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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

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


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

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