C {z}k,d implies f = Sk,d f ∈ As (S0 , E ),

S0 = S(d, ±, ρ), ± > sπ. With S1 = S(d + 2π, ±, ρ) we see that S0 and S1

˜ ˜ ˜

are already a normal covering, and S0 © S1 is a sector of opening larger

than π(s ’ 2) and bisecting direction d + π. So one direction of the proof

in both cases is completed. The opposite directions, however, follow from

2

Remark 9 (p. 117).

We now show that k-summable series can always be decomposed as a

sum of such series with just one singular direction:

ˆ ˆ

Theorem 41 Let k > 0 and f ∈ C {z}k be given, and let f have m ≥ 2

singular directions in any half-open interval of length 2π. Then

ˆˆ ˆ

f = f1 + . . . + fm ,

ˆ

where each fj ∈ C {z}k has exactly one singular direction.

ˆ

Proof: Let the singular directions of f be 0 < d1 < . . . < dm ¤ 2π, and

ˆ

de¬ne d0 = dm ’ 2π. For 1 ¤ j ¤ m, de¬ne fj = Sk,d f , for dj’1 < d < dj .

Then fj is holomorphic in a sectorial region Gj with bisecting direction

(dj’1 + dj )/2 and opening dj ’ dj’1 + π/k. Moreover, fj (z) ∼1/k f (z) in

ˆ

=

Gj . With G0 = Gm e’2πi and f0 (z) = fm (ze2πi ) in G0 , de¬ne ψj (z) =

fj (z) ’ fj’1 (z) in Gj’1 © Gj , 1 ¤ j ¤ m. For arbitrary aj ∈ Gj’1 © Gj ,

Remark 9 (p. 117) implies that CHaj ψj is k-summable in all directions

but dj’1 modulo 2π. As in the proof of Theorem 40 one can show that

m

ˆ ˆ

f ’ j=1 CHaj ψj = f0 converges; hence is k-summable in every direction.

ˆ ˆ ˆ

De¬ning fj = CHaj ψj , 1 ¤ j ¤ m ’ 1, and fm = f0 + CHam ψm , the proof

2

is completed.

Exercises:

1. Show that if f ∈ E {z}k has exactly one singular ray d0 , then f (z 2 )

ˆ ˆ

is in E {z}2k and has exactly two singular rays d0 /2 and π + d0 /2.

7.4 Functions with a Gevrey Asymptotic 121

∞

2. Let f (z) = 0 “(1 + n/k) z 2n , k > 0. Show f ∈ E {z}2k , determine

ˆ ˆ

ˆ ˆ

the singluar rays of f , and explicitly write f as a sum of formal series

which each have one singular ray.

ˆ ˆˆ

3. Choose f ∈ C [[z]]s so that g = S (B1/s f ) is a rational function. Show

ˆ

that a decomposition of f into series, having exactly one singular ray

each, can be achieved through a partial fraction decomposition of g.

7.4 Functions with a Gevrey Asymptotic

The following result can be thought of as a generalization of the well-

known characterization of removable singularities. As we shall see in the

exercises below, it is very useful in showing that some functions have a

Gevrey asymptotic.

Proposition 18 Let s > 0, any sector S and any function f , holomorphic

in S, be given. Then f ∈ As (S, E ) is equivalent to the existence of a normal

covering S0 , . . . , Sm , with S0 = S, and functions fj , holomorphic in Sj ,

0 ¤ j ¤ m, with f0 = f and fm (z) = f0 (ze’2πi ), z ∈ Sm , so that all fj are

bounded at the origin, and

fk’1 (z) ’ fk (z) ∈ As,0 (Sk’1 © Sk , E ), 1 ¤ k ¤ m.

ˆ

Proof: If f ∈ As (S, E ), let f = J(f ), and choose any normal covering

S0 , . . . , Sm so that S0 = S and the opening of S1 , . . . , Sm’1 is less than or

equal to sπ. Then Proposition 10 (p. 73) shows existence of fj ∈ As (Sj , E )

with J(fj ) = f , 1 ¤ j ¤ m ’ 1, and with fm (z) = f0 (ze’2πi ) we obtain

ˆ

fk’1 (z) ’ fk (z) ∈ As,0 (Sk’1 © Sk , E ), 1 ¤ k ¤ m. Conversely, if Sj and fj

are as stated, de¬ne ψj = fj ’ fj’1 , 1 ¤ j ¤ m, and (with aµ ∈ Sµ’1 © Sµ

˜

and Sj as in Theorem 39)

j m

(CHaµ ψµ )(ze2πi )

gj (z) = (CHaµ ψµ )(z) +

µ=1 µ=j+1

˜ ˜

for z ∈ Sj and 0 ¤ j ¤ m. Then Theorem 39 shows gj ∈ As (Sj , E )

and J(gj ) = g , independent of j. Moreover, h(z) = fj (z) ’ gj (z) is also

ˆ

independent of j, and single-valued and bounded at the origin. Therefore,

the origin is a removable singularity of h, and consequently fj = h + gj ∈

˜

As (Sj , E ), and even in As (Sj , E ), in view of Exercise 1 on p. 72, because

˜ 2

Sj and Sj only di¬er in their radius.

ˆ

Let f ∈ E {z}k,d be given. Then, there exists a sector S of opening larger

than π/k in which Sk,d f (z) ∼1/k f (z) holds. Using the above proposition,

ˆ= ˆ