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.