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to hold. Then we apply the integral operator Tq , integrating in a di-
rection close to dq , then Tqв€’1 , integrating close to dqв€’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 deп¬Ѓned by
application of the integral operator Tjв€’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 inп¬Ѓnity
and has the correct growth, so that the next operator applies.

Л†
2. Show that to f в€€ 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-
Л†в€’1 Л†
Л†в€’1
Л†
ries fq = Tq в—¦ . . . в—¦ T1 f . The others are successively obtained as
fjв€’1 = Tj fj , 1 в‰¤ j в‰¤ q, integrating in a direction close to dj . From
Theorem 27 (p. 91), conclude

fj (z) в€јsj fj (z) = (Tjв€’1 в—¦ . . . в—¦ T1 f )(z) in Gj ,
Л†в€’1 Л†
=Л† Л† 0 в‰¤ j в‰¤ q в€’ 1,

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

Л†Л† Л† Л†
3. For f1 , f2 в€€ E {z}T ,d , show that f1 + f2 в€€ E {z}T ,d , and we have
Л† Л† Л† Л†
ST ,d (f1 + f2 ) = ST ,d f1 + ST ,d f2 .

Л†
4. Let ej (z) be the kernel of Tj , and let f в€€ E [[z]]. For a natural
Лњ
number p, consider the operators Tj with kernels ej (z) = p ej (z p )
Лњ
Л†
of corresponding orders Оєj = pОєj , and set g (z) = f (z p ). Show that
Лњ Л†
f в€€ E {z}T ,d if and only if g в€€ E {z}T ,d , with d = pв€’1 d. If this is so,
Л† Лњ
Л† ЛњЛњ
show (ST ,d f )(z) = (S Лњ Лњ g )(z 1/p ), wherever both sides are deп¬Ѓned.
Л† Л†
T ,d

Л† Л†
5. For f в€€ E {z}T ,d , deп¬Ѓne fj and fj as in Exercise 2. For each j,
Л†
1 в‰¤ j в‰¤ q в€’ 1, show: fj в€€ E {z}T (j),d(j) , with T (j) = (Tj+1 , . . . , Tq ),
Л†
d(j) = (dj+1 , . . . , dq ), and ST (j),d(j) fj = fj .

6. For k = (1/Оє1 + . . . + 1/Оєq )в€’1 , show E {z}k,dq вЉ‚ E {z}T ,d , and
Л† Л† Л†
ST ,d f = Sk,dq f for every f в€€ E {z}k,dq .

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

Lemma 19
Л†
(a) Let f в€€ E {z}T ,d and j, 1 в‰¤ j в‰¤ q, be given. Then there exists
Л† Л†
Оµ > 0 so that f в€€ E {z}T ,d and ST ,d f = ST ,d f hold for every
Лњ Лњ
Лњ Лњ Лњ Лњ
d = (d1 , . . . , djв€’1 , dj , dj+1 , . . . , dq ) satisfying |dj в€’ dj | < Оµ, 2Оєj |dj в€’
Лњ
djв€’1 | в‰¤ ПЂ, and 2Оєj+1 |dj+1 в€’ dj | в‰¤ ПЂ.
Л†
(b) Let f в€€ E [[z]] be given, and let d = (d1 , . . . , dq ) be admissible with
Лњ Лњ Лњ Лњ
respect to T . Then d = (d1 , . . . , dq ) with dj = dj + 2ПЂ, 1 в‰¤ j в‰¤ n,
Л†
is also admissible with respect to T , and f в€€ E {z}T ,d if and only if
f в€€ E {z}T ,d . If so, then (ST ,d f )(z) = (ST ,d f )(ze2ПЂi ), for every z
Л† Л† Л†
Лњ Лњ
where either side is deп¬Ѓned.
Let T = (T1 , . . . , Tq ) and d = (d1 , . . . , dq ), admissible with respect to T ,
Лњ
be given, and assume q в‰Ґ 2. With 1 в‰¤ ОЅ в‰¤ q, deп¬Ѓne T = (T1 , . . . , Tqв€’1 ) in
case ОЅ = q, resp. = (T1 , . . . , TОЅв€’1 , TОЅ в€— TОЅ+1 , TОЅ+2 , . . . , Tq ) otherwise, and
Лњ Лњ
d = (d1 , . . . , dОЅв€’1 , dОЅ+1 , . . . , dq ). It is easily seen that then d is admissible
Лњ
with respect to T .
ЛњЛњ
Lemma 20 For q в‰Ґ 2 and T , d, resp. T , d as above, we have E {z}T ,d вЉ‚ ЛњЛњ
Л† Л† Л†
E {z}T ,d , and for every f в€€ E {z}T ,d we have (ST ,d f )(z) = (ST ,d f )(z)
ЛњЛњ ЛњЛњ
wherever both sides are deп¬Ѓned, which holds at least for z with |z| suп¬ѓciently
small and |d1 в€’ arg z| в‰¤ ПЂ/(2Оє1 ).

Л†
Proof: Let f в€€ E {z}T ,d be given. In case ОЅ = q, we п¬Ѓnd that g =
ЛњЛњ
в€’1 Л†
в€’1
Л† Л†
S в—¦ Tq в—¦ . . . в—¦ T f is entire and of exponential growth at most Оєq in every
1
в€’1
sector of inп¬Ѓnite radius. Hence we conclude that g = Tq g = S в—¦ Tqв€’1 в—¦ . . . в—¦
Лњ
Л†в€’1 Л†
T1 f , where integration in the integral operator may be along any ray.
This completes the proof in this case. In case 1 в‰¤ ОЅ в‰¤ q в€’ 1, let g be as
above, and deп¬Ѓne g = TОЅ+2 в—¦ . . . в—¦ Tq g; in particular, let g = g for ОЅ = q в€’ 1.
Лњ Лњ
Then from Lemma 18 (p. 160) we п¬Ѓnd that TОЅ в€— TОЅ+1 g = TОЅ в—¦ TОЅ+1 g , and
Лњ Лњ
2
this is all we need to show.

ЛњЛњ
Exercises: For T , d, T , d as above, we deп¬Ѓne the projection ПЂОЅ by
ЛњЛњ
ПЂОЅ (T , d) = (T , d).
ЛњЛњ
1. For 1 в‰¤ ОЅ1 < . . . < ОЅp в‰¤ q (1 в‰¤ p в‰¤ q в€’ 1), let (T , d) = ПЂОЅ1 в—¦
. . . в—¦ ПЂОЅp (T , d). Show E {z}T ,d вЉ‚ E {z}T ,d . Moreover, show for every
ЛњЛњ
Л† Л† Л†
f в€€ E {z}T ,d that (ST ,d f )(z) = (ST ,d f )(z) wherever both sides are
ЛњЛњ ЛњЛњ
deп¬Ѓned.
j
2. Show E {z}kj ,dj вЉ‚ E {z}T ,d , for kj deп¬Ѓned by 1/kj = 1 1/Оє ,
Л† Л†
1 в‰¤ j в‰¤ q. Moreover, for every f в€€ E {z}kj ,dj show (Skj ,dj f )(z) =
Л†
(ST ,d f )(z) wherever both sides are deп¬Ѓned.
164 10. Multisummable Power Series

Л† Л†
3. Assume f в€€ E {z}T ,d в€© E [[z]]1/kqв€’1 . Show that then f в€€ E {z}T ,d , for
ЛњЛњ
ЛњЛњ
(T , d) = ПЂq (T , d).

10.4 The Main Decomposition Result
Let integral operators T = (T1 , . . . , Tq ) of respective orders Оє1 , . . . , Оєq , and
a multidirection d = (d1 , . . . , dq ), admissible with respect to T , be given,
and deп¬Ѓne k = (k1 , . . . , kq ) as in Exercise 2 on p. 163. It follows from
Л†
this and Exercise 3 on p. 162 that fj в€€ E {z}kj ,dj , 1 в‰¤ j в‰¤ q, implies
Л† Л†
fj в€€ E {z}T ,d . In this section we will show in case of Оєj > 1/2,
f=
Л†
1 в‰¤ j в‰¤ q, that every f в€€ E {z}T ,d is obtained in this fashion. To do so, we
use the following lemma:
Лњ Лњ
Lemma 21 Let real numbers k > k > 0 be given, so that k > 1/2, and
Лњ
deп¬Ѓne Оє by 1/Оє = 1/k в€’ 1/k. Let T be any integral operator of order Оє. For
Л† Лњ Л† Л†Л† Л† Л† Л†
h в€€ E {z}k,d , for some real number d, deп¬Ѓne f = T h. Then f = f1 + f2 ,
ЛњЛњ
Л† Л†
where f1 в€€ E [[z]]1/Оє and f2 в€€ E {z} Лњ. k,d

Л†
Proof: By deп¬Ѓnition, h = Sk,d h is holomorphic in a sectorial region G =
ЛњЛњ
G(d, О±), with О± > ПЂ/k, and h(z) в€ј1/k h(z) in G. Let S = S(d, О±, ПЃ) be a
Лњ Лњ = ЛњЛ† ВЇ ЛњЛњ Лњ
ВЇ
Лњ
closed subsector of G with 2ПЂ > О± > ПЂ/k, and let Оі denote the positively
Лњ
ВЇ
oriented boundary of S. Decomposing Оі = Оі1 + Оі2 , where Оі1 is the circular
part of Оі, let
1 h(w) ЛњЛњ Лњ
z в€€ S(d, О±, ПЃ), j = 1, 2.
hj (z) = dw,
wв€’z
2ПЂi Оіj

Then h = h1 +h2 by CauchyвЂ™s Formula, and h1 is holomorphic at the origin.
Therefore, h2 (z) = h(z)в€’h1 (z) в€ј1/k h2 (z) in S(d, О±, ПЃ), and h2 (z) = h(z)в€’
= ЛњЛ† ЛњЛњ Лњ Л† Л†
Л† Л†
h1 (z), where h1 (z) is the power series of h1 (z), and hence converges. So
Л† Л†Л† Л† ЛњЛњ
Л†
f1 = T h1 в€€ E [[z]]1/Оє . Moreover, h2 remains holomorphic in S = S(d, О±),
Л†
tending to zero as z в†’ в€ћ in S. So Theorem 27 (p. 91) implies f2 (z) =
(T h2 )(z) в€ј1/k f2 (z) = (T h2 )(z) in a sector of opening larger than ПЂ/k
Л† Л†Л†
=
Лњ Л†
and bisecting direction d. Hence by deп¬Ѓnition f2 в€€ E {z} Лњ. 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 deп¬Ѓne kj
j
by 1/kj = 1 1/Оє , 1 в‰¤ j в‰¤ q. If all Оєj > 1/2, for 1 в‰¤ j в‰¤ q, then for
q
Л† Л† Л† Л†
f в€€ E {z}T ,d we have f = j=1 fj , with fj в€€ E {z}kj ,dj , 1 в‰¤ j в‰¤ q, and
q
Л† Л†
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 в‰Ґ 2. From Exercise 5 on p. 162
Л†в€’1 Л†в€’1 Л†
Л† Лњ
we conclude that h(z) = Tqв€’1 в—¦ . . . в—¦ T1 f в€€ E {z}Оєq ,dq . Set k = kq , k =
Оєq , then the number Оє in Lemma 21 equals kqв€’1 . Applying the lemma
Лњ Л† Л† Л† Л†
with d = dq and T = T1 в€— . . . в€— Tqв€’1 , we п¬Ѓnd f = f1 + f2 , with f2 в€€
Л† Л†
E {z}k,dq вЉ‚ E {z}T ,d , and f1 в€€ E [[z]]1/kqв€’1 . Since f2 в€€ E {z}T ,d , we have
Л†
f1 в€€ E {z}T ,d в€© E [[z]]1/kqв€’1 , hence according to Exercise 3 on p. 164 we
Л† Лњ
Лњ
п¬Ѓnd f1 в€€ E {z}T ,d , with T = (T1 , . . . , Tqв€’1 ), d = (d1 , . . . , dqв€’1 ). Applying
ЛњЛњ
Л† 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 )
Лњ Лњ Лњ Лњ
and T = (T1 , . . . , Tq ) of integral operators with Tj and Tj having the same
order Оєj > 0, for 1 в‰¤ j в‰¤ q. Then E {z}T ,d = E {z}T ,d for every multidi-
Лњ
Л† = S Лњ f for every f в€€ E {z}T ,d =
Л† Л†
rection d satisfying (10.1), and ST ,d f T ,d
E {z}T ,d .
Лњ

Proof: Use Exercise 4 on p. 162 to see that we may assume Оєj > 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:

вЂў 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 deп¬Ѓne Оє1 , . . . , Оєq by Оє1 = k1 , 1/Оєj = 1/kj в€’ 1/kjв€’1 ,
2 в‰¤ j в‰¤ q, so that in turn kj are as in Theorem 50.
Л†
вЂў 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
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