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Proof: Follows from Theorems 18, 20 in Section 4.5, and Exercise 2 on
2
p. 72.

ˆ ˆ
Theorem 35 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

2
Proof: Follows from Theorem 19 (p. 71).

ˆˆ ˆ
Theorem 36 For ¬xed, but arbitrary, k > 0 and d, 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.

Proof: The ¬rst statement follows from the previous theorem and the fact
that for a Banach algebra E every series g1 can be identi¬ed with the
ˆ
element of L(E , F){z}k,d corresponding to the mapping g2 ’ g1 g2 . For
ˆ ˆˆ
the second one, use Exercise 2 to conclude that (Sk,d f )’1 is de¬ned on
ˆ
a sectorial region with opening larger than π/k, then apply Theorem 21
2
(p. 72).
Roughly speaking, the results in this section may be easily memorized
as saying that that E {z}k,d has the same algebraic properties as E {z},
the space of convergent power series. What these properties are in detail
depends on whether E is an algebra, or has a unit element. In particular, we
ˆ
obtain that C {z}k,d is a di¬erential algebra, and f ∈ C {z}k,d is invertible
if and only if it has a nonzero constant term.
104 6. Summable Power Series

Exercises:
1. Let E be a Banach algebra with unit element e. Show that every
a ∈ E with e ’ a < 1 is invertible and

’1
(e ’ a)n .
a =
n=0


2. Under the assumptions of the previous exercise, let f be holomorphic
in a sectorial region G, and assume that f0 = lim f (z) (for z ’ 0 in
˜
G) exists and is invertible. Show that then a sectorial region G ‚ G
exists, whose opening is smaller than, but arbitrarily close to, that of
˜
G, so that for z ∈ G all values f (z) are invertible.

3. Let f ∈ E {z}k,d have zero constant term. Show that z ’1 f (z) ∈
ˆ ˆ
E {z}k,d , and Sk,d (z ’1 f (z)) = z ’1 (Sk,d f )(z).
ˆ ˆ

ˆ
4. Let E be a Banach algebra, and let f ∈ E {z}k,d formally satisfy the
n-th order linear di¬erential equation
n
aj (z)y (j) = 0,
j=0

with coe¬cients aj (z) holomorphic near the origin. Show that f =
ˆ
Sk,d (f ) then is a solution of the same equation.
5. Let E be a Banach space whose elements are functions of one or sev-
eral variables ω = (ω1 , . . . , ων ) on some domain „¦ ‚ C ν . Moreover,
assume that the norm on E is such that the mapping f ’ f (ω), for
¬xed ω ∈ „¦, is a continuous linear functional on E ; e.g., this is so if
ˆ
f = sup„¦ |f (ω)|. For f (z) = fn z n ∈ E [[z]] we then have that all
coe¬cients fn are functions of ω, and we can consider the mapping
ˆ ˆ
f (z) ’ fω (z) = fn (ω) z n from E [[z]] to C [[z]], for every ¬xed
ˆ ˆ
ω ∈ „¦. Show: If f is k-summable in direction d, then so is fω . Hence,
the above mapping maps E {z}k,d into C {z}k,d , for every ω.




6.4 De¬nition of k-Summability
In view of Lemma 10 part (c), we will from now on identify directions d that
di¬er by integer multiples of 2π, despite the fact that regions still are con-
sidered on the Riemann surface of the logarithm. From Lemma 8 part (b)
ˆ
we see that the set of directions d for which some f is k-summable is always
open. Examples show that the complement of this set can be uncountable;
6.4 De¬nition of k-Summability 105

ˆ
however, in most applications one encounters series f being k-summable
in all directions d but (after identi¬cation modulo 2π) ¬nitely many direc-
ˆ
tions d1 , . . . , dn . Whenever this is so, we simply call f k-summable. The
ˆ
directions d1 , . . . , dn then are called the singular directions of f , provided
ˆ
they occur. For the set of all k-summable series f we write E {z}k , and
it may sometimes be convenient to de¬ne E {z}∞ = E {z}, the set of all
convergent series.
It is immediately clear that Theorems 34“36 remain correct if we replace
E {z}k,d by E {z}k , and the identities for the respective sums then hold for
every but ¬nitely many d (modulo 2π).
ˆ
Lemma 10 part (a) says that Sk,d f is independent of d, as d varies in an
interval not containing singular directions. At a singular direction, however,
the sum will change abruptly, as can be seen from the following proposition.
ˆ
Proposition 12 For ± < d0 < β, k > 0, assume that f is k-summable in
all directions d ∈ (±, β), d = d0 . For some d1 , d2 with ± < d1 < d0 < d2 <
β, |d1 ’ d2 | < π/(2k), assume
ˆ ˆ
(Sk,d1 f )(z) = (Sk,d2 f )(z),

ˆ
for all z where both sides are de¬ned. Then f is k-summable in direction
d0 .

ˆ
Proof: For 1 ¤ j ¤ 2, the functions fj = Sk,dj f are holomorphic in
sectorial regions Gj = G(dj , ±j ) of opening ±j > π/k. Owing to our as-
sumptions, these sectors overlap on the Riemann surface of the logarithm,
and within their intersection the functions are equal. Hence f = f1 = f2 is
holomorphic in G1 ∪ G2 , and f (z) ∼k f (z) there. Obviously, G1 ∪ G2 con-

tains a sectorial region of opening larger than π/k and bisecting direction
ˆ 2
d0 , so f is k-summable in direction d0 .
If a formal power series is k-summable in every direction, then the next
proposition shows convergence. In the terminology of summability theory
one can therefore say: Absence of singular rays is a Tauberian condition
for k-summability.
ˆ
Proposition 13 Assume f ∈ E {z}k has no singular direction; in other
ˆ ˆ ˆ
words, f ∈ E {z}k,d for every d. Then f ∈ E {z}, i.e., f (z) converges for
su¬ciently small |z|.

ˆ
Proof: According to Lemma 10 parts (a), (c), the function f = Sk,d (f ) is
independent of d and single-valued; i.e., f (ze2πi ) = f (z) for every z. From
ˆ
Proposition 7 (p. 65) we then obtain f ∈ E {z}. 2
ˆ ˆ
Suppose f ∈ E {z}k is given, and d0 is a singular direction of f . Then
ˆˆ
for g = S (Bk fk ) two things can happen: Either g(u) is singular for some
106 6. Summable Power Series

u0 with arg u0 = d0 , so that holomorphic continuation along arg u = d0
breaks down at this point; or otherwise, since directions d close to d0 are
not singular, g(u) is holomorphic in a sector S(d0 , µ), for some µ > 0, but
is not of exponential growth at most k along arg u = d0 . This can occur,
according to Exercise 5 on p. 102, but only if g(u) is of in¬nite exponential
order along arg u = d0 ; this can be seen from the following application of
Phragmen-Lindel¨f™s principle:
o
ˆ
Proposition 14 For k > 0 and ± < d0 < β, assume that f is k-summable
˜
in direction d for every d = d0 , d ∈ (±, β). Moreover, for some k > k
ˆˆ
assume that g = S (Bk f ) can be holomorphically continued along arg u = d0
˜ ˆ
and is of exponential growth at most k along arg u = d0 . Then f is k-
summable in direction d0 .

ˆˆ
Proof: In S = S((± + β)/2, β ’ ±) we have that g = S (Bk f ), along
every ray arg u = d but for d = d0 , is of exponential growth at most k. An
application of Phragmen-Lindel¨f ™s principle (p. 235) then shows that the
o
2
same holds for d = d0 .
From Exercise 6 on p. 102 we learn that for ¬xed d a divergent series
ˆ
f may very well be k1 - and k2 -summable in direction d. This changes
drastically, if we require the same for all but ¬nitely many directions at the
same time, because then it converges:

Theorem 37 For k1 > k2 > 0 we have

E {z}k2 © E {z}k1 = E {z}k2 © E [[z]]1/k1 = E {z}.

ˆ
Proof: Trivially E {z} ‚ E {z}k2 © E {z}k1 ‚ E {z}k2 © E [[z]]1/k1 . If f ∈
ˆˆ
E {z}k2 © E [[z]]1/k1 , then g = Bk2 (f ) is entire and of exponential growth
ˆ
at most k everywhere, with 1/k = 1/k2 ’ 1/k1 . Therefore, Proposition 14
ˆ
implies that f , with respect to k2 -summability, cannot have any singular
2
directions, hence converges because of Proposition 13 (p. 105).
The previous theorem can be rephrased as saying that k-summable series
of Gevrey order strictly smaller than 1/k necessarily converge. Thus, we
found another Tauberian condition for k-summability! In fact, the proof
shows that being of Gevrey order strictly smaller than 1/k is a su¬cient
condition for absence of singular directions with respect to k-summability.
So Proposition 14 is the stronger result!

Exercises:
ˆ
1. Investigate k-summability of the formal power series f in the Exer-
cises in Section 6.2.
6.5 General Moment Summability 107

ˆ
2. For k1 > k2 > 0, let f ∈ E {z}k2 ,d © E [[z]]1/k1 , for every d ∈ (±, β),
ˆ
± < β. Prove f ∈ E {z}k1 ,d for every d ∈ (± ’ π/(2k), β + π/(2k))
with 1/k = 1/k2 ’ 1/k1 .

ˆ “(» + n) z n is 1-summable,
3. For Re » > 0, show that f (z; ») = 0
its 1-sum being the function
∞(d)
w»’1 ’w/z
’»
f (z; ») = z e dw,
1’w
0

for d = 2kπ, k ∈ Z.
ˆ ˆ
4. For complex » = 0, ’1, ’2, . . . and f (z; ») as above, show f (z; ») ∈
E {z}1 .
ˆ fn z n ∈ E {z}1 and complex ±, β with β = 0, ’1, . . .,
5. For f (z) =
show
(±)n n
ˆ z ∈ E {z}1 .
f (z; ±, β) = fn
(β)n

6. Show that the generalized hypergeometric series

(±1 )n . . . (±p )n z n
F (±1 , . . . , ±p ; β1 , . . . , βq ; z) =
(β1 )n . . . (βq )n n !
n=0

(with βk = 0, ’1, ’2, . . .) is 1-summable for p = q + 2, q ≥ 0.




6.5 General Moment Summability
In Section 5.5 we have introduced general pairs of integral operators with
features similar to those of Laplace and Borel operator. Here we shall show
that each such pair de¬nes a summability method, and that all methods

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