For more details, see Ramis™s discussion of such abstract summation pro-

cesses in [229].

In particular, the requirement that products be summed to products is

rarely satis¬ed for general summation methods, but we shall see that k-

summability does have all the required properties, making it an ideal tool

for treating formal solutions of ODE “ except that it is not powerful enough

to sum all formal series arising as solutions of ODE!

A trivial example of a summation method satisfying all the requirements

listed above is as follows: Let X = E {z} and take S = S , mapping each

convergent power series to its natural sum. Clearly, this method is too

weak in that it applies to convergent series only. Constructing a summa-

bility method suitable for formal solutions of ODE may also be viewed

as ¬nding a way of extending S to larger di¬erential algebras, since an-

other natural requirement to make is that for convergent power series a

summation method produces the natural sum.

6.1 Gevrey Asymptotics and Laplace Transform 99

6.1 Gevrey Asymptotics and Laplace Transform

In what follows we again consider a given Banach space E . We have seen

in Section 4.7 that the mapping J : As (G, E ) ’ E [[z]]s , owing to Wat-

son™s Lemma, is injective for sectorial regions G of opening more than sπ,

ˆ

meaning that given a formal series f (z), there can be at most one function

f ∈ As (G, E ) with f (z) ∼s f (z) in G. The following result will be used

=ˆ

in the exercises below to show that in this case the mapping cannot be

surjective, except for the trivial case of dim E = 0.

ˆ ˆˆ

Theorem 33 Let s = 1/k > 0 and f ∈ E [[z]]s , hence g (u) = (Bk f )(u)

ˆ

converges for |u| su¬ciently small. Set g = S g ; thus g(u) is holomorphic in

ˆ

some neighborhood of the origin. Then, for every ¬xed real d the following

two statements are equivalent:

(a) There exists a sectorial region G = G(d, ±) with ± > sπ, and f ∈

ˆ

As (G, E ) with f = J(f ).

(b) There exists a sector S = S(d, µ) for su¬ciently small µ > 0 so that g

admits holomorphic continuation into S and is of exponential growth

not more than k there.

Moreover, if either statement holds, then f = Lk g follows.

Proof: Follows immediately from Theorems 22“24 in the previous chapter.

2

According to the theorem, existence of f as in (a) is linked to the holo-

morphic continuation of g plus its behavior when approaching in¬nity. In

ˆ

principle, (b) can be veri¬ed for a given formal series f , and if satis¬ed,

then the function f can be computed via Laplace transform.

Existence of power series that cannot be continued beyond their disc of

convergence is well known; one such example is contained in the exercises

below.

Exercises: Let s > 0 be given, and set k = 1/s.

∞ n

1. For g (z) = 0 z 2 , show that g has radius of convergence equal to

ˆ ˆ

one. Moreover, show that g = S g cannot be holomorphically contin-

ˆ

ued beyond the unit disc.

ˆ

2. For g as above, set f = Lk g . Use Theorem 33 and the previous

ˆ ˆ

ˆ

exercise to conclude that no f ∈ As (G, C ) with f = J(f ) can exist if

the opening of G exceeds sπ. Hence the map J : As (G, E ) ’ E [[z]]s ,

for E = C , is not surjective. Generalize this to arbitrary E with

dim E > 0.

100 6. Summable Power Series

6.2 Summability in a Direction

ˆ ˆ

Let k > 0, d ∈ R and f ∈ E [[z]] be given. We say that f is k-summable

in direction d, if a sectorial region G = G(d, ±) of opening ± > π/k and

ˆ ˆ

a function f ∈ A1/k (G, E ) exist with J(f ) = f . If this is so, then f ∈

E [[z]]1/k follows from results in Section 4.5, and Watson™s Lemma (p. 75)

ˆ

guarantees uniqueness of f ∈ A1/k (G, E ). In view of Theorem 33, f is k-

ˆˆ

summable in direction d if and only if g = Bk f converges and g = S (ˆ) is

ˆ g

˜

holomorphic and of exponential growth at most k in a sector S = S(d, µ), for

some µ > 0. If this is so, we call the function f = Lk g, integrating along rays

ˆ ˆ ˆˆ

close to d, the k-sum of f in direction d and write f = Sk,d f = Lk —¦S —¦ Bk f .

ˆ

Remark 8: If we are given a series f ∈ E [[z]]1/k whose k-summability in

direction d is to be investigated, and if we then want to compute its sum,

we are presented with the following problems:

ˆˆ

1. The function g(z), locally given by the convergent series Bk f , has to

be holomorphically continued into a sector S(d, µ), for some µ > 0,

which typically will be small.

2. In this sector, we have to show that g is of exponential growth not

larger than k.

3. We have to compute the integral Lk g.

At ¬rst glance, one might think that item 1 might be the major problem.

However, there are explicit methods for performing holomorphic continu-

ation, once we know that g is holomorphic in S(d, µ). Moreover, we shall

˜ ˜

show in Lemma 8 that we may even replace k by k < k, with k ’ k su¬-

ˆ˜ ˆ

ciently small; then the sum of Bk f will be an entire function, so that item 1

is no problem at all. This entire function, in general, will have too large an

exponential growth in all directions but arg u = d. So in a way, the main

3

di¬culty lies in the problem of verifying item 2.

As we shall show, this summation method has all the properties listed

at the beginning of the chapter, plus several additional ones, and we begin

with proving some which are direct consequences of the de¬nition:

Lemma 8 For every ¬xed k > 0 the following holds:

ˆ ˆ

(a) Let f be convergent. Then for every d, the series f is k-summable in

ˆ ˆ

direction d, and (Sk,d f )(z) = (S f )(z) for every z where both sides

are de¬ned.

ˆ

(b) Let f be k-summable in direction d, and let µ > 0 be su¬ciently

ˆ ˜ ˜

small. Then f is k-summable in all directions d with |d ’ d| < µ, and

ˆ ˆ

(Sk,d f )(z) = (Sk,d f )(z) for every z where both sides are de¬ned.

˜

6.2 Summability in a Direction 101

ˆ

(c) Let f be k-summable in direction d, and let µ > 0 be su¬ciently

ˆ ˆ

small. Then f is (k ’µ)-summable in direction d, and (Sk’µ,d f )(z) =

ˆ

(Sk,d f )(z) for every z where both sides are de¬ned.

Proof: For the ¬rst statement, use f (z) = (S f )(z) ∼1/k f (z) in S, for

ˆ ˆ

=

every k > 0 and every sector S of su¬ciently small radius. To prove the

second, use the de¬nition and observe that for a sectorial region G(d, ±)

˜

of opening ± > π/k and µ < ± ’ π/k, there exists G(d, ±µ ) ‚ G(d, ±)

˜

with d as above and ±µ > π/k. Finally, for (c) use that ± > π/k implies

± > π/(k ’ µ) for small µ > 0, and “(1 + N/k)/“(1 + N/(k ’ µ)) ’ 0 as

N ’ ∞. 2

The following lemma shall prove useful in later chapters where we will

study multisummablility.

ˆ

Lemma 9 Let f be k1 -summable in direction d, let k > k1 , and de¬ne k2

ˆˆ

by 1/k2 = 1/k1 ’ 1/k. Then g = Bk f is k2 -summable in direction d, and

ˆ

ˆ

Sk2 ,d g = Bk (Sk1 ,d f ).

ˆ

Proof: Follows immediately from Theorem 23 (p. 80) and the de¬nition

2

of k1 - resp. k2 -summability in direction d.

On one hand, for k < 1/2 the k-sum of formal series is analytic in sectorial

regions of opening more than 2π, which forces us to consider such regions

on the Riemann surface of the logarithm. On the other hand, statement (c)

of the next lemma shows that k-summability does not distinguish between

directions di¬ering by multiples of 2π.

Lemma 10

ˆ

(a) Let f be k-summable in direction d, for every d ∈ (±, β), ± < β, and

ˆ ˆ

¬xed k > 0. Then (Sk,d1 f )(z) = (Sk,d2 f )(z) for every d1 , d2 ∈ (±, β)

and every z where both sides are de¬ned.

ˆ

(b) Let f be kj -summable in direction d, 1 ¤ j ¤ 2, with k1 > k2 > 0

ˆ ˜

and some ¬xed d, then f is k1 -summable in all directions d with

˜ ˆ ˆ

2 |d ’ d| ¤ π (1/k2 ’ 1/k1 ), and (Sk1 ,d f )(z) = (Sk2 ,d f )(z) for every

˜

z where both sides are de¬ned.

˜ ˆ

(c) For d = d + 2π, k-summability of f in direction d is equivalent to

k-summability of f in direction d, and (Sk,d f )(z) = (Sk,d f )(ze’2πi )

ˆ ˜ ˆ ˆ

˜

for every z where both sides are de¬ned.

Proof:

ˆˆ

(a) The function g = S (Bk f ) is holomorphic and of exponential growth

not more than k in S((± + β)/2, β ’ ±). Hence f = Lk g ∼1/k f in a

ˆ

=

102 6. Summable Power Series

region which, for every d ∈ (±, β), contains a sector of opening larger

ˆ

than π/k and bisecting direction d. So Sk,d f = f , for every such d.

(b) From the de¬nition of k2 -summability in direction d we conclude ex-

istence of f with f (z) ∼1/k2 f (z) in G(d, ±) for some ± > π/k2 . Thus,

ˆ

=

g = Bk1 f is holomorphic and of exponential growth not more than k1

in S(d, ± ’ π/k1 ), and the assumption of k1 -summability in direction

d implies that g is holomorphic at the origin. Hence statement (b)

follows.

(c) Use Exercise 4 on p. 72.

2

Exercises: Always assume k > 0.

∞

ˆ “(1+n/k) z n is k-summable in every direction

1. Show that f (z) = 0

d = 2jπ, j ∈ Z.

ˆ ˆˆ

2. Assume that f is chosen so that g = S (Bk f ) is a rational function.

Let d be so that no poles of g lie on the ray arg u = d, and show that

ˆ

then f is k-summable in direction d.

∞

ˆ n

3. Show that f (z) = 0 “(1 + n/k) z /“(1 + n) is k-summable in

directions d with (2j + 1/2)π < d < (2j + 3/2)π, j ∈ Z.

ˆ

4. For f as in Exercise 3, let d be such that (2j’1/2)π < d < (2j+1/2)π,

ˆ

j ∈ Z. Show that f is (resp. is not) k-summable in direction d, if k ≥ 1

(resp. 0 < k < 1).

ˆ ˆˆ ˆ

5. Assume f so that g = S (Bk f ) is as in (4.2) (p. 63). Show that f is

k-summable in all directions d = 2jπ, j ∈ Z, and is not k-summable

in the remaining directions.

∞

ˆ “(1 + 2n) z n /“(1 + n)

6. Show that f (z) = 0

(a) is 1-summable in every direction d ∈ [’π, π) but one (which?).

(b) is 1/2-summable in every direction d with (2j + 1/2)π < d <

(2j + 3/2)π, j ∈ Z.

6.3 Algebra Properties

ˆ

Let k > 0 and d be given, and let E {z}k,d denote the set of all f that are

k-summable in direction d. The following theorems are direct consequences

from the results on Gevrey asymptotics in Section 4.5, together with the

de¬nition of k-summability in direction d.

6.3 Algebra Properties 103

ˆˆ ˆ

Theorem 34 For ¬xed, but arbitrary, k > 0 and d, let f , g1 , g2 ∈ E {z}k,d

be given. Then we have

g1 + g2 ∈ E {z}k,d , Sk,d (ˆ1 + g2 ) = Sk,d g1 + Sk,d g2 ,

ˆ ˆ g ˆ ˆ ˆ

d

ˆ ˆ ˆ

f ∈ E {z}k,d , Sk,d (f ) = (Sk,d f ),

dz

z z z

ˆ ˆ ˆ

f (w)dw ∈ E {z}k,d , Sk,d f (w)dw = (Sk,d f )(w) dw.

0 0 0

Finally, if p is a natural number, then

ˆ ˆ ˆ

f (z p ) ∈ E {z}pk,d/p , Spk,d/p (f (z p )) = (Sk,d f )(z p ).