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. 37
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let p ∈ N with p kq > 1/2, and let f ∈ E {z}k,d . Then we can ¬nd µ, ρ > 0
such that to every φ ∈ R there exists a function f (z; φ), holomorphic in
Sφ = S(φ, µ, ρ), such that the following holds:
170 10. Multisummable Power Series

(a) f (z; φ) ∼1/kq f (z) in Sφ , for every φ ∈ R.
ˆ
=
(b) For every φ1 , φ2 with |φ1 ’ φ2 | < µ, i.e., Sφ1 © Sφ2 = …, and every
j = 1, . . . , q we have the following: If φ1 , φ2 ∈ Ij , then f (z; φ1 ) ’
f (z; φ2 ) ∼1/kj’1 ˆ in Sφ1 © Sφ2 .
0
=

(c) f (z; φ) = f (ze2pπi ; φ + 2pπ) in Sφ , for every φ ∈ R.

Proof: Using Theorem 51, we see that we may, without loss of generality,
restrict ourselves to cases with kq > 1/2, so that we may take p = 1.
q
ˆ ˆ ˆ
In this situation, Theorem 50 implies f = j=1 fj , with fj ∈ E {z}kj ,dj .
q
ˆ ˆ
Hence f (z) = (Sk,d f )(z) = j=1 fj (z), with fj = Skj ,dj fj holomorphic
ˆ ˜
and asymptotic to fj of Gevrey order kj in Sj = S(dj , ±j , ρ), ρ > 0,
˜
±j > π/kj . Taking µ > 0 so small that φ ∈ Ij implies Sφ ‚ Sj , we de¬ne
q
f (z; φ) = j=1 fj (z; φ), with fj (z; φ) = fj (z) if φ ∈ Ij , while for φ ∈ Ij
we take any fj (z; φ) ∼1/kj fj (z) in Sφ . Since φ and φ + 2π cannot both
ˆ
=
be in Ij , we can in particular arrange fj (z; φ) = fj (ze2πi ; φ + 2π). Using
k1 > . . . > kq and I1 ‚ I2 ‚ . . . ‚ Iq , one can check that (a)“(c) hold. 2
The next result is, in a sense, the converse of the above one:
Proposition 25 Let k = (k1 , . . . , kq ) and d = (d1 , . . . , dq ) be admissible,
with kq > 1/2, and let I0 = [dq ’ π, dq + π]. For µ, ρ > 0 and φ ∈ I0 , assume
existence of f (z; φ), holomorphic in Sφ = S(φ, µ, ρ), such that the following
holds:
(a) f (z; φ), in the variable z, is bounded at the origin, for every φ ∈ I0 .
(b) For every φ1 , φ2 ∈ I0 with |φ1 ’ φ2 | < µ we have the following: If
φ1 , φ2 ∈ Ij for some j, 1 ¤ j ¤ q, then f (z; φ1 ) ’ f (z; φ2 ) ∼1/kj’1 ˆ 0
=
∼1/k ˆ
in Sφ1 © Sφ2 . If φ1 or φ2 is not in Iq , then f (z; φ1 ) ’ f (z; φ2 ) = q 0
in Sφ1 © Sφ2 .
(c) f (z; dq ’ π) = f (ze2πi ; dq + π) in Sφ0 .
ˆ ˆ
Then there exists a unique f ∈ E {z}k,d with f (z; d1 ) = (Sk,d f )(z) in Sd1 .

Proof: Take any partitioning φ0 = dq ’ π < φ1 < . . . < φm’1 < φm =
φ0 + 2π of the interval I0 with φj ’ φj’1 < µ, 1 ¤ j ¤ m, and such
that all the boundary points of all intervals Iν occur in the set of φj . For
aj = ρ exp[i(φj + φj’1 )/2], with 0 < ρ < ρ, de¬ne
˜ ˜

gj = CHaj (f (·; φj ) ’ f (·; φj’1 )), 1 ¤ j ¤ m.

Then gj (z) ≡ 0 whenever φj , φj’1 ∈ I1 . Otherwise, let ν be taken maxi-
mally so that φj or φj’1 is not in Iν , i.e., φj ¤ dν ’ π/(2kν ) or φj’1 ≥
dν + π/(2kν ). Then Proposition 17 (p. 116) implies gj (z) ∼1/kν gj (z) in
ˆ
=
10.7 Applications of Cauchy-Heine Transforms 171

˜˜˜ ˜
S(dj , ±j , ρ), with dj = π + (φj + φj’1 )/2, ±j = 2π + µ ’ (φj ’ φj’1 ),
˜
˜˜˜
gj = CHaj (f (·; φj ) ’ f (·; φj’1 )). For φj ¤ dν ’ π/(2kν ), S(dj , ±j , ρ) con-
ˆ
tains a sector with bisecting direction dν and opening more than π/kν , so
gj ∈ E {z}kν ,dν , and (Skν ,dν gj )(z) = gj (z). In case φj’1 ≥ dν + π/(2kν ), a
ˆ ˆ
sector of the same opening, but with bisecting direction dν ’2π, is contained
˜˜˜
in S(dj , ±j , ρ); hence again gj ∈ E {z}kν ,dν , but this time (Skν ,dν gj )(z) =
ˆ ˆ
2πi m
gj (ze ). Consequently, g = j=1 gj ∈ E {z}k,d . De¬ne
ˆ ˆ

j m
gµ (ze2πi ), 0 ¤ j ¤ m, z ∈ Sφj © Sφj’1 .
fj (z) = gµ (z) +
µ=1 µ=j+1


Then (Sk,d g )(z) = fj (z), for every j with φj ∈ I1 . Moreover, h(z) =
ˆ
f (z; φj ) ’ fj (z) can be seen to be independent of j, holomorphic and
ˆ
single-valued for 0 < |z| < ρ, and bounded at the origin. So h = J(h)
˜
ˆˆˆ ˆ
is convergent; hence f = h + g ∈ E {z}k;d , and Sk,d f = f (·; φj ), for every j
with φj ∈ I1 . 2
With help of the above two propositions, it is now easy to deal with
products of multisummable series:
ˆ ˆ
Theorem 52 Let E , F both be Banach spaces, and let f ∈ E {z}k,d , T ∈
L(E , F){z}k,d , ± ∈ C {z}k,d . Then
ˆ

ˆˆ ˆˆ ˆ
ˆ
T f ∈ F{z}k,d , Sk,d (T f ) = (Sk,d T ) (Sk,d f ),
ˆˆ ±ˆ ˆ
± f ∈ F{z}k,d , Sk,d (ˆ f ) = (Sk,d ±) (Sk,d f ).
ˆ

Proof: Without loss in generality, let kq > 1/2. Let T (z; φ), resp. ±(z; φ),
ˆ
ˆ
resp. f (z; φ) be the functions corresponding to T (z), resp. ±(z), resp. f (z)
ˆ
by means of Proposition 24. The products T (z; φ)f (z; φ) and ±(z; φ)f (z; φ)
then can be checked to have the same properties, and therefore Proposi-
2
tion 25 can be used to complete the proof.
As for the case of q = 1, this shows that E {z}k,d is an algebra, in the case
of E being a Banach algebra, and we can also characterize the invertible
elements:
ˆˆ ˆ
Theorem 53 Let f , g1 , g2 ∈ E {z}k,d be given. If E is a Banach algebra,
then
g1 g2 ∈ E {z}k,d , Sk,d (ˆ1 g2 ) = (Sk,d g1 ) (Sk,d g2 ).
ˆˆ gˆ ˆ ˆ
ˆ
Moreover, if E has a unit element and f has invertible constant term, then

f ’1 ∈ E {z}k,d , Sk,d (f ’1 ) = (Sk,d f )’1 ,
ˆ ˆ ˆ

wherever the right-hand side is de¬ned.
172 10. Multisummable Power Series

Proof: The ¬rst statement follows from Theorem 52, since every element of
E can be regarded as a linear operator on E . For the second one, let kq >
1/2; otherwise, make a change of variable z = wp with su¬ciently large
p ∈ N. Then, let f (z; φ) be as in Proposition 24. Making ρ and µ smaller if
needed, we may arrange that g(z; φ) = f (z; φ)’1 exists for every z ∈ Sφ . It
then follows by means of Proposition 25 that g(z; d1 ) = (Sk,d g )(z) in Sd1 ,
ˆ
ˆ’1 .
for some g . Since g(z; φ) f (z; φ) ≡ 1, this shows g = f 2
ˆ ˆ

As we have shown above and in Section 10.5, multisummability in a
direction satis¬es the same rules as k-summability. It is obvious that The-
orems 51 and 53 also hold with E {z}k instead of E {z}k,d . However, note
that the Main Decomposition Theorem on p. 164 does not generalize to
E {z}k , as is shown in one of the exercises below. Whether it admits gener-
alization when we consider decompositions into sums of products of series
in E {z}kj seems to be unknown “ such decompositions hold, however, for
formal solutions of systems (3.1) (p. 37), as was shown in Theorem 43
(p. 134).
Note that the properties of multisummability shown in this chapter, to-
gether with Theorem 43 (p. 134), enable us to conclude that every formal
fundamental solution of every system (3.1) (p. 37) is multisummable. The
same is true even for formal solutions of nonlinear systems, but we shall
not prove this here.

Exercises:

1. For ψ(z) = (1 ’ z)’1 exp[’z ’1 ], show that f (z) =
ˆ fn z n , with
0

1/2
ψ(w)w’n’1 dw,
fn = “(1 + n/2)
0


is in E {z}k , k = (2, 1), and determine all singular multidirections.

ˆ ˆ ˆˆ
2. Let f = f1 + f2 , fj ∈ E {z}kj , j = 1, 2, with k1 > k2 > 0, and let
ˆˆ
ˆ
g = S (Bκ2 —¦ Bκ1 f ). Let d = (d1 , d2 ) be singular of level 2, and de¬ne
h± by application of Lκ2 to g, with integration along arg z = d2 ± µ,
with small µ > 0. Show that ψ = h+ ’h’ is holomorphic in the sector
S = S(φ, π/κ2 ).

ˆ ˆ
3. Show that f ∈ E {z}k , k = (2, 1), exists, which is not the sum of f1 ∈
ˆ
E {z}2 and f2 ∈ E {z}1 . Why does this not contradict Theorem 50?

4. Show that Exercise 5 on p. 104 generalizes to multisummability.
10.8 Optimal Summability Types 173

10.8 Optimal Summability Types
ˆ
It is common to say that a formal power series f is multisummable, if we
ˆ
can ¬nd an admissible k so that f ∈ E {z}k . The following theorem shows
that there always exists an optimal choice for k.
˜ ˜ ˜˜
Theorem 54 Let k = (k1 , . . . , kq ) and k = (k1 , . . . , kq ) both be admissible,
ˆ
and assume f ∈ E {z}k © E {z}k . Then the following holds:
˜

˜ ˜˜ ˆ
(a) If {k1 , . . . , kq } © {k1 , . . . , kq } = …, then f converges.
˜ ˜˜ ˆ ˆˆ
(b) In case {k1 , . . . , kq }©{k1 , . . . , kq } = {k1 , . . . , kq } = …, assume without
ˆ ˆ ˆˆ ˆ ˆ ˆˆ
loss of generality k1 > k2 > . . . > kq (> 0), so that k = (k1 , . . . , kq )
ˆ
is a type of multisummability. Then f ∈ E {z}k . ˆ


¯ ¯ ¯¯
Proof: Let k = (k1 , . . . , kq ) be the type of multisummability corresponding
˜ ˜˜
to the union {k1 , . . . , kq } ∪ {k1 , . . . , kq }. From Proposition 23 we learn that
ˆ ˆ
f ∈ E {z}k , and there, f has no singular directions of level j whenever
¯
¯ ˜ ˜˜
kj ∈ {k1 , . . . , kq } © {k1 , . . . , kq }. Hence Proposition 22 resp. Exercise 2 on
2
p. 169 complete the proof.
˜
Under the assumptions of Theorem 54, let d = (d1 , . . . , dq ), resp. d =
˜ ˜˜ ˜
(d1 , . . . , dq ), be admissible and nonsingular with respect to k, resp. k. In
ˆ ˆ ˆ
case (a), we then have (Sk,d f )(z) = (Sk,d f )(z) = (S f )(z), for every z
˜˜
where the ¬rst two expressions are both de¬ned. In case (b), assume in
˜ ˜
addition that kj = kν implies dj = dν , for all j, ν with 1 ¤ j ¤ q, 1 ¤ ν ¤ q .
˜
ˆ ˆ ˜˜ ˆ ˜˜
De¬ne d so that kj = kν (= kν ) implies dj = dν (= dν ). Then Exercise 1
ˆ ˆ ˆ
on p. 163 implies (Sk,d f )(z) = (Sk,d f )(z) = (Sk,d f )(z), for every z where
˜˜ ˆˆ
all three expressions are de¬ned. This shows the following: While di¬erent
ˆ
choices of multidirections in general produce di¬erent functions (Sk,d f )(z),
ˆˆ
it is not possible by choosing a di¬erent summability type k to produce even
more functions as corresponding sums.

Exercises:
∞ 2πi/p
ˆ , with integer p ≥ 2, de¬ne
n
1. For f (z) = n=0 fn z and µ = e

ˆ ˆ
nn
g (z) = f (zµ) = n=0 fn µ z . Show that f multisummable implies g
ˆ ˆ
multisummable, and the corresponding optimal types concide.

ˆ ˆ
2. For f and p as above, de¬ne fj (z) = n=0 fnp+j z n , j = 0, . . . , p ’ 1.
ˆ ˆ
Show that f is multisummable if and only if fj is multisummable for
every j = 0, . . . , p ’ 1. What about the corresponding optimal types?
3. Given k1 > . . . > kq > 0, and de¬ning κj accordingly, we say that
a subset of Rq contains almost all multidirections d = (d1 , . . . dq ),

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. 37
( 61 .)



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