<<

. 35
( 61 .)



>>

to hold. Then we apply the integral operator Tq , integrating in a di-
rection close to dq , then Tq’1 , integrating close to dq’1 , and so forth.
162 10. Multisummable Power Series

The assumptions made are such that these operators all apply. In
particular, the inequalities (10.1) assure that the function de¬ned by
application of the integral operator Tj’1 is analytic for values of the
variable close to the origin, having argument close to dj . By assump-
tion, this function then can be holomorphically continued to in¬nity
and has the correct growth, so that the next operator applies.

ˆ
2. Show that to f ∈ E {z}T ,d there corresponds a tuple (f0 (z), . . . , fq (z))
of functions. The function fq (z) is the sum of the convergent se-
ˆ’1 ˆ
ˆ’1
ˆ
ries fq = Tq —¦ . . . —¦ T1 f . The others are successively obtained as
fj’1 = Tj fj , 1 ¤ j ¤ q, integrating in a direction close to dj . From
Theorem 27 (p. 91), conclude

fj (z) ∼sj fj (z) = (Tj’1 —¦ . . . —¦ T1 f )(z) in Gj ,
ˆ’1 ˆ
=ˆ ˆ 0 ¤ j ¤ q ’ 1,

q
where sj = j+1 1/κν , and Gj is a sectorial region with bisecting
direction dj+1 and opening larger than π/κj+1 . The function f0 then
ˆ ˆ ˆ
is the T -sum of f , and in particular f0 (z) = f (z).

ˆˆ ˆ ˆ
3. For f1 , f2 ∈ E {z}T ,d , show that f1 + f2 ∈ E {z}T ,d , and we have
ˆ ˆ ˆ ˆ
ST ,d (f1 + f2 ) = ST ,d f1 + ST ,d f2 .

ˆ
4. Let ej (z) be the kernel of Tj , and let f ∈ E [[z]]. For a natural
˜
number p, consider the operators Tj with kernels ej (z) = p ej (z p )
˜
ˆ
of corresponding orders κj = pκj , and set g (z) = f (z p ). Show that
˜ ˆ
f ∈ E {z}T ,d if and only if g ∈ E {z}T ,d , with d = p’1 d. If this is so,
ˆ ˜
ˆ ˜˜
show (ST ,d f )(z) = (S ˜ ˜ g )(z 1/p ), wherever both sides are de¬ned.
ˆ ˆ
T ,d


ˆ ˆ
5. For f ∈ E {z}T ,d , de¬ne fj and fj as in Exercise 2. For each j,
ˆ
1 ¤ j ¤ q ’ 1, show: fj ∈ E {z}T (j),d(j) , with T (j) = (Tj+1 , . . . , Tq ),
ˆ
d(j) = (dj+1 , . . . , dq ), and ST (j),d(j) fj = fj .

6. For k = (1/κ1 + . . . + 1/κq )’1 , show E {z}k,dq ‚ E {z}T ,d , and
ˆ ˆ ˆ
ST ,d f = Sk,dq f for every f ∈ E {z}k,dq .




10.3 Elementary Properties
The following lemma lists properties of multisummability that are direct
consequences of the de¬nition, and the proof can be left to the reader:
10.3 Elementary Properties 163

Lemma 19
ˆ
(a) Let f ∈ E {z}T ,d and j, 1 ¤ j ¤ q, be given. Then there exists
ˆ ˆ
µ > 0 so that f ∈ E {z}T ,d and ST ,d f = ST ,d f hold for every
˜ ˜
˜ ˜ ˜ ˜
d = (d1 , . . . , dj’1 , dj , dj+1 , . . . , dq ) satisfying |dj ’ dj | < µ, 2κj |dj ’
˜
dj’1 | ¤ π, and 2κj+1 |dj+1 ’ dj | ¤ π.
ˆ
(b) Let f ∈ E [[z]] be given, and let d = (d1 , . . . , dq ) be admissible with
˜ ˜ ˜ ˜
respect to T . Then d = (d1 , . . . , dq ) with dj = dj + 2π, 1 ¤ j ¤ n,
ˆ
is also admissible with respect to T , and f ∈ E {z}T ,d if and only if
f ∈ E {z}T ,d . If so, then (ST ,d f )(z) = (ST ,d f )(ze2πi ), for every z
ˆ ˆ ˆ
˜ ˜
where either side is de¬ned.
Let T = (T1 , . . . , Tq ) and d = (d1 , . . . , dq ), admissible with respect to T ,
˜
be given, and assume q ≥ 2. With 1 ¤ ν ¤ q, de¬ne T = (T1 , . . . , Tq’1 ) in
case ν = q, resp. = (T1 , . . . , Tν’1 , Tν — Tν+1 , Tν+2 , . . . , Tq ) otherwise, and
˜ ˜
d = (d1 , . . . , dν’1 , dν+1 , . . . , dq ). It is easily seen that then d is admissible
˜
with respect to T .
˜˜
Lemma 20 For q ≥ 2 and T , d, resp. T , d as above, we have E {z}T ,d ‚ ˜˜
ˆ ˆ ˆ
E {z}T ,d , and for every f ∈ E {z}T ,d we have (ST ,d f )(z) = (ST ,d f )(z)
˜˜ ˜˜
wherever both sides are de¬ned, which holds at least for z with |z| su¬ciently
small and |d1 ’ arg z| ¤ π/(2κ1 ).

ˆ
Proof: Let f ∈ E {z}T ,d be given. In case ν = q, we ¬nd that g =
˜˜
’1 ˆ
’1
ˆ ˆ
S —¦ Tq —¦ . . . —¦ T f is entire and of exponential growth at most κq in every
1
’1
sector of in¬nite radius. Hence we conclude that g = Tq g = S —¦ Tq’1 —¦ . . . —¦
˜
ˆ’1 ˆ
T1 f , where integration in the integral operator may be along any ray.
This completes the proof in this case. In case 1 ¤ ν ¤ q ’ 1, let g be as
above, and de¬ne g = Tν+2 —¦ . . . —¦ Tq g; in particular, let g = g for ν = q ’ 1.
˜ ˜
Then from Lemma 18 (p. 160) we ¬nd that Tν — Tν+1 g = Tν —¦ Tν+1 g , and
˜ ˜
2
this is all we need to show.

˜˜
Exercises: For T , d, T , d as above, we de¬ne the projection πν by
˜˜
πν (T , d) = (T , d).
˜˜
1. For 1 ¤ ν1 < . . . < νp ¤ q (1 ¤ p ¤ q ’ 1), let (T , d) = πν1 —¦
. . . —¦ πνp (T , d). Show E {z}T ,d ‚ E {z}T ,d . Moreover, show for every
˜˜
ˆ ˆ ˆ
f ∈ E {z}T ,d that (ST ,d f )(z) = (ST ,d f )(z) wherever both sides are
˜˜ ˜˜
de¬ned.
j
2. Show E {z}kj ,dj ‚ E {z}T ,d , for kj de¬ned by 1/kj = 1 1/κ ,
ˆ ˆ
1 ¤ j ¤ q. Moreover, for every f ∈ E {z}kj ,dj show (Skj ,dj f )(z) =
ˆ
(ST ,d f )(z) wherever both sides are de¬ned.
164 10. Multisummable Power Series

ˆ ˆ
3. Assume f ∈ E {z}T ,d © E [[z]]1/kq’1 . Show that then f ∈ E {z}T ,d , for
˜˜
˜˜
(T , d) = πq (T , d).



10.4 The Main Decomposition Result
Let integral operators T = (T1 , . . . , Tq ) of respective orders κ1 , . . . , κq , and
a multidirection d = (d1 , . . . , dq ), admissible with respect to T , be given,
and de¬ne k = (k1 , . . . , kq ) as in Exercise 2 on p. 163. It follows from
ˆ
this and Exercise 3 on p. 162 that fj ∈ E {z}kj ,dj , 1 ¤ j ¤ q, implies
ˆ ˆ
fj ∈ E {z}T ,d . In this section we will show in case of κj > 1/2,
f=
ˆ
1 ¤ j ¤ q, that every f ∈ E {z}T ,d is obtained in this fashion. To do so, we
use the following lemma:
˜ ˜
Lemma 21 Let real numbers k > k > 0 be given, so that k > 1/2, and
˜
de¬ne κ by 1/κ = 1/k ’ 1/k. Let T be any integral operator of order κ. For
ˆ ˜ ˆ ˆˆ ˆ ˆ ˆ
h ∈ E {z}k,d , for some real number d, de¬ne f = T h. Then f = f1 + f2 ,
˜˜
ˆ ˆ
where f1 ∈ E [[z]]1/κ and f2 ∈ E {z} ˜. k,d


ˆ
Proof: By de¬nition, h = Sk,d h is holomorphic in a sectorial region G =
˜˜
G(d, ±), with ± > π/k, and h(z) ∼1/k h(z) in G. Let S = S(d, ±, ρ) be a
˜ ˜ = ˜ˆ ¯ ˜˜ ˜
¯
˜
closed subsector of G with 2π > ± > π/k, and let γ denote the positively
˜
¯
oriented boundary of S. Decomposing γ = γ1 + γ2 , where γ1 is the circular
part of γ, let
1 h(w) ˜˜ ˜
z ∈ S(d, ±, ρ), j = 1, 2.
hj (z) = dw,
w’z
2πi γj

Then h = h1 +h2 by Cauchy™s Formula, and h1 is holomorphic at the origin.
Therefore, h2 (z) = h(z)’h1 (z) ∼1/k h2 (z) in S(d, ±, ρ), and h2 (z) = h(z)’
= ˜ˆ ˜˜ ˜ ˆ ˆ
ˆ ˆ
h1 (z), where h1 (z) is the power series of h1 (z), and hence converges. So
ˆ ˆˆ ˆ ˜˜
ˆ
f1 = T h1 ∈ E [[z]]1/κ . Moreover, h2 remains holomorphic in S = S(d, ±),
ˆ
tending to zero as z ’ ∞ in S. So Theorem 27 (p. 91) implies f2 (z) =
(T h2 )(z) ∼1/k f2 (z) = (T h2 )(z) in a sector of opening larger than π/k
ˆ ˆˆ
=
˜ ˆ
and bisecting direction d. Hence by de¬nition f2 ∈ E {z} ˜. 2
k,d
We are now ready to prove the main decomposition result [17, 19]:
Theorem 50 (Main Decomposition Theorem) Let T = (T1 , . . . , Tq )
and d = (d1 , . . . , dq ), admissible with respect to T , be given, and de¬ne kj
j
by 1/kj = 1 1/κ , 1 ¤ j ¤ q. If all κj > 1/2, for 1 ¤ j ¤ q, then for
q
ˆ ˆ ˆ ˆ
f ∈ E {z}T ,d we have f = j=1 fj , with fj ∈ E {z}kj ,dj , 1 ¤ j ¤ q, and
q
ˆ ˆ
ST ,d f = j=1 Skj ,dj fj .
10.4 The Main Decomposition Result 165

Proof: We proceed by induction with respect to q: In case q = 1, the
statement holds trivially, hence assume q ≥ 2. From Exercise 5 on p. 162
ˆ’1 ˆ’1 ˆ
ˆ ˜
we conclude that h(z) = Tq’1 —¦ . . . —¦ T1 f ∈ E {z}κq ,dq . Set k = kq , k =
κq , then the number κ in Lemma 21 equals kq’1 . Applying the lemma
˜ ˆ ˆ ˆ ˆ
with d = dq and T = T1 — . . . — Tq’1 , we ¬nd f = f1 + f2 , with f2 ∈
ˆ ˆ
E {z}k,dq ‚ E {z}T ,d , and f1 ∈ E [[z]]1/kq’1 . Since f2 ∈ E {z}T ,d , we have
ˆ
f1 ∈ E {z}T ,d © E [[z]]1/kq’1 , hence according to Exercise 3 on p. 164 we
ˆ ˜
˜
¬nd f1 ∈ E {z}T ,d , with T = (T1 , . . . , Tq’1 ), d = (d1 , . . . , dq’1 ). Applying
˜˜
ˆ 2
the induction hypothesis to f1 , we see that the proof is completed.
The last theorem has the following important consequence:
Corollary to Theorem 50 Suppose that we have tuples T = (T1 , . . . , Tq )
˜ ˜ ˜ ˜
and T = (T1 , . . . , Tq ) of integral operators with Tj and Tj having the same
order κj > 0, for 1 ¤ j ¤ q. Then E {z}T ,d = E {z}T ,d for every multidi-
˜
ˆ = S ˜ f for every f ∈ E {z}T ,d =
ˆ ˆ
rection d satisfying (10.1), and ST ,d f T ,d
E {z}T ,d .
˜

Proof: Use Exercise 4 on p. 162 to see that we may assume κj > 1/2, in
which case the assertion follows directly from Theorem 50, Exercise 2 on
2
p. 163 and Exercise 3 on p. 162.
In view of the above corollary, we shall from now on use the following
terminology:

• A tuple k = (k1 , . . . , kq ) will be called a type of multisummability, or
simply, admissible, provided that k1 > k2 > . . . > kq > 0. Given such
a k, we always de¬ne κ1 , . . . , κq by κ1 = k1 , 1/κj = 1/kj ’ 1/kj’1 ,
2 ¤ j ¤ q, so that in turn kj are as in Theorem 50.
ˆ
• Given a type k of multisummability, we shall say that f (z) is k-
summable in an admissible multidirection d, if there exists a tuple

<<

. 35
( 61 .)



>>