1

(T ’ f )(u) = u» E(w) w’»’1 dw,

2πi γ

with γ as in Hankel™s formula (p. 228). Hence T ’ f equals u» times

a constant. Using that T is the inverse operator, we conclude that

this constant equals 1/m(»). In particular, this shows m(u) = 0 for

Re u > 0. Moreover, we have the following integral representation for

the reciprocal moment function:

1 1

E(w) w’u’1 dw,

=

m(u) 2πi γ

Compare this to Hankel™s formula for the reciprocal Gamma function

(B.10) (p. 228), and note that the integral also converges for u on the

imaginary axis.

Exercises: In the following exercises, let e(z), E(z) be a given pair of

kernel functions of order k > 1/2, and let ± > 0.

1. For e(z; ±) = z ± e(z), ¬nd E(z; ±) so that e(z; ±), E(z; ±) are a pair

of kernel functions of order k.

2. For 0 < ± < 2k, de¬ne e(z; ±) = e(z 1/± )/±, resp.

w±’1

1

E(z; ±) = E(w) dw. (5.16)

w± ’ z

2πi γ

Show that e(z; ±) and E(z; ±) are a pair of kernel functions of order

k/±. Compare (5.16) to the integral representation of Mittag-Le¬„er™s

function (p. 233). Moreover, show that for the corresponding opera-

tors T and T± one has T —¦ s± = s± —¦ T± , with s± de¬ned on p. 78.

5.6 Kernels of Small Order 89

3. Under the same assumptions as in the previous exercise, de¬ne

’± dz

E((u/z)± ; ±) f (z)

g(u) = ,

2πi z

γk („ )

for f ∈ H(G, E ) continuous at the origin, and G a sectorial region

of opening larger than π/k. Show that g = T ’ f .

5.6 Kernels of Small Order

In the previous section we restricted ourselves to kernels and corresponding

operators of order k > 1/2. Here we generalize these notions to smaller

orders.

Kernel Functions of Small Orders

A C -valued function e(z) will be called a kernel function of

˜

order k > 0 if we can ¬nd a pair of kernel functions e(z), E(z)

˜

˜

of order k > 1/2 so that

˜ ˜ z ∈ S(0, π/k).

e(z) = e(z k/k )k/k,

˜ (5.17)

From Exercise 2 on p. 88 we conclude that if a pair of kernels of some

˜

order k > 1/2 exists so that (5.17) holds, then there exists one for any such

˜

k. In particular, if k happens to be larger than 1/2, then we can choose

˜

k = k, hence e(z) is a kernel function in the earlier sense. Moreover, to

verify that e(z) is a kernel function of order k, we may always assume that

˜

k = p k, for a su¬ciently large p ∈ N. This then implies the following

characterization of such kernel functions:

For arbitrary k > 0, e(z) is a kernel function of order k, if and only if it

has the following properties:

• The function e(z) is holomorphic in S+ = S(0, π/k), and z ’1 e(z)

is integrable at the origin. Moreover, for every µ > 0 there exist

constants c, K > 0 such that (5.10) holds.

• For positive real z = x, the values e(x) are positive real.

∞

• For some p ∈ N with p k > 1/2, the function Ep (z) = 0 z n /m(n/p)

is entire and of exponential growth not more than p k. Moreover, in

the sector S(π, π(2 ’ 1/(pk))) the function z ’1 Ep (1/z) is integrable

at the origin.

Let a kernel function e(z) of order k with 0 < k ¤ 1/2 be given. Then we

de¬ne the corresponding integral operator T as in (5.13). The de¬nition of

90 5. Integral Operators

T ’ , however, cannot be given as in (5.14): While we can de¬ne an entire

function E(z) by means of (5.12), this function does not have the same

properties as for k > 1/2. Therefore, we de¬ne the operator T ’ as follows:

˜

Take any k > 1/2, then by de¬nition, resp. Exercise 2 on p. 88, there exists

˜

a pair e(z), E(z) of kernel functions of this order, such that (5.17) holds.

˜

Abbreviate ± = k/k and set f = s± f , for f ∈ A(k) (S, E ) and s± as on

˜ ˜

˜

p. 78. Then f ∈ A(k) (S, E ), for suitable S, and the functions h = T f and

˜ ˜ ˜

˜ ˜˜ ˜ ˜

h = T f are related by h = s± h. In view of this relation between T and T ,

we now de¬ne:

• Let a kernel function e(z) of order 0 < k ¤ 1/2 be given. Choose

˜ ˜

k > 1/2 and let e(z), E(z) and ± be as above. For a sectorial region

˜

G = G(d, ±) of opening larger than π/k, and any f ∈ H(G, E ) that

is continuous at the origin, we de¬ne with γk („ ) as in the de¬nition

of Borel operators on p. 80:

’1 dz

(T ’ f )(u) = E((u/z)1/± ) f (z) .

˜

2±πi z

γk („ )

For this de¬nition to give sense, we have to show that the right-hand side

˜

does not depend upon the choice of k. This, however, follows directly from

Exercise 3 on p. 89.

This de¬nition of T ’ has the following consequence: For f (z) = s± f and

˜

g = T ’ f , resp. g = T ’ f we obtain g (u) = s± g. This relation and the

˜˜

˜ ˜

corresponding one for T which was shown above can be formally stated as

s± —¦ T ’ = T ’ —¦ s± .

˜ ˜

s± —¦ T = T —¦ s± , (5.18)

In particular, this shows that T maps a power un to m(n) z n , while

T ’ is inverse to T , at least for such powers. This motivates the de¬nition

of formal operators T , T ’ , acting on formal power series f by means of

ˆ

ˆˆ

termwise application of T , resp. T ’ . Consequently, T and T ’ are inverse

ˆ ˆ

to one another.

Exercises: In the following exercises, let ej (z), 1 ¤ j ¤ 2, be two kernel

functions of the same order k > 0, and let mj (u) denote the corresponding

moment functions.

∞ n

1. For k(z) = 0 m1 (n) z /m2 (n), show that the power series has

positive, but ¬nite, radius of convergence.

2. For k > 1/2 show that the function k(z), de¬ned above, can be holo-

morphically continued along all rays except for the positive real axis.

3. For 0 < k ¤ 1/2, let p ∈ N be so that p k > 1/2. De¬ne kp (z) =

∞ n

0 m1 (n/p) z /m2 (n/p). Conclude from the previous exercise that

kp (z) can be holomorphically continued along all rays except for the

positive real axis. Use this fact to show the same for k(z).

5.7 Properties of the Integral Operators 91

5.7 Properties of the Integral Operators

In this section, we consider ¬xed operators T, T ’ of some order k > 0. From

the de¬nition of these operators in case of k ¤ 1/2, especially (5.18), one

can see that in some of the proofs to follow it su¬ces to consider k > 1/2.

Using the corresponding formal operators de¬ned above, we show that

Theorems 22 and 23 in Sections 5.1 resp. 5.2 immediately generalize to the

operators T, T ’ :

Theorem 27 Let f ∈ A(k) (S, E ), for k > 0 and a sector S = S(d, ±),

and let g = T f be given by (5.13), de¬ned in a corresponding sectorial

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

=ˆ

ˆˆ

s2 = 1/k + s1 , and let g = T f . Then

ˆ

g(z) ∼s2 g (z)

=ˆ in G.

Proof: By assumption, for every δ > 0 there exist constants c, K > 0 such

that rf (u, n) ¤ c K n “(1+s1 n) for every u with |d’arg u| ¤ ±’δ, |u| ¤ 1

and every n ≥ 0. As was seen before, this implies fn ¤ c K n “(1 + s1 n),

for every n ≥ 0. Using that f (u) is assumed to be of exponential growth

not more than k, it is easily seen that (making c, K larger if needed)

k

rf (u, n) ¤ c K n “(1 + s1 n)eK|u| ,

for |d ’ arg u| ¤ ± ’ δ and arbitrary |u|. Since the operator T maps

un rf (u, n) to z n rg (z, n), one can complete the proof estimating the in-

2

tegral in a standard manner.

Theorem 28 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 T 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

(T ’ f )(u) ∼s2 (Tˆ f )(u) in S.

’ˆ ˜

=

Proof: Observe that T ’ maps z n rf (z, n) to un rT ’ f (u, n), and estimate

2

as in the proof of Theorem 23 (p. 80).

The ¬rst inversion theorem of Section 5.3 also generalizes to the situation

of arbitrary integral operators T and T ’ , as we show now.

Theorem 29 Let G(d, ±) be a sectorial region, and let k > π/±. For f ∈

H(G, E ) which is continuous at the origin, let

g(u) = (T ’ f )(u), u ∈ S = S(d, ± ’ π/k).

92 5. Integral Operators

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

˜ ˜

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

˜ ˜

˜

z ∈ G © G.

f (z) = (T g)(z),

Proof: For k > 1/2, the proof has been given in property 4 (p. 87). For

smaller values of k, use the de¬nition of T , T ’ given in Section 5.6 in terms

of operators T , T ’ of order k > 1/2, together with (5.18).

˜

˜˜ 2

To prove the analogue to the second inversion theorem of Section 5.3 is

harder, owing to the fact that we have to use di¬erent rays of integration

in the integral operator T in order to cover the path of integration γk („ )

used in the integral representation for T ’ . Here, we give a proof that in

the case of T ™s being a Laplace operator is considerably di¬erent from the

proof of Theorem 25 (p. 83).

Theorem 30 For a sector S = S(d, ±) of in¬nite radius and k > 0, let

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

f = T ’ g follows.

Proof: Without loss of generality, assume that d = 0 and k > 1/2; the

latter assumption can be made owing to (5.18). Let β± be the following

path: from the origin along the ray arg z = ±(µ + π)/(2k) to a point of

modulus r, and then on the circle of radius r to the real axis, with ¬xed

µ, r > 0. Then, for 0 < δ < ± we may, for r and µ su¬ciently small and

t on the path β± , represent T f = g by the integral (5.13) with „ = ±δ.

Doing so, we de¬ne for positive real u

∞(±δ)

’1 du dz

f± (t) = E(t/z) e(u/z) f (u)

2πi uz

0

β±

∞(±δ)

du

= f (u) k± (t, u)

,

u

0

’1 dz

k± (t, u) = E(t/z) e(u/z) .

2πi β± z

By de¬nition of the paths β± we conclude that f+ ’f’ = T ’ g. To evaluate

the kernels k± (t, u), we make a change of variable z = 1/w, giving

’1 dw