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two steps: First we apply the Splitting Lemma, decoupling the system
into two one-dimensional systems. Then we apply Theorem 11 (p. 52)
to both of these systems. From Exercise 1 on p. 55 we conclude that
we are able to take N = 1 and M = r’1 = 0. This shows that there is
an HLFFS for which the transformed system has a coe¬cient matrix
Solutions 251

of the form B(z) = zA0 + diag [a, d], and then F (z) = I + 1 Fn z ’n
ˆ
can be computed directly as follows: Denoting the entries in the ¬rst
column of Fn by fn and gn , they satisfy the recursions
n fn = ’b gn , (»1 ’ »2 ) gn+1 = (n + d ’ a) gn + c fn , n ≥ 1,
with g1 = c/(»1 ’ »2 ). Setting ± + β = d ’ a, ± β = ’b c, we obtain

fn = (»1 ’»(β)nn!
(±)n

n
2)
, n ≥ 1.
(1+±)n’1 (1+β)n’1 
gn = c (»1 ’»2 )n (n’1)!

Similarly, one can obtain the other two entries of Fn .
In case (b), use a scalar exponential shift to eliminate », then use a
constant transformation T that commutes with A0 and produces a
new matrix A1 for which the (new) element c vanishes. Then, use the
transformation
b1/2 ’b1/2
1
x = diag [1, z /2] x
˜
1 1
and the change of variable z = w2 to obtain a system of the form
treated in case (a).

Section 4.1:
1. Note that by de¬nition of sectorial regions it su¬ces to prove the
statement for sectors.
2. Use the previous exercise to see that it su¬ces to deal with k = 1 and
„ = 0, in which case the inequalities describe a circle with center on
the positive real axis, touching the origin, resp. the right half-plane
in case of c = 0.
3. Let I = (a, b); then the union has opening ± + b ’ a and bisecting
direction (b + a)/2.

Section 4.2:
z
1. Use [f (z) ’ f (0)] z ’1 ’ f (0) = z ’1 (u) ’ f (0)] du.
[f
0

2. Conclude that di¬erentiability at the origin is equivalent to f (z) =
f (0)+z f (0)+z h(z), with h(z) continuous at the origin and h(0) = 0.
Then, use Cauchy™s integral formula for f (z).
3. Use the fact that, according to the previous exercises, di¬erentiability
at the origin is equivalent to continuity of the derivative at the origin.
4. Justify termwise integration of the power series expansion of ezt .
252 Solutions

Section 4.3:

6. Proceed as in Exercise 2 on p. 41 to show that E [[z]]s is closed with
respect to multiplication. Moreover, use (B.13) to derive that E [[z]]s
is closed with respect to derivation.

7. Compare Exercise 3 on p. 41.

8. Use (B.13) (p. 231).


Section 4.4:
z ’w
1. Observe 1 ’ e’z = e dw and estimate the integral.
0

5. Use the previous two exercises.

6. Use Proposition 8 (p. 66).


Section 4.5:

2. Use Exercise 1.

3. Use (B.13) (p. 231).
N ’1 N ’m’1
5. Observe z m f (z) ’ fn’m z n = z m (f (z) ’ fn z n ) and
n=0 n=’m
use (B.13) (p. 231).

6. To show that every element of As,m (G, C ) is invertible, observe that
the number of zeros of f (z) in an arbitrary closed subsector of G is
¬nite.
ˆ
7. Take any f ∈ E [[z]]s ∀s ≥ 0, and observe Ritt™s theorem.


Section 4.6:

2. For (a), use the same arguments as in the proof of Proposition 7
(p. 65) to show gn ¤ cn for every n ≥ 0. For (b), use the integral
representation for the Gamma function to show
∞(d)
k
z N rf (z, N ) = z ’k uN rg (u, N ) e’(u/z) duk
0

and estimate as usual.

4. For x ≥ 0, ¬nd the maximum of xn e’x and use Stirling™s formula.
Solutions 253

6. Abbreviate s = 1/k. By assumption there exist c, K > 0 so that
¯
f (z) ¤ c (|z|K)n “(1 + sn) for every n ≥ 0 and z ∈ S. According to
Stirling™s formula the same holds, for di¬erent constants c, K, with
“(1 + sn) replaced by nsn . For small |z|, the term (|z|K)n nsn ¬rst
decreases, then increases with respect to n and is approximately min-
imal for n with |z|K es ns ¤ 1 < |z|K es (n + 1)s . For this n we have
˜ ’k
(|z|K)n nsn ¤ e’sn ¤ ce’K|z| , with su¬ciently large constants
˜
˜˜
c, K, independent of z.

7. Use the previous exercise.


Section 4.7:

2. Let G = {z ∈ G : ze2πi ∈ G}. Conclude that f (z) ’ f (ze2πi ) ∼s ˆ in
˜ =0
˜ and then use Watson™s Lemma and Proposition 7 (p. 65).
G


Section 5.1:

1. Make a change of variable (u/z)k = x in (5.1). The formula holds
whenever the integral exists, which is certainly the case for » = 0.

2. Make a change of variable (u/z)k = x in (5.1) and estimate f .

3. Substitute uk = ak + tk , conclude that f (t) = f ((ak + tk )1/k ) is
˜
of exponential growth at most k on the line arg t = „ , and use the
previous exercise.


Section 5.2:

1. Use (B.10) (p. 228).

2. Estimate the integral for g = Bk f in terms of the length of the path
of integration times maximum of the integrand, occurring along the
circular part of γ(„ ), to obtain g(u) ¤ c exp[Kρκ1 + (|u|/ρ)k ], with
ρ being the radius of the circular part. Then, observe that ρ can be
selected, depending upon |u|, so that the right-hand side is minimal.

3. Make a change of variable z ’k = w, both in (5.4) and (5.3).


Section 5.3:

2. For real t > 0, de¬ne
∞(±„ )
du
±
f (u) e’uz
h (z) = ,
u’t
0
254 Solutions

with su¬ciently small „ > 0. Use Theorem 22 (p. 79) to show that
h± (z) is holomorphic in a region G± containing the straight line
z(y) = x0 + iy, with su¬ciently large x0 (¬xed) and ’∞ < y <
∞, and h+ (z(y)) (resp. h’ (z(y)) tends to zero as y ’ ’∞ (resp.
y ’ +∞). For z(y) as above, check that h± (z(y)) can be represented
by an integral as above, but integration along the positive real axis,
with a small detour about t, say, along a half-circle, in the upper,
resp. lower, half-plane. Using Cauchy™s integral formula, show
h’ (z(y)) = h+ (z(y)) + 2πif (t) e’tz(y) .
From this “connection relation,” conclude
lim h± (z(y))etz(y) = “2πi f (t).
y’±∞

Then, observe that the left-hand side of (5.5) can be seen to equal
(2πi)’1 [h+ (z(’y))’h+ (z(y))]. From these identities, conclude (5.5).

Section 5.4:
N ’1
1. Observe n=1 (ws /z)n = (wsN z 1’N ’ws ) (ws ’z)’1 , and use (B.10)
(p. 228) and (B.14) (p. 232).
2. Observe (B.10) (p. 228). Can you show that the asymptotic expansion
is of Gevrey order s by estimating the integral representation for
Es (z; N )?

Section 5.5:
1. Compute the moment function corresponding to e(z; ±) and de¬ne
E(z; ±) analogous to (5.12). Then, derive the integral representation
1 dw
w’± E(w)
E(z; ±) = ,
w’z
2πi γ

for |z| < 1, and discuss the behavior for z ’ ∞.
3. Analyze the behavior of E(z; ±) as z ’ ∞, following the same argu-
ments used in the proof of Lemma 6 (p. 84), and then proceed as in
the proof of Theorem 26 (p. 84).

Section 5.6:
2. Show the integral representation
∞(„ )
du
k(z) = e1 (u/z) E2 (u) ,
u
0

with 2k|„ | < π, and discuss convergence of the integral, using the
behavior of E2 (u) as u tends to in¬nity.
Solutions 255

3. Use the fact that the moment functions mj (u/p) correspond to the
kernels p ej (z p ), which are of order kp. Moreover, show
p’1
kp ([z e2jπi ]1/p ),
p k(z) = 0 < arg z < 2π.
j=0



Section 5.8:
3. Use Lemma 7 (p. 94).
4. Stirling™s formula (B.12) (p. 229) implies that the radius of conver-
gence equals 1. Holomorphic continuation follows from Exercise 3 on
p. 90.

Section 6.1:
ν m
1. Show the functional equations g(z) = ν=0 z 2 + g(z 2 ), m ≥ 1.
m’1

Use this to show that g(r) cannot have a limit as r ’ 1’. Then,
conclude that g(z) cannot have a radial limit at any one of the points
zjm = exp[2jπi2’m ], for arbitrary j ∈ N.
2. If E contains at least one a = 0, multiply all the terms of the power
ˆ
series f (z) by a.

Section 6.2:
1. Observe (S (Bk f ))(u) = (1 ’ u)’1 .
ˆˆ

5. Use Exercises 4“7, p. 63.

6. For part (a), show (1 ’ 4z)’1/2 = “(1 + 2n) [“(1 + n)]’2 z n , in
n=0
the disc |z| < 1/4.

Section 6.3:
1. Show that (e ’ a)n is a Cauchy sequence, and hence de¬nes some
b ∈ E . Then, verify a b = b a = e.
2. Show ¬rst that it su¬ces to consider the case of f0 = e. Then, con-
¯
clude for every closed subsector S of G with su¬ciently small radius
¯
that e ’ f (z) < 1 for every z ∈ S, and apply the previous exercise.

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( 61 .)



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