closed subintervals of (d ’ ±/2, d + ±/2), but may go o¬ to in¬nity when

„ approaches the boundary values. Consequently, a change of „ essentially

means a rotation of the region G, corresponding to holomorphic continua-

tion of g. Therefore, g(z) is independent of „ and holomorphic in the union

of the corresponding regions G(„, π/k), which is a, in general bounded,

sectorial region G(d, β), with β = ± + π/k. For short, we write g = Lk f

and call Lk the Laplace operator of order k. We sometimes also say that

g = Lk f is the Laplace transform of order k of f .

One immediately sees that g(z) being the Laplace transform of order

k of f (u) is equivalent to g(z 1/k ) being the Laplace transform of order 1

of f (u1/k ), in appropriate sectorial regions. Introducing a mapping s± , for

± > 0, by (s± f )(z) = f (z ± ), this property of Laplace operators can be

conveniently stated as saying that, slightly more generally,

L±k —¦ s± = s± —¦ Lk , (5.2)

for every k > 0.

5.1 Laplace Operators 79

The reason for the factor z ’k in front of the integral representation of

g(z) is that we want the Laplace transform of a power u» to be equal

to “(1 + »/k) z » , as is to be shown in Exercise 1. Accordingly, termwise

∞

ˆ

application of Lk to a formal power series f (u) = 0 fn un produces the

∞ ˆˆ

formal power series g (z) = 0 fn “(1 + n/k) z n , which we denote by Lk f .

ˆ

ˆ

The operator Lk will frequently be called the formal Laplace operator of

order k.

The following theorem shows that Laplace operators link very naturally

with the notion of Gevrey asymptotics, making them the perfect tool for

what will follow.

Theorem 22 Let f ∈ A(k) (S, E ), for k > 0 and a sector S = S(d, ±), and

let g = Lk f be its Laplace transform of order k, de¬ned in a corresponding

sectorial region G = G(d, ± + π/k). For s1 ≥ 0, assume f (z) ∼s1 f (z) in

=ˆ

ˆˆ

S, take s2 = 1/k + s1 , and let g = Lk f . Then

ˆ

g(z) ∼s2 g (z)

=ˆ in G.

Proof: For arbitrarily ¬xed a = ρei„ ∈ S, we may split the path of integra-

tion in (5.1) into the line segment from 0 to a and the ray from a to in¬nity,

thus de¬ning two functions g1 (z), g2 (z). From the exercises on p. 74, to-

gether with the ones below, we conclude g1 (z) ∼s2 g (z), g2 (z) ∼1/k ˆ in

=ˆ 0

=

∼s g (z) in G(„, π/k), owing to Theorem 18

G(„, π/k). This implies g(z) = 2 ˆ

(p. 71). Since G is the union of these regions as „ varies in the interval

(d ’ ±/2, d + ±/2), the proof is completed. 2

Let A(k) (S, E ) = A(k) (S, E ) © As1 (S, E ), i.e., the set of all holomorphic

s1

E -valued functions on S that are of exponential growth not more than k and

have a Gevrey asymptotic of order s1 at the origin. The previous theorem

then says that Lk maps A(k) (S, E ) into As2 (G, E ), and J —¦ Lk = Lk —¦ J. ˆ

s1

In the following sections we show that Lk is in fact bijective and its inverse

equals Borel™s operator.

Exercises:

1. Prove that the Laplace transform of u» , with complex », Re » > 0,

equals “(1 + »/k) z » . Discuss for which other » this statement holds

as well.

2. For f ∈ A(k) (S, E ), k > 0, show that g = Lk f is bounded at the

origin.

3. For f ∈ A(k) (S, E ), k > 0, take a = ρei„ ∈ S and de¬ne

∞(„ )

’k

z ∈ G(„, π/k).

f (u) exp[’(u/z)k ] duk ,

g(z) = z

a

Show g(z) ∼s ˆ in G(„, π/k) for s = 1/k.

=0

80 5. Integral Operators

5.2 Borel Operators

Assume that „ ∈ R, k, r > 0, and 0 < µ < π are given. Let γk („ ) denote the

path from the origin along arg z = „ + (µ + π)/(2k) to some z1 of modulus

r, then along the circle |z| = r to the ray arg z = „ ’ (µ + π)/(2k), and back

to the origin along this ray. In other words, the path γk („ ) is the boundary

of a sector with bisecting direction „ , ¬nite radius, and opening slightly

larger than π/(2k), and the orientation is negative. The dependence of the

path on µ and r will be inessential and therefore is not displayed.

Let G = G(d, ±) be a sectorial region of opening ± > π/k. For every „

with |„ ’ d| < (± ’ π/k)/2, we may choose µ and r so small that γk („ ) ¬ts

into the region G. If f (z) is holomorphic in G and bounded at the origin,

we de¬ne the Borel transform of f of order k by the integral

1

z k f (z) exp[(u/z)k ] dz ’k ,

(Bk f )(u) = (5.3)

2πi γk („ )

for u ∈ S(„, µ/k). Observe that for these u the exponential function in the

integral decreases along the two radial parts of the path, so that the integral

converges absolutely and compactly and represents a holomorphic function

of u. The integral operator Bk de¬ned in this manner will be referred to as

the Borel operator of order k.

From Cauchy™s theorem we conclude that Bk f is independent of µ and

r, and a change of „ results in holomorphic continuation of Bk f . So we

¬nd that Bk f is also independent of „ , and therefore is holomorphic in the

sector S(d, ± ’ π/k). By a simple change of variable, one can show that, in

analogy to (5.2),

s± —¦ Bk = B±k —¦ s± , k, ± > 0.

An exercise below shows that termwise application of Bk to a formal

∞

ˆ

power series f (z) = 0 fn z n produces the formal power series

∞

ˆˆ

(Bk f )(z) = fn z n /“(1 + n/k),

0

ˆ

and Bk is called the formal Borel operator.

As for Laplace operators, we show that Borel operators also “respect”

Gevrey asymptotics:

Theorem 23 Let G = G(d, ±) be an arbitrary sectorial region, let f be

holomorphic in G, and for s1 ≥ 0 assume f (z) ∼s1 f (z) in G. Let k > 0

=ˆ

be such that ± > π/k, so that Bk f is de¬ned and holomorphic in S =

S(d, ± ’ π/k). De¬ne s2 by s2 = s1 ’ k ’1 if 1/s1 < k, resp. s2 = 0

otherwise. Then

(Bk f )(u) ∼s2 (Bk f )(u) in S.

ˆˆ

=

5.2 Borel Operators 81

Proof: For „ close to d and γk („ ) in G, we have for su¬ciently large

c, K > 0 that |rf (z, N )| ¤ c K N “(1 + s1 N ), for every N ≥ 0 and every

z ∈ γk („ ). With g = Bk f we ¬nd that

uN rg (u, N ) = Bk z N rf (z, N ) (u), u ∈ S(„, µ/k).

Breaking γk („ ) into three pieces, the two radial parts and the circular arc

of radius r, and estimating the resulting three integrals in the usual way,

¯

we ¬nd for N ≥ 0 and arbitrarily ¬xed u ∈ S(„, µ/(2k), ρ), ρ > 0

|uN rg (u, N )| ¤ c K N (2π)’1 “(1 + s1 N ) (I1 + I2 + I3 ),

with

r

xN ’1 exp[’ˆ (|u|/x)k ] dx,

I1 , I3 ¤ k I2 ¤ c rN exp[(|u|/r)k ],

c ˜

0

for suitable constants c, c > 0, independent of u, N and r. It su¬ces to

ˆ˜

consider large N ; the ¬nal estimate then holds automatically for all N ,

with possibly enlarged constants. For su¬ciently large N , however, we may

choose r = |u| (k/N )1/k , implying I2 ¤ c |u|N (k/N )N/k eN/k and, after

˜

k

substituting y = c (|u|/x) ,

ˆ

∞

y ’1’N/k e’y dy ¤ |u|N c’1 (k/N )1+N/k .

I1 , I3 ¤ |u| c N N/k

ˆ ˆ

ˆ

cN/k

˜˜

Using Stirling™s formula, this shows, with suitably large C, K > 0, inde-

pendent of u and N ,

˜˜

|rg (u, N )| ¤ C K N “(1 + N s1 )/“(1 + N/k).

Another application of Stirling™s formula and the fact that closed subsectors

¯

of S can be covered by ¬nitely many sectors of the form S(„, µ/(2k), ρ)

2

completes the proof.

In case 1/s1 ≥ k, Theorem 23 says that Bk f is holomorphic at the origin,

ˆˆ

and Bk f is its power series expansion. If 1/s1 > k, one can see, either from

ˆˆ

the ¬nal estimate in the proof or by looking at Bk f and estimating its

coe¬cients, that Bk f is an entire function of exponential growth not larger

than (k ’1 ’ s1 )’1 . For a corresponding statement on the growth of Bk f in

case 1/s1 ¤ k, see the next theorem in the following section.

Exercises:

1. For f (z) = z » , » ∈ C , show (Bk f )(u) exists and equals u» /“(1+»/k),

for every k > 0.

82 5. Integral Operators

2. Under the assumptions of Theorem 23, let G be a sector of in¬nite

radius, and assume f ∈ A(κ1 ) (G, E ), for some κ1 > 0. Show that

then Bk f ∈ A(κ2 ) (S, E ) for 1/κ2 = 1/k + 1/κ1 .

3. For G(d, ±), f , k and „ as in (5.3), and su¬ciently large c > 0,

show that the curve βy , parameterized by z(t) = ei„ (c + it)’1/k ,

’y ¤ t ¤ y, is contained in G for arbitrary y > 0. Use Cauchy™s

integral theorem to conclude for arg u = „ that

1

z k f (z) exp[(u/z)k ] dz ’k = (Bk f )(u).

lim (5.4)

y’∞ 2πi βy

5.3 Inversion Formulas

The following two theorems say that, roughly speaking, Bk is the inverse

operator of Lk , considered as mappings between certain spaces of analytic

functions. In the ¬rst theorem we shall not worry about the exact region

where Lk (Bk f )(z) will be de¬ned, compared to that of f (z). Instead, we

show Lk (Bk f )(z) = f (z) for su¬ciently many z, so that the Identity Theo-

rem in Exercise 3 on p. 230 allows to conclude that they both are the same

analytic function.

Theorem 24 Let G(d, ±) be a sectorial region, and let k > π/±. For any

f ∈ H(G, E ) that is bounded at the origin, let

u ∈ S = S(d, ± ’ π/k).

g(u) = (Bk f )(u),

Then g(u) is of exponential growth not more than k in S, so that (Lk g)(z)

˜ ˜

is holomorphic in a sectorial region G = G(d, ±), with π/k < ±, and

˜ ˜

˜

z ∈ G © G.

f (z) = (Lk g)(z),

Proof: Split γk („ ) into three pieces, as in the proof of Theorem 23. Then

the integrals over the radial parts uniformly tend to zero as u ’ ∞ in

S(„, µ/(2k)). In the integral over the circular arc, expand exp[(u/z)k ] into

the exponential series and integrate termwise to see that this integral is an

entire function of u of exponential growth not more than k. Altogether, this

shows Bk f of exponential growth not more than k in S(„, µ/(2k)). Varying

„ , we ¬nd the same in S.

To prove f = Lk g, it su¬ces to do so for z with arg z = d and |z|

su¬ciently small. For such z, we may insert the integral for Bk f into that

5.4 A Di¬erent Representation for Borel Operators 83

for Lk g, both with „ = d. Interchanging the order of integration and then

evaluating the inner integral gives

’k wk’1 f (w)

(Lk —¦ Bk f )(z) = dw.

wk ’ z k

2πi γk (d)

The function Fz (w) = wk’1 f (w)(wk ’ z k )’1 can be seen to have, in the

interior of γk (d), exactly one singularity, namely, at w = z, this being a

pole of ¬rst order with residue f (z)/k. Since γk (d) has negative orientation,

2

the Residue Theorem in Exercise 3 on p. 224 completes the proof.

Theorem 25 For a sector S = S(d, ±) of in¬nite radius and k > 0, let

f ∈ A(k) (S, E ) and de¬ne g(z) = (Lk f )(z), z ∈ G = G(d, ± + π/k). Then

u ∈ S.

f (u) = (Bk g)(u),

Proof: Using (5.4) for the representation of the Borel operator, ¬x y and