m
DIFFERENTIAL OPERATORS AND
DIFFERENTIAL EQUATIONS OF INFINITE ORDER WITH CONSTANT COEFFICIENTS RESEARCHES IN CONNECTION WITH INTEGRAL FUNCTIONS OF FINITE ORDER
BY
Dr P. C. SIKKEMA The University of Gronbngen
/11.50 cloth P3.50
P. NOORDHOFF N.V. - 1953 - GRONINGEN - DJAKARTA
PREFACE This book originated from the study of H. Muggli's thesis (cf. references at the end of this book). During this study it appeared to me that all Muggli's theorems could be generalized or improved upon. With the aid of my results, which were deduced under very general conditions, a generalization of a theorem of J. M. Whittaker's
could also be proved. In it Whittaker's notion "asymptotic period" has been replaced by a more general one. Moreover, an improvement upon another theorem of his was possible. My researches have been published in the present writer's Groningen thesis, which appeared in 1953. However, it seemed to me, that it might be of some use to publish them also in the form of a monograph since up to now there
did not exist a book which gives an exact treatment of problems lying in the field of linear differential operators and differential equations both of infinite order with constant coefficients. The present book is an almost unaltered reprint of the author's thesis mentioned above. The only difference between the text of the
book and the thesis lies in the fact that a few pages of the thesis, where results closely related to the problems treated are briefly mentioned, have, not been reprinted since the proofs of them have been omitted in the thesis. For a detailed survey of the contents the reader is referred to the Introduction. I have prefaced this Introduction by a Preparatory Chapter where a summary is given of those notions and properties from the theory of integral functions and of the ordinary differential equations of finite order that are used in this book. I am most grateful to Prof. C. S. Meijer, who has read the manuscript and has made many useful comments. In conclusion, I should like to thank Noordhoff's publishing-office for their valuable help in all matters concerning the publishing of the book. Groningen, September 1953
P. C. SIKKEMA
CONTENTS
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Page 3
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17
Necessary and Sufficient Conditions
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39
PREPARATORY CHAPTER .
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INTRODUCTION .
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CHAPTER
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CHAPTER
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Properties of the Function h(x) = F(D) -* y(x)
CHAPTER III. CHAPTER IV.
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V.
REFERENCES .
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79
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Further Investigation of the Function . . . . . . . h(x) = F(D) -> y(x)
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On the Differential Equation
F(D) - y(x) = h(x) CHAPTER
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149
Chapters III and IV Continued
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180
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222
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The issue of this work has partly been made possible by a grant of tb a Netherlands Organization for Pure Research (Z.W.O.).
PREPARATORY CHAPTER Where in this work we speak of a function this function is nearly always an integral function. It seems therefore desirable to summarize
in this Preparatory Chapter those properties of integral functions that we use in this work. The proofs of these properties will mostly be omitted, since the reader can find them in excellent text-books published in this field. The most important of them are Pringsheim [3] '), Valiron [3], Titchmarsh [1] and Bieberbach [1]. THE NOTION INTEGRAL FUNCTION. A one-valued function /(z) that
is defined in a region G of the complex z-plane, is said to be analytic
in G if it is differentiable at every point of G. Such a function 1(z) is said to be an integral function of z if the region G contains every point of the complex z-plane. A function 1(z) defined for all values of z is called a bounded function if there exists a positive constant M, such that I 1(z) I < M holds for all values of z. From this definition it follows that a function
/(z), which is identically equal to a constant is bounded. Moreover, such a function is apparently an integral function. Conversely, if an integral function 1(z) is bounded, it follows from a well-known theorem
of Liouville 2) that it is then identically equal to a constant. An integral function 1(z) can be expanded into a power series of ascending powers of z, 1(z) = anzn, whose radius of convergence n=o
R is equal to 00 3). If a function 1(z) can be expanded into a power series of ascending powers of z, 1(z) _
anzn, whose radius of con-
vergence R is positive, then that function 1(z) is analytic in the region determined by I z I < R 4). Now, we see that if R = oo, the function
1(z) is analytic in the whole complex z-plane; then it is an integral function. 1) Numbers in brackets refer to the references at the end of this work. 2) Cf. Titchmarsh [1] p. 85. 3) Cf. Titchmarsh [1] p. 83. 4) Titchmarsh [1] p. 69. 3
The case with R = oo will certainly occur if only a finite number of the coefficients a,, differs from zero, i.e. if the function 1(z) is a polynomial. Such a function is also said to be a rational integral function. If R = oo and if there are infinitely many of the coefficients a different from zero, then /(.-) is called a transcendental integral function. Examples of transcendental integral functions are the functions ez, sin Z, Ze-2z2, ee:.
(1)
If we examine these four functions more closely, we see at once
that, if z - x + iy moves from z = 0 to z = oo along the positive real axis, the absolute values of these functions, which are equal to ex, I sin x 1, xe-2x2,
ee:
respectively, present a very different behaviour. The function ex increases indefinitely and so does the function eex, but the latter function increases much more rapidly than the first. The function sin x I remains finite, it oscillates and has no limit if x tends to infinity. The function xe-2x2 also remains finite; its limit for x -* 00 does exist and this limit is equal to zero. The picture changes completely if z moves from z = 0 to z = oo i along the positive purely-imaginary axis. The absolute values of the functions (1) are then equal to the following four functions of y 1, .2(ev
-
e-v), ye2j2, eeos v,
respectively. From this we see that if z moves from z = 0 to z = 00 i along the positive purely-imaginary axis, the functions ez and ee.
remain finite, whereas the absolute values of the functions sin z and ze-2z2 increase indefinitely. However, the absolute value of the latter function increases much more rapidly than that of the first function.
Thus we see two phenomena occurring at the same time. In the first place we see, that the absolute values of two different integral functions f(z) and g(z) may present a different behaviour if the variable
z moves along the same continuous curve to z = 00. In the second place we see that it may happen also that the absolute value of the same integral function 1(z) presents a different behaviour if z moves along two different continuous curves to infinity. ORDER OF AN INTEGRAL FUNCTION. If we examine the functions
ex and ze-2z2 once again, there are more things that strike us. In fact,
if we take the circle with the centre at z = 0 and with radius r, then for all sufficiently large values of r this circle has a point where the 4
absolute value of the function ze- 2 exceeds the maximum that the absolute value of the function e-' attains on that circle. Such a phenom-
enon always presents itself with two integral functions of different order. We shall now communicate what is meant by saying that an integral function is of a certain order. Definition A. If with an integral function /(z) there exists no finite real constant A, such that the relation (2)
I f(z)
I S era
holds for all sufficiently large values of I z I = r, then the function f(z) is said to be an integral function of infinite order. If with an integral function 1(z) there does exist a finite real constant A, such that relation (2) holds for all sufficiently large values of I z I = r, then the function 1(z) is said to be an integral function of finite order. If the integral function 1(z) is of finite order and if it is not identically equal to a constant, then the set of all values A with the property that relation (2) holds for all sufficiently large values of I z I = r, possesses a lower bound t,; this number a is non-negative 5). Then e is called the order of the integral function 1(z) or: f(z) is said to be an integral function of the order Lo.
An integral function 1(z) which is identically equal to a constant is said to be of the order zero.
This notion of order of an integral function is due to Pringsheim [1, p. 129]. If not otherwise explicitly stated we shall from now on always assume
that the integral functions occurring in this work are of finite order.
There exist integral functions 1(z) with the property that the number a belongs to the set of all numbers A for which (2) holds. However, this is not the case with all integral functions. Yet it follows from definition A that with every positive e there exists a number K
(in general dependent of e), such that the function /(z) satisfies the inequality I f(z) 15 Ker
+e
5) Proof. If this lower bound should not exist, or if it should be negative if
it did exist, then there would exist a negative number A, such that there would be - oo < A < 0 and < A < 0, resp., while the integral function &) would satisfy the relation I &) I S er4 for all sufficiently large values of I z I = r. Then there would exist a positive number M, such that the relation I 1(z) I < M would hold for all values of z. The integral function 1(z) would then be a bounded function and therefore it would be identically equal to a constant.
Since this contradicts the assumption that f(z) should not be identically equal to a constant, the assertion is true. 5
for all values of z, while in case e > 0 there exists with not a single positives a number L, such that the function /(z) satisfies the inequality I /(z)
Lere-8 I
for all values of z. Clearly the function ec' is of infinite order. The functions ez and ze-2z2 are both of finite order, the first being of the order 1, the latter of the order 2. Any polynomial is an integral function of the order zero,
as may be proved in the following way: If v(z) is a polynomial, then there exists with every positives a number ro(s), such that v(z) satisfies the inequality I v(z) I <e'
for all values of I z I = r > ro(e). Hence the order P is ; 0. As the order is not negative we have P = 0. However, polynomials are not the only integral functions that are of the order zero. With the help of the formulae (6) and (15) it may be proved that for instance the transcendental integral function zn
g(z)=Inin 00
is an integral function of the order zero. THE NOTION TYPE. If we compare the two functions e2z and e3z, it
appears, though they are both of the order 1, that on every circle with z = 0 as centre the maximum of the absolute value of the function e3z is nevertheless greater than the maximum of the absolute value of the function e2z. Accordingly for a great number of researches
in which the maximum M(r) of the absolute value of an integral function on a circle with z = 0 as centre and radius r plays a role, the notion order is not sufficient to characterize precisely the rapidity
of growth of M(r) as r -> oo. In order to be able to do this more accurately one uses the notion type of a certain order. This notion is also introduced by Pringsheim [2, p. 264-265]. If with a non-constant integral function /(z) of the (finite) order there exists a finite real constant B, such that /(z) satisfies the relation (3)
11(z)
I
esr'
for all sufficiently large values of I z I = r, it is easy to prove 6), 6) This may be done in a similar way as was done in the case of the order. See footnote 6). 6
that the set of all numbers B having this property has a lower bound
y and that this lower bound is Z 0. Definition B. I/ the lower bound y just mentioned is positive, then the integral function 1(z) is said to be of the normal type of the order n. If one wants to state at the same time the magnitude of the type, one says
that the integral function 1(z) is of the normal type y of the order Lo.
If y = 0, the function f(z) is said to be of the minimum type of the order a 7).
If there exists no finite real constant B, such that relation (3) holds for all sufficiently large values of I z I = r, then the function 1(z) is said to be of the maximum type of the order 8).
This notion of type has not been introduced for integral functions of infinite order nor for integral functions that are identically equal to a constant 9).
We have already seen that the order of an integral function is at least equal to zero. If the order is equal to zero and if the function
is not identically equal to a constant, it may be proved in a very simple way that the function is of the maximum type of the order zero 10).
CONNECTION WITH THE COEFFICIENTS OF THE POWER SERIES EX00
PANSION. If 1(z) _
a given integral function of finite order n=0
then naturally the question arises to give a relation between the order p of the function and the sequence of the coefficients a0, a1, a2, .... To answer this question one can prove the following three properties, in which the behaviour for large values of n of the absolute values of the coefficients plays a role 11\
') One does not speak of the normal type zero, but of the minimum type. 8) One uses the expression maximum type and not normal type oo. 9) For integral functions of infinite order a related notion is introduced, viz. the notion function-type. See Blumenthal [1]. We do not need this notion function-type here. 10) In fact, there cannot exist a finite real constant Al, such that we have M for all sufficiently large values of I z I = Y. If there should exist I /(z) such a constant the function 1(z) would be a bounded function, hence it would be identically equal to a constant. This contradicts the assumption. 11) For a proof of the properties P I and P 2 we refer to Pringsheim [3] Bd II. 2, §§ 87, 88 and 89. Since, as far as I know, property P 3 is not to be found in one of the well-known text-books, I shall give a proof of it in footI
note 13). 7
P i. If the function J (z) _ I anzn is an integral function of the order e, n=o
then the formula 12)
1
(4)
1
lim{n!e}61 a I}+ = 0 n-.00
holds for every positive 6. This is also true for e - 0.
If e > 0, then the formula 1
1
(5)
lim sup {n !Q-4 a 1 } 7i = 00 n-.00
holds for every positive 6 that is less than Q.
P 2. 1/ to the expression 1(z) _
anzn there exists a finite positive n=0
number e, such that the relation (4) holds for every positive 6 and that moreover the relation (5) holds for every positive 6 that is less than then 1(z) is an integral function of the order e. P 3. If the numbers an are such that the formula 1
(6)
lim {n ! 4 I a , , 1} = 0 n-.0o 00
holds for every finite positive constant A, then 1(z)
function of the order zero 13).
anzn is an integral n=o
00
Between the type of an integral function 14) 1(z) _ I anzn of the n=0
order a and the sequence of the coefficients ao, a1, ... there also exists a relation in which the behaviour for large values of n of the 12) In order to keep the formulae surveyable we shall always write it!' instead of (n!)° if there is no fear of misunderstanding. 13) Proof. Since formula (6) holds for every positive constant A, it certainly holds for A = 1 and, as a consequence, there exists a positive number
B not depending on n, such that we have n! (
a,,
! S B for it = 0, 1, ....
From this it follows that 1(z) satisfies the inequality ! 1(z) I S Be for every finite value of z. Hence f (z) is an integral function of finite order. If it should be
of the order e > 0, then relation (5) would hold for every positive 6 less than o and consequently for 6 = e. Therefore we would have 2
1
limsup{n!e !an!}n = on. n-.oo
However, this contradicts the fact that relation (6) holds for every positive constant A and therefore also for A = ?. Hence the function 1(z) is an integral e
function of the order zero. 14) Here the case in which the function 1(z) is identically equal to a constant is left out of consideration, because for this case we did not introduce the notion of type. 8
coefficients plays a role. In fact, conditions are known that are necessary and sufficient for an integral function of the order e to be of the minimum type, of the normal type, of the normal type y and of the maximum type respectively, of that order o. In those conditions the behaviour for large values of n of the absolute values of the numbers a0, a1, ... occurs. The conditions referred to are stated in the following properties 15) : 00
P 4. If the function 1(z)
n=0
anzn is an integral function of the order
> 0, then a necessary and sufficient condition that the function /(z) should be of the minimum type of that order o, is, that the numbers an should satisfy the relation 1
1
(7)
lim {n! 0 I an I }n = 0. 00
P 5. 1/ the function 1(z) _ I a,,zn is an integral function of the order n=0
Lo > 0, then a necessary and sufficient condition that the function 1(z) should be of the normal type of that order o, is, that the numbers an should satisfy the relation 1-
1
(8)
0 < lim sup {n! Q I an j}n < oo. oc
P 6. 1/ the function 1(z) =
anzn is an integral function of the order n=0
o > 0, then a necessary and sufficient condition that the function f (z) should be of the normal type y of that order o, is, that the numbers an should satisfy the relation 1
1
(9)
1
lim sup {n ! e I an I } n = (oy) e . n-9 00 00
P 7. If the function f(z) _ n=0
anzn is an integral function of the order
o > 0, then a necessary and sufficient condition that the function f(z) should be of the maximum type of that order o, is, that the numbers an should satisfy the relation 1
(10)
1
limsup{n!e Ianl}n =oo.
For the sake of completeness we remark that in this Preparatory Chapter we have already noticed that a non-constant integral function
of the order zero is always of the maximum type of the order zero. 16) For the proofs of (7), (9) and (10) we refer to Pringsheim [3] Bd II. 2, §§ 87, 88 and 89. The proof of (8) may be given in a very simple way with the help of (9). 9
We have already seen that formula (4) holds if the function Co
a"z" is an integral function of the order e. Then there cer-
1(z) = n=0
tainly exists a positive constant C, such that we have 1
lim{n!°Ia"I}n=0.
(1I)
1l- 00
In fact, formula (4) holds for every positive S, so that, if e > 0, formula (11) holds for every number C that satisfies the inequality
0 < C < 1 , while (11) holds for every positive number C if
0.
P
Conversely, if there exists a finite positive number C', such that Co
formula (11) with C = C' holds, then 1(z) = I a"z" is an integral n=0
function of finite order as may easily be seen in the following manner. There are two possibilities. The first possibility is, that formula (11)
not only holds for C = C' but for every positive number C. Then /(z) has the property P 3. Therefore in this case /(z) is an integral function of the order zero. The second possibility is, that formula (11) holds for C = C', but not for every positive number C. Then it can be proved quite simply 16) that there exists a positive number 1
e 5 C , such that relation (4) holds for this number a and for every positive 6, while relation (5) holds for this number p and for every 6 that satisfies the inequality 0 < 6 < e. Then 1(z) possesses the property P 2. Consequently it is now an integral function of the order Q. Thus we see that in both cases 1(z) is an integral function of finite order. 16) We may e.g. furnish the proof by dividing all positive numbers into two classes. The first class will contain all those positive numbers C for which
(11) holds. This class is not null because the number C' belongs to it. The second class will contain all those positive numbers C for which (11) does not hold. According to the assumption made that (11) does not hold for every positive number C, this class is neither null. Every number of the first class is less than every number of the second class. In fact, if a number C" of the first class should exceed a number C"' belonging to the second class so that formula (11) should not hold for C = C"', then (11) would certainly not
hold for C = C", which contradicts the fact that C" belongs to the first class. Hence this way of dividing the positive numbers into two classes is a Dedekind section which defines one positive number. Let this number be e
As the number C' belongs to the first class, it is clear that we have C' S -; e
10
From the above it follows in the first place, that if we wish to examine whether 1(z) _
anzn is an integral function of finite order,
00
n=0
it is enough to investigate whether there exists a positive number C with the property (11). Then we see that the following property holds:
P 8. 1/ 1(z) is defined by 1(z) = I anzn and i f to the numbers an n=0
there exists a positive constant C, such that the formula
lim{n!c and}IL=0 n- m
holds, then f(z) is an integral function of finite order, which order does
not exceed C Now three examples will be examined. EXAMPLE 1.
F(z)n=2= -
z"--_1_
(0 < a < oo).
(log n)n n!
Here we have as = al = 0,
an = ---I 1
(n = 2, 3, ... ).
(log n)nni
We see immediately that (11) holds for every number C that satisfies
the inequality 0 < C < 1 The function F(z) is therefore of finite a
I
.
1
hence e S C,. From the definition of the number-- it follows that formula (11) Q
1 . Hence (4) holds for every positive 6. Besides it follows from the definition of the number 1 , that formula (11) does not e holds for every number C <
hold for any number C >
1 . As a consequence, formula (5) holds for every
S with 0 < 6 < e. In fact, if there should exist a number SI with 0 < 81 < e, such that (5), with 6 replaced by b,, does not hold, then the left-hand member of (5), with 6 replaced by J8 would be equal to zero. Then formula (11) would
. As e - Jd, < e, this number C would belong to the hold for C = I second class, so that formula (11) cannot hold for this number C. The assumption that (5) should not hold for a number 8, with 0 < 6 < e leads therefore to a contradiction. Hence (5) holds for every 5 with 0 < 8 < Q. II
order. In order to calculate the order o of this function we determine the number o such that in this special case formula (4) holds for every positive number S, i.e. that we have 1 (12) lim in ! e ° IF I
n-.oo
(log n) n n ! o
for every positive 6. Moreover we determine the number o such that in this special case formula (5) holds for every 6 satisfying the ine-
quality 0 < 6 < e, i.e. that we have 1
1
1
lim sup is !
(13)
00
n-. oo
(log n) n n ! a
for every number 6 with 0 < 6 < o. It is easy to see that formulae (12) and (13) both hold if we take = a. In fact, we have for every positive 6
+s - v <0, I
so that
limIn la+a n- oo
1
°_--_ --n=0 (log n n J
.
Moreover, we have for every number 6 that satisfies the inequality 0 < 6 < Q, the formula
a-b - - >0. I
(14)
1
If we put the left-hand member of (14) equal to It and if we use the well-known formula 17)
n! = e-n nn V'2nn (1 + o(1)),
(15)
we see that we have _1
-1
WO-d a-
1
(log n) n
1
I e-Nn n"'(2.-m)-2_0 (log n) n
+ o(1))
f1 n
n" " e '`(2 rn)2n (I + o(1)). log n
17) See for example Valiron [2] § 8!. This formula also follows naturally from the well-known Stirling formula F(z) = e-'z1#V'2r (1 0(1)), valid for large values of ! z J with ! arg z ! i - d. For the latter formula .
we refer to Whittaker-Watson [1], sections 12.33 and 13.6. 12
Becapse of It > 0, the latter expression tends to 0o if n increases indefinitely. Hence formula (13) is correct if we take e = a, even if we replace lim sup by lim. This means that F(z) has the property P 2 with e = a. Therefore F(z) is an integral function of the order a. Moreover we see that, because of 1
lim n! a n;°°
1
1
1- n = 0, (log 91) n n
1
a
formula (7) holds in this special case. Hence the function F(z) an integral function of the minimum type of the order a.
is
EXAMPLE 2.
a"zn
(0
G(z)=2:co - ---i n-0 ylia
Here we may prove in a similar way as in example I that the function G(z) is an integral function of finite order and that this order is equal
to a. For this function G(z) formula (8) is valid since we have I a In
1
lim sup n! a
'
1
°°
1
n= I a j.
nIa
Hence the function G(z) is of the normal type of the order a. Finally, by means of formula (9) we determine the type of the function G(z). 1
From (ay) a = I a I it follows y = v I a Ia, so that the function G(z) is an
integral function of the normal type 1 I a Ia of the order a. EXAMPLE 3.
H(z) =
°°
(log n)"zn
n=1
yjla
1-
(0 < Cr < oo).
The function H(z) is also an integral function of finite order and this order is equal to a. For this function H(z) we have lim sup . n! ny00
Q (log n)" 1 n
= 00
n!a
Hence it follows from (10) that H(z) is an integral function of the maximum type of the order a. 13
LINEAR DIFFERENTIAL EQUATIONS OF FINITE ORDER. Let /(z) be a
polynomial of degree n z 0: /(z) = anzn + a,,,_1zn-1 + ... + ao
(a,, - 0).
As is well-known 18) this polynomial is called the characteristic (or generating) polynomial of the linear differential expression a"_1y("-1)(x)
any(")(x) +
+ ... + aoy(x)
Now this differential expression is represented symbolically by ... + ao) - y(x) 1(D) - y(x) _ (anDn + an-1Dn-1 +
If 1(z) and g(z) denote two polynomials of degree Z 0, then we can form the two linear differential expressions of which j (z) and g(z) are the characteristic polynomials. For these two differential expressions, which we may now symbolically represent by I (D) -> y(x) and g(D) - y(x), it is easy to see that we have
1(D) - y(x) + g(D) - y(x) = {/(D) + g(D)} -. y(x) If a differential expression is given of which the characteristic polynomial is the product of two polynomials 1(z) and g(z), then that differential expression may be represented by {/(D)g(D)} -*y(x) and it is quite easy to prove that (16) {f (D)g(D) } - y(x) = I (D) - {g(D) -> y(x)} = g(D) -* {/(D) - y(x) }.
Now let a polynomial k(z) be written in the form 19) m
k(z) _ fl (z µ=1
in which m, v1, ... , vdenote positive integers while A1, ... , A, are distinct (complex) numbers. In connection with formula (16) we then obtain for m Z 1 e.g. m
(17)
k(D) - y(x) = { ff (D - A,,)"µ} -* y(x) µ=1
{ F1 (D - AuY"}
Particularly, we have for m = 1 k(D) -> y(x) = {(D - Al)"1} -. y(x) = 1
{(D - A1)" - y(x) }.
{(D - Al)"1 - y(x)}.
From this it may be deduced that the totality of all solutions of 18) In this section not a single proof is given. For proofs the reader is referred to Kamke [1] Nrs 127, 133, 134, 137.
19) For m = 0 we have k(z)- 1. 14
the homogeneous linear differential equation of order n.- with constant coefficients
k(D) -. y(x) = 0
is given by
y(x)
=
p,,(x)e-', µ=1
in which for ,u = 1, ... , m p,,(x) denotes an arbitrary polynomial in x the degree of which does not exceed v,, - I, with finite (complex) coefficients.
In this work we shall also encounter inhomogeneous linear differential equations of order n with constant coefficients. In the cases considered here the right-hand member of such a differential equation is always an integral function. Let such a differential equation
be written in the form m
(18)
y(x) = r(x),
(D
{ µ=1
where therefore r(x) denotes an integral function. Now if u(x) is some solution of (18) it can be proved that every other solution of this equation may be denoted by m
Y(x) = u(x) +
q,(x)ex"x u=1
There q,(x) denotes for = 1, ..., m a polynomial depending on the function y(x) of which the degree does not exceed v,, - I. The totality of all solutions of the differential equation (18) may be written in the form m
y(x) = u(x) + Y, p,,(x)e*µx,
(19)
p=1
where denotes for It = 1, ... , m a polynomial of which the degree does not exceed v,, - 1 with finite indetermined (complex) coefficients.
Moreover it may be proved that the function VIA
Y(x) = s(x) _
Y. aYeez"x 'U=1 Q=1
x
J (o -
t)1
1
a aptr(t)dt
0
is a solution of the differential equation (18) if the numbers ape are furnished by the following decomposition into partial fractions ?n
n
a
v
t
Obviously the function s(x) is an integral function 20). 20) This statement is not proved in Kamke [1]. However, it can easily be
shown.
15
The totality of all solutions of the equation (18) may then be written
in the form (19) with u(x) = s(x), i.e. in the form y(x) = s(x)
µ=1
pµ(x)e"',
where the polynomials p, (x) (,u = 1, ..., r) are mentioned above.
From this it follows that every solution of the equation (18) is an integral function.
From (19) it appears that we may also write the totality of all solutions of the differential equation (18) in the form (20)
y(x) = u(x) + cleA1x + ... + clx"1-1e'1x + c,, +1+1eA2x eA2x + .. . + cY1Tg2xti.,-1eA2x +
... + cnxv--'ea,"x, err.
where cl, ... , cn denote arbitrary (complex) constants and n = Finally, we give the following definition:
Definition C. If we speak of a set of oo" (p
a'1;.
x=1
1) functions y(x) we
mean a set of all functions y(x) with the property that there exist p constants
c1, ... , c,,, such that
y(x) = 90) + c19,1(x) + ... + cvTv(x) Here 92(x), q1(x), ..., gq,,(x) denote given integral functions of which the p functions ¢21(x), ..., T,(x) are linearly independent.
From definition C and from (20) we see that the number of solutions of the differential equation (18) amounts to oo'n.
16
INTRODUCTION
Before giving in this Introduction a survey of the aim of this work and its most important results we give a number of definitions. They will enable us to bring our results clearly to the fore. Moreover they
will appear to be very useful in formulating the theorems and in proving ours. In the definitions
I
to 14 the letter x always denotes a complex
variable.
Definition z. We shall say that an integral function I (x) is an integral function not exceeding the minimum type of the order a (a > 0), if one of the following two cases presents itself : 1.
I(x) is of the minimum type of the order a,
2. I (x) is of the order aw, where co < a 1).
Definition z. We shall say that an integral function l(x) is an integral function not exceeding the normal type of the order a (a > 0), if one of the following three cases presents itself : 1.
2.
f (x) is of the normal type of the order a, I(x) is of the minimum type of the order a,
3. I (x) is of the order w, where o) < a 1).
Definition 3. We shall say that an integral function I (x) is an integral function not exceeding the normal type A of the order a (a > 0), if one of the following four cases presents itself : 1.
2. 3. 4.
I (x) is of the normal type A of the order a, I(x) is of the normal type t of the order a, where t < A, I (x) is of the minimum type of the order a, I (x) is of the order co, where w < a 1).
') Here the case where the function /(x) is identically equal to a constant, is included. 17 2
Definition 4. We shall say that an integral function f(x) is an integral function not exceeding the normal type, less than A, of the order a (a > 0), if one of the following three cases presents itself : 2.
I (x) is of the normal type t of the order a, where t < A, I(x) is of the minimum type of the order a,
3.
I(x) is of the order co, where av < a 1).
1.
Definition 5. We shall say that an integral function I (x) is an integral function not exceeding the maximum type of the order a (a > 0), if one of the following four cases presents itself : 1. I (x) is of the maximum type of the order a, 2. I (x) is of the normal type of the order a, 3. I(x) is of the minimum type of the order 4. I (x) is of the order co, where co < a 1).
a,
Definition 6. We shall say that an integral function I (x) is an integral function not exceeding the maximum type of the order zero, if one of the following two cases presents itself : 1.
I(x) is of the maximum type of the order zero,
2. I (x) is identically equal to a constant.
Definition 7. We shall say that two non-constant integral functions of finite order are of the same kind, if they are of the same order and both either of the minimum type, or of the normal type and moreover of the same normal type, or of the maximum type of this order.
Of two constant (integral) functions we shall say that they are of the same kind. Definition 8. If the function I (x) is an integral function of the order and if the function g(x) is an integral function of the order a, then we say that the function g(x) is of the same kind, at most as the function I (x), if one of the following cases r, 2a, 2b, 2c, 2d, 2e, 21, 3a and 3b presents itself : 1.
o > a,
2. o = a 0 0, a) f(x) and g(x) are both of the minimum type, b) c)
I (x) is of the normal type, g(x) of the minimum type,
f(x) is of the normal type y, g(x) of the normal type r, where y Z r,
d) f(x) is of the maximum type, g(x) of the minimum
type, 18
e)
f(x) is of the maximum type, g(x) of the normal type,
f) 3.
f(x) and g(x) are both of the maximum type,
o = a = 0, a) I (x) and g(x) are both constant functions, b)
I (x) is a non-constant function, g(x) is a constant function.
Instead of the expression "the function g(z) is of the same kind at most as the function I(x)" we also use the phrase "the function f(x) is of the same kind at least as the function g(x)".
Definition 9. If the function I (x) is a non-constant integral function of the order P and if the function g(x) is an integral function of the order a, then we say that the function g(x) is of a lower kind than the function I (x), if one of the cases r, 2a, 2b, 2c, 2d and 3 presents itself : 1.
P > a,
a 0 0, a) I (x) is of the normal type, g(x) of the minimum
2.
type, b)
I (x) is of the normal type y, g(x) of the normal type r, where y > z,
c)
I (x) is of the maximum type, g(x) of the minimum type,
d) f(x) is of the maximum type, g(x) of the normal type,
a = 0, I (x) is a non-constant function, g(x) is a constant
3.
function.
Instead of the expression "the function g(x) is of a lower kind than the function /(x)" we also use the phrase "the function I (x) is of a higher kind than the function g(x)". 00
Definition io. If F(z) _
n=0
anzn and if the function y(x) is an integral
function of x, then by F(D) -> y(x) we understand the expression 00
= any(n)(x). n=0 00 F(D) _ anDn we call a differential operator.
h(x)
n=0
If we wish to indicate expressly that only a finite number of the numbers
an is different from zero, then we speak of a differential operator of finite order. If we wish to indicate expressly that infinitely many of the numbers an are different from zero, then we speak of a differential operator
of infinite order. 19
The series
we shall call the generating power series of the n=0
00
differential operator F(D) _ 7 anDn. n=0
Definition II. If the function y(x) is an integral function of x and 0 any(n)(x) converges for every finite value of x, then we if the series n=0 anDn is applicable shall say that the differential operator F(D) _ CO
n=o
to the function y(x).
0
If the series I any"n)(x) does not converge for every finite value of x, n-0 then we shall say that the differential operator F(D) is not applicable to the function y(x).
Definition 12. By an ordinary linear differential equation with constant coefficients for the function y(x) we mean an equation of the form (1)
0
any(n)(x) = h(x),
n=o
where the numbers an (n = 0, 1, ...) denote given constants of which at least one differs from zero, and h(x) is a given integral function. For the sake of brevity such a differential equation will often be called a differential equation. In view of definition zr we shall often write the differential equation
(s) in the form F(D) -> y(x) = h(x). If the function h(x) is not identically equal to zero, the differential equation (i) is said to be an inhomogeneous differential equation. If, on the contrary, the function h(x) is identically equal to zero, the differential equation (z) is said to be a homogeneous differential equation. The numbers an are called the coefficients of the differential equation.
Definition 13. If only a finite number, but at least one, of the coefficients an differs from zero, we say that the differential equation (z)
is of finite order. I/ aN 0 0, but an = 0 for every positive integer n Z N + 1, the differential equation (i) is said to be of order N. In the special case N = 0 the differential equation reduces to the algebraic equation aoy(x) = h(x) for the unknown function y(x). If infinitely many of the coefficients an differ from zero, we say that the differential equation (r) is of infinite order.
Definition 14. A function y(x) = rt(x) will be called a solution of the differential equation (r) if and only if the following four conditions are satisfied: 20
1.
the function rt(x) is an integral function. 00
anDn is applicable 3) to the 2. the differential operator 2) F(D) = n=o function rt(x). 3. the function F(D) -> ij(x) is an integral function. 4. the integral function F(D) -* n(x) is identical with the function h(x).
Now we proceed to give a survey of the aim and the most important results of this work.
Evidently each differential operator of finite order
anDn + a.._1Dii-1 + ... + ao
(n Z 0)
has the property that it is applicable to all integral functions of finite order y(x) 4).
An important difference between the properties of a differential operator of finite order and those of a differential operator of infinite 00
order F(D) _
n=0
anDn is, that the latter differential operator is
not always 5) applicable to all integral functions of finite order y(x). E.g. the operator
F(D) _
Dn n=0
is not applicable to the integral function y(x) = ex which is of the order 1,
because the series
-
do
n=0 dxn ex does not converge for every finite (complex) value of x (see definition 11).
Often it is necessary to know whether a differential operator 0c F(D) = anDn is applicable to a given integral function of finite n-0
order y(x). Accordingly, various authors 6) who occupy themselves 2) See definition 10. 3) See definition 11. 4) Of course such a differential operator is also applicable to every integral function of infinite order. 5) There also exist differential operators of infinite order that are applicable to all integral functions of finite order y(x). Such an operator is e.g. 00 an
F(D)
--DI, where a denotes an arbitrary (complex) number. For
n=0yyl
we have
00
an
F(D) -s y(x) ="=0 I it! y(`)(x) = y(x + a). 6) See e.g. Ritt [1], Perron [1], Hilb I'll, Pblya [1], Valiron [1J, Sheffer [1]. 21
with differential equations of the type we consider here (comp. definition 12), make use of conditions for the numbers an that are sufficient for such a differential operator to be applicable to a function y(x). They do not trouble themselves about the question whether the condition they use is also necessary for a differential operator 00
F(D) = anD", they consider, to be applicable to the function. Now it n=o appears to be extremely difficult to give a condition that is both 00 necessary and sufficient for a differential operator F(D) _ I anDn to 0
be applicable to one single given integral function of finite order y(x). It is easier to find a condition that is necessary and sufficient for CO
a differential operator F(D) =
anDn to be applicable to all functions n=0
y(x) belonging to a certain more or less extensive class of integral functions of finite order. The first to deduce such conditions for specified classes of integral functions of finite order is Muggli [1]. In his paper, which forms the starting-point of this work, he considers the following two') classes,: a) the class of all integral functions the order of which does not exceed a number a, where a Z 1 8). b) the class of all integral functions not exceeding the normal type r of the order 1. Sometimes Muggli replaces this class by the class of all integral functions not exceeding the minimum
type of the order 1 (comp. Theorem B). With respect to the class mentioned under a) Muggli proves the following theorem: 00
Theorem A.
The differential operator F(D) _
n=o
anDn is
a--
plicable to all integral functions y(x) the order of which does not exceed
a number a, where a > 1, i f and only i f the numbers an satisfy the condition i
i
lim sup I a,, N' ogn < eQ
_l
9L-
7) The other two classes that Muggli considers are the class of all analytic functions and the class of all integral functions. As these two classes are very extensive they are considerably less interesting for our purpose than the classes mentioned under a) and b). Accordingly, in this work they will be left out of consideration. 8) In our terminology this is the class of all integral functions not exceeding the maximum type of the order a, where a 1 (comp. definition 5). 22
anzn is an In case a > I this condition expresses that F(z) _ 7&-0 a
integral function the order of which is less than
a - 1
In case a = I this condition expresses that F(z) is an integral function of finite order.
With respect to the applicability of the differential operator F(D) to all functions mentioned under b) Muggli proves: Theorem B.
anDn is ap-
The differential operator F(D) _ 0
plicable to all integral functions respectively not exceeding the normal type r of the order 1, and not exceeding the minimum type of the order 1, if and only if the numbers an satisfy respectively the condition 1
(2)
1
lim sup I a,, In < ---
r
fl-'oo
and 1
lim sup I a,,
(3)
' < oo.
jj_ov 00
Condition (2) expresses that F(z) =
a function that is
analytic for I z i a r. Condition (3) means that F(z) is a function that is analytic at z = 0 9).
In chapter I we generalize and improve upon Muggli's results con-
tained in the theorems A and B. In the proofs of our theorems in chapter I we use a method which bears a slight resemblance to that used by Muggli to prove the theorems A and B. There we do not start from the two classes considered by Muggli and mentioned under a) and b), but from four classes of integral functions of finite order. These four
classes are all much less comprehensive than the classes a) and b) considered by Muggli. They are :
u) the class of all integral functions y(x) of the normal type r of the order a, where 0 < a < oo. the class of all integral functions y(x) of the minimum type of the order a, where 0 < a < oo. 9) The phrase "a function F(z) = E00 anzn is analytic for n=0
IzI
r or at
z = 0" means, as usual, that the radius of convergence of the power series CO
E an.;n is greater than s or greater than zero respectively. u=0 23
the class of all integral functions y(x) of the maximum type of the order a, where 0 < a < oo. a) the class of all integral functions y(x) of the maximum type of the order zero. With respect to each of these four classes we deduce a condition for the numbers a that is necessary and sufficient for the differential
operator F(D) _ I0.a,,Dn to be applicable to all integral functions =0
y(x) belonging to that class. Moreover we prove that the differential operator F(D) _
anDn n=0
is not only applicable to all integral functions that belong to the class in question, but also to all integral functions that are of a lower kind than the functions of this class 10), if the numbers a,, fulfil the necessary and sufficient condition deduced with respect to that class. So we deduce in theorem 1 a condition for the numbers a,, that is 00
necessary and sufficient for the differential operator F(D) =
anDn n=0
to be applicable to all integral functions y(x) belonging to the class mentioned under a), i.e. to all integral functions y(x) of the normal type r (0 <,r < oo) of the order a (0 < a < oo). This condition is that the numbers a,, are such that the expression CO
(4)
G(z) =
zn,
n=0 n!a
which is formed with these numbers a,,, defines an integral function
not exceeding the normal type, less than (ar)
i
of the order
1.
This condition is particularly interesting if a satisfies the inequality
0 < a < 1. In fact, the remark following theorem I tells us that in this case there exist differential operators F(D) that are applicable to all integral functions y(x) of the normal type r of the order a, but whose generating power series
00
anzn does not define an analytic
n=0
function; it diverges for every value of z
0. However, this divergence
cannot be "too rapid" since the expression G(z) in (4), which is associated with this power series, has the property stated. Such "divergent" operators that are applicable to all integral functions of the normal type r of the order a do not exist if a Z 1. In fact, if a > I 10) The functions belonging to this class are naturally all of the same kind (see definition 7). 24
a"z" is to say
it follows from theorem I that the function F(z) _ n=0
the least analytic at z = 0. In Muggli's paper mentioned the function
F(z) _
0
a"z" is always to say the least analytic at z = 0. Ac-
cordingly, "divergent" operators do not occur in Muggli's paper. In the theorems 2, 3 and 4 we successively deduce a condition for the numbers a" that is necessary and sufficient for the differential 00
operator F(D) =
a"D" to be applicable to all integral functions n=0
y(x) that belong to the class of integral functions mentioned above under #), y) and b) respectively. Particularly, if in the cases mentioned
under fl) and y) the number a satisfies the inequality 0 < a < 1, then as in the similar case mentioned under a), there exist differential 00
operators F(D) whose generating power series
a"z" diverges in
every point z 0 0. Nevertheless these differential operators are applicable to all functions y(x) belonging to the classes mentioned under fi) and y) resp., with 0 < a < 1. In the case mentioned under 6) there also exist differential operators F(D) that are applicable to all functions y(x) belonging to the class mentioned under 6) and whose a"z" only converges for z = 0.
generating power series n=0
For these last three cases the existence of such "divergent" operators was already guaranteed by theorem 1, as may be seen from the proof
of the remarks that we subjoined to the theorems 2, 3 and 4. Our theorem 1 is a generalization of Muggli's theorem B, so far as it concerns all integral functions not exceeding the normal type r of the order 1. In fact, theorem 1 occupies itself with all integral functions of (or not exceeding) the normal type r of the order a, where a denotes
an arbitrary number of the interval 0 < a < oo. Moreover, theorem I
is for each a with 0 < a < oo an improvement on theorem B,
which only deals with case a = 1. In fact, in theorem 1 we give with respect to the numbers a" a condition for the differential operator 00
F(D) _ =0
a"D" to be applicable to all integral functions belonging
to the relatively narrow class of all integral functions that are of the normal type r of the order a. Theorem 1 and the subjoined footnote
tell us next, that the same condition also holds for the applicability to all integral functions that belong to the considerably more extensive class of all integral functions not exceeding the normal type r of the order a. On the other hand, in theorem B Muggli deduces a condition 25
for the applicability of a differential operator F(D) -
to all n= o
integral functions not exceeding the normal type r of the order 1. Therefore he at once starts from the class of all integral functions not
exceeding the normal type r of the order 1 and not from the much narrower class of all integral functions of the normal type z of the order 1 as we do in theorem I for each value of a with 0 < a < oo. Similarly theorem 2 is a generalization of and an improvement on Muggli's theorem B so far as it concerns all integral functions not exceeding the minimum type of the order 1. Similarly theorem 3 is a generalization of and an improvement on Muggli's theorem A. In his paper already mentioned Muggli proceeds to examine the expression
h(x) = F(D) - y(x),
assuming that the conditions of theorems A and B are fulfilled. He deduces the following three theorems, the first of which deals with the case where the order of the integral function y(x) is not less than 1. This theorem reads like this: Theorem C. If the function y(x) is an integral function of the order 1, and if the numbers a satisfy the condition mentioned a, where a
in theorem A, then h(x) = F(D) -k y(x) is an integral function the order of which does not exceed a.
The second theorem deals with the case where the integral function y(x) does not exceed the normal type -r of the order 1 and the minimum
type of the order 1 respectively. It runs as follows: Theorem D. If the function y(x) is an integral function not exceeding the normal type z and the minimum type of the order I respectively and if the numbers a satisfy the conditions (2) and (3) respectively, mentioned in theorem B, then h(x) = F(D) -> y(x) is an integral function not exceeding the normal type z and the minimum type of the order I respectively.
In case the function y(x) is an integral function the order of which is less than 1, this theorem states it only roughly since it then only 26
tells us of the function h(x) that this function does not exceed the minimum type of the order 1. For functions y(x) of an order less than I Muggli improves this statement in the third theorem. In this theorem he takes as a condition for the numbers an the condition (3) of theorem B. However, this condition admits of no "divergent" operators since 00
the associated expression F(z) _
anzn is a function that is analytic n=o
at z = 0 (see theorem B). This third theorem is Theorem E. If the function y(x) is an integral function of the order a, where a < 1, and if the numbers an satisfy the condition (3) mentioned in theorem B, then h(x) = F(D) -> y(x) is an integral function the order of which does not exceed a.
In chapter II we generalize and improve upon the theorems C, D and E. Among other things in this chapter we deduce the theorems
7, 8, 9 and 10. In these theorems the function y(x) is an arbitrary integral function that belongs respectively to the classes of integral functions mentioned above under a), fl), y) and (S).
In the proofs of his theorems C and D Muggli determines an inequality for the functions I y(n)(x) I (n = 0, 1, ...) and so obtains an
inequality for the function CO
h(x) = F(D) ' y(x)
any(n)(x).
n=o
The proof that Muggli gives of his theorem E is essentially based 00
on the assumption that the function F(z) _ anzn is analytic n=o at z = 0. It is clear that this method of proving fails completely if also differential operators are admitted whose generating power 00
series I anzn diverges for every z n=O
0. Moreover, the method to
prove theorems C and D cannot be of any use in our considerations as the results that may be obtained by it are not accurate enough.
The method we use in proving the theorems 7-10 is based upon the expansion of the functions y(n)(x) (n = 0, 1, ...) into a power series in ascending powers of x. This method appears to be also extremely fruitful in the chapters III and IV. In theorem 7 the function y(x) is an integral function of the normal type r of the order a > 0, while the numbers an occurring in the dif27
have the property mentioned in
ferential operator F(D) = I 0
theorem 1 (so, if 0 < a < 1, a class of differential operators is admitted 00
converges only for z = 0). whose generating power series I n=0 Then it appears that in the study of the properties of the function h(x) = F(D) -* y(x) for a we have to distinguish two cases, viz. 0 < a S 1 and a > 1. In fact, in the first case the function h(x) is an integral function of the same kind at most as the function y(x). There-
fore this is also true for 0 < a < 1 and particularly also, if the differential operator F(D) belongs to the class of "divergent" operators then admitted. In the second case, so where a > 1, the situation is different. If the expression G(z) in theorem 7 (or theorem 1) now defines
an integral function not exceeding the minimum type of the order 1, then the function h(x) is still an integral function of the same kind at most as the function y(x). On the contrary, the function h(x) may
be of a higher kind than the function y(x), if the expression G(z) mentioned defines an integral function of the normal type y of the 1
order I (where y < (ar) -a). If in this case the function h(x) is of a higher kind than the function y(x), then it appears that the order of the function h(x) equals the order of the function y(x). However, the type of the function h(x) then exceeds the type of the function y(x). Nevertheless, the type of the function h(x) cannot exceed a certain bound then. This bound is to be found in (II, 29) 11) and we prove that it cannot be replaced by a smaller number. In theorem 8 and 9 respectively the function y(x) is an integral function of the minimum type and maximum type respectively of the order a > 0. There the numbers a,,, occurring in the differential 00
operator F(D)
a,,D", have the property mentioned in theorems n=0
2 and 3 respectively. So, if a satisfies the inequality 0 < a < 1, again a class of "divergent" operators is admitted. The latter is also the case in theorem 10, where the function y(x) is an integral function of the maximum type of the order zero and the numbers an have the property mentioned in theorem 4. In each of the three cases that come up for discussion in theorems
8, 9 and 10 it appears, that the function h(x) = F(D) -* y(x) is an integral function of the same kind at most as the function y(x). ") If in this work we refer to a formula occurring in another chapter then that in question before the number of this formula we place the indication of that chapter. E.g. (II, 29) means formula (29) of chapter II; (P, 8) means formula (8) of the Preparatory Chapter. 28
In chapter III we restrict ourselves to those integral functions y(x) that do not exceed the minimum type of the order I and that are not identically equal to a constant. So these functions have the property to be either of the minimum type of the order 1 or of an order less than 1. For these functions y(x) we have shown in chapter II
that, under certain restrictions with respect to the applicability of the differential operator F(D), the function h(x) = F(D) y(x) is an integral function of the same kind at most as the function y(x). We
formulated these results in a surveyable manner in theorem 11. In chapter III we shall considerably improve on these results. In fact, we prove that, if the function y(x) is a transcendental integral function
not exceeding the minimum type of the order 1, then the integral function h(x) is of the same kind as the function y(x). For instance, if the function y(x) is an integral function of the normal type r of the order a (then we now have 0 < a < I since the function y(x) does not exceed the minimum type of the order 1 (see definition 1)), we prove that the function h(x) is likewise an integral function of the normal type r of the order a. So this is also true if the differential 00
operator F(D) _ a certain class of operators in this case admitted, since a satisfies the inequality 0 < a < 1, whose generating power series diverges for z
. 0.
In theorem 15 we shall also examine the case where the function y(x) is a non-constant rational integral function, or identically equal
to a constant. So in chapter III we see that a transcendental integral function y(x), not exceeding the minimum type of the order 1, is transformed
into a function h(x) that is of the same kind as the function y(x) by every differential operator F(D) _ that is applicable CO
n=0
to all integral functions that are of the same kind as the function y(x). Now the question arises whether, conversely, to a given transcendental integral function h(x) not exceeding the minimum type of the order 1 there can be found an integral function y(x) that is of the same kind as the function h(x) and has the property of being transformed into the function h(x) by a given differential operator F(D) that is applicable
to all integral functions of the same kind as the function h(x). In chapter IV we answer this question in the affirmative. We even prove that with every transcendental integral function h(x) not exceeding the minimum type of the order I and with every differential operator F(D) that is applicable to all integral functions of the same kind as the function h(x), there can be found an integral function y(x) of the 29
same kind as the function h(x) that by the operator F(D) is transformed into the function h(x). Moreover we prove that there does not exist a function y(x) of a lower kind than the function h(x) that is transformed
into the function h(x) by the operator F(D).
In theorem 16 we assume that the function h(x) is an integral function of the normal type r of the order a (0 < a < 1), while the differential operator is constructed in such a way that the numbers an have the property that the expression -
n=0
G(z)
an
zn,
,n
formed with them, defines an integral function not exceeding the 1
normal type, less than (ar) a, of the order 1. So particularly, as we have already seen, a class of "divergent" operators is also admitted. Moreover we assume that at least one of the numbers an differs from zero. Then it appears that the differential equation
F(D) - y(x) = h(x) has a solution y(x) = rj(x) that is an integral function of the same kind at most as the function h(x). Besides we prove that this function 17(x)
is even of the same kind as the function h(x). We go on to show that this function fi(x) is uniquely determined, if ao - 0. In case ao = ... = a9_1 = 0, a, 0 (p 1) there are ooD such solutions fi(x) 12). In the theorems 17, 18 and 19 we similarly assume that the function
h(x) is an integral function of the minimum type of the order a (0 < a 1), an integral function of the maximum type of the order a (0 < a < 1) and a transcendental integral function of the maximum type of the order zero, respectively.
In theorem 19 we besides consider the case where the function h(x) is a non-constant polynomial or identically equal to a constant. So we also consider as a special case that of the homogeneous differential equation, viz. where h(x) = 0. So we see that in chapter TV we prove that the differential equation F(D) -± y(x) = h(x),
where F(D) and the integral function h(x) satisfy certain conditions,
has a solution y(x) that is of the same kind as the function h(x). 12) For the p functions with the help of which these oop solutions are formed,
see the remark that follows the proof of theorem 21 (page 2)5). 30
Meanwhile it is not true that this differential equation may not have solutions of a higher kind than the function la(x). A quite simple example, where we do find this, is the differential equation of finite order (D - 1) --> y(x) = cos -,/x.
(5)
So in this case h(x) = cos -,/x, which function is an integral function of the normal type I of the order 1 13). Besides, F(D) = D - I so that
ao = - 1, a1 = 1, a,, = 0 (n 2). Hence the conditions of theorem 16 are satisfied, if we take v = 2, r = 1. Then this theorem tells us that this differential equation has one and only one solution y(x) = rl(x)
not exceeding the normal type I of the order I and, moreover, that this function is of exactly the normal type I of the order Now from the Preparatory Chapter we know that the totality of all solutions of (5) is given by 4.
y(x) _ 7(x) + cex,
where c denotes an arbitrary (complex) constant. Then, obviously, every solution belonging to c
0 is of the normal type I of the order 1
and therefore of a higher kind than the function fi(x). An example 14) of a differential equation of infinite order, which just as equation (5) has one and only one solution that is an integral function not exceeding the normal type 1 of the order 4, is furnished by (6)
j(D
l
- 1)._o 1 Di J - y(x) = cos x it.
1.
This equation also comes under theorem 16. It has one and only one solution that is an integral function of the same kind (at most) as the function cos Vx + 1, i.e. of (not exceeding) the normal type 1 of the order J. Now we assert that this solution is the function )(x)
that we mentioned in the first example. If the function y(x) is an integral function of finite order, we may conveniently deduce from theorem 3 that the differential operator D" (D-1)Z n=o n!
n-1 n=1
n!
19) This follows namely from cos Vx _ }(ei `'X_ + e-' fix) and from the de-
finitions A and B. ") See the remark following theorem 21 for another example of a differential equation that has solutions of a higher kind than the function h(x). 31
is applicable to the function y(x). Then we have {(D
- 1)
,' O It!
+
-* y(x)
1
-- Dn
-
I
91=1
Y(x) + 2;
n
1
n!
ycni(x)
I
Y(x)
= y'(x + 1) - y(x + 1).
So, if the differential equation (6) has a solution y(x) _ fi(x), that is an integral function of finite order, then for this function .(x) we have therefore
0'(x + 1) - O(x + 1) = cos x + 1. Hence the function O(x) satisfies the differential equation (5). The latter has one and only one solution that is an integral function of (not exceeding) the normal type I of the order 1, viz. the function n(x). Then this function rl(x) is also the only solution of the differential
equation (6) with this property. As far as we know, Muggli is the only author who, within the frame of
chapter IV, has found some results of a somewhat general character. In his paper already mentioned he proves 15) Theorem F. If h(x) is an integral function not exceeding the minimum type of the order 1, if the numbers a,, satisfy the condition (3) of 00
theorem B and if F(D) _
n=0
a,Dn, then the differential equation
F(D) - y(x) = h(x)
(7)
has a solution that is an integral function not exceeding the minimum type of the order 1. Theorem G. If h(x) is an integral function of the order a, where a < 1, if the numbers a,, satisfy the condition (3) of theorem B and if 00
F(D) _
n=0
anDn, then the differential equation
F(D) -* y(x) = h(x)
has a solution y(x) = yo(x) that is an integral function of the order a. 16) Contrary to theorem G theorem F restricts itself to the communication that there exists a solution with the property desired. 32
The most general solution of the order a has the form y0(x) + P(x). There P(x) denotes a polynomial of degree m - I (m Z 1) with arbitrary coefficients, where m is the multiplicity of the zero z = 0 of the function F(z). If m = 0, then P(x) - 0; so then there is just one solution with the property mentioned.
It is obvious that our theorems 16-19 of chapter IV not only constitute a considerable generalization but also a considerable improvement of the Muggli theorems F and G. To begin with, Muggli's
theorems do not admit "divergent" operators, of which precisely for 0 < a < 1 we admit a certain class also. Moreover theorem F tells us only that the differential equation (7) has a solution that does not exceed the minimum type of the order 1. Therefore this theorem
tells us little, since it does not indicate how large the number of such solutions is. Besides, according to this theorem, it might be possible
that such a solution is of a lower kind than the function h(x) if the latter function is of the minimum type of the order 1; also it might be possible that such a solution is of a higher kind or of a lower kind than the function h(x) if the order of this function is less than 1. All this holds similarly for theorem G, which only tells us of the function yo(x) that it is of the same order as the function h(x). Hence the function might very well not be of the same kind as the function h(x), e.g. it might be of the minimum type, of the normal type A, with
0 < A < r or r < A < oo, or of the maximum type of the order a, if the function h(x) is of the normal type r of the order a. In chapter V we continue the researches we began in the chapters III and IV. Now we first start from a given integral function y(x) of the normal type r of the order 1 and we suppose that the differential 00
operator F(D) _ I anDn satisfies the condition of theorem I with n=0 a = 1, the numbers an not all being equal to zero. Then it follows from theorem 7 that the function (8)
h(x) = F(D) - y(x)
is an integral function of the same kind at most as the function y(x). In theorem 20 we first state under what conditions the function (8) is of the same kind as the function y(x). Then we give in theorem 20 conditions under which the function (8) is of a lower kind than the function y(x). In this last case we state that then the function h(x) is of the same kind as a function ul(x) that we define in assertion 3 of theorem 20. In theorem 21 we, conversely, assume that the function h(x) is a 33 3
given integral function of the normal type r of the order 1. The dif00
ferential operator F(D) _ I a,,Dl is also given and it has the property m=o
that the condition of theorem 1, with a = 1, is satisfied, the numbers a not all being equal to zero. Then theorem 21 gives an affirmative answer to the question whether the differential equation F(D)
y(x) = h(x)
has a solution y(x) = 71(x) that is an integral function of the same kind at most 16) as the function h(x). This answer was already known to Sheffer 17) who, as we do, also gives the number of solutions with this property. We prove more than Sheffer, however, viz. we show that the
solutions found are not only all of the same kind at most, but even all of the same kind as the function h(x). Next, in theorem 22, we deal with the homogeneous differential
equation F(D) -- y(x) = 0, whereupon we state the connection that exists between the solutions of the inhomogeneous equation F(D)- y(x) = h(x) in case ao = 0, and those of the associated homogeneous equation.
In the latter part of chapter V we show that our theorems 12, 13, 14, 15 and 20 enable us to prove a generalization of a theorem of J. M. Whittaker 18). We formulate this generalization in
theorem 24.
In the Whittaker theorem in question there occurs the notion "asymptotic period" of an integral function of finite order. This notion "asymptotic period" is a generalization of the notion "period" of an integral function. As is well-known, the number a (a 0) is
16) In the remark that follows theorem 21 we give an example of a differential equation of the type considered here, that has a solution of a higher kind than the function h(x). 17) Sheffer [ 1]. Our method of proving theorem 21 shows resemblance to that
of this author. 18) J. M. Whittaker [1] p. 86. There also results are stated concerning integral functions of finite order that are of the maximum type of the order 1 or of an order exceeding 1. Just before we were going to press it appeared from the Mathematical Reviews vol. 13 (1952) p. 929, that Gelfond [1] has generalized the notion asymptotic period for the latter functions. We had no access to Gelfond's paper. Here we shall not enter into Whittaker's and Gelfond's results, since we restrict ourselves to integral functions not exceeding the normal type of the order 1. 34
called a period of the integral function y(x), if for every finite (complex)
value of x Y(x + a) - y(x) = 0.
Now Whittaker calls a number P an asymptotic period of an integral function of finite order y(x), if this number P has the property
that the order of the integral function O(x) defined by (9)
O(x) = Ax + fl) - Y(x),
is less than the order of the function y(x). It is clear that every period of an integral function of finite order is also an asymptotic period of this function. However, the converse statement does not always hold as may be seen from the following example. We define the integral function fi(x) by fi(x) = e2 + x.
Obviously this function is of the order 1, while the function O(x) = (x + 2nci) - fi(x) = 2ni is an integral function of the order zero. The number
= 27ri therefore
is an asymptotic period of this function $(x). Nevertheless, this number jI is not a period of the function $(x). Now Whittaker proves the following theorem 18): Theorem H. If the function y(x) is an integral function not exceeding the normal type of the order 1 and if A denotes the set of asymptotic Periods of the function y(x), then the following holds: 1.
If the integral function y(x) does not exceed the minimum type of the order 1, then A is null.
2. If the integral function y (x) is of the normal type of the order 1,
then either A is null, or A is the set of the points U, where A denotes
a constant 0 0 and k = ± 1, + 2, .. . In theorem 24 we start, like Whittaker in his theorem H, from an integral function y(x) not exceeding the normal type of the order 1. However, we do not consider the set A, but the set B of all numbers
w (co 0 0) that have the property that the integral function (10)
k(x) = (ewn - 1) - y(x) = y(x + co) - y(x)
is of a lower kind than the function y(x). 35
If a number i is an asymptotic period of a function y(x) not exceeding the normal type of the order I, then fi is not only an element of the set A, but also an element of the set B, since the function (9) is of a lower kind than the function y(x) (see definition 9). From this it follows that the set A of asymptotic periods of an integral function not exceeding the normal type of the order 1, considered by Whittaker, is a sub-set of our set B. Conversely, if to is an element of our set B, then this number co is not always an asymptotic period of the function y(x). In fact, if the function (10) is of a lower kind than the function y(x), but of the same order as the function y(x), then the number w is indeed an element of the set B, but not of the set A. So we see that the definition of our numbers co is a generalization of the notion asymptotic
period, defined by Whittaker. Therefore, our theorem 24 is a generalization of Whittaker's theorem H. An example of a function y(x) that has the property that there exists a number w belonging to the set B but not to the set A, is furnished by the function
y(x)=e'-x- ex+x, which is an integral function of the normal type 2 of the order 1. Now, if we take co = 7ri, we have
k(x) = y(x + ni) - y(x) = 2ex + ni. This function k(x) is of the normal type I of the order I and hence (according to definition 9) of a lower kind than the function y(x). So this number ni is an element of the set B. Since the functions y(x) and k(x) are both of the order 1, this number ni is not an asymptotic
period. Hence in this case the set A is a proper sub-set of the set B. According to theorem 24 our set B consists in this special case of the
numbers kni (k = ± 1, ± 2, ...) 19). It is easy to show that in this case the set A consists of the numbers 2kni (k = ± 1, + 2, ...). Our theorem 23, in which we summarize those results of the theorems
12, 13, 14, 15 and 20, which we need in order to prove theorem 24, is in a certain sense a very far-going generalization of theorem 24 and therefore of Whittaker's theorem H.
For in theorem H it must be examined whether a differential operator of the special form Yn e''n-I=I--Dn
n=1 U.
19) In fact, the number fi, occurring in theorem 24, is equal to 2. 36
where y denotes an arbitrary constant -A 0, transforms an integral function y(x) not exceeding the normal type of the order 1, into a function
(x) = (In_1 Yn.D1)
- y(x) = y(x + y) - y(x),
the order of which is less than the order of the function y(x). 00
In theorem 23 we admit every differential operator F(D) _ n=0
that is applicable to all functions that are of the same kind as the function y(x) and we indicate under what conditions the function h(x) = F(D) - y(x) is of a lower kind than the function y(x). Moreover, we indicate under what conditions the function h(x) is of the same kind as the function y(x).
Finally in chapter V we prove an improvement on another theorem due to J. M. Whittaker [1, p. 22], viz. the theorem I to be mentioned below, in as far as in that theorem the integral function f(x) does not
exceed the normal type of the order 1. This theorem I is, in turn, an improvement on a theorem that originated with Guichard [1] and that was also proved by Appell [ I ] and Hurwitz [ 11. The Guichard's
theorem reads like this: If the function f(x) is an integral function, then there exists an integral function g(x) with the Property
g(x ± 1) - g(x) = f(x). Whittaker improved upon this theorem as follows:
(11)
Theorem I. If the function f(x) is an integral function of finite order t,, then there exists an integral function, likewise of the order for which formula (ii) holds.
Our theorem 26 deals with the difference equation Y(x + w) - y(x) = h(x), (12) where the function h(x) is an arbitrary integral function not exceeding the normal type of the order 1. In case this function h(x) is not identically equal to a constant 0, theorem 26 tells us a.o. that equation (12) has a solution y(x) = rt(x) that is an integral function of the same
kind at most as the function h(x) and that, apart from an additive constant, is uniquely determined. Moreover, theorem 26 tells us that this function n(x) is even of the same kind as the function h(x). So 37
our theorem 26 is clearly an improvement on Whittaker's theorem I, in so far as in this theorem the integral function f(x) does not exceed the normal type of the order 1.
To prove our theorem 26 we need a part of the results of our theorems 16, 17, 18, 19 and 21. These results needed we summarize in theorem 25 which, in fact, is a generalization of theorem 26. On account of the very general character of theorem 25 - as we know a certain class of "divergent" operators is also admitted, if the order of the given integral function h(x) is less than i - it is a far-going generalization of Whittaker's theorem I, in so far as in this theorem the integral function I (x) does not exceed the normal type of the order 1.
38
CHAPTER I
NECESSARY AND SUFFICIENT CONDITIONS
Necessary and sufficient conditions for a differential operator 0 F(D) _ a,iDn to be applicable to all integral functions that are of the same type of a fixed finite order will now be derived. In this chapter we first give thirteen lemmas. Then we shall deduce in theorem I a condition that is necessary and sufficient for a dif00
ferential operator F(D) = 7, a,,Dn to be applicable to all integral =o
functions y(x) of the normal type r of the order a > 0. After that we give in theorem 2 and theorem 3 a condition that is necessary and
sufficient for a differential operator F(D) _06 I anD" to be apn=0
plicable to all integral functions y(x) of the minimum type of the order a > 0, and to all integral functions y(x) of the maximum type of the order a > 0 respectively. Theorem 4 occupies itself with a condition that is necessary and sufficient for a differential operator 00
F(D) _
a0Dn to be applicable to all integral functions y(x) that 0
are of the maximum type of the order zero. Finally, theorem 5 gives a condition that is necessary and sufficient for a differential operator
F(D) _
0
a0Dn to be applicable to all integral functions y(x) that f==0
are of the normal type of the order a > 0.
Lemma i. I/ the function I (x) is an integral function of the order a, then (1)
lim {n!a+i fl
f(n)(x) J}n
=0
00
for every finite value of x and for every positive 6. 39
If a > 0, then fInI(x) J}+ = 00
lim sup (n!Q-6
(2)
n-.00
for every finite value of x and for every 6 satisfying 0 < 6 < a. Proof. As the function I (x) is an integral function of the order a, the function g(1=) = I (x + ) is for all finite values of x an integral function of , which is also of the order a. With the help of the well-known expansion 00
g(" (0)
00
g() = n=o
(3)
(n'(x)
n
it follows from the formulae (P, 4) and (P, 5), with an = e = a, that the assertions of this lemma are correct. Lemma 2. zero, then
{(n)x) !-\and n!
If the function I (x) is an integral function of the order lim {n!A I /((x) 1}7n = 0
(4)
n-00
for every finite value of x and for every finite real number A. Proof. For A > - I the assertion follows immediately from formula (I) of lemma 1, with a = 0. Then as a matter of course it holds for A S - I so that formula (4) is correct for every finite value of x and for every finite real number A.
Lemma 3. If the function 1(x) is an integral function of the normal type, of the normal type A, and of the minimum type, respectively, of the order a > 0, then for every finite value of x the following formulae hold respectively (5)
1 -1
0 < lim sup {n!a
-I-
f(n)(x) 1}n < oo,
n-.oo
1_1
lim sup {n!u
(6)
1
I f(n'(x) 1}7 _ (a),)a,
n-.o0
1_1
lim {n!
(7)
0.
n-CO
Proof.
We again make use of the function
which we introduced
in the proof of lemma 1. Then formulae (5), (6) and (7) follow from (3) combined respectively with (P, 8), (P, 9) and (P, 7), with an = n!-1 I f(n)(x) 40
1,
Lemma 4. If the function f(x) is an integral function not exceeding the normal type of the order a > 0, then
0 S lim sup {n!a -1 I f(n)(x) I)n < oo
(8)
nyco for every finite value of x.
Proof. According to definition 2 there are three possibilities with respect to the function I(x): The first possibility is that the function I (x) is of the normal type of the order a. Then formula (5) of lemma 3 holds. Hence formula (8)
is now correct. The second possibility is that the function I (x) is of the minimum type of the order a. Now formula (7) of lemma 3 holds. Formula (8) is now also correct. The third and last possibility is that the function f(x) is of the order
w < a. Now we choose a number 6 satisfying 0 < 6 < a - w. For this number 6 we have 1
co+a
9)
1
a
According to lemma I we have now for every finite value of x and for every positive 6, and therefore also for the number 6 which was chosen by us, 1
Jim {n! aye -` f (n)(x)}21 = 0.
(10)
n-boo
From (9) follows -1 I f(n)(x) I}n Z {n !Q -1 I f(-)(x)
n
and from this formula and from (10) we see that 1 -1
Jim {n!a
1
I f(n)(x) I}"= 0.
n-t oo
This means that formula (8) also holds now. Lemma g. If the function I (x) is an integral function not exceeding the normal type A of the order a > 0, then (11)
0
Jim sup {n! a -1
I /(n) (X)
I}n
(al)1
n-.oo for every finite value of x.
Proof. According to definition 3 there are four possibilities with respect to the function f (x) : 41
The first possibility is that the function f (x) is of the normal type 2 of the order a. In this case formula (6) of lemma 3 holds, so that (11) is correct.
The second possibility is that the function I (x) is of the normal type t of the order a, with t < A. In this case lemma 3 tells us that formula (6) holds if in it we replace A by t. Hence we have urn sup {n!__1 I f"">(x) I}n = (at) n-. 1
1
and, as t < A, we have (at) u < (aA)
so that formula (11) is also
correct now.
The third possibility is that the function I (x) is of the minimum type of the order a, while the fourth and last possibility is that the function I (x) is of the order ai < a. The proof of the validity of formula
(11) for these two possibilities runs quite analogous to the proof of formula (8) of lemma 4 for the second and third possibilities mentioned
in that proof. Lemma 6. If the function I (x) is an integral function not exceeding the normal type, less than A, of the order a > 0, then (12)
0
lim sup {n!u
I /(">(x) 1) n < (ad) a
n-. 00
for every finite value of x.
Proof. According to definition 4 there are with respect to the function I (x) three possibilities : The first possibility is that the function I (x) is of the normal type t
of the order a, where t < A. In this case the proof of formula (12) runs analogous to the proof of formula (11) of lemma 5 for the second possibility mentioned in that proof. The second possibility is that the function I (x) is of the minimum
type of the order a, while the third and last possibility is that the function I (x) is of the order w < a. The proof of the validity of formula (12) for these two possibilities runs analogous to the proof of formula
(8) of lemma 4 for the second and third possibilities mentioned in that proof. Lemma 7. If the function I (x) is an integral function not exceeding the minimum type of the order a > 0, then (13)
lim {n!a
for every finite value of x.
42
-1
f(")(x) 1}n = 0
Proof. According to definition I there are two possibilities with respect to the function I(x): The first possibility is that the function I (x) is of the minimum type of the order a, while the second possibility is that the function f(x) is of the order co < a. The proof of the validity of formula (13) for these two possibilities runs analogous to the proof of formula (8) for the second and third possibilities mentioned in that proof.
Lemma 8. If the function I (x) is an integral function not exceeding the maximum type of the order a > 0, then lim {n! a
(14)
-8-1
I f (n)(x) I}n
=0
n-. eo
for every finite value of x and for every positive e. Proof. According to definition 5 there are four possibilities with respect to the function I(x): The first, second and third possibilities are that the function I (x)
is of the maximum type, of the normal type, and of the minimum type, respectively, of the order a. Then the function I (x) is an integral
function of the order a > 0. Hence we have, according to formula (1) of lemma 1, for every finite value of x and for every positive S 1
(15)
lim {n!a+a-1 I f(n,(x) I)n = 0. n-boe
From this formula it follows that (14) holds for every finite value of x and for every positive E. The fourth and last possibility is that the function I (x) is of the order
w < a. Then it follows from formula (1) of lemma 1, in which a is replaced by to, that we have for every positive 6 I
I
Jim {n!w
I f(n)(x) I}n = 0.
Since co < a, it follows that in this case formula (15) also holds. Now formula (14) is also correct for every positive E. Lemma q. 1/ a and A denote two positive numbers and if the numbers a (n = 0, 1, ...) are such that the following formulae hold respectively 1
(16)
0
(17)
0 s lim sup {n!
1
lim sup {n! a I an I } n < oo, 1
1
I an
11n
1
S (aA)a, 43
1
1
1
0 S lim sup {n!a I an In < (aA) a,
(18)
n_+oo 1-
-1
lim fit! a I an 1}, = 0,
19)
n-+oo
then f(x)
n=o
is an integral function not exceeding the normal
type of the order a if formula (Z6) holds ; an integral function not exceeding the normal type A of the order a if formula (17) holds ; an integral
function not exceeding the normal type, less than A, of the order a if formula (z8) holds; and an integral function not exceeding the minimum type of the order a if formula (ig) holds. Proof.
If we put 1
1
lim sup {n!a an 1}n = 1,
(20)
n-+oo
then there are for this number l two possibilities, viz. a) b)
l = 0, l > 0, where
0
(21)
if formula (16) holds, 0 < l S (aA) a
(22)
if formula (17) holds, and
0 < l < (a2)
(23)
if formula (18) holds.
ad a) Now we have. 1
urn (n!u n- M
1
a,, I}
=0
and from this formula it follows that I (x) has the property P 8, with
C = a- Hence f(x) is an integral function of finite order and this 1
.
order does not exceed a. If the function I (x) is of the order a, it follows
from (P, 7) that it is of the minimum type of the order a. Consequently the function f(x) is either of the minimum type of the order a, or its order is less than a. So according to definition I the function I (x) does not exceed the minimum type of the order a if formula (19) holds. ad b). Now we use formula (21) if (16) holds, formula (22) if (17) holds and formula (23) if (18) holds. 44
In each of these three cases there exists to the number Z one and only one positive number t, which satisfies 1
(at) 4 = 1.
(24)
From (20) and from (21), (22) and (23) respectively, it follows that we have for every positive 6 I_
i
lim{n!ala.J}n=0, and for every a, that satisfies 0 < 6 < a, i
1
lim sup {n!aa J an J}%i = 00. n-.00
This means that I (x) possesses the property P 2. Consequently I (x) is
an integral function of the order a and, because of (20), (24) and (P, 9) with e = a, r = t, it is an integral function of the normal type t of the order a. So, in case where (16) holds, the function I (x) is an integral function of the normal type of the order a. In case where (17) holds,
it follows from (22) and (24) that the number t does not exceed A. Hence the function I (x) is now an integral function of the normal type t of the order a, where t 5 A. In case where (18) holds, it follows from (23) and (24) that the number t is less than A. The function /(x) is now an integral function of the normal type t of the order a, where t < A. In case where formula (16) holds, it follows from ad a) and ad b) that I (x) is an integral function, which is either of the normal type of the order a, or of the minimum type of the order a, or of an order which is less than a. According to definition 2 in each case the function /(x) does not exceed the normal type of the order a.
In case where formula (17) holds, it follows from ad a) and ad b) that /(x) is an integral function, which is either of the normal type A of the order a, or of the normal type of the order a, where this type is less than A, or of the minimal type of the order a, or of an order which is less than a. According to definition 3 in each case the function /(x) does not exceed the normal type A of the order a.
In case where formula (18) holds, it follows from ad a) and ad b) that fix) is an integral function, which is either of the normal type of the order a, where this type is less than A, or of the minimum type of the order a, or of an order which is less than a. According to definition 4 in each case the function f(x) does not exceed the normal type, less than A, of the order a. 45
Lemma zo. I/ the number a is positive and if there exists to the numbers an (n = 0, 1, ...) a positive number 6, such that lim sup {n!a
0
(25)
+e
1 an 1}n < oo,
n-.eo 00
anxn is an integral function, the order of which is less
then f(x)
than a.
n-0
Proof.
If formula (25) holds, we have for e = J8 1
lim {n ! a n-00
(26)
1
a I)- - 0.
We write (26) in the form lim{n!
1+as a
1
n->oo
from which we see that I (x) possesses the property P 8 with C =
I + as
a Consequently I (x) is an integral function of finite order, which order does not exceed a
Because of S > 0, we have e > 0 and
a
1 + as
1 +- ae < a. Hence the order of the function I (x) is less than a. Lemma i z. If the number a is positive and if, for every positive S, the numbers an satisfy the relation lim {n!a -°1 an I In = 0,
(27)
n-+00 00
then f(x) =
n=0 not exceed a.
is an integral function, the order of which does
As formula (27) holds for every positive S, it is also correct
Proof.
for a number b = 81 satisfying 0 < S1 < -I -. We now write formula a (27) with d = bl in the form 1-ad,
1
lim {n!-'a- Ian 1}+L = 0. n-.oo
From this formula it follows that I (x) possesses the property P 8 with
C=
I -a6,-. Hence I (x) is an integral function of finite order, which a
order does not exceed
46
a
I - a61
I
As we may choose 81 arbitrarily in the interval 0 < 61 < - and a
/(x) does not depend on the choice of Sl in this interval, the order of the function f(x) does not exceed a = a. lim a,-.o 1 - a61 Lemma 12. C o n d i t i o n s: The numbers d (n = 0, 1, ...) are all Z 0 and they are such that for every positive 6 the formula 1
lim sup {-n!" d,,} = oo
(28)
n .oo
holds.
possesses a sub-sequence Assertion: The sequence (k = 0, 1, ...) which has the following three properties : 1. all numbers d,,k are > 0; 1
2.
lim
-.t exists 1) ;
k-. o0
3.
for every positive number 6 is 1
(29)
lim {n,!' d,, k} nk = 00. k-.oo
Proof.
With respect to 1
I = lim sup
n
n-.oo
we consider two cases, viz, a) 0 < l 5 oo, and fl) l = 0. In case a) the sequence possesses a sub-sequence {dnk}, so that dnk > 0 for k = 0, 1, ... and 1
lim (d,yk) hk = 1. k-.oo
As we now have l > 0, it is clear that for this sequence (29) holds for every positive number 6. In case fl) l = 0; so we have
formula
1
(30)
lim
n = 0.
n- CO
Now it follows from formula (28) that the sequence has the property that infinitely many of its elements satisfy the inequality do > n!_ k, 1) Even if this limit is oo, we say that it exists. 47
where k denotes an arbitrarily chosen, fixed positive integer. In fact,
if this should not be so, there would exist a positive integer Nk, such that for every it 1,'k the element d would satisfy 1
d
k.
n!
Consequently 1
1
1
(n ! i.,
{n !
1
1
k) it = 1
for every is Z Nk. Hence formula (28) was not correct for S = --k which contradicts the fact, that (28) holds for every S > 0. We now construct a sequence {dnk} as follows: In the sequence the element is the element with the smallest
subscript, for which do > n!-1. If k Z I the element dnk is in the sequence the element with the smallest subscript > nx_1, for which d,, > n! k+1. We now assert, that the sequence {dnk}, constructed in this way, possesses the three properties mentioned in the lemma. It is clear
that the numbers dnk are all positive. We also know that for the sequence formula (30) holds. This formula then also holds for each infinite sub-sequence and consequently also for the sequence In order to show that formula (29) holds for every positive number S, we choose the positive number S arbitrarily and keep it. Let h be that positive integer that satisfies
(31)
Then we have for each positive integer k
It
1
1
i > (nk!)
d,tx > (nk!)
h.
Consequently 1
{nk!a
1
{nk!+
1
'k}>tk
(k Z h).
Because of (31) it follows from this formula, that for the fixed number b I
lim {nk!8 k-. Co
dnk},zk =
co.
As we may choose the positive number d arbitrarily, lemma 12 is
proved.
Lemma 13. C o n d i t i o n s: The numbers d,, (n = 0, 1, ...) are 0 and they are such that for every finite real number v, not de-
all
48
pending on n, the formula (32)
lim sup {n!'d}W
_ 00
n- W
holds.
Assertion: The sequence
possesses a sub-sequence
(k = 0, 1, ...) which has the following two properties : 1. all numbers d,,,k are > 0; 2. for every finite real number v, not depending on n,,, is I
(33)
d,,,,F-., = oo.
lim {n,.!
From the conditions of this lemma it follows, that, for every fixed non-negative integer k, the sequence contains infinitely many elements, which satisfy the inequality d > n!k. In Proof.
fact, if this should not be so, there would exist a positive integer Nk, such that for every n Z Nk the element do would satisfy the inequality d S n!k. As a consequence formula (32) would not hold for v = k, which contradicts the fact, that (32) holds for every finite real number v
not depending on n. Now we construct a sequence
as follows: In the sequence {dn} the element d,,0 is the element with the smallest
subscript, satisfying d > WO = 1. For k Z 1 the element dnk is in the sequence the element with the smallest subscript > nk_I, for which is d,, > n!k. Now this sequence {d.flk} has the two properties mentioned above. Evidently all elements dnk are > 0. Now let v be an arbitrarily, but fixed, chosen positive number. Let then h be that positive integer which satisfies
h - 1 S v < h.
(34)
Then we have for every positive integer k z h and consequently I
I
{nk!-P
}nk
(k Z h).
From this and from (34) we see that for the fixed number v we have I
lim
oo.
k_oo
As we may choose the positive number v arbitrarily, we thus find that formula (33) holds for every positive number Y. Then it holds a fortiori for every finite real number v not depending on n.
A necessary and sufficient. condition for the differential
Theorem z.
00
operator F(D) = 7, anDn to be applicable to all integral functions y(x)
n0
that are of the normal type z of the order a > 0, is, that the numbers an are such that (35)
G(z)
_
ai zn n=0 yyla
defines an integral function not exceeding the normal type, less than i
(o r) a , of the order 1.
I/ the numbers an have this property, then the differential operator F(D) is not only applicable to all integral functions y(x) of the normal type r of the order a > 0, but also to all integral functions y(x) not exceeding the normal type r of the order a 2).
In case a > 1 the necessary and sufficient condition mentioned above is equivalent to the condition that the numbers an are such that 00
the generating power series
anzn defines an integral function not n=o
exceeding the normal type, less than (a - 1) (ra°)11-° of the order
or
a - 1
In case or = I the condition mentioned above is equivalent to the condition that the numbers an are such that the generating power series 00
Y. anzn defines a function which is analytic for jzI S r 3). n=0 REMARK 4).
In case 0 < a < I there exist differential operators
2) This means therefore, that the condition mentioned that is necessary and
sufficient for the differential operator F(D) to be applicable to all integral functions y(x) of the normal type r of the order a, is sufficient for the differen-
tial operator F(D) to be applicable to all integral functions y(x) that do not exceed the normal type r of the order or. However, this condition is not only sufficient but also necessary for the differential operator F(D) to be applicable to all integral functions y(x) that do not exceed the normal type r of the order a. This follows immediately from the fact that this condition is necessary for the applicability to all integral functions belonging to a sub-class of the class of all integral functions y(x) not exceeding the normal type r of the order a, namely the sub-class of all integral functions y(x) of the normal type r of the order a. 3) See footnote °) on page 21. 4) The proof of what we assert in this remark follows on the proof of theorem 1. 50
F(D) =
a,Dn that are applicable to all integral functions y(x) of n=o
00
the normal type z of the order a, but whose generating power series
anzn
n=0
does not define an analytic function; in fact, this power series diverges for every value of z 0. From this it follows that in this case it is not possible to give analogous equivalent conditions as in the cases a > 1 00
and a = 1. However, the power series
anzn does not diverge "too n=0
rapidly" since the expression G(z) in (35), associated with this power
series, has the property stated in theorem 1. For each value of a with 0 < a < I and for each value of z with 0 < z < oo there exists a non-enumerable infinite set of such "divergent" operators. Proof of theorem r. First we shall prove that the condition is a sufficient one. We therefore assume that the expression (35) defines
an integral function not exceeding the normal type, less than (or)-", of the order 1. For this function G(z) it follows from lemma 6, in l
which I (x) is replaced by G(z), .l by (az) a and a by 1, that we have l
l
0 5 lim sup I G(")(z) In < (az) o
(36)
n-> o0
for every finite value of z. For z = 0 formula (36) gives, because of G(n)(0) = rltl- aa" (37)
a
lim sup {n! l
0
n
az
l
n-* 00
If the function y(x) is an integral function of the normal type z of the order a > 0, lemma 3 gives (38)
lim sup {n! Q - l I y(n)(x) 1}n = (az)
1
n-.00 for every finite value of x.
Now it follows from (37) and (38) that we have 1
(39)
_
l
l
lim sup I any(n)(x) In < (az) a (az) Q = 1. n-+ 00 00
Consequently the series n=U
any(")(x) converges for every finite value
of x so that the condition is a sufficient one.
Now we shall prove that the condition is also a necessary one. 51
Let the differential operator ¢(D) =
n=0
anDn be of such a nature that
the numbers an have the property that the expression G(z) in (35) does not define an integral function not exceeding the normal type, 1
less than (a-r) a, of the order 1. Then we have lim sup {n!
(40)
1-1a
1
1
J an J)n Z (vrr) a.
n-00
In fact, if formula (40) should not hold, we would have (37). Then it would follow from lemma 9 (in which we replace an by ann! _
_
00
1
1
and a by 1) that G(z) _ ann ! zn is an integral 0 _ function not exceeding the normal type, less than (oa)a, of the A by (ar)
1
order 1. Since this is not true formula (40) holds. From (40) it follows that the sequence a0, a,, ... contains an infinite sub-sequence a,,, anl, ..., having the following two properties: 1. all numbers ank are different from zero, 2. the formula (41)
Jim {nk!
I_1 a
k- o0
_1
1
J ank J}nk Z (or)
holds. To each of the numbers ank of the sequence {ank} there corresponds a number cnk, which will be defined as follows:
If in (41) the sign of inequality holds, that is to say if _1 (42)
_1
1
lim {nk!1 k- w
ank J}nk > (oT)
then nk
(43)
(ar)
C nk
nk!
(k = 0, 1, ...);
1_, a Jan,
if in (41) the sign of equality holds, that is to say if (44)
then (45)
00
1_1_
Jim {nkr k-. 00
cnk= 1
I
1
a I an, J}nk = (a r)
I a
(k=0, 1, ...).
From this definition of the numbers cnk it follows that the series
k=0 C,,
diverges. This is at once clear in the case in which formula (44)
holds so that Cnk is defined by (45). In the case in which formula (42) 52
holds it follows from (42) that there exists a positive integer K, having
the property that for all positive integers k Z K the following inequality holds {nkl
a
I
ank
I}nk > (ar) a.
From this formula and from (43) we see that the numbers cnk satisfy the inequality 0 < cnk < 1 for all positive integers k Z K, so that the series
is now also divergent.
k-0 Cnk
By means of the numbers cnk and ank we now define the numbers bnk as follows: 1
bnk =
(46)
Mk
(k = 0, 1, ... ).
! Cnkank
With the help of these numbers bnk we form the expression 00
(47)
?1 (x) =
bnkx"k.
k=0
From (46) we see that we then have I
bnk I
Mk
a
nk . Cnk
ank I
-
1
{nk!1
a Cnk 1 ank I} 1.
and from this formula and from respectively (43) and (44), (45) together it follows that we now obtain i (48)
i
i
lira {nk I o I bnk I } nk = (ar) o k-.oo
As a consequence of (48) we have for every positive a i
i
lira {nk!o±o I bnk 1}nk = 0 k-+oo
and for every 6 which satisfies 0 < 6 < a lim {nk!-aI bnk I}nk = 00.
kyao
This means that the expression 71(x) in (47) has the property P 2
with o = a. It defines therefore an integral function of the order a. Then it follows from (48) and from (P, 9) with e = a and y = r, that the function -q(x) is an integral function of the normal type r of the order a.
53
To this function n(x) we now apply the differential operator O(D) _
anDn. Then we find for x = 0 n=0
{
00
00
n,r!anlbnk =
k=0
n=0
---
(because of (46)).
k=0 Cnk
As we have already proved that this series diverges, it follows from
definition 11 that the differential operator is not applicable to the function fi(x). This means that the differential operator #(D) is not applicable to all integral functions that are of the normal type r of the order a > 0. Hence the condition is also a necessary one. We now suppose that the function y(x) is an integral function not exceeding the normal type t of the order a > 0. Then it follows from lemma 5 with I (x) = y(x), a = a, ?. = t, that for every finite value of x we have y(n)(x) 1}n 5 (at)a .
lim sup {n!a
(49)
n-.00
If the numbers an have the property mentioned in the theorem, then formula (37) holds and from this formula and from (49) we see that in this case formula (39) holds also. This means that the differential CO
operator F(D) = T, anDn is also applicable to all integral functions n=0
not exceeding the normal type t of the order a > 0. We next suppose a > 1. If the numbers an are such that the expression
(35) defines an integral function not exceeding the normal type, less _
than (at)
I
,
then formula (37) holds. We write this formula in the form 1
1
o
n
0 S lim sup jn! a-i
(50)
_
nr m
We now put
1
< (ar)
an
= v. Then we have, as we see after a brief calculation, (a - 1) (ta°) I
(51)
a-1
a
Ca-1
_
1
(at)
V
°
Consequently we may write (50) in the form I
I
(_)-
a- I
(52)
0 S Jim sup in! Q i J an I 1 n
<
fl-±00
Now it follows from lemma 9, with a = 54
a
a-1 a
a-1
and 2 = v, that the
power series
n=0
a, zn defines an integral function not exceeding the
normal type, less than v, of the order
a - 1
Conversely, if the power series I anzn defines an integral function n=0
F(z) not exceeding the normal type, less than v, of the order
a
a- I'
where v is given by formula (51), then according to lemma 6, with I (x) = F(z), A = v and a =
a
1
, we have for every finite value of z
-1 0
lim soup {n! o
a
1
I F(-)(z) 1}' < (a
v)Q
-1 Q
-1
For z = 0 this formula gives formula (52) in virtue of Fn)(0) = n!an.
On account of (51) formula (50) follows from (52). Consequently, formula (37) holds. This means, as we showed in the reasoning that followed formula (40), that the expression G(z) in (35) defines an 1
integral function not exceeding the normal type, less than (6a) of the order 1. With this the statement concerning case a > I is proved.
Finally we suppose a = 1. If the numbers an are such that the expression (35) with a = 1, defines an integral function not exceeding
the normal type, less than z-1, of the order 1, then formula (37) with a = 1 holds, i.e. we have 1
(53)
lim sup I a,, I n < z-1 n-oo
Then it is clear that we have (54)
lim inf n-.00
Ianln
>z
and from this it follows that the radius of convergence of the power 00 series I anzn exceeds z 5). Consequently this power series defines n=0
a function that is analytic for I z
z. 00
6) As is well-known the radius R of a power series E Cnzn is given by the formula n=n (A)
R = lim inf
-
1
. ICnln From this formula it is easy to deduce that also
(B)
R=_
1..
lira sup I cn In n-00 Cf. Pringsheim [3], Bd IL 1, p. 244. 55
00
anxn defines a function that is
Conversely, if the power series n-0
analytic for I z 15 r, its radius of convergence exceeds T. Then it follows from (A) that formula (54) holds. Hence (53) holds and this means that (37) with a = t is valid. From this we may deduce with the help of the reasoning that followed formula (40), that the expression G(z)
in (35) with a = 1, defines an integral function not exceeding the normal type, less than r-1, of the order 1. This proves theorem I conclusively.
1-
Proof of the remark following theorem i.
We take an = n! 0 l,hn,
(ar)_
where 0 < f <
.
Then we have
lanl lim A--.00
In! -iyZ f
_1
1
n = p < (ar) 0,
o
whereupon it follows from lemma 9 (formula (18)) with a = I and 1
A = (ar)
that in this special case the expression G(z) in (35) defines 1
an integral function not exceeding the normal type, less than (or) of the order 1. Since the differential operator 00
F(D) =
1
_
n!Q
p^D-
n=°
fulfils the conditions of theorem 1, it is applicable to all integral functions of the normal type r of the order a. Its generating power series 00
1
00 n! a -1 finzn is on account of 0 < a < 1 divergent for every value of n=o
z 0 0. It is clear that for 0 < a < 1 there exists a non-enumerable infinite set of such "divergent" operators. To see this one need only replace the coefficient of D° in the differential operator just mentioned by an arbitrary constant.
Theorem z.
A necessary and sufficient condition for the differential 00
operator F(D) _
a,,D' to be applicable to all integral functions y(x) =o
that are of the minimum type of the order a > 0, is, that the numbers a7, are such that (55)
G(z) _ 2 and Z. n=o n!a
defines an integral function not exceeding the normal type of the order 1. 56
If the numbers an have this property, then the differential operator F(D) is not only applicable to all integral functions y(x) of the minimum type of the order or > 0, but also to all integral functions y(x) not exceeding the minimum type of the order a 6). In case a > I the necessary and sufficient condition mentioned above is equivalent to the condition that the numbers an are such that the ge00
anzn defines an integral function not exceeding the
iterating power series n=0
a
normal type of the order
a
1
.
In case a = I the condition mentioned above is equivalent to the condition that the numbers an are such that the generating power series
0 I anzn defines a function that is analytic at z = 0').
n=0
In case 0
REMARK 8). 00
F(D) = 7, anD' that are applicable to all integral functions y(x) of the n=0
Co
anzn
minimum type of the order a, but whose generating power series n=0
does not define an analytic function; in fact, this power series diverges for every value of z 0 0. See the similar remark following theorem 1.
Proof of theorem z. First we shall prove that the condition is a sufficient one. We therefore assume that the expression (55) defines an integral function not exceeding the normal type of the order 1. For this function G(z) it follows from lemma 4 where I (x) is replaced by G(z)
and a by 1, that we have 1
0 S lim sup I G(n)(z) In < 00
(56)
n-Foo
for every finite value of z. For z = 0 formula (56) gives, because of 1
G(n)(0) = n!1 a am (57)
0
1im sup {n!1
1 an !}n < oo.
n-. oo
If the function y(x) is an integral function of the minimum type of 9) That the condition mentioned is not only sufficient but also necessary for a differential operator F(D) to be applicable to all integral functions y(x) that do not exceed the minimum type of the order a(a > 0), may be proved in an analogous way as in the similar case in theorem 1. See footnote 2). 7) For the notion "analytic at z = 0" see footnote 9) on page 21. 8) The proof of this assertion follows the proof of theorem 5. 57
the order a > 0, lemma 3 gives (58)
-1
lim {n! -a
I y(11) (x) I}n = 0
n-00
for every finite value of x.
Now it follows from (57) and (58) that we have 1
lim I any(n>(x) In = 0.
(59)
n-+00 W
Consequently the series
a,,y(n)(x) converges for every finite value of n=o
x so that the condition is a sufficient one.
Now we shall prove that the condition is also a necessary one. 00
anDn be of such a nature no that the numbers an have the property that the expression G(z) Let the differential operator #(D) =
in (55) does not define an integral function not exceeding the normal type of the order 1. Then we have (60)
lim sup {n!
I an
11n
= oo.
In fact, if formula (60) should not hold, formula (57) would be true. Then it would follow from lemma 9 (in which we replace an by ann! 00
1
and a by 1) that G(z) _ 7, ann!
Zn is an integral function of at most
n=o
the normal type of the order 1. Since this is not true formula (60) holds.
From (60) it follows that the sequence ao, a1, ... contains an infinite sub-sequence an., an,, ... (no Z 2) having the following two properties: 1. all numbers ank are different from zero, 2. the formula (61)
lim {nk!
1_.1
k 00
a
1
I ank I}n'k = 00
holds.
To each of the numbers ank of the sequence {ank} we associate a number cnk. The choice of these numbers cnk depends on the be-
haviour for large values of k of the numbers nk!1 namely have either a) the property that the formula
numbers nk!1 (62)
a I ank I
lim {nk!1 k-00
holds for every positive e, 58
-` I ank I} nk = 0 1
a
I ant I. These
or
b) the property that formula (62) does not hold for every positive e. Now we define the numbers cnk as follows:
In case a) we take
(k = 0, 1, ...);
cnk = 1
(63)
while in case b) we take (log nk)"k
cnk=-- 1 --
(64)
(k = 0,1,...).
a I ank
nk!
In case a) it follows immediately from the definition of the numbers cnk and from (62) that we have _
(65)
hm {nk!1 k-.oo
1
1
a _ECna
a.., I}nk = 0
for every positive e. And since the numbers positive it follows from (65) that we have lim {nk!
(66)
1-ia -e
cnk I ank I}
cnk
and
I ank I
are all
nk = Op 1
k-e00
for every positive e. In case b) it follows from (64) 1_ 1 _6
(67) fn,!- a
1
cnk
I
1
e
ank i}nk = {nk!-` (log nk) 1k}nk = nk! nk log Mk-
Consequently formula (65) also holds in case b) for every positive
Since no Z 2, we have cnk > 0 for k = 0, 1, ..., so that from (65) and (67), in connection with (P, 15), it follows that formula (66) also holds in case b).
We now assert that it follows from the definition of the numbers cnk that the series (68)
k=0 Cnk
diverges. This is at once clear in case a) considered above, where cnk = 1. In case b) considered above where, as we know, formula (62) does not hold for every positive e, the numbers cnk are given by (64).
This means that we may write the series (68) in the form 1--1
a
00
(69)
k=0
nk!
I ank
(log nk)'"k
59
Since formula (62) does not hold for every positive s, there exists a positive eo, such that we have
I-
lim sup {nk!
° -2eo
I}nk > 0.
I
k-.
Obviously we then have lim sup {nk !
1-
1
°
-BU
I } nk = 00
Ic o°
and, as a consequence, we certainly have I} = 00.
lim sup {nk!1 Q -e0 I
(70)
k-+oo
Now it follows from (70) that the inequality k! -n,!
1- 1a -e o
I>1
I
is satisfied for infinitely many values of k. For those values of k we have therefore 1
nk! 1- ° I a., I
>nk!eo
and from this formula we see that the inequality nk 1
1
__I °
I ank I > (log
nk)'nk
holds for infinitely many values of k. From this it follows, in connection with (P, 15), that the series (69) and therefore also the series (68) diverges. By means of the numbers and a,,,, we now define the numbers b
nk
=
I
(k=0,1,...).
nk !C. k a, k
With these numbers
we form the expression 00
(72)
77(x) 0
In case a) it follows from (71) and (63) that we have 1
(73)
1
' °° I' bIlk b I}nk
(
1 -I
I
1
}Jf yyZ1
c.n k
I
I ank
Now we see from (61) that we have
I
Ink _ {nkl J
I- I lim {nk! k-oo 60
a.., I}
1 nk
= 0.
1-
_
1
a
I a nk
I}
_1
Ilk .
From this formula and from (73) it follows that the formula 1
1
lim Ink- I bnk I}nk = 0
(74)
k'-O0
holds. By virtue of (71) we further have in case a) l_
I
I
1
bnk I}nk = Ink 11
fnk 1 a
cnk I ank I}
a
1
nk
and from this formula and from (66) it follows that +e
lim {,n,!
(75)
bnk I}nk = 00
k-oo
holds for every positive e. In case b) it follows from (71) and (64) that 1
1
{nk!a I bnk I}nk = {nkI 1 - "a cnk I ank
_
1
nk = (lognk) -1
As a consequence, formula (74) holds also in case b). Besides we have in case b) {nk!a
8 I bnk
1
s
I}nk = nk. nk (lognk)-I,
from which we see that in this case also formula (75) holds for every positive E.
Thus we see that for the coefficients bnk of the power series in (72)
formula (74) holds both in case a) and in case b). Consequently in both cases formula 1
(76)
1
lim Ink! a+a I bnk I}'nk = 0 k-+oo
holds for every 6 z 0 and therefore certainly for every b > 0. Moreover, we saw that both in case a) and in case b) formula (75) holds for every positive e. Consequently we have in both cases for every number 6 that satisfies the inequality 0 < 6 < a, 1
(77)
1
lira {nk!a-e I bnk I}nk = 00. k-*oo
Now it follows from (76) and (77) that the expression in (72) has the property P 2 with @ = a, and this means that this expression defines an integral function of the order a. Then it follows from the formulae
(74) and (P, 7) that the function n(x) is an integral function of the minimum type of the order a. 61
To this function 77(x) we now apply the differential operator 00
(D) _ 7, a1Dn. We then find for x = 0 n=0 00
00
00
ann(n)(x)}x=0 = 2: nk!a,Lkbfyk 12: k=0 k=0 n-0
1
(because of (71)).
Cnk
As we have already proved, this series diverges so that, according to definition 11, the differential operator ¢(D) is not applicable to the function 71(x). This proves that the differential operator c(D) is not applicable to all integral functions of the minimum type of the order a > 0. Consequently the condition is also a necessary one. We now suppose that the function y(x) is an integral function not exceeding the minimum type of the order a > 0. Then it follows from lemma 7, with I (x) = y(x) and a = a, that we have lim {n!a - i I y(n)(x) I}n = 0
(78)
n-.00 for every finite value of x.
If the numbers an have the property mentioned in the theorem, then formula (57) holds. From this formula and from (78) it follows that formula (59) now also holds. This means that the differential 00
operator F(D) =
n=0
anDn is also applicable to all integral functions
y(x) that do not exceed the minimum type of the order a > 0. We next suppose a > 1. If the numbers an are such that the expression (55) defines an integral function not exceeding the normal type of the order 1, then formula (57) holds. We write this formula in the form 1
1
f
(79)
0 S limsupln!Q al
an
Jf In
< oo.
n-.00
Then it follows from lemma 9 with a =
a-a 1 that n=0 anzn defines
an integral function not exceeding the normal type of the order 00
Conversely, if the power series
a- I
anzn defines an integral function n=0 a F(z) not exceeding the normal type of the order , then we have, a according to lemma 4, with /(x) = F(z) and a = a , for every a-1 finite value of z (80) 62
0 S lim sup {n!
1-1 I F('(z) j}n < oo.
For z = 0 this formula gives formula (57) in virtue of F(n)(0) = n!an.
-I
Lemma 9, in which we replace an by ann! Q and a by 1, then says that the expression (55) defines an integral function not exceeding the
normal type of the order 1. Herewith the statement concerning the case a > 1 is proved. Finally we suppose a = 1. If the numbers an are such that the expression (55) with a = 1, defines an integral function not exceeding the normal type of the order 1, then formula (57) with a = I holds, i.e. we have 1
0
(81)
urn sup a, n-00
This clearly shows that 1
lim inf
(82)
n-+oo
I
1>
0
an in
and from this it follows, in connection with formula (A) on p. 55, that 00
the radius of convergence of the power series I anzn is positive. This n=0
power series defines therefore a function F(z), that is analytic at
z=0.
00
anzn defines a function that is
Conversely, if the power series n=0
analytic at z = 0, its radius of convergence is positive. Then it follows from (A) that formula (82) holds. Hence (81) holds and this means
that formula (57), with a= 1, is valid. From lemma 9, in which we replace an by n!-'an and a by 1, now follows that the expression (55),
with a = 1, defines an integral function not exceeding the normal type of the order 1. This proves theorem 2 conclusively. Theorem 3.
A necessary and sufficient condition for the differential 00
operator F(D) = Y, anDn to be applicable to all integral functions y(x) n=0
that are of the maximum type of the order a > 0, is, that the numbers an
are such that (83)
G(z)
°°
n-0
a
n zn 1
n!Q
defines an integral function the order of which is less than 1. If the numbers an have this property, then the differential operator F(D) is not only applicable to all integral functions y(x) of the maximum 63
type of the order a > 0, but also to all integral functions y(x) not exceeding
the maximum type of the order a > 0 9). In case a > I the necessary and sufficient condition mentioned above is equivalent to the condition that the numbers a are such that the generating power series 2,
defines an integral function the order of
n=0
a
which is less than - -
a - I
In case a = I the condition mentioned above is equivalent to the condition that the numbers an are such that the generating power series
defines an integral function of finite order. n=0 REMARK 10).
In case 0 < a < 1 there exist differential operators
00
that are applicable to all integral functions y(x) of
F(D) _ 11-0
the maximum type of the order a, but whose generating power series 00
n-0
does not define an analytic function; in fact, this power series
diverges for every value of z - 0. See the similar remark following theorem 1. Proof of theorem 3. First we shall prove that the condition is a necessary one. We therefore assume that the expression (83) defines an integral function the order of which is less than 1. Then this function G(z) has the property P I so that, if its order is equal to a (so 0 5 ,u < 1), we have (84)
lm ln!a+a
a
ait }nl - 0
ytia
1J
for every 6 > 0. If the function y(x) is an integral function of the maximum type
of the order a > 0, then of course it is an integral function of the order a > 0, so that from formula (1) of lemma 1 follows that (85)
lim {n!a
I y(n)(x) }n = 0
for every finite value of x and for every positive e. 9) That the condition mentioned is not only sufficient but also necessary for a differential operator F(D) to be applicable to all integral functions not
exceeding the maximum type of the order a(a > 0), may be proved in a way analogous to a similar case in theorem 1. See footnote 2). 10) The proof of this assertion follows the proof of theorem 5. 64
Now it follows from (84) and (85) that we have 1 + 1 -L-1
1
(86)
lim {n!A+a - a
a
1
I any(n)(x) I}n = 0
n+oo
for every finite value of x and for every choice of the positive numbers
6ande. If in this formula we take s = 6, then we see that (87)
any(n)(x) }ii = 0
lim {n!N+a n-oo
holds for every finite value of x and for every positive value of 6. Since the number ,u satisfies the inequality 0 S ,u < 1, it is possible to choose the positive number 6 in such a way that (,u + 6)(6 + 1) = 1. If we have thus chosen the number 6 and if we substitute this number 6
in (87), then we find that we have (88)
lim I any(n)(x) I- = 0 n-. o0 00
for every positive value of x. Hence the series I any(n)(x) converges n=0
for every finite value of x. So the condition is a sufficient one. Now we shall prove that the condition is also a necessary one. Let
a,Dn be of such a nature that
the differential operator O(D) = n-0
the numbers an have the property that the expression G(z) in (83) does not define an integral function the order of which is less than 1. Then we have for every positive 6 (89)
lim sup
{n!1+e- a I
an I}n = 00.
n-.00
In fact, if formula (89) should not hold for every positive 8, there would exist a positive number 61 with the property that (90)
0 S lim sup
{n!1+a1-a I
an I}n
< 00
n-,. 00 ann!_
would hold. Then lemma 10 (where we replace an by
6 by 61
and a by 1) tells us that G(z) _
00
_
1
ann! azn
n=0
is an integral function the order of which is less than 1. Since this is not true we see that (90) does not hold. So (89) is valid. _
From (89) and lemma 12 with do = n!1
1
a I an I
it follows that the 65 5
sequence of numbers ao, al, ... contains an infinite sub-sequence ano, an,, ... (no Z 2) possessing the following three properties: 1. all numbers ank are different from zero, 2. lim {nk!
a I and I Ink exists,
k3. the formula (91)
a
lim {nk!
I
k- w
ank I}nk = oo
holds for every positive number 6. To each of the numbers ank we now associate a number cn The choice of these numbers cnk depends on the behaviour for large values
of k of the numbers nk!1 an0,
i o I ank I.
From property 2 of the sequence
it follows that the following two cases may present
themselves : a)
lim {nk!
(92)
1--
I ank I}nk = 0,
k_oo
b)
urn {nk!
(93)
I ank j}nk > 0.
k-.oo
Now we define the numbers c, as follows:
In case a) we take
(k = 0, 1, ...) ;.
cnk = 1
(94)
while in case b) we take (95)
1
(k = 0, 1, ... ).
cnk = nk!
a I ank I
(log nk) nk
In case a) it follows from the definition (93) of the numbers c,2 that the series 1
(96) C=--o cnk
diverges. In case b) we choose a number l as follows : If the left-hand member of (93) is finite, then we put l equal to that left-hand member. If the left-hand member of (93) is infinite, then we put l equal to an arbitrarily chosen finite positive number. From the manner in which we chose the number 1, it follows that there exists a positive integer 66
K, such that we have
1--a I ank 1
{nk!
for every positive integer k
1
I}nk > 1-1
K.
So for kzK _
nk!1
(97)
1
a
I>
I
Besides there exists a positive integer L with the property that for every positive integer k L the inequality 11 (log nk) > 1 holds. Let M denote the larger of the two numbers K and L. Consequently
(97) certainly holds for every positive integer k za M. From (95) and (97) follows for every positive integer P > M
--1
1
nk!1
2;
'57
>
a
k=0
k=0 Cnk M-1
k=0
+
I ank I (log nk)nk = M-1
P
p
1
{il (1og nk)}"k > k=M
+ k=M
1:=0
nk!1
a
I ank I (log nk)nk +
k=0
1.
k=-4f
From this it follows that in case b) the series (96) also diverges. By means of the numbers cnk and ank we now define the numbers bnk as follows
--
1
bnk =
(98)
(k = 0, 1, ...).
nk !cnka,, k
With these numbers bnk we form the expression 00
21(x) =
(99)
k=0
bnkxnk
In case a) it follows from (98) and (94) 1
(100)
{nk!o
1-1
1
I bnk
1
nk = nkla
1
_ nk = {nk!
1-1
__1
I ank I} nk.
a
I a nk
From this formula and from (92) follows that in case a) we have 1
1
lim {nk!a I bnk I}nk = 00.
(101)
k-.oo
Besides in case a) we see from (100) that we certainly have for every positive b {nk! 1a
1 -d I bnk I}nk 1 = {nk! 1+d- a I ank i}
1
nk
and from this and from (91) we conclude that (102)
lira {Mk! a -° I bnk I } k = 0 k-.m 67
holds for every positive S. So we have 1
1
lim {nk!a+
(103)
b,bk I}nk
=0
k-.oo
for every positive e. In case b) it follows from (98) and (95) {nk!Q I bnk I}nk = jnk!a
(104)
Cnk I a,, k
II 1
a cnk I aRk I}
_ {nk I
nk = log nk,
so that in case b) formula (101) also holds.
Besides in case b) it follows from (104) that certainly for every positive b 1
{nk! a
-a
1
I bnk I )nk = nk!
E
nk log nk,
so that in case b) formula (102) also holds for every positive S. Hence in case b) formula (103) is valid for every positive e.
So we see that both in case a) and in case b) for the coefficients bnk of the power series (99) the formulae (101) and (103) hold. With the help of the same reasoning we used in the proof of theorem 2 11) we now can prove that the expression (99) defines an integral function of the order a. Then it follows from formula (101) and from formula (P, 10) that the function n(x) is an integral function of the maximum type of the order a.
To this function n(x) we now apply the differential operator 00
O(D) = I anDn. We then find for x = 0 n=O 00
oa
00
{ S` a,,(nI(x)}r_0 =Ink' ankbnk = I n=O
k=0
k=0
1
(because of (98)).
Cnk
We have already proved that this series diverges so that, according to definition 11, the differential operator O(D) is not applicable to all integral functions of the maximum type of the order a > 0. Consequently the condition is also a necessary one. We now suppose that the function y(x) is an integral function not exceeding the maximum type of the order a. Then it follows from lemma 8 with I (x) = y(x), a = a, that formula (85) holds for every finite value of x and for every positive e. If the numbers an have the property mentioned in the theorem, 11) There we started from the formulae (74) and (75) and by way of (76) and (77) we showed that (72) defines an integral function of the order a. 68
then formula (84) holds. From this formula and from (85) it follows that formula (88) is valid as we saw in the beginning of the proof of
0 theorem 3. This means that the differential operator F(D) _ I a,D' n=0
is also applicable to all integral functions y(x) that do not exceed the maximum type of the order a > 0. Finally we suppose a 1. If the numbers an are such that the
expression (83) defines an integral function the order of which is less than 1, then formula (84) holds for every positive b. Now if we
put w = I - 2e then, since 0 S ,u < 1, we have 0 < e S J. So llm{n!1-2c+e n-
a
I a. 11-n = 0
holds for every positive b.
If we take 6 = s then it follows from this formula that
Ianl}n = 0.
urn {n !T-
(105)
n-00
Now 1
I -{-e
1 -E
1 -E2
> 1 +e;
from this and from (105) follows (106)
lim {n!1
a
I an I}n = 0.
If a > 1, it follows from (106) by application of lemma 10, with or
a = a _ 11 , b = s, that the power series function the order of which is less than 00
00
anzn defines an integral
n=0
a
a- I
Conversely, if the power series I anzn defines an integral function n=0
F(z) of the order e, where
a
1
,
then F(z) has the property P 1,
so that formula (P, 4) holds for every positive b. So we obtain lim 93-+00
n! a+o +
--all
0
n!a
for every positive b and from this it follows that for every positive 69
6' we have lim
(107)
1+1-a'
n! e
1
a
n= 0.
1
n-.oo
Since t) <
1
l an l
a
n! a 1
we have the inequality - -}-
1
> 1. Now if we put
N
1
- + - = I + 77, then -q > 0 and from (107) it follows that the a formula
i
Em n!i+o-a'
0
n-.oo
nIa
holds for every positive 6'. If in this formula we take S' = .'171, we consequently have a
lim n!1+}°
I
n- oo
i
I
0 < Co.
I
nta 1
From this, according to lemma 10, where we replace an by n! Q an,
a by 1
and S by -,q, it follows that the expression G(z) in (83)
defines+an integral function the order of which is less than
1
1+4?
Hence certainly the order of the function G(z) is less than 1.
Herewith the statement of theorem 3 relative to case a > 1
is
proved.
If a = 1, we write (106) in the form llm {n!2Je n-oo
a. j)n = 0.
+}6 I
From this, according to lemma 10, where we replace a by
2 ,
6 by Js,
E
00
it follows that the power series I anzn now defines an integral function n=0
the order of which is less then
?
.
And as e > 0 we have
8
2
< oo
so that this function is of finite order. 00
Conversely, if the power series 71 anzn defines an integral function n=o
of finite order and if this order is equal to Q, then this function has the property P 1 so that formula (P, 4) holds for every positive a and therefore also for b = 1. Consequently we have lim I n (e+f
n-.oo
70
+1
an I I L= n!
J
0.
If we write this formula in the form 912 +
lim 1n! Q+1
an I
1
2(Q+1)
n-0
n!
n-,oot
and if we apply lemma 10, where we replace a,, by n!-'a,,, a by
and S by
,
1
2(P + 1)
P+1 P + 3/2
then we see that the expression G(z) in (83) with
a = I defines an integral function the order of which is less than P + 1 Hence certainly the order of the function G(z) is less than 1. + 3/2 This proves theorem 3 conclusively. Theorem 4.
A necessary and sufficient condition for the differential 00
operator F(D) _
to be applicable to all integral functions y(x) of n=0
Me maximum type o l the order zero, is, that there exists a finite real number v, depending neither on n nor on z, with the property that (108)
G(z)-zn n=0 n!" a
defines an integral function the order of which is less than 1. REMARK 12).
a,Dn that
There exist differential operators F(D) = n=0
are applicable to all integral functions y(x) of the maximum type of the order zero, but whose generating power series00I anzn does not define an n=o
analytic function; in fact, this power series diverges for every value of 00
z r 0. However, the power series
n=0
does not diverge "too
rapidly", since there exists a finite real number v, depending neither on n nor on z, with the property that the expression G(z) in (108) associated with this power series has the property stated in theorem 4.
There exists a non-enumerable infinite set of such "divergent" operators. Proof of theorem 4. First we shall prove that the condition is a sufficient one. We therefore assume that there exists a finite real
number v with the property that the expression G(z) in (108) defines an integral function the order of which is less than I. Let the order 32) The proof of this assertion follows the proof of theorem S. 71
of this function G(z) be equal to ,u so that 0 S u < 1. The function G(z) has the property P 1, i.e. the formula `Jan J}n = 0
lim {n! µ+a n-oo
(109)
holds for every positive number S. So formula (109) also holds for S = I -,u, so that also lim {n!1-° I an J}n = 0.
(110)
n- oo
If the function y(x) is an integral function of the maximum type of the order zero, then lemma 2 tells us that 1
(111)
J}n = 0
lim {"!A I n-roo
for every finite value of x and for every finite real number A.
As formula (111) holds for every finite real value of A, it also holds for A = v - I and then it follows from (110) and from (111),
with A = v - 1, that 1
Jim I any(n)(x) J n = 0 n-ioo 00
for every finite value of x. Hence the series I any(n)(x) converges n= 0
for every finite value of x so that the condition is a sufficient one. Now we shall show that the condition is a necessary one. Let the
differential operator O(D) =
0
n=0
anDn be of such a nature that the
numbers an have the property that with these numbers an there does not exist a finite real number v, neither dependent on n nor on z, such that the expression (108) defines an integral function the order of which is less than 1. From this it follows that 1
(112)
lim sup {n!-" J an J}n = co,
however large the number v is chosen. In fact, if this should not be so, there would exist a positive number /L, such that 1
lira sup {n!-µ J an J}n < oo. Then (113)
llm n-.oo
72
{n!_µ-1
1
J an }n
0.
From (113) follows lim
r
nj2 ayn 1= 0.
From this formula we see that the expression °° n=O
a
H(z)"
!µ+3 zn
has the property P 8, with C = 2 and an replaced by
a !,+3.
So this
expression H(z) defines an integral function the order of which does not exceed 1. This proves that there did exist a real number v, viz. v =,u + 3, with the property that the expression (108) associated with the differential operator O(D) defines an integral function the order
of which is less than 1. This contradicts the assumption we made with respect to the operator q(D). Hence formula (112) holds for every finite value of v. From (112) and lemma 13 with d,, _ a,, I follows that the sequence
a0, al, ... contains an infinite sub-sequence a"o, anl, ... possessing the following two properties: 1. all numbers a"k are different from zero, 2. for every finite real number v, not dependent on nk, we have i
lim {nk! ° I ank I) nk = oo. k-oo
From these two properties we see that the formula
nk=0
(114) k->oo
holds for every finite real number v, not dependent on nk. With the help of these numbers ank we now define the numbers b"k as follows (115)
bnk =
nk 1 a"k
(k = 0, 1,
... ).
With these numbers b,,k we form the expression 00
b"kink
n(x)
k=0
and we shall prove that this expression defines an integral function of the maximum type of the order zero. To that end we shall first show that q(x) has the property P 3. 73
Let A denote an arbitrarily chosen positive number. Then from (115) follows
_
1
{nk!-4
(116)
I
b,,,, I}nk = {n,E!i-A I ank I}
1
k.
Since formula (114) holds for every real number v, it holds also for
v = A - I and then from (114), with v = A - 1, and from (116) we obtain for every positive number A 1
lim {nk!A I bnk I}nk = 0. k-,co
So r7(x) has the property P 3. Hence 77(x) is an integral function of the order zero. As the numbers ank are all finite the numbers b,,k
are all 0 0, so that the function 77(x) is not identically equal to a constant. Consequently it is an integral function of the maximum type of the order zero. To this function r7(x) we apply the differential operator 00 O(D) _ a1 Dn. We then find for x = 0 n=0
00
00
00
nk!ankbnk=
{I anr7(n)(x)}x=0 =
1
(because of (115)).
k=0
k=0
n=0
Since this series diverges this differential operator q(D) is, according to definition 11, not applicable to the function r7(x). So it does not apply to all integral functions of the maximum type of the order zero. This means that the condition is also a necessary one. This proves theorem 4. Theorem 5.
A necessary and sufficient condition for the differential 00
operator F(D) =
anDn to be applicable to all integral functions y(x) n=0
of the normal type of the order a > 0, is, that the numbers an are such that (117)
G(z)
a
n-0 yila
zn
defines an integral function not exceeding the minimum type of the order 1.
If the numbers an have this property, then the differential operator F(D) is not only applicable to all integral functions y(x) of the normal type
of the order a > 0, but also to all integral functions not exceeding the normal type of the order a 13). 13) That the condition mentioned is not only sufficient but also necessary
for a differential operator F(D) to be applicable to all integral functions not exceeding the normal type of the order a > 0, may be proved in an analogous way as in the similar case in theorem 1. See footnote 2). 74
In case a > 1 the necessary and sufficient condition mentioned above is equivalent to the condition that the numbers a.n are such that the geCO
anzn defines an integral function not exceeding
nerating power series n=O
the minimum type of the order
a a - 1
In case a = 1 the condition mentioned above is equivalent to the con00
anzn dition that the numbers an are such that the generating power series n=0 defines an integral function. REMARK 14).
F(D) _
In case 0 < a < 1 there exist differential operators
anDn that are applicable to all integral functions y(x) of the n=0
normal type of the order a, but whose generating power series
anzn n=0
does not define an analytic function; in fact, this power series diverges for every value of z 0 0. See the similar remark following theorem 1.
Proof of theorem 5. First we prove that the condition is a sufficient one. We therefore assume that the expression (117) defines an integral function not exceeding the minimum type of the order 1. If A denotes an arbitrarily chosen finite positive number, then, according to the definitions 4 and 1, this integral function G(z) is an integral function not exceeding the normal type, less than A, of the
order 1. So, according to definition I with r =
1 a-°, the differential Cr
operator F(D) is applicable to all integral functions not exceeding I
the normal type - A-° of the order a. Since we may choose the positive a
number 2 arbitrarily, the differential operator F(D) is applicable to all integral functions not exceeding the normal type of the order a. This proves the second assertion of theorem 5 at the same time. Now we shall show that the condition is also a necessary one. Let 00
the differential operator ¢(D) = I anDn be an operator of such a n-0
nature that the numbers an have the property that the expression (117) does not define an integral function not exceeding the minimum type of the order 1. Now we choose a finite positive number u. With respect to the expression (117) there are now two possible cases, viz. a) this expression defines an integral function of the normal type v of the order 1 where 0 < v <,u, and fl) this expression does not define an 11) The proof of this assertion follows the proof of theorem 5. 75
integral function not exceeding the normal type, less than u, of the order 1. In case a) G(z) is not an integral function not exceeding the normal type, less than v, of the order 1, as follows from definition 4.
Then from theorem 1, with i = 1 v-°, follows that the differential a operator O(D) is not applicable to all integral functions of the normal
type
1a v-° of the order a. So it is certainly not applicable to all in-
tegral functions of the normal type of the order a. In case P) it follows
from theorem 1, with z =
I a
,u-°, that the differential operator O(D) I
is not applicable to all integral functions of the normal type --,u-° a of the order a. So it neither applies in this case to all integral functions
of the normal type of the order a. Hence the condition is also a necessary one. Next we assume a > 1. If the numbers an are such that the expres-
sion (117) defines an integral function not exceeding the minimum type of the order 1, then from lemma 7, with I (x) replaced by G(z) and a by 1, it follows that we have for z = 0 (118)
lim {n!1 Q I an 1}n = 0. n-.co
From this and from lemma 9 (formula (19)) with a = -
v.--,
a - I
it follows that the power series
anzn defines an integral function n=0
a
not exceeding the minimum type of the order --- . 00
a- I
Conversely, if the power series 7, anzn defines an integral function n=0 a F(z) not exceeding the minimum type of the order then from ' a lemma 7, with a = a and I (x) = F(z), it follows for z = 0, that 1
a-1
formula (118) holds. From this and from lemma 9 (formula (19)), with I _ ' °° ' - replaced by I and an by n! ° an, follows that G(z) _ n! o anzn n=0 a is an integral function not exceeding the minimum type of the order 1.
Finally we assume a = 1. If the numbers an have the property that the expression (117), where a = 1, defines an integral function not exceeding the minimum type of the order 1, then formula (118)
with a = I holds, so that (119) 76
limIa,, In = 0.
Then I
(120)
lim i n~00 Ian ln
= 00
and from this it follows, in connection with formula (A) of page 55, that Co
anz" is 00. Hence this the radius of convergence of the power series power series defines an integral function. n=0 Co
anz" defines an integral function,
Conversely, if the power series n=o
then formula (120) holds. Then formula (119) also holds i.e. formula (118) with a = I is valid. This means that the expression (117) with a = I defines an integral function not exceeding the minimum type of the order 1. This proves theorem 5.
Proof of the remarks that follow the theorems 2, 3, 4 and 5. In the cases occurring in the theorems 2, 3 and 5 let the number a be equal to a number al, where 0 < al < 1. Then there exists a
number p such that al < P < 1. Besides, let r denote a positive number. Now it follows from the remark following theorem 1 that there exists a non-enumerable infinite set of differential operators
F(D) = anDn whose generating power series anz" diverges n=0 n=0 for every value of z 0 0 and that nevertheless are applicable to all integral functions of the normal type r of the order Lo. Then, according
to theorem 1, these differential operators are also applicable to all integral functions not exceeding the normal type r of the order Lo. As this is the case with the integral functions that are mentioned in the
remarks following the theorems 2, 3, 4 and 5 (see definition 3), the stated differential operators are also applicable to all those functions. CONCLUDING REMARK. From the preceding we see that we proved the theorems 1, 2, 3 and 4 independently of each other, but that we deduced the condition mentioned in the first paragraph of theorem 5 from theorem 1. This is not to say that between the theorems 1 -5 there does not
exist another connection than the one just mentioned between theorem
5 and theorem 1. For instance we may deduce the assertion mentioned in the second paragraph of theorem 3 from theorem I and the first assertion of theorem 3. In fact, if the expression (83) defines an integral function 77
the order of which is less than 1, then, according to definition 4, for every value of r satisfying 0 < r < oo G(z) certainly is an integral I
function not exceeding the normal type, less than (ar) a , of the order 1. Then, according to theorem 1, the differential operator F(D) certainly is applicable to all integral functions not exceeding the normal type r of the order a. But since to r we may assign all finite positive values,
the differential operator F(D) is applicable to all integral functions not exceeding the normal type of the order a. As, according to the first paragraph of theorem 3, the differential operator F(D) is also applicable
to all integral functions of the maximum type of the order a, it is, in connection with the definitions 2 and 5, applicable to all integral functions not exceeding the maximum type of the order a.
We can prove besides that the condition mentioned in the first paragraph of theorem 2 is a sufficient condition for the differential Co
operator F(D) = Z
n=0
to be applicable to all integral functions
of (or not exceeding) the minimum type of the order a > 0, by using theorem 1. In fact, if G(z) defines an integral function not exceeding the normal type of the order 1, it follows from the definitions 2 and 1, that for this function G(z) there exist the following two possibilities: a) it is an integral function of the normal type of the order 1, fl) it is an
integral function not exceeding the minimum type of the order I. In case a) let G(z) be an integral function of the normal type A of the
order 1. Then, according to definition 4, it is certainly an integral function not exceeding the normal type, less than 2A, of the order 1. In case fl), according to definition 4, G(z) is quite certainly an integral function not exceeding the normal type, less than 2A, of the order 1,
with an arbitrary choice of 2A lying in between 0 and oo. Now if G(z) is an integral function not exceeding the normal type, less than 2A, of the order 1, then theorem 1 tells us that the differential operator F(D) is applicable to all integral functions not exceeding the normal 1
type - (2A) of the order a. So it positively is applicable to all ina tegral functions of (or not exceeding) the minimum type of the order a.
78
CHAPTER II
PROPERTIES OF THE FUNCTION h(x) = F(D) -, y(x) From the Preparatory Chapter we know that, if y(x) is an integral function of the order a > 0, it is either of the minimum type of the
order a, or of the maximum type of the order a, or of the normal type of the order a; in the last case it is of the normal type r of the order a, where r is equal to the number y as given in definition B. If y(x) is an integral function of the order zero and if it is not identically
equal to a constant, then it is of the maximum type of the order zero. For each of these cases we deduced in chapter I a condition 1) that is necessary and sufficient for a differential operator F(D) = 7, a,,Dn n-0
to be applicable to all integral functions that are of the same kind 2) as the function y(x). If, in one of these cases, a differential 00
operator F(D) = 7, a,,Dn satisfies the condition bearing on this n=0
case, then this operator is certainly applicable to the function y(x) so that, according to the definition of applicability (definition 11), 00
the series I any(n)(x) converges for every finite value of x. From 9L=0
this it follows that 00
h(x) = F(D) -- y(x) = I a.ny(n)(x) n=0
is a function that is defined for every finite value of x. Of this function h(x) we shall deduce some properties in this chapter
and in chapter III. For a differential operator we first take the special operator O(D) = e". Then theorem 6 shows that, if the function I (x) satisfies certain conditions, there exists for the function h1(x) = e" 2 . AX) a simple integral representation. From this integral representation it 1) See the theorems 2, 3, 1 and 4 respectively. 2) See definition 7 in the Introduction for the notion "of the same kind". 79
is easy to see that in this case the function la(x) is an integral function. This leads us to the problem whether the function h(x) is still an integral function, if the choice of the differential operator is less special. This
is one of the problems that are the subject of this chapter. While studying this problem we shall suppose that the non-constant integral function y(x) of finite order has the power series exCo
pansion y(x) = 0o
F(D) _
bmx-. We also suppose that the differential operator
m=0
anDn satisfies the condition mentioned in that theorem of n=0
chapter I that refers to functions of the same kind as this function y(x) (one of the theorems 1-4). So then the function h(x) already mentioned is defined for every finite value of x. Obviously we have for n = 0, 1, .. and for every finite value of x 00
{1)
b,,,m(m - 1)...(m - n + l)xm-n}
a,yl")(x) = a1, m=0
-x
00
(n + A) !
A-0
At
=anI bn+x
A
so that
Now we consider the series
{ I anbn+x I n=0 1x=0
(3)
(n
+.t
A
) i I X I,}
Because of (1) this series dominates the series 00
2, anyln>(x) n=0
(4)
for every finite value of x.
Since none of the terms in the repeated series in
i f'i
(3)
is negative,
we have (5)
I anbn+A I
(n i
' )I X I } _
°°
xlx {
I anbn+a I (n + A)!).
We shall examine the convergence of the repeated series in the right-hand member of (5), treating successively the cases where the
integral function y(x) is of the normal type r of the order a > 0 (theorem 7), of the minimum type of the order a > 0 (theorem 8), of the maximum type of the order a > 0 (theorem 9) and of the 80
maximum type of the order zero (theorem 10). In this examination it will appear in each of these cases, that the repeated series in the right-hand member of (5) converges for every finite value of x. Then the series (3) also converges for every finite value of x. So, if R denotes an arbitrary finite positive number, the series (3) converges for I x I = R and the terms of this series are not dependent on arg x. As the series (3)
dominates the series (4) then, according to the Weierstrass test for uniform convergence, the series (4) converges uniformly on the set of the points x for which I x I = R. Moreover, since each of the functions
any"n)(x) (n = 0, 1, ...) is an integral function, its power series expansion in the last member of (1) certainly converges for J x 15 R. On account of Weierstrass' well-known double-series theorem 3) we then have for every value of x, satisfying I x I < R, n=0
(n. ... ) !
00
00
an {
L
.
00
bn+x - -------I -- xx r
1=0
J
xA
co
x=0 A. n=0
anbn+x (n + A) !}.
As the positive number R may be chosen arbitrarily, this last formula holds for every finite value of x. From this and from (2) it then follows that the function h(x) may be expanded into a power series converging for every finite value of x. Hence the function h(x) is an integral function and its power series expansion is h(x) =
(6)
{I anbn+x(n + A) !}. A=O A !
After having proved theorem 6 we treat the cases mentioned above for the function y(x). After having shown for one of these cases that
the function h(x) is an integral function, we shall in each case immediately investigate, whether in that case the function h(x) is of the same kind as the function y(x) or not. In the proof of theorem 6 we shall use the following lemma 4) : 3) See e.g. Caratheodory [1] p. 205. 4) Cf. Bromwich [1] Art. 176,B. From the proof of the theorem formulated
there it appears that it is Bromwich's intention to assume in that theorem 00
00
that the series E I f(u) I and not the series E fn(u) converges uniformly n=0 n=0 on the interval 0 S u S b. Our lemma 14 is in so far slightly more general than the theorem given by Bromwich in Art. 176,B, that in lemma 14 the functions q(u) and fn(u) (n = 0, 1, ...) are complex functions, while Bromwich assumes that these functions are real. Bromwich does not say this in formulating the theorem stated, but from the proof he gives of that theorem it appears that he assumes
that the functions mentioned are real. 81 6
Lemma 14. b denotes an arbitrary finite positive number. Let the following conditions be satisfied: 1. The functions qq(u) and fn(u) (n = 0, 1, ...) are complex functions
of the real variable u, that are defined in every point of the interval
OSu<00.
2. The function T(u) is continuous in every point of the interval
oSu<00.
3. The products 4p(u) f n(u) (n = 0, 1, ...) are integrable in the proper u b. sense of Riemann over the interval 0
4. The integrals
fT(U)fn(U) I du
(n=0,1,...)
0
are all convergent.
5. The series I dur
I
0
converges. 6. The series
I I fn(u) I du
n=0
converges uniformly on the interval 0 S u S b.
Assertion:
j f
p(u)fn(u)du J=
J"1(u) :j fn(u) , du. 0
0
.) the real and the imaginary part then it appears that we may prove lemma 14 in a similar way as Bromwich proved the theorem mentioned in Proof.
If we take of the functions qq(u) and 1,,(u) (n = 0, 1,
. .
footnote 4). In order to spare space we refer to the proof of that theorem.
The following lemma will be used in proving theorems 7-10: Lemma 15. If the integral function k(x) for every finite value of x satisfies the inequality (7)
82
I k(x) I S
x=0
Ixix
-
A1e
where a, P and p denote Positive numbers that do not depend on x, then the function k(x) is an integral function not exceeding the normal type 1
fil
of the order Q.
e
Proof.
The expression
(8)
defines an integral function of the normal type 1 f1Q of the order a s). e
This means (see definition B) that to every positive number 8 there belongs a positive quantity L that is not dependent on , such that for every finite value of the inequality Le(.Q1
IfOI
holds. From this formula and from (8) and (7) it follows that the func-
tion k(x) satisfies the inequality
Ik(x)I In connection with definition A it follows from this that the in-
(9)
tegral function k(x) is of finite order and that its order does not exceed
e. If its order is equal top we see from (9), in connection with definition B, that the function k(x) is either of the minimum type of the order p, or of the normal type x of the order e, where x satisfies the inequality
0<x5
pe +e. e
The function k(x), however, does not depend on a and as we may choose the positive number e arbitrarily and hence arbitrarily near to zero, x satisfies the inequality
0<x<
(10)
1
fle.
LO
So the function k(x) is either of the normal type x of the order e, where x satisfies (10), or of the minimum type of the order e, or of an order less than e. So, according to definition 3, it does not exceed
the normal type
1
PQ of the order e.
e
°) One may take in example 2 of the Preparatory Chapter (p. 11) G(z)
n=,t, a=fi, a=e.
83
If the function I (x) is an integral function not ex-
Theorem 6.
ceeding the normal type b of the order 2 and if a denotes a complex number with the property
41aIb<1,
(11)
then the differential operator O(D) = e"D2 is applicable to the function I (x) and then the formula 6) (12)
'
a"DZ -* I (x) =
f e_U2f (x + 2u la)du
n
-00
holds. The function a `D2 -> I (X) is an integral function.
Proof. If a = 0 formula (12) becomes the identity I (x) = I (x) and since I (x) is an integral function the theorem in this case is proved.
Now we assume a 0 0. First we shall prove that the operator O(D) = eaD2 is applicable to the function f (x), by showing that in this special case the condition of theorem 1 is satisfied.
To that end we consider the expression 0o
G(z) =2;
(13)
n=o
a --zn' n!
n
where a2n = a (n = 0, 1, ...) and aln+ =0 (n = 0, 1, ...). So we n. can write this expression in the form G(z) =
(14)
1
an
"0
n!(2n)!}
z. 2n
Now, according to formula (P, 15), we have
urn (2n)! (2n)}in
urn
(2n)! n!(2n)!}}ln
= (2 1 a I)}.
while
2n+1 = 0. lim jI(2n + 1) ! a2n+1 (2n+1)!}1 n-ool I
Now it follows that (15)
lim sup lJn! n-.oo
1
a
"
1-
(n = (2 1 a I)'.
n!# J
6) For the argument of -,/a in (12) we may take both ,>: arg a and I arg a + on,
as appears from substituting u = - v. 84
It is easy to prove 7) now that (14) and consequently (13) defines an integral function of the normal type (2 1 a I)+ of the order 1. It follows from (11) that the inequality (2b)-1
(2 1 a I)1 <
is valid so that, according to definition 4, the integral function G(z) does not exceed the normal type, less than (2b)-4, of the order 1. From this we see that we may apply theorem I with a = 2, r = b, an
a2n = n!i , a2n+I = 0 (n = 0, 1, ...). Hence the differential operator O(D) = eaD' is applicable to all integral functions not exceeding the normal type b of the order 2 so that it certainly is applicable to the function &). Then it follows from the definition of applicability (definition 11) that the series an
converges for every finite
f(2n)(x)
n O n. value of x.
Now an
an f(2n)(x)
(2n) !
n!
(2n)!
n!
(2n) !
°°
22n
r e-"'u2n du
(n
= 0, 1,
-
),
-00
we have for every finite value of x (18)
elD' -+ f(x)
an f(2n)(x) n-O n. 1
°e
22nanf(2n)(x)
7
(2n)'
°O
ll-// f e-"'u2ndu I.
J
We shall now prove that in the last member of (18) we may interchange the order of integration and summation. To that end we put for a fixed value of x
9,(u) = e", fn(u) _
(2u)
2n
,/ (Ln)
(x)
(n
= 0, 1, ...)
(2
and we assert that these functions satisfy the conditions of lemma 14. 7) This may be done in the same way as in example 2 of the Preparatory Chapter (page 11). 85
Obviously the conditions 1, 2, 3 and 4 of lemma 14 are fulfilled. Now we shall show that condition 5 is satisfied too. Since the function I (x) is an integral function not exceeding the normal type b of the order 2, it follows from lemma 5, with A = b and
a = 2, that for every finite value of x we have {n!-}
Jim sup
I f(n)(x) I}n S (2b)}.
n-.oo
Consequently i
lim sup
{(2n)!-}
1 f(2n)(x) I}-2n 5 (2b)}
n- m
and also i
lira sup {(2n) !-} I /(2n) (X) I}n 5 2b. n-boo
From this last formula it follows that to every positive number e there exists a finite positive quantity K, dependent on E and on x,
but not on n, such that (19)
(n = 0, 1, ...).
I f(2n)(x) 15 K (2b + e)n (2n)!}
Now we choose the number e such that it satisfies the inequality 1
0 <e <
(20)
2
a1 Ib
which is possible because of (11) and we form the series n=0
(I a I (2b + e))n
(2n)!* n!
Applying to this series the well-known d'Alembert test of convergence, we see that because of (20) this series converges. From this and from (19) now follows that the series aa nom=0
f(2n)(x)
n.
1
converges for every finite value of x. Then, according to (16) and (17), this is also the case with the series 0o
r
22nan,(2n)(x)
l
(2n)!
,.oo
l
I' o0
This means that condition 5 of lemma 14 is also satisfied. Finally we shall prove that the last condition of lemma 14 is also 86
fulfilled. Let b denote an arbitrary finite positive number. Then we shall show that the series (2u) 2nanf(2n)(x)
I
0o
(21)
(2n)!
n=O
converges uniformly on the interval 0 S u S b. According to a well-known formula of Cauchy we have (22)
f(2n)(x) 15
I
(2n)RM{R}
(n
2ft
= 0, 1, ... ).
There R denotes a finite positive number and M{R} is the greatest value that I 1(t) I attains if t runs along the circle with x as the centre and radius R. Now we take R = 4b I a 1;. Then it follows from (22) (2n) ! M{4b I a I }}
I f (2n)(x)
(n
(4b)2n I a In
I
= 0, 1, ... )
and from this formula we see that we have for every finite fixed value of x and for every value of u of the interval 0 S u S b (2b)2n I a 1 n(2n) !M{4b I a
(2u)2nanf(2n)(x) (2n) !
I
I}}
-
(2n) ! (4b) 2n I a In
S
Consequently the series
M{4b I a 11} 00
sn
-
n=0
M{4b I a I}}. 22n
is a dominating series of the
series (21) on the interval 0 S u S b. Since this dominating series converges and since its terms are not dependent on u, the series (21) converges uniformly on the interval 0 S u 5 b. So the last condition of lemma 14 is also fulfilled.
Therefore we may apply lemma 14 which tells that in the last member of (18) we may interchange the order of integration and summation. This means that we have eaD2 - /(
1- r e_Uf 00 (2u)2nat/(2n)(x) } 1/n
J
00
In
(2n1)
du
!
for every finite value of x. 0 Now the sum I in the right-hand member of this formula is equal to 8)
n=0
8) The two sums in the first line following here are certainly convergent for every finite value of x and of u since they are the Taylor expansions of the two integral functions in the second line following here. For the first line following here see also footnote 6). 87
2u.%/a)n
(2u,%/a)n
2ui/a) + /(x - 2u-,/a)}. Hence .00
eaD$
- 1(x) =
e-"8 {/(x + 2u i/a) + I (x - 2u AM)) du.
2
J
%/.-r -00
From this formula (12) immediately follows, provided that the integral in (12) converges for every finite value of x. The latter will now be proved. From the datum that the function /(x) is an integral function not exceeding the normal type b of the order 2 it follows, in connection
with the definitions 3, A and B, that to every positive number 6 there exists a quantity L(8), not dependent on x, such that the inequality (23)
1/(x)
! S L(6)e(b+a) I x 12
holds for every finite value of x.
Now we choose the number b such that 4(b + 28) I a I < I. (24) According to (11) this is certainly possible. Then in (23) we replace x by x + 2u i/a 9) which leads to I /(x + 2u 1/a) I
L (S)e(t+e){ 1
X12 } 4 I xuv/a' + 41 a I I u12).
From this it follows that if we make a fixed choice for x, there exists a positive quantity M(8) with the property that we have for every finite value of u (25)
I f(x+2uVa) 15 M(d)e0+2e) 4 1alIu:8
As we have chosen the number S in such a way that it satisfies there exists a positive number n, such that
(24)
4(b+26) 1 a I < 1 -27.
Naturally this number 27 does not depend on x nor on u. Then it follows from this and from (25) that, if x is chosen fixed, the inequality (26)
I /(x + 2u a) I S M(6)e(i-a) u I-
0) For arg Va we may choose either j arg a or j arg a + a. This makes no difference for the right-hand member of the inequality following here. 88
holds for every finite value of u. So we have for every finite value of u I e-u2f(x + 2u-,/a) 15 M(b)e-°u', from which the convergence of the integral in the right-hand member of (12) follows. So formula (12) is proved. That the function e°12 --> I (X) is an integral function, we may
prove quite simply with the help of formula (12) by using a wellknown theorem 10) on the analytic nature of functions represented by an infinite integral. Herewith theorem 6 is proved conclusively. Now we proceed to treat the cases for the integral function y(x) mentioned at the beginning of this chapter.
First we assume that the integral function y(x) is of the normal type r of the order a > 0. For this case the following theorem holds: Theorem 7. Let the function y(x) be an integral function of the normal type r of the order a > 0.
A ssertions: 1.
If the numbers a have the property for the expression
(27)
a1
G(z) _
zn
n-0 n1 a
to define an integral function not exceeding the normal type, less than 1
(ar) a, of the order 1, and if the differential operator F(D) is defined by 00
F(D) = Y,
then the function
n=0
h(x) = F(D) - y(x)
is an integral function. 2.
If the number a satisfies the inequality 0 < a 5 1, then this
function h(x) is an integral function of the same kind at most as the function y(x). 3. If the number a satisfies the inequality a > 1 and if the expression G(z) in (27) defines an integral function not exceeding the minimum type
10) See e.g. Whittaker-Watson [1] 5.32. Clearly the conditions (i), (ii) and
(iii) stated there are satisfied. That the condition (iv) is also fulfilled may easily be shown, by proving first that the function
dx
f(x) is of the same kind
at most as the function f(x). This immediately follows from the remark subjoined to lemma 18 (see page 121). After that an inequality of the form 8 (26) may be deduced for the function /(x -1- 2u Va). ax
89
of the order 1, then the function h(x) is also an integral function of the same kind at most as the function y(x). 4. If the number a satisfies the inequality a > I and if the expression G(z) in (27) defines an integral function of the normal type y of the order 1,
where y < (or) - Q , then the function h(x) is an integral function not exceeding the normal type (28)
r(1 - do-1)1 °
of the order a, where
d = y(a'r) 5. If, in the case of assertion 4 the function h(x) is an integral function of the normal type u of the order a, then this number It satisfies the condition a
(29)
0 < ,u S r(1 - dQ-i )
the upper bound, occurring in this inequality, cannot be replaced by a smaller number. REMARK. We point out the fact that there exists an important difference between the results that theorem 7 gives in the assertions 2 and 3
and the one it gives in assertion 4. For, theorem 7 tells its that in the cases mentioned in assertions 2 and 3 the function h(x) is an integral function of the same kind at most as the function y(x). In other words, in the cases mentioned in the assertions 2 and 3 the theorem excludes the possibility that the function h(x) is of a higher kind 11) than the function y(x). In the case mentioned in assertion 4, however, theorem 7 does not exclude the possibility that the function h(x) is of a higher kind than the function y(x). It is true that in this case the order of the function h(x) cannot exceed the order of the function y(x), yet if the order of the function h(x) equals the order of the function y(x), i.e. equals a, and moreover the function h(x) is of the normal
type ,u of the order a, the number (28) is an upper bound to ,u (see also definition 3) and this upper bound is greater than r. So if, in the case mentioned in assertion 4, the function h(x) is of the normal type,u of the order a, theorem 7 does not exclude the possibility that the inequality ,u > r holds. But with respect to the present case the theorem contains 11) On the basis of the definitions 8 and 9 the reader may easily convince himself of the fact that the function h(x) is of a higher kind than the function y(x) if it is not of the same kind at most as the function y(x). 90
still more, viz. that it does happen indeed that the inequality u > r holds. This follows from the last assertion of theorem 7, which a.o. tells us that the upper bound (29) for the number 4u cannot be replaced by a smaller number. In the proof of theorem 7 we shall give an example 12) where this number,u is equal to the upper bound which is given by formula (29) for this example.
Before furnishing the proof of theorem 7 we shall give the following lemma, which will be used in proving theorem 7. The proof of theorem 7 follows the proof of this lemma.
Lemma r6.
If the numbers a, Co and c satisfy the conditions
a>0, w>0 and
0
(30)
then 00
lim a-m nm0
Proof.
(n+a m
cn\
a=(I-am) C-1
First we shall deduce an inequality for the sum
S-Zcn
Cnla'n-F a,
n-0
which obviously exists and is finite.
If we put (31)
then u,, > 0 and
/ cn(nn a)- =un
(n=0,1,...)
\ 00
(32)
S = 71 un. n=0
Let us suppose now that the number a satisfies the inequality (33)
a > 2 (c-
This is permitted since a tends to infinity. From the data concerning c and cu follows that then certainly a > 0. In an orthogonal Cartesian coordinate system with coordinates and q we draw, for a fixed value of a satisfying (33), the curve with 12) See page 106. 91
the equation 13)
(s Z 0).
+ a )W
77 = c` I
(34)
This curve we shall call curve I. On curve I lie the points (n, un) (n = 0, 1, ... ). In order not to disturb the proof here we shall show at the end of this proof that in one and only one point of this curve the tangent to this curve is parallel to the s-axis, if the points of the curve for which = 0 and = oo are left out of consideration. Let the abscissa of that point be equal to C. We shall also show there that between the points of the curve for which E = 0 and _ ' the slope of the curve is positive and that between the points for which = c and = oo the slope is negative. Then it is clear that we have if I S n _<
(35)
un >
(36)
un < 1[n_1 if n
14),
+ 1.
Besides curve I we consider curve II of which the equation is -}- a)°'
(37)
( Z 1).
Curve II arises from curve I by translating every point of this last curve along the distance + 1 parallel to the E-axis. By equating the right-hand members of (34) and (37) we see that curves I and II, apart from the point for which = oo, have one and only one point in common. Of this point the abscissa is equal to
(38)a
a
c'-1
From (33) follows Sa > 2. From the manner in which curve II arises
from curve I it is easy to see that the inequality
a-1
(39)
holds.
13) If f > 0, 0 5 y
we understand by + 1)
\
the expression
r(/J
f(y + 1) r(f - Y + 1) 14) Since the number a to be defined in (38) satisfies the inequality a > 2,
it follows from (39) that C > 1, so that there exists at least one number n
1, viz. n = 1, for which (35) holds. 92
Each point (n - 1, u1) (n = 1, 2, ...) of curve I has the property lies then that the point (n, u,,_1) lies on curve II. This point (n, and only then also on curve I, if it coincides with the point of inter-
section of curves I and II, i.e. if n =Q. If this is the case then un-1 = un. Therefore, if s,, is equal to an integer n (n Z 3 because of Q > 2), then there exists one and only one value for n for which un-1 = u,,. Let this value of n be N, so that uN_1 = UN. From the definition of the number N, from (39) and from (35) follows (40)
uN_x > u_,_x_1 for A = 1, .. ., N - 1;
from this definition, from (39) and (36) follows (41)
uN+x < uN+x-1 for A= 1, 2, ....
Now it follows from (40) and (41) that un < u.,, for every value
of n:74-N-1 and :N.
If $,, is not equal to an integer
3, then it follows from the above,
that we always have un_1 u (n = 1, 2, . . .). In this case the integer lying in between Q - 1 and E. we shall call N. We assert that in this case we have un < it., for n N. In the first place is u,,. > "N+1. In fact,
` N1UN - uN+1 = C N(N+alw
c 'v+I(N+1+a N+1 (NN a)° 1
1
(N + 1
N+
Now it follows from the definition of the number N that
N>Q-1=
to
i
a)°'}
(because of (38))
C W- 1 and from this it is easily seen that 1-c(N+I+a)°'>0,
N+ 1 so that it,, - UN+1 > 0. We may prove quite analogously the inequality UN > u_v_1, by using the fact that also N < EQ. Moreover it is obvious that (42)
N-I <$,-I
and from this, from (39) and (35) follows uN-x > UN_x_I for A = 1, ... ,
N - 1. So (43)
(A=O,...,N-1).
UN_x > uN-x-1
From (42), (39) and (36) together with the inequality uN > uN+1,
just proved, it follows that now also formula (41) holds. Then it follows from (41) and (43) that we now have u,, < UN for every value N - 1 and of n N and therefore certainly for every value of n N.
Both in the case where Q is equal to an integer (Z 3) and in that where , is not equal to an integer, it follows from the definition of the number N that we may write (44)
where the number 0 depends on a, but always satisfies the inequality OZ. 0 < 1. (45) Then from
(44) and (31), with n = N, follows
(46)
UN
= CSQ-B
+ v tt -0a
Sa
o-0
Now it follows from (31), (44) and (38) UN!A UV.{A-1
N+A±a _ N+A
`
- c{
}
a - 0(1 - c a-0(c_
1
Q-B+A+a " 6+A + A(1 - cw) 1
-1)+A(c-W-1)
If we now put
c w- 1 =q,
(47)
then we have q > 0, as follows from the data concerning c and w, and
(48)
4--0+-g
uN+x UN+x-1
l
q+ I
q+ I
a - q9+qA
0)
A
-
(A= 1 , 2 ,
...).
1
Now, if, with a fixed choice of a satisfying (33), A runs through the
sequence of the positive integers from I to oo, then we see that the quotient (48) for A -> oo has a limit-value and that we have (49)
lim
x-oo uN+x-1 94
= (q ±
Since q > 0, the inequality (q + 1) -" < (lq + I)-' holds and from this and from (49) we see that the formula (50)
< (Jq + 1)
°
holds for all sufficiently large values of A.
Substitution of (48) in (50) gives after a short reduction (51)
(Jq+ 1)(a - 4+ j- 0 +
4..+_1 A)
So this inequality is satisfied for all sufficiently large values of A and from this and from the fact that 0 has the property (45), it follows
that (51) is certainly satisfied by all positive integers A satisfying the inequality (52)
(Jq+ 1)(a + q+ 1 A) < a - q+qA'
which arises from (51) by replacing in the left-hand member of (51)
the number 0 by zero and in the right-hand member of (51) the number 0 by 1. The positive integers satisfying (52) are all positive integers for which the inequality (53)
A>(a+2)(1+
1
91
holds. If L denotes the smallest positive integer that is larger than right-hand member of (53), then (53) and hence also (51) and (50) are satisfied for every integer A Z L. In the following way we shall construct a dominating series of the series
UN + u.v+1 + ... + "N I-L + uV+r,+l + ... . In the first place we replace each of the L terms uv+1, .., UN+ L by UN. From (41) it follows that, owing to this, each of these terms is replaced by a larger quantity, viz. u,`-, which exceeds each of these terms in magnitude. Moreover, for v I we replace the terms by uN(1 + q)-w'. This causes each of these terms to be replaced by a larger number as follows from (41) and from the fact, just proved, that (50) holds for every positive number A L. Then 15) 00
13) Since q > 0 and to > 0 the series Ii (1 + ,Jq)-°" certainly converges. r=1
95
(54)
'UN + UN+1 +... < (L + I)uN +uN
+jq)-wr
r=1
(1 00
{L + 1 +(1 r=1
+44)-wr}uN.
As we have already seen, the inequality u,, < u,,v holds for every value
of n 0 N - I and 0 N, while uN_ 1 S u N-, so that 18) u0 + u1 + ... + UN-1 < Nun-. (55) From this and from (44), (45), (38) and (47) it follows a
u0 + u1 + ... + uN-1 < -'tN =
(56)
UN.
q
Now, because of (54) and (56), we have (57)
N -I
00
n=0
n= S
2; un=Y,
n=0
a q
r=1
J
where
A = --- + L + I +
(58)
q
iq)-wr
(1 +
r=1
From the definition of the number L we see that
L
(59)
(a + 2) C 1 +
1 J + 1. q
From this and from (58) we see that quantity A satisfies the inequality (60)
A
2
+4+
q
2
+
,
r=1
q
(1 + jq)
-wr .
00
From (57) and from the fact that I u > uN and from (32) we n=0
have for the sum S the following inequality (61)
UN < S < AuN,
where A satisfies (60). This is the inequality wanted. 16) Because of
, > 2 it follows from (44) and (45) that N Z 2, so that
in (55) the sign of equality cannot hold. 96
Now we shall show that (62)
lim {UN) 71.
a-.oo
exists and we shall determine this limit. From (46) follows 1
{uN}a = c
a
r
C
eQ _Q{ r(a- 0 + a + I) lQ
As from (38) and (47) it follows that 1
Sa + a =
C
=1 q
a
and
------ a, we have therefore q
_
rt/c (63)
.
r(a- 0 + 1) r(a + 1)
+a - 0
{UN} a= C q
w
1
a-0+11
q
a
r a -0+1)r(a+1)
a0
q
Applying the well-known Stirling's formula 17) to the three rfunctions occurring in the right-hand member of this formula, we find after an elementary calculation that (62) exists indeed and that we have 1
(64)
1
lim {uN}a = (1 - cam a-.oo
As the three members in (61) are all positive, it follows from (61)
that the inequality 1
(65)
1
1
1
{uN}a < S- < A
{uN}a
also holds. Because of (58) and (60), we see that 1
lim {A}a = 1 a-.o*
and from this formula, from (64) and (65) it follows that 1
1
lim Sa = (1 - cOj U- W
Now we have yet to prove that in one and only one point of curve I with equation (34) the tangent to that curve is horizontal, if the points
for which = 0 and = oo are left out of consideration. 17) See the formula for r(z) in footnote 17) on page 10. 97 7
From (34) it follows that '1,()
=
d
d log r!()
=d d{ loge f wlobI'($+a+ 1)
1) -wlogr(a+1)} 1) -V($+ 1)},
=log c+CO where d
1' (z) =
dz
log r(z)-
z 0, it follows from 0 if
Since 71(E) is finite for every finite value of
this formula that then and only then (66)
v( + a +
+ 1) = log (c- W ).
The right-hand member of this formula is positive because of C-
1
w>l.
From (33) it appears that a + 1 > c w From this inequality and from the fact that for > 0 the function ,V(E) is strictly increasing (this follows immediately from formula 18) 00
(67)
) _ Y, ( + n)-2 > 0
1V (
(5 > 0)),
It-0
we see that we have _
(68)
1
V(a + 1 ) - 1V(l) > i (c w) - V(1).
We shall now first show that the inequality _
1
1
tV(1) > log (C-
(69)
is correct.
This formula will be proved as soon as we have shown that the formula TV(s) - V(1) > log 8
(70)
holds for every j > 1. In fact, formula (69) follows from it for fi = c- w . If 14 > 0 we have 19) a-u - e-9u
(71)
TV(s) -'V(1) = OV(A)
+C
f 0
e-11
du,
18) See e.g. p. 80 in the Bohmer's nice book [1]. In this book a diagram of the function w(x) for real values of x is also given. 1e) See Whittaker-Watson [1], p. 260, ex. 16. 98
where C denotes Euler's constant ; moreover we have
20)
n0 e-u - e-s"
log 8 =
(72)
.I
U
- du.
0
We also have for /3 > 1 oo a-p - e-#u ' ° e-7L - e-P'u
f J 0
1 -e-"
du -
du
u 0
=
I
(e -u - e-flu)
I
1
11-e-"--- - u- du > 0
0
and from this and from (71) and (72) it follows that formula (70) and therefore also formula (69) is proved. Then it follows from (69) and (68) that (73)
+V(a + 1) - p(1) > log (c
i).
From this we see that the function (74)
V( + a + 1) - t ( + 1),
occurring in the left-hand member of (66), exceeds the right-hand member of (66) if = 0. Besides, it follows from
v($+a+ 1)-v(E+
1)
(0<0<1)
and from (67), that the function (74) is positive for every $ Z 0 and that it tends to zero if $ -* oo. Moreover, it follows from (67) W
1 ) -v"(E+
n=0
+a+ 1 +n)-2
-I 00
n=0
1 +n)-2 <0,
so that for Z 0 the function (74) is continuous, positive and strictly decreasing and that it tends to zero if $ -* oo. From this it follows in connection with (73) that there exists one and only one value of $, lying in between 0 and oo, with the property that formula (66) holds. This value of $ we called . Moreover we see for 0 < $ < C that
V( + a + 1) -TV(E+ 1) >log(c 20) See Whittaker-Watson [1], p. 116, ex. 6. 99
From this formula it follows, in connection with the formula for
n'() that
0 if 0 < $ < C, because for these values of
we
rl(e)
0. Similarly we have '() < 0 if C < $ < oo. have Herewith lemma 16 is proved conclusively. REMARK.
In the above proof the inequality (53) is important. 00
This inequality enables us to deduce for the series
n=0
u,, a dominating
series which is suitable for our purpose, viz. the series (57). Proof of theorem 7. From theorem it follows that the differential operator F(D) is applicable to the function y(x), so that the function h(x) = F(D) -+ y(x) is defined for every finite value 1
of X.
00
Since the function y(x) _ I bmxm is an integral function of the m0 normal type r of the order a > 0, it follows from (P, 9) that for the coefficients bm the formula 1
1
1
lim sup {m!
(75)
b,,,}+n = (oT)
holds.
If the numbers a have the property for the expression (27) to define an integral function not exceeding the normal type, less than 1
(ar) Q , of the order 1, then lemma 6 tells us, if in it we replace /(x) I
by G(z), A by (ar) Q and a by 1, that the formula 1
(76)
1
0 S lim sup ( G"n)(z) In'- < (ar)^ o n-00
holds for every finite value of z. 1
Because of G(2)(0) = n!1 aa,, formula (76) gives for z = 0 1
0 5 lim sup (n!
1
1
n-, 00
1
1
limsup {n! 1 QIan IJn =v,
then it follows from (77) that we have _
(79) 100
_
1
1 a 1) i < (ar) a .
a
1
05 v <(ar) Q.
Now we define the number r1 by (80)
From (75), (80) and (78) follows that to every positive E there belongs a positive number K = K(s), neither dependent on m nor on n, such that (r, L Elm (m=0,1,...) K ------ 1 (81) I bm I m!Q
and at the same time (y+E)n
(82)
(n=0,1,...).
n!' Now we investigate the convergence of the repeated series in the right-hand member of (5), i.e. the series xIa I
A=O
A.
{I I n=o
A ) ! }.
To that end we consider the series
+)!
'x Ia 0 (v + E)n(r1 + E)n+A
Ky
I
Because of (81) and (82) this series dominates the series stated in (5) for every finite value of x. This dominating series we transform into Kz
(i1
+ E) 1
Ix Ix
j 1
(v + E)"(Zl + E)^
ll
1
n
/
1
Putting
(v + e)(r, + e) = c,
(83)
we may write this last series in the form cn(n+A) 1-
(tl+E)xlx
(84)
K2 A=U
In=o
n
Now we shall show that this repeated series converges for all sufficiently small values of e and for every finite value of x. In the first place it follows from (79) and (80) that we have (85)
vi1 < 1. 101
Now we choose the number a in such a way that it satisfies the inequality - vri 0<e<1+.v+ri I
(86)
From the definition of the numbers v and ri follows v z 0 and ri > 0 and from this, from (85) and (86) we conclude that the number e just chosen is less than 1. Then we see from this and from (83) and (86)
that
c=yri+e(v+ri)+e2
(87)
so that, because of 0 < a < oo, =p(yt + A/
C,
-
00
(88)
(n + A
) - (1
n
n=oCn
1
- c)x+i
21).
From this formula it follows that the series K2
ao
/ri + e).
1-c.o 1-c
(89)
Ixx
10
dominates the series (84) for every finite value of x. And since the latter series in turn dominates the repeated series in the right-hand member of (5), the series (89) dominates the repeated series in the right-hand member of (5). Since the series in (89) converges for every finite value of x, the series in (84) and therefore also that in (5) converges for every finite value of x. Moreover we have 00
(90)
X 1;' I
i- {2: I an.b.+a I (n + A)!)
SK2
A=O
lai
(
l_i
n=p
As we showed in the two paragraphs following formula (5), it follows from the convergence of the repeated series in the right-hand member of (5) that h(x) is an integral function and that it has the power series expansion (6). Herewith assertion I of theorem 7 is proved.
Now we consider the case where 0 < a S 1. Consequently we
have 1 - 1 S 0, so that we may replace the inequality (88) by the a
21) In the left part of formula (88) the sign of equality occurs only for
d=0.
102
following, the right-hand member of which does not depend on A:
-
/+A\1 Qi
n=0
cnl
n
n=p c"
1
I- C
From this formula and from (90) follows that now 00
(91)
Ixl,
00
i- {
I anon+x
(n + A)!} S
K2
Ix12
00
1-
(ti
C
+
E)
Ai
1
.
In connection with formula (6), it follows from this that in the case now considered the integral function h(x) satisfies the inequality K2
h(x)I
° S1-c
A=O
zlxI
(t1+e) i
.
Ai z
Hence, according to lemma 15 with a = K
1-c
(3 = tl + e and
,
= a, the function h(x) is an integral function not exceeding the normal type
(r1 + e)° of the order a. Since the function h(x) does
not depend on E and since we may choose the number e arbitrarily in
the interval (86), the function h(x) is an integral function not ex1
ceeding the normal type 4p of the order P, where 4p = lim --- (r + e)°, -
8-.0
i.e.
or
= v t' = t (because of (80)). According to definitions 3 and 8
the function h(x) therefore is of the same kind at most as the function y(x). Herewith we have also proved assertion 2 of theorem 7. Next we consider the case where a > 1 and the expression (27) defines an integral function not exceeding the minimum type of the order 1. Then lemma 7, with I (x) replaced by G(z) and a by 1, tells us
that we have for z = 0 lim{n!
=0,
n-+o°
so that in this case the number v defined in (78) is equal to zero. Now, if in (82) and (83) we replace v by zero and if we choose a
number a satisfying the inequality (86) with v = 0, then formula (87) with v = 0 holds. In this case the inequality (88) also holds and so does (90) with c defined by (83) with v = 0. As a consequence,
the function h(x), possessing the power series expansion (6), now satisfies the inequality I x Ix co tl + E /( I S I h(x) 1 - Etl - E2 A=O \ 1 - Etl - -2 / A!K2
A
1
103
K2
i-
According to lemma 15, with a =
tl -!-
I - Eti - E2 E2' 1 - EE function not exceeding and e = a, the function h(x) is now an integral the normal type
1(
it + E
a \I -EZl-E2
}Q
of the order a. Since the function
h(x) does not depend on E and since we may choose the number E arbitrarily in the interval (86) with v = 0, the function h(x) is an integral function not exceeding the normal type a of the order a, because of
lim e-+0
- (.._. T, + .8. _)a = Or \I -erl-82
T' = T. Cr
So in this case it is an integral function of the same kind at most as the function y(x). This proves assertion 3 of theorem 7. Now we consider the case where a > I and the expression (27) defines an integral function of the normal type y of the order 1, where
y < (ai) ° Then lemma 3, with /(x) replaced by G(z), a by I and 2 by y, tells us that we have for z = 0 (92)
lim sup {n!
1- 1
a,,
1
= y,
so that now the number v defined in (78) has the value y. In the formulae (82) and (83) we now replace v by V. From (80) and (92) it follows that formula (85) with v = y holds. Now we choose a number E satisfying the inequality (86) with v = y and then we see that formula
(87) with v = y is valid. Hence in this case we also have c < 1. Now formula (90) also holds and in connection with this formula we form the expression c"
nn
4-0 \ Because of the fact that 0 < c < I and a > 1, hence I
1
111
a
> 0,
we may apply 22) lemma 16, with a = A and w = I - --, to this expression.
a
22) Here we do not entirely use the result of lemma 16, since here A does not increase continuously, but runs through the sequence of the positive integers.
104
Consequently
limc 00
(n
!
Tn
.-Y°° n=0
A` 1 1v
1 a1
1
JI
1-a
a
-co-1) a
=(1
From this formula it follows that corresponding to the number e, chosen by us, there exists a positive quantity Q, not dependent on A, such that _ 1-_ as +e)z(1 -c°Q') as (A = 0, 1, ...) cn(n n=0
n
From this formula and from (90) it follows, in connection with (6),
that in the present case the function h(x) satisfies the inequality a
o°
(rl + e) (1 + e) 1 - c °
h(x) I S KZQ
1-all a
t
x
Il 1
'
ta where c = (y + e)(r1 + e). Now it follows from lemma 15, with a a, that the intea = KZQ, _ (r1 + e)(1 + e)(1 ca-') ° and .1=0
I.
J
-
1
a
- cQi
e)a
e)a
gral function h(x) does not exceed the normal type (rl +
(1 + a
)1_a
of the order a. However, the function h(x) does not depend on e and since we may choose the number a in the interval (86), with v = y, arbitrarily, the function h(x) is an integral function not exceeding the normal type a of the order a, where (1
V= lim
e-+0 a
(r1
e)° (1
+E)'(1 -C a=
Putting d equal to the value of c for s - 0, so d = yr1 = y(ar) (because of (80)) we find
This proves assertion 4 of theorem 7.
It remains to prove the last assertion of theorem 7. The first part of this assertion follows immediately from assertion 4 of theorem 7 and definition 3. In order to prove the second part we start from the function
(b > 0), y(x) = e'-2 which, according to definitions A and B, is of the normal type b of
the order 2. 105
On account of theorem 6 the differential operator (a > 0, 4ab < 1) ¢(D) = eaj/2 is applicable to this function and because of the same theorem we
have the formula
f°°
1
(93)
1/A
_
_bx2
e-u°eI,rx+2u%'u)Zdu
eb22 = ,
1
= ._.. -el-4ab, V1 - 4ab
in which the square-root has to be taken positive. So in this special case the function h(x) is equal to the function in the last member of (93). According to definitions A and B it is an
integral function of the normal type u =
---b 1
4ab
of the order 2.
In this case the function G(z) is defined by (14) and the number y is equal to (2a)1, as is easy to see from formula (15) with I a = a. Then
d = y(ar)
(2a)l (2b)' = (4ab)t.
The upper bound to the number u, occurring in (28), becomes in this special case
r1-
d°-1
1-Q
= b(l
-
(4ab)4
-2)-1 =
b
1 -4ab
and this is exactly the value of ,u. Hence in this special case the upper bound is attained so that it cannot be replaced by a smaller number. This proves theorem 7. REMARK.
Later on we shall make use of the following assertion,
which at this moment can be proved easily: If the function y(x) is an integral function not exceeding the normal type r of the order a where a satisfies the inequality 0 < a s 1, and if 0 the differential operator F(D) is defined by F(D) _ T, aaDn, where the numbers a are such that the expression "-0 °
a
G(z) _ Y, _ --i ` z" n=0
n!-
defines an integral function not exceeding the normal type, less than (ar) (94)
of the order 1, then the function
k(x) = F(D) a y(x)
is an integral function not exceeding the normal type r of the order a. 106
To prove this assertion we first observe that from theorem 1 it follows that the differential operator F(D) is applicable to the function y(x). Moreover, we see that from lemma 5, with n = m, 2 = r, a = a and I (x) = y(x), for x = 0 it follows that for the numbers bm formula holds. Then in almost the same way as (75), with = replaced by we did in the proof of the assertions 1 and 2 of theorem 7 that the function h(x), occurring there, is an integral function not exceeding the normal type -r of the order a, we may prove that the function k(x) in (94) is an integral function not exceeding the normal type r of the order a. Theorem 8. Let the function y(x) be an integral function of the minimum type of the order a > 0. If the numbers an have the property that the expression G(z) =
(95)
0O
a "`
9L=o ni
i zn
defines an integral function not exceeding the normal type of the order I 00
and if the differential operator F(D) is defined by F(D) _ n=O
then the function
h(x) = F(D) - y(x) is an integral function of the same kind at most as the function y(x). Proof.
From theorem 2 it follows that the differential operator
F(D) is applicable to the function y(x), so that the function h(x) = F(D) -> y(x) is defined for every finite value of x. 00
Since the function y(x) =
bmxm is an integral function of the m=0
minimum type of the order a > 0, it follows from (P, 7) that for the coefficients bm the formula (96)
lim fin!
I
I-
I bm j}1n = 0
m-.oo
holds.
If the numbers a have the property that the expression (95) defines an integral function not exceeding the normal type of the order 1, then lemma 4 tells us, if in it we replace f(x) by G(z) and a by 1, that we have for every finite value of z (97)
0 S lim sup I G(n)(z)
< 00.
n-. 00
107
Because of G("'(0) = n!1
-
1
an, formula (97) gives for z = 0 the
formula 1
1
0 - Jim sup {n!1
(98)
I a I}n < oo.
a
n-.oo
If we put _
1
Jim sup {n!1
(99)
1
J a I}n = y,
n-oo
then it follows from (98) that y satisfies the inequality
0Sy
(100)
From (96) and from (99) and (100) follows that corresponding to every positive number e there exists a positive number M = M(e),
neither dependent on m nor on n, such that (101)
(m=0,1,...)
IbmlaM-"'
M!-
and at the same time a. I S M
(102)
(Y + e)n
(n = 0, 1, ...).
.
n!1-
Now we shall investigate the convergence of the repeated series in
the right-hand member of (5). To that end we consider the series (103)
8I
M2
x-
X
Il
n+
Co
to-0
en(Y + e)n
1-
1
Cr
As follows from (101) and (102) this series dominates for every finite value of x the series in the right-hand member of (5). Now we choose the
number e such, that for y > 0 it satisfies the inequality
0 <e
(104)
<min(1 4Y
,
)
and for y = 0 the inequality
0<e<- 2. 1
(105)
Then, if y > 0, 1
s(Y+e)=ey+e2 < 4YY+()2and, if y = 0, 1
e(Y + e) = e2 < 108
2
2=
so that both for y > 0 and for y = 0
<
e(y +
.
From this it follows that for every finite value of x the series
M2 I 00
ex l
I
f
I
2n 1
\n nn+ 1)
dominates the series (103) and therefore also the series in the righthand member of (5). As this dominating series is identical with the series 2M2
xll
(2e)a I
.
and the latter, because of 0 < a < oo, converges for every finite value of x, then this is also the case with the repeated series in the right-hand member of (5) and we have the inequality (106)
1
A=O
Ix
X I
Al
-I I anon+x I (n + A)!j < 2M2 f (2s)a {n=0 a=o 00
x1I I
A
Because of the convergence of the series in the right-hand member of (5) the function h(x) is an integral function. Moreover, for every finite value of x it satisfies the inequality 00
I h(x) I < 2M2
(2e)a I x1lz x=0
,
,tta
as follows from (106) and (6). So, according to lemma 15, with a = 2M2,
/1 = 2e and i = a, the function h(x) is an integral function not ex-
ceeding the normal type
(2e)° of the order a. Since the function
h(x) does not depend on e and since we may choose the number e in
the interval (104), if y > 0, and in the interval (105), if y = 0, arbitrarily, the function h(x) is an integral function not exceeding the minimum type of the order a, because of lim
1 (2e)° = 0. Hence it is
Q-.° a
an integral function of the same kind at most as the function y(x), as follows from definitions I and 8. Thus theorem 8 is proved.
REMARK. Now we can easily furnish the proof of the following assertion : 109
is an integral function not exceeding
If the function y(x) = Y,
m=0
the minimum type of the order a > 0 and if the differential operator
F(D) is defined by F(D) =
anD", where the numbers an have the n=0
property mentioned in theorem 8, then the function (107)
k(x) = F(D) - y(x)
is an integral function not exceeding the minimum type of the order a.
In fact, in the first place it follows from theorem 2, that the differential operator F(D) is applicable to the function y(x). Besides, it follows from lemma 7, with n = m, a = a and I (x) = y(x), and with x = 0, that for the coefficients bm formula (96) holds. Then we may prove in quite the same way as in theorem 8, that the function k(x) in (107), like the function h(x) occurring in theorem 8, is an integral function not exceeding the minimum type of the order a. Theorem 9.
a>0.
Let the function y(x) be of the maximum type of the order
If the numbers an have the property that the expression (108)
a G(z)= izn
n=on!o
defines an integral function the order of which is less than I and if the CO
differential operator F(D) is defined by F(D) _
n=0
anDn, then the
/unction
h(x) = F(D) -. y(x) is an integral function of the same kind at most as the function y(x).
From theorem 3 it follows that the differential operator F(D) is applicable to the function y(x) so that the function h(x) is defined for every finite value of x. Proof.
00
As the function y(x) _
b,,,x"^
is an integral function of the
order a > 0, it follows from (P, 4) that the coefficients bm are such that for every positive 8 we have (109)
lim{m!'
FbmI}no=0.
If the numbers an have the property, that the expression (108) defines an integral function the order of which is equal to e, then 110
_
1
that for every
it follows from (P, 4), with an replaced by an n! positive number 17 we have 1
1
lim{n!g+
(110)
1
Ianl}n = 0.
n-.00
Since formula (110) holds for every positive number n it also holds, if we replace q by the number 6 occurring in (109). Hence for every positive 6 not only formula (109) is correct, but also the formula 1
(111)
1
1
an l}n = 0.
lim {n!Q+a 5L- 00
From (109) and (111) it follows that corresponding to an arbitrarily chosen positive number E there exists a positive number N = N(e, 6) depending on e and on 6, but not on m nor on n, such that
(m=0,1,...)
N
1 bnj
(112)
and at the same time
anlSN
(113)
En 1
n!Q+a
(n=0,1,...).
1
a
-
Now we shall investigate the convergence of the series in the righthand member of (5). To that end we consider the series a n+ 1- (n + A) ! 00 n _x N2 (114) --
(I A=o
%t!
i -ia . n=o nIQya
(n + ,).-
Because of (112) and (113) this series dominates the series in the right-hand member of (5) for every finite value of x. We transform the series (114) into the series - 8A x iA
115)
oo
N2
d=0
a -a
0
e2n
n+ it
1 -1-8
yZ1QTa
1-
}
a +8 (. 1
Now we choose the numbers 6 and e such that they satisfy the
inequalities (116)
0<6 < min 11-e
12+e'
and (117)
11
a1
0 <E < 1
respectively.
111
I From (116) follows 6 < -+
2
P
< 1 < 1, so that certainly 62 < 6.
e
If we make use of this it follows from (116) that I
(118)
a+a
1-e-d(e+2) > -1-8= I -(e+e6+6+62) +a >0. a+a
Besides, it also follows from (116), that
1 - 1 +b < 1
(119)
v
--1a + 1a = 1.
Because of (118) and (119) the series Ez
ll,x la E 2n
N2
(120)
'°
-°
a' 0
l
n
dominates the series (115) for every finite value of x. Therefore it dominates for every finite value of x the series (114) and hence also the series (5). Now, according to (120), we may write the series (117)
in the form a
N2 1
E2
(
la
f
Alo -a
E2)
from which, because of (116), we see that the series (120), which dominates the series in the right-hand member of (5), converges for every finite value of x. Hence this is also the case with the series (5) and we have 00
(121) 1
1x12 .-
--.
N2
o,
- {I I anb,,+a I (n + A)!} <
a
E
Ix
1 - E2)
1 - E2 a-_°
la
o -a
Since the series in the right-hand member of (5) converges we see that the function h(x) is an integral function and from (121) and (6) it follows that for every finite value of x it satisfies the inequality N2
I hh(X)
0o
E
I < 1--- Ez a-°
a
- E2 Nz
Ix
la
1-6'
Alo E
= .i , and-- = - 6, - Ez s2 a e the integral function h(x) does not exceed the normal type Because of lemma 15, with a =
1
(li)(e)o
112
1
1
of the order
= --d 1
a
1
ab .
So it certainly is an integral function
the order of which does not exceed
a
1 -a8
However, the function h(x) does not depend on 6 and since we may choose the number 6 in the interval (116) arbitrarily, we see that the order does not exceed lim -
a
a-o 1 -a8
= a. So the integral function
h(x) does not exceed the maximum type of the order a (definition 5). Hence, according to definitions 5 and 8, it is of the same kind at most as the function y(x).
This proves theorem 9. REMARK. Now we can easily demonstrate the following assertion: If the function y(x) is an integral function not exceeding the maximum type of the order a > 0 and if the differential operator F(D) is defined 00
by F(D) _
where the numbers a,, have the property mentioned n
0
in theorem 9, then the function
k(x) = F(D) - y(x)
(122)
is an integral function not exceeding the maximum type of the order a.
In fact, in the first place it follows from theorem 3 that the differential operator F(D) is applicable to the function y(x). Besides, it follows from lemma 8, with n = m, a = a, e = 6 and f(x) = y(x), and with x = 0, that for the coefficients bm formula (109) holds for every positive 6. Then we may prove in quite the same way as in theorem 9 that the function k(x) in (122), just as the function h(x) occurring in theorem 9, is an integral function not exceeding the maximum
type of the order a. Theorem io.
Let the function y(x) be an integral function of the
maximum type of the order zero.
If the numbers a have the property that there exists a finite real number v, depending neither on n nor on z, such that the expression (123)
a-Zn n G(z)_In=o n!"
defines an integral function the order of which is less than 1 and if the 00
differential operator F(D) is defined by F(D) = 2, a1,D", then the n=0
113 8
unction
h(x) = F(D)
y(x)
is an integral function of the same kind at most as the function y(x). Proof. From theorem 4 it follows that the differential operator F(D) is applicable to the function y(x) so that the function h(x) is defined for every finite value of x. Co b.x° is an integral function of the Since the function y(x) _ m=o
order zero, it follows from (P,4) with e = 0, that for every positive number A the formula lim {m ! " I
(124)
b,,, l } 719 = 0
,,,-.oo
holds.
If the numbers a have the property that there exists a finite real number v, depending neither on n nor on z, such that the expression (123) defines an integral function of the order Q, then it follows from that for every positive number (P,4), in which we replace a by b we have lim {n!e+e
I a J} n = 0.
n-.oo
Because of P < I this formula also holds for S = I - g, so that 1
lim {n!1
(125)
I a I}n = 0.
n-.oo
From (124) and (125) it follows that to every positive s there exists
a finite positive number R = R(E, A), depending neither on m nor on n, such that
E"'
(126)
WiA
(m=0,1, ...))
and at the same time En
Ia.IsR n !'
(127)
(n = 0, 1,
... ).
Now we investigate the convergence of the series in the right-hand member of (5). To that end we consider the series (128)
E
A-0 114
A
E2n
z
RZ A-
I
n-0 n!l-°(n
}- ) - (n +
A)! .
Because of (126) and (127) this series dominates the series in the right-hand member of (5) for every finite value of x. We transform the series (128) into 00
(129)
R2
I
I
f 00
+ Y
2n
z-n
Now we choose the number A such that it satisfies the inequality A > max (v, 1).
(130)
Also, we choose the number e such that the inequality
0<e<1 holds. Then we have A > v and also A > I so that for the sum i
n=0
in (129) the following inequality holds: /n A\'-A 00 00 E2n
o n! A v
1
n
1
<E2nn =O
1- E2
As a consequence, the series
Ixlz
R2 z
e2 A=O
A!A
dominates the series (129) for every finite value of x. So for every finite value of x it dominates the series (128) and therefore also the series in the right-hand member of (5). And since, because A > 1, this dominating series converges for every finite value of x, the series in the right-hand member of (5) does so too and we have (131)
Ixlz A-0
A!
Rz-
00
( n=O
I anon+zI
- ez
1 - e2 x=o
Ixlz A!A
Because of the convergence of the right-hand member (5) the function h(x) is an integral function and because of (131) and (6) it satisfies the inequality
Ix
R2
I h(x) I < 1 -e2
According to lemma 15, with a =
A=O
R2
1-e2
A!A
a and
P-A 1
the
1
integral function h(x) does not exceed the normal type Aed of the
order A . Then it appears from definition 3 that the order of the 115
function h(x) is not greater than
I
A
.
As the function h(x) does not
depend on A and as we may choose the number A in the infinite interval (130) arbitrarily, we see that the function h(x) is an integral function of the order zero, because of lim
A
= 0. Hence, according to
definition 8, it is of the same kind at most as the function y(x). This proves theorem 10. CONCLUDING REMARK.
In lemma 19 of chapter III we shall make 00
use of the fact that the series I, any(n)(x) converges uniformly on and n=0
in each circle C with the centre in x = 0, if the function y(x) is an integral function not exceeding the minimum type of the order and the numbers an have the property that the differential operator 1
F(D) = anDn is applicable to all integral functions that are of n=0 the same kind as the function y(x). That this is indeed so is clear from the following: We consider the case where the function y(x) is an integral function
of the normal type i of the order a (0 < a < 1), while the numbers an have the property mentioned in theorem 1 (and hence in theorem 7).
We have already seen that for every finite value of x the series (3) dominates the series (4). Now it follows from (5) and (91) that the series (3) with I x I = R has a finite sum. Moreover, the series (3) with I x I = R dominates the series (3) with I x I S R. Then the series 00
I any(n)(x) converges uniformly on the domain I x 5 R.
n=0
If the function y(x) is an integral function either of the minimum type of the order a (0 < a S 1), or of the maximum type of the order a
(0 < a < 1), or of the maximum type of the order zero, and if the numbers an have the property mentioned in theorems 2, 3 and 4 respectively (and so in theorems 8, 9 and 10 respectively) then in a similar way we may refer to the inequalities (106), (121) and (131).
116
CHAPTER III
FURTHER INVESTIGATION OF THE FUNCTION h(x) = F(D) -+ y(x)
In this chapter we shall occupy ourselves with functions y(x) of which the order is less than 1 and with functions y(x) of the minimum type of the order 1. This means therefore 1) that we now consider functions y(x) that do not exceed the minimum type of the order 1.
The aim of chapter III is to improve considerably the results of chapter II in so far as they bear upon the functions that we shall now consider. These results of chapter II may be thus recapitulated: Theorem zz. Conditions: 1. Let the function y(x) be an integral function of either
a) the normal type a of the order a (0 < z < oo, 0 < a < 1), or b) the minimum type of the order a (0 < a S 1), or c) the maximum type of the order a (0 < a < 1), or d)
the maximum type of the order zero.
00
2. The differential operator F(D) is defined by F(D) _ Y, n=o
where in cases a), b), c), d) mentioned above the numbers a have the following property :
In case a) that the expression
G(z) =,, a l
(1)
n=o
zn
n,
defines an integral function not exceeding the normal type, less
than (aa) of the order I; In case b) that the expression (r) defines an integral function not exceeding the normal type of the order 1 ; In case c) that the expression (z) defines an integral function the order of which is less than 1; ') See definition
1.
117
In case d) that there exists a finite real number v, not dependent on n nor on z, such that the expression
H(z) =° 7 an n z"
(2)
defines an integral function the order of which is less than 1. A s s e r t i o n: In cases a)-d) the expression h(x) = F(D) -* y(x) defines an integral function of the same kind at most as the function y(x). Proof.
That in cases a) -d) the assertion is true follows immediately
from theorems 7, 8, 9 and 10 respectively.
So theorem 11 tells us that in cases a)-d) the integral function h(x) is of the same kind at most as the function y(x). Now it is the purpose of chapter III to show that - with one exception in the cases mentioned the function h(x) is even of the same kind as the function y(x), provided that in the differential operator F(D) not all coefficients a are equal to zero; the exception in question comes up for discussion in theorem 15 and it bears upon a case where the function h(x) is of a lower kind than the function y(x).
In order to prove this we shall treat these four cases separately and we shall do it in theorems 12, 13, 14 and 15 respectively.
The case where all coefficients a in the differential operator are equal to zero, will be left out of consideration all the time. For in this
non-interesting case the function h(x) is identically equal to zero. Then it is of the same kind as the function y(x) if the function y(x) is identically equal to a constant (see definition 7) ; it is of a lower kind than the function y(x), if the latter is not identically equal to a constant (see definition 9). Before we proceed to treat the first case, viz. that, where the function y(x) is an integral function of the normal type z of the order a
(0 < a < 1), we prove four lemmas. Lemma 17.
C o n d i t i o n s: The infinite sequence of real numbers co, Cl,
(3)
.. .
has the following three properties:
c, Z0
a) b)
for infinitely many values of n is c > 0,
C)
lim Cn = 0. n-. 00
118
(n = 0,1,...),
The sequence
1o, a1, .. .
(4)
is formed by all those positive integers At, arranged in ascending order that have the property that in the sequence (3) the number cA, (l = 0, 1, ... ) is only followed by numbers that are smaller than the number cA, 2).
A s s e r t i o n: For every finite real non-negative number b we have 1
1.
Jim sup {;t,!° cA'}7A9 = Jim sup x-.oo
1-.00
Proof. From the definition of the numbers X10, A1, ... it follows that, if in the sequence (3) there occurs a number between the numbers cA, and cA,+, (l Z 0), which therefore are consecutive numbers in the sequence c4, cA1, . . ., then this number is not greater than cA,+1 In fact, if this number, occurring between cA, and cA,+1, should be greater than cA,+1, then between cA, and cA,+1 in the sequence (3) there would occur a number 3) the suffix of which is not to be found in the sequence (4) while this number does have the property that in the sequence (3)
it is only followed by numbers that are smaller than the number in
question. But this contradicts the assumption that the sequence (4) is formed by all those positive integers that have the property stated. So if l is a non-negative integer with the property that in the sequence (3) between c,,, and cAl+1 there occurs a number c,,, then (n = ),-f-1, ..., AL")(5) C. C CA'+1
Now it follows from property c) that there exists a non-negative integer L, such that for every l Z L the inequality cA,+1 <
1
holds. Then, if l Z L and if in the sequence (3) between c,,, and c,,,+, 2) From the properties a), b) and c) of the sequence (3) it follows easily that the number of such positive integers x, is infinite. In fact, if the sequence 7.0, 1.1, ... should contain only a finite number of elements (or possibly none
at all), then this would mean that there would exist a positive integer N, such that for every n Z N the number c would have the property that it is followed by at least one number not smaller than this number c,,. From would contain infinitely many numbers this it follows that the sequence with m z N. that are not less than a positive number cm of the sequence Because of property b) such a number c,,, certainly exists. Hence lim sup c,, Z cm > 0, which contradicts property c). "'_°° 3) This number is not necessarily equal to the number occurring between the numbers cA, and cA,+1 mentioned in the preceding sentence, if in the sequence (3) between ca, and cA,+1 there occurs more than one number. 119
there occurs a number c,,, it follows from (5), because of n < Al+1, I
1
1
<
(Cn) n S (C21+)
(Cxl+)xl+l
(Al + 1 S n < A111)
This means that of the numbers 1
1 S n S 1+1)
(C.) n
(6) 2
the number (cx,+1)A`+1+1 is the greater. If in the sequence (3) between the
numbers cA, and c2t+l there occurs no number, then this last assertion is obvious, since in that case the sequence (6) contains only one number, viz. 1
z++ .
Now it follows from what we have just found that for every finite real non-negative number 8, the number Cat+1}zt+,
is the greater of the numbers (Al + 1 S n S 21+1).
{n!dcn}n
From this it follows without difficulty that the following formula holds : 1
1
lim sup {(A,+ !)o cxl+1}xl+ = lim sup {n!°
1
lim sup {n!°cn}n, n-.oo
l-+oo Z1+1 Sn 51.1+I
l-+oo
so that lemma 17 is proved. Lemma i8. C o n d i t i o n s: p denotes a fixed non-negative integer. The function gi(n) is defined for n. = 0, 1, .. , and it has the following two properties :
fi(n) is finite and 0 0 for n = 0, 1,
...
1
1.
In 00
Moreover, the expression M(z)
of finite order. Assert n s:
anzn defines an integral function 7t=0
1.
The expression 1V (z) = G, n=o
defines an integral function. 120
2. If the function M(z) is a Polynomial of degree k > 0 and if, moreover,
P z k, then the integral function N(z) is of a lower kind than the function M(z). 3.
The integral function N(z) is of the same kind as the function
M(z) in the following three cases :
a) M(z) is a polynomial of degree zero, 9) M(z) is a polynomial of degree k > 0 and, moreover, 0 S P S k-1, y) M(z) is a transcendental integral function of finite order. 4. The integral function N(z) is an integral function of the same kind
at most as the function M(z).
REMARK. We obtain a special case of lemma 18 if we take (n + P)! Then 00
N(z) = I (n + P) (n + p - 1) ... (n + 1) n=O
and so N(z)
dd vv
M(z).
Proof of lemma i8. First we consider the case where M(z) is a polynomial of degree k > 0 and p z k. From the Preparatory Chapter we know that M(z) is an integral function of the maximum type of the order zero. In this case N(z) is identically equal to a constant, viz. equal to 9?(0)ak if P = k, and equal to 0 if P k + 1. So the function N(z) is an integral function, which, according to definition 9, is of a lower kind than the function M(z). Next we consider the case where M(z) is a polynomial of degree zero.
So then M(z) = ao, so that N(z) = g7(0)ao if P = 0, and N(z) - 0 if p z 1. Therefore the function N(z) is an integral function and as the functions M(z) and N(z) are both identically equal to a constant, the function N(z) is, according to definition 7, of the same kind as the function M(z). Now let the function M(z) be a polynomial of degree k > 0, while
P satisfies the inequality 0 S p s k - 1. Then it is clear that N(z) is a polynomial of degree k - P > 0 so that it is an integral function. In this case the functions M(z) and N(z) are both polynomials of a posi-
tive degree. Therefore they are both integral functions of the maximum type of the order zero so that, according to definition 7, the function N(z) is of the same kind as the function M(z). Finally we assume that the function M(z) is a transcendental 121
integral function of finite order. Let its order be equal to p. Then with respect too the following two cases are possible: p = 0 and 0
lim {n!A I an l}n = 0
(7)
n-*oo
holds for every finite positive value of A. Now
{n!A I (n)an+v }n =
(8)
(n + p)!' I q (n)an+D I [
I a(n) I
{(n + p) ! Ian+v 1 }n+P 1 i n J
L
'
{(n + p) ... (n + 1)}A .
[{(n + p) ... (n+
n
1)}A
Besides, it follows from (7) that we also have i
lim {(n +
P)!-4
1
an+,
1} n+P = 0.
9L-, o0
From this formula and from (8) we see that for n -> oo the limit of the first factor in the last member of (8) exists and is equal to zero. Moreover, for n -+ oo the limit of the second factor in the last member of (8) also exists, as follows from the datum, and this limit is equal to 1. Then for n -* oc the limit of the last member of (8) exists as well and
this limit is equal to zero. From this it follows that the formula i
lim {n!A 19'(n)an+, I}n = 0 n-+oo
holds for every finite positive value of A.
This means that the numbers T(n)an+ , (n = 0, 1, ...) have the property P 3 so that N(z) is an integral function of the order zero. Besides, since the function M(z) is a transcendental integral function, infinitely many coefficients of its power series expansion are different from zero. Consequently, in the power series expansion of the function N(z) infinitely many coefficients T(n)an+9 are different from zero, since T(n) 0 0 for n = 0, 1, .... So the function N(z) is also a transcendental integral function of the order zero and hence of the maximum type of the order zero. Then, according to definition 7, it is of the same kind as the function M(z).
In the second case is 0 < p < oo. Now the function M(z) has the property P 1 so that formula (P,4) with an = an holds. So we have 122
for every positive S
an1}n=0,
urn {n! n->oo
while for every positive S that is less than e, i
i
limsup{n!e-al
and}n=oo.
n-.oo
Now we can prove quite easily 4) that for every positive S also i i iim {n! T p(n)an+v I }n = 0, n-* oo
while for every positive S that is less than e, 1
lim sup {n! QQ-a I T(n)a.+9 }n = oo. n-ioo
Consequently, N(z) has the property P 2 so that N(z) is an integral function of the order e. Now if the function M(z) is an integral function of the minimum type of the order e, then formula (P,7) with an = an holds. Then we may prove without any difficulty that formula (P,7) is also valid if in it we replace an by q)(n)an+,. Hence the function N(z) is an integral function of the minimum type of the order e. If the function M(z) is an integral function of the normal type y or of the maximum type of the order e, then the function N(z) is also an integral function of respectively the normal type y and the maximum type, of the order e, as we may prove similarly by using formula (P,9) and (P, 10) respectively. According to definition 7 the function N(z) is in each case
of the same kind as the function M(z). Herewith the proof of assertions 1, 2 and 3 is furnished. Now, in connection with definitions 7, 9 and 8, assertion 4 is evident.
Lemma i9. Conditions: The integral function y(x) satisfies one of the conditions a), b), c), d) mentioned in theorem rx under i. The differential operator F(D) is defined by F(D) =
anDn, where n=0
the numbers an have that property mentioned in theorem ii under 2. that bears upon the case mentioned under z. where the function y(x) belongs.
4) We use the formula that arises from (8) by replacing A by by
1
e - a
respectively.
1
e+b
and
123
Moreover, ao = ... = a9_1 = 0, aD F(D) = D'O(D), where
0 (p Z 1) and we put
00
O(D) = 2; an+vD". n=0
Assertions:
1.
have that property mentioned
The numbers
in theorem iz under 2., with an replaced by an+D, that bears upon the case mentioned under i. where the function y(x) belongs. 2. If u(x) = c(D) a y(x) is either a transcendental integral function, or a polynomial of degree not less than p + 1, or identically equal to a constant, then h(x) = F(D) -* y(x) is an integral function of the same kind as the function u(x). 3. 1/ u(x) is a polynomial of degree g where 1 5 g S p, then h(x) is an integral function of a lower kind than the function u(x).
Proof.
First we assume that the function y(x) comes under one
of the three cases a), b) or c). Now we shall prove that the expression
G1(z)zn nIa a
(9)
00
n=0
defines an integral function of the same kind at most as the integral function defined by the expression G(z) in (1). To that end we write the expression (9) in the form
{(n+P)!} n=0
n.
a
(n+. p)to a
and we apply lemma 18 with an = i (n = 0, 1, ...) and with 1 nia (n+p) (n = 0, 1, ... ). This function (p(n) obviously T(n) _ n! satisfies the conditions of this lemma. Then assertion 4 of lemma 18 tells us that G1(z) is an integral function of the same kind at most as the function G(z). If the function y(x) comes under case a) or case b) or case c) then, according to definition 4, definitions 8 and 2, and definition 8 respectively, the expression G1(z) defines an integral function
not exceeding the normal type, less than (ar)- a, of the order an integral function not exceeding the normal type of the order
1, 1
and an integral function the order of which is less than 1, respectively. This means that assertion I holds if the function y(x) comes under one of the three cases a), b) and c). 124
Now we assume that the function y(x) comes under case d). Then, in a similar way as in case of the expression G1(z) considered above, we may prove that the expression
v
H1(z) = 00 an n=0 Ii.
zn
defines an integral function of the same kind at most as the expression H(z) in (2). Then, according to definition 8, the expression H1(z) defines
an integral function of an order less than 1 so that assertion 1 now also holds.
Now we proceed to prove assertions 2 and 3. From theorem 11 it follows that the function h(x) = F(D) -- y(x) is an integral function which, in each of the four cases a)-d), is of the same kind at most as the function y(x). Hence the function h(x) is certainly an integral function of finite order and we have 0.
(10) h(x) = {DP¢(D)}- y(x) = {I00an+PDn+P}- Y(x) _
an+PY(n+P)(x).
n=0
n-0
Moreover, from assertion 1 just proved and from theorems 1, 2, 3 and 4 respectively it follows that the differential operator O(D) is applicable to the function y(x). According to definition 1 in each of the cases a) -d) this function y(x) is an integral function not exceeding the minimum type of the order 1. Then, with F(D) replaced by q(D),
the concluding remark of chapter II tells us that the series 00
an+PY(n)(x)
n=0
converges uniformly on and in every circle with centre at the origin. Then we have in every point x that lies in the finite part of the complex x-plane and for !5 = 0, 1, ... b) DP -- {2: an+PY1n1(x)} n=0
-
an+PY(n+v)(x)
n=0
Hence 00
(1 I)
DP --> u(x)
an+PY(n+P)(x).
n=0
Then it follows from (10) and (11), that we have h(x) = DP-* u(x) and so h(x) = u(P)(x).
6) Cf. Titchmarsh [1] p. 95. 125
Now, if u(x) =
0*
c"xn, then
n-0 CO
h(x) =
n=0
(n + p) ... (n +
1)c"+,x".
Then, applying lemma 18 with the subjoined remark, we see that assertions 2 and 3 are also true. This proves lemma 19.
The following lemma is due von Koch [1, p. 353-354]. In this chapter we shall only use it to prove theorem 15. Lemma 20. 1/ to the following system o l infinitely many linear equations with infinitely many unknowns
(12)
x1 + 12x2 + C13x3 + C14x4 + ... = dl = d2 x2 + C23x3 + C24x4 + .
x3+C34x4+
=d3
there exists a finite positive number M with the following two properties 00
1+2: IChk 12 5 M2
a)
(h = 1, 2, ...),
k=h+1
where M does not depend on h, 1
b)
lim sup
E d,,
1
Ih<
M
h-oo
then system (12) has a solution {xk} that satisfies the condition 1
(13)
1
urn sup Xk V< k+a,
M
and this solution is the only one of system (12) that satisfies (13).
Now we proceed to treat the case where the function y(x) is an integral function of the normal type z of the order a (0 < a < 1). For this case the following theorem holds:
Theorem 12. Conditions: The function y(x) is an integral function o l the normal type z 0/ the order a (0 < a < 1). The differential 00
operator F(D) is defined by F(D) _ 2, a"Dn where the numbers an n=0
are not all equal to zero. 126
A s s e r t ti o n: If the numbers an have the property that the expression G(z) =
(14)
a " zn
n=0
1
n!Q
defines an integral function not exceeding the normal type, less than
(at), of the order 1, then the integral function h(x) = F(D) -± y(x) is of the same kind as the function y(x). Proof. From theorem 7 it follows that h(x) is indeed an integral function.
Now we first assume ao - 0. Let the integral function y(x) have 00
In this power series
the power series expansion y(x) = I
expansion there are infinitely many coefficients bm different from zero; 'n-0 in fact, the function y(x) is not a polynomial, since its order is positive.
As the function y(x) is an integral function of the normal type r of the order a (a > 0) the coefficients bm have the property that, because of formula (P,9), with an = bm, p = a and y = r, the formula 1
1
(15)
1
lim sup {m!a I bm }m = (ar)a yn-*0O
holds.
Since the expression (14) defines an integral function not exceeding 1
the normal type, less than (at)
of the order 1, it follows from 1
lemma 6, with f(x) replaced by G(z), ,1 by (ar) a and a by 1, and with
z = 0, that we have (16)
lim sup {n!
1_
1
a
_
1
1
I an 1}n < (at) a.
n-00
With respect to the left-hand member of (16) there are now two possibilities :
The first possibility is the one where this left-hand member is _
1
positive, but less than (ar) Q ; the second possibility is the one where
this left-hand member is equal to zero. In the first case we put the number y equal to this left-hand member, i.e. (17)
y = lim sup {n!1
an I}'W
n- 00 127
so that with a view to (16) we have 0 < y < (UT)
(18)
In the second case we choose the number y such that it satisfies the inequality (18). With the help of the number y we define the numbers An in both cases as follows n! 1
(19)
° an = Any-
(n = 0, 1, ...).
Then
Ao=ao
(20)
0.
If the left-hand member of (16) is positive, formula (17) holds and from this and from (19) it follows that then we have i
lim sup I An I n= 1. n-b oo
If the left-hand member of (16) is equal to zero, then y satisfies (18) and from this formula and from (19) we see that in this case i
Jim1AnIn=0.
n-. -
Hence in both cases we certainly have 1
lim sup I An In S 1.
(21)
n-.oo
From formula (18) we see that we may choose a positive number s
such that 1
y{(ar) a + e} < 1.
(22)
Now we define the numbers Bn,(e) as follows 1
(23)
yn,!a bm
1
= Bm(s) &T) a + e}"
Putting (24)
(at) a + e = r1(e)
and 1
(25) 128
T1(0) = (ar)
(m = 0, 1, ...).
we see that from (15), (23), (24) and (25) it follows that
lim sup I B.(e) I n = ri(O) < 1.
(26)
In chapter II we have seen (formula (II, 6)) that
h(x) _ 1=0 A! {I an. bn+z (n + A)!) Because of (19), (23) and (24) it follows from this formula that
h(x) _
z=o A!
(In=0n! ° _
00
-
A) !
Q.B.+,t(--)(Yti(O)n( +
(ri(e))z A! a
(F)(r1(e))"+x
° Bn+x
ln=0
(n + 1)!)
A)1
°}
Putting for A = 0, 1, .. . (27)
DA(e) = L.t AnBn+A(e)(Yrl(e))n ( n n=0
+ /1
/
a
then it follows from the last but one formula that
h(x) _
(28)
X' x=o
(ri(e))z Da(e) i . Al
Now we shall prove that
i Jim sup {(rl(e))x I Di(e) I} x = r1(0).
(29)
A-oo
From theorem 7 we already know that the function h(x) is an integral function of the same kind at most as the function y(x), i.e.,
according to definitions 8 and 3, that the function h(x) does not exceed the normal type r of the order a. Then lemma 5, with I (x) = h(x),
A = r and a = a, and with x = 0, tells us that, because of (ri(e))Da(s)A!=
i °
which follows from (28), we have the formula
lim sup {(ri(e)) I Dx(e) I) a 5 (ar) a = r1(0) x-.oo
(because of (25)).
Now if we shall have proved that also (30)
Jim sup {(rl(e))'t I De(s) IF" Z rl(0),
then formula (29) is correct. Therefore we now prove formula (30). 129 9
To begin with it follows from (26) that lim I Bm(E) 1 = 0.
Moreover, it follows from definition (23) of the numbers Bm(e) that infinitely many of the numbers Bm(e) are different from zero, because infinitely many of the numbers bm are not equal to zero. Then we have for infinitely many values of m IB(E)I>0.
Let the sequence 4, 2i, ... be formed by all those positive integers A, arranged in an ascending order, that have the property that in the sequence
(31)
{ B0(E) 1.
the number I B,1t(e)
I
I B1(E) I, ...
(1 = 0, 1, ...) is only followed by numbers that
are smaller than this number I Bat(e) I. Then lemma 17, applied with c = I Bm(e) I and S = 0, tells us that the formula 1
1
lim sup I Bit(e) I _ = lim sup I Bm(e) I m
(32)
1-.C0
holds. Moreover, it follows from (27), where we replace A by A,r and from
the property of the number A,, that in the sequence (31) the number I Bat(e) I is only followed by numbers, that are smaller than I Bat(e) I,
that we have forl=0, 1, ... 6) IDa,(e)I
(33)
(n+Az
00
z I AOBx,(e) I R
n
00 r
AoBat(e) I-
n=1
I A. I I Bn+a,(E) I (Yzl(e))n {
n+ t 1- a 1
00
>IAOB1,(e)I-IBa,(e)IIIAnf(yz1(e))n \ n=1
1
n
/
°) All the series occurring in (33) are convergent, because the series in the last but one line of (33) converges as will be proved in the paragraph that follows formula (34). 130
00
> I AOB1,(E) I- I BA,(--)
I
1
n-1
1
At
I A. I (Yil(E))" 1
a
/
(because a < 1)
= I B1,(--)
111
AO -1
B1,(e)
1 n=1
I An I (Yt1(E))ttl
I ao I - - K(e)
I
(1 + Aa)°
(because of (20)),
-1
if we put K(e) =
(34)
n=1
I A. I (vil(e))".
Now because of (21), (24) and (22) the series in the right-hand member of (34) converges, while its sum K(e) does not depend on At.
From this it follows, since 0 < a < 1 and so
1 - 1 > 0, Cr
lim---K(e)1
= 0.
1
1
Jim sup I DA,(--) IT, 1-i00
lim sup I I BA,(--) 11'
I ao I - -
K()1 -1 } 1
(1+At)
l
= Jim sup I B11(e) 11` Jim S I ao I t
:
1-+00
K(ei
(1 +At)a
1111
,IT
1
1
= lim sup I BA,(--) 1-2-1 = lim sup I
In
n-i o0
1-100
(because of (32)).
From this and from (26) we see that the formula 1lim Z-+oo sup
rim
-L
I Dl,(--) h
i1( ) 1
Besides, since the sequence {I D1t(e) I a, }
(l = 1, 2, ...) is a sub-
1
sequence of the sequence {I D1(e) I A } (A = 1, 2, ...) we have 1
1
lim sup I D1(e) I 1 Z lim sup I D1,(e)
11` ,
i-, 00
131
so that also 1
lim sup I DA(s) I
zl( )
a-1W
From this formula it follows at once that formula (30) is correct. Consequently formula (29) is proved. As 'r1(0) > 0 it follows from the formulae (29) and (28), that the
the formulae (P, 4) and (P, 5), with n = 1, P = a and an = (r (E)) 1Da(E)
(35)
R!°
hold respectively. for every positive 6 and for every positive S, less than a. This means that the (integral) function h(x) has the property P 2 with P = a. Hence it is of the order a. Besides it follows from (29) that for the function h(x) formula (P, 9), with an
1a (z1(0))° hold. So the function h(x) is an integral function of the normal type 1 (r1(0))° _ (ar) ° = r defined by (35), P = a and y =
Cr
a
of the order a. Therefore, according to definition 7, the function h(x) is of the same kind as the function y(x). Hence, theorem 12 is proved for the case where a° 0 holds. Now we suppose a° = 0. As we have assumed that not all numbers
an are equal to zero there exists a positive integer p, such that
a°= ... =a,-1=0, a,
0.
Then we put F(D) = DPO(D), W
where ¢(D) = n=°
an+9Dn.
Since the numbers a have the property mentioned in theorem 11 under 2. bearing upon case a) mentioned under 1. the function y(x) belongs, according to lemma 19 (assertion 1) the coefficients an+D have the same property, with an replaced by an+,,. This means that the expression an
G1(z)
1
Zn
n-0
defines an integral function not exceeding the normal type, less than
(or) of the order 1. Moreover, since in O(D) the coefficient of D°, i.e. a, differs from zero, it follows from theorem 12, with F(D) replaced by O(D) and so an 132
that the function u(x) _ O(D) - y(x) is an integral function by of the same kind as the function y(x). Hence, certainly the function u(x) is a transcendental integral function. Then it follows from lemma
16 (assertion 2) that the function h(x) is of the same kind as the function u(x). Therefore it is also of the same kind as the function y(x). Herewith theorem 12 is proved conclusively. Next we shall consider the case where the function y(x) is an integral
function of the minimum type of the order a (0 < a 5 1). For this case the following theorem holds: Theorem 13. C o n d i t i o n s: The function y(x) is an integral function of the minimum type of the order a (0 < a S 1). The differential 00
anD", where the numbers an n-0 are not all equal to zero. A s s e r t i o n: I/ the numbers an have the property that the expression
operator F(D) is defined by F(D) =
(36)
G(z) = I -a n=O n!
i
zn
defines an integral function not exceeding the normal type of the order 1, then the integral function h(x) = F(D) --* y(x) is of the same kind as the lunction y(x). Proof. From theorem 8 it follows that h(x) is indeed an integral function.
Now we first assume a0 = 0. Let the integral function y(x) have 00
the power series expansion y(x) _
bmxm. Then infinitely many ne=0
of the coefficients bm differ from zero, since the order a of the function y(x) is positive.
Since the function y(x) is an integral function of the minimum type of the order a (a > 0) the coefficients bm have the property that, because of (P, 7) with an = b0, and e = a, the formula .
(37)
1
lim {m! a I bm j} n' = 0 1)L-+co
holds.
Besides, the function y(x) has the property P 1, so that formula (P, 5) with n replaced by m, an by bm and P by a, holds, i.e. we have for every positive 6 (38)
limsup
=o0
n'-*o0
133
Since the expression (36) defines an integral function not exceeding
the normal type of the order 1, it follows from lemma 4, with 1(x) replaced by G(z) and a by 1, and with z = 0, that (39)
an I}n < oo.
0 5 lim sup {n!1
With respect to the expression (40)
(n!1_
1
Q I an I}n
lim sup n-±oo
there are two cases possible. The first case is that where this expression is positive and finite; the second case is that where this expression is
equal to zero. In the first case we put the number
equal to this
expression, i.e. (41)
=limsup(n!1 ° Ianl}n, n-.oo
so that with a view to (39) we have
0 < C < oo.
(42)
In the second case we choose the number 4 such that it satisfies (42). With the help of the number C in both cases we define the numbers
An as follows (43)
n!
1-1 ° an = Ann
(n
= 0, 1, ...).
Then (44)
AO = ao
0.
Now, if the expression (40) is positive, then formula (41) holds and from (43) it then follows that we have Iim sup I An I n = 1. ft- W
If the expression (40) is equal to zero, then C satisfies (42) and from this and from (43) we see that in this case 1
limIA.in =0. So in both cases certainly
(45)
urn sup A,, 1T n-,.oo
134
51.
Then with the help of the numbers C we define the numbers B. by (46)
(m=0, 1,...).
j
M
There a denotes a positive number that has been chosen so large that IAnI (47)
an
n=1
This is certainly possible, since we have assumed ao numbers An satisfy (45).
0 and since the
Then from (37), (46) and (42) it follows that 1
limIBm[m = 0 m-ioo
and certainly that
limIBmI=0. nt-+ 00
Besides it follows from (46)
m!°IB,nI =m!abmI (m=0, 1,...) and from this, from (38) and (42) we see that for every positive b 1
lim sup {m!° I B,,, IF-- = oo.
(48)
In-00
In chapter II we have seen (formula (II, 6)), that we have {nI anon+z (n + A)!)-
h(x) = AI -0
Because of (43) and (46) it follows from this that h(x) =
g
1
n=
C A nBn+t
(n + 1) !
1
n !1 - ° (ab)n+l (n + A). a 00
_
A
-Bn+xn + al
n
n
a's
1
(ac)x 4!o
l
l j
Putting An
(49)
E.4
( = n=0 7 nBn+x a"
n
+ 2 1n
>
o
(2 = 0, 1, ... ),
it follows from the last but one formula that (50)
h(x) =
00
Ex
xz
A-0 (aC)
la 135
Now we shall prove that for every positive 6 1
1-Ca
lim sup
(51)
Ez
°
I
A = oo.
r)
From definition (46) of the numbers Bit follows that infinitely many of the coefficients B,,, differ from zero because infinitely many are not equal to zero. So then for infinitely many of the numbers
values of m we have B. > 0. Let the sequence AO, Al,
. . .
be formed by all those positive integers
A,, arranged in an ascending order that have the property that in the sequence
IBOI, 1B1I, ... the number I BA, I (1 = 0, 1, ...) is only followed by numbers that are smaller than this number I BA, I. Then it follows from lemma 17, applied with c = I B1, that for every positive S (52)
1
(53)
lim sup {A,!° I BA, I)r: = lim sup {m!° I B,a, 1}. I-00
M-00
Moreover, it follows from (49), with A replaced by A,, and from the
property of the number A, that in the sequence (52) the number I BA, I
BA, I, that for
is only followed by numbers that are less than
1=0, 1, ...
7)
(54)
I Ea, I _
oo A n
-a I A OBA, I --
z AOBA,I-
2
.+ AL1
B,,+A=
an n-1 a
Cn
(n
An I
a
n=1
0o
---IBn+A,I
I A.
z IAoBA,l-1BA,I n=1
I
a an
z1AOBA,I-IBA,1 iA"1 n=1 an = I BA, I
{I
a0 1n=1 -' i-an 00
n
I
i
Cn
n
a
)
n
Ai\1-
J _.u
because 1
1501 a /
(because of (44)).
7) All the series occurring in (54) are convergent, since the series in the last member of (54), i.e. the series in the left-hand member of (47), converges. 136
From this it follows, in connection with (47), that
(1=0, 1, ...).
I
I
From this inequality it follows that for every positive 6 Jim sup {A,!° I EA, I}xt Z lim sup {A,!6 I Ba, I
I ao I}z`
l_o0
= lim sup {A,!° I Bxt I}xt = lim sup {m!° I Bm I}nt m-.00
1-.- oo
(because of (53)). Besides, since the sequence {I EA,
(1 = 0, 1, ...) is a sub-sequence of the sequence {I E., I} (A = 0, 1, .. .), we have for every positive S t
1
Jim sup {A!° I Ez I}x Z lim sup {A,!' I Eat l}Tt A-00
1-.oo
and from this, in connection with the last but one inequality and (48), we conclude that formula (51) is correct.
Moreover, we already know from theorem 8 that the function h(x) is an integral function of the same kind at most as the function y(x). According to definitions 8 and 1, it is therefore an integral function
not exceeding the minimum type of the order a. Then, for x = 0 lemma 7, applied with I (x) replaced by h(x) and a by a, tells us because of i _x . A!1
(as follows from (50)),
a
that (55)
lim
A!°-
t-.oo
Ex I
I
-)
= 0.
a
C (ad)z A ! a
From this and from formula (51), written in the form 1
lim sup A ! a
I
lEA
+a
(
i
I
' = oo,
A!--
which is valid for every positive 6, it follows that the function h(x) has the property P 2. So it is of the order a. Moreover, it follows from (55) that formula (P, 7), with n = A,
e = a and
EA (aC), A ! a
holds, so that the function h(x) is of the minimum type of the order a. 137
So, according to definition 7, it is of the same kind as the function y(x). Hence theorem 13 is proved in case ao z 0. Now we assume ao = 0. In this case we can furnish the proof in literally the same way as in the similar case in the proof of theorem 12, provided that in the latter proof we replace the expression "case a)" by "case b)" and the expression "the normal type, less than (or)
by "the normal type". Herewith theorem 13 is proved conclusively.
Now we shall treat the case where the function y(x) is an integral
function of the maximum type of the order a (0 < a < 1). For this case we have Theorem 14. C on d i t i o n s: The function y(x) is an integral
function of the maximum type of the order a (0 < a < 1). The dif00
ferential operator F(D) is defined by F(D) = T, a,Dn, where the numbers n=o a,, are not all equal to zero. A s s e r t i o n: I/ the numbers a,, have the property that the expression
G(z) =
(56)
-
n=0
-i zn n I-1
defines an integral function the order of which is less than 1, then the integral function h(x)=F(D)-> y(x) is of the same kind as the functiony(x).
Proof. From theorem 9 it follows that h(x) is indeed an integral function.
First we assume ao
0. Let the integral function y(x) have the
power series expansion y(x) = !n=o
b,,,x". Then of this power series
expansion infinitely many coefficients are different from zero because the order of the function y(x) is positive.
Since the function y(x) is an integral function of the maximum type of the order a (a > 0) the coefficients bm have the property that, according to formula (P, 10) with a = bm and e = a, the formula (57)
lim sup {m! a I bm JIM = 00 9n-.oo
holds.
Besides, the function y(x) has the property P 1, so that formula (P, 4), with n replaced by m, an by bm and e by or, is valid, i.e. for every
positive e we have (58)
lim {m!a -f I bm l}m = 0. m-. oo
138
Let the order of the integral function G(z) defined by (56) be equal to e < 1. This function G(z) has the property P 1, with a replaced by a
i so that for every positive 6
(59)
' Ia
lim
!i+a
n- co
J
i.._
n' = 0.
n!
Now we choose a number 0 such that it satisfies
0<0< 1 -1, if 0<<1; and
0<0<00, if e=0. Then it follows from (59) that for the number 0 thus chosen the formula (60)
n!+e' ai n = 0
lim n-.oo
n!a
holds. In fact, if a satisfies 0 < e < 1, then 1 + 0 < there exists a positive number S for which I
we may take simply S =
1+B
I
e+6
1e
.
Consequently
= I + 0. If
0,
.
With the help of the number 0 we now define the numbers A,, in the following manner: (61)
n!1
at,=At,
+e
(n=0,1,...).
Then (62)
AO = ao 0 0,
while from (61) and (60) it follows that 1
(63)
lim I A. I n = 0. n-.oo
Besides, since 0 > 0 and 0 < a < 1, so - - 1 > 0, we may choose a number il such that it satisfies the inequality (64)
0 < 27 < min (o,
--1
For the number 17 thus chosen formula (58) holds, if in it we put 139
e equal to ij, because (58) is valid for every positive a and hence also
for e = . Therefore we have lim {m!a -" I b,,, 11m = 0.
(65)
ni-> 00
With the help of the number 17 we now define the numbers Bm by 1
m!0 -"b,,, = B.
(66)
(m = 0, 1, ...).
For these numbers Bm we have, with a view to (65), 1
limIB.mllrt ..-0,
and so certainly
lim)BmI=0. ,n-.00
From (66) and (57) it follows that for the numbers Bm the formula 1
(67)
lim sup {m!" I Bm I}gin = 00 9n- 00
also holds.
In chapter II we have seen (formula (II, 6)) that h(x) _ Y.
X-1
-
- {Y. a,,bn+x(n + A)!}-
;M0 A. n=O
Because of (61) and (66), it follows from this that xx
o0
AnBn+x
h(x)
n
nt
(n + A)1 }
-+e (n + A)!a1 0 00
n
AnBnl (n + A}1 -
x
2-0 AlQ
J
ln= n!e`'' \
n
1
/
Putting A}1- a +" AnBn+x (n + n=o n!e-" n it then follows from the last but one formula that
(68)
cc
b(Z)
2: Ar a=o
i X,
a
`
Aio
Now we shall prove that the formula 1
(69)
lim sup {A!" I MA I} A-.0o
is correct.
140
= 00
(A=0, 1,
+o
From definition (66) of the numbers B. it follows that infinitely many of the numbers Bm differ from zero, since infinitely many of the numbers bm are not equal to zero. So then for infinitely many values of m the inequality I B. I > 0 is satisfied.
Let the sequence AO, A1, ... be formed by all those positive integers At, arranged in an ascending order that have the property that in the sequence
IBOIB11,... the number I B,, I (l = 0, 1, ...) is only followed by numbers that are smaller than this number I B,, I. Then it follows from lemma 17, applied with c = I Bm 1, that for every positive 6 1
1
lim sup {At!6 I B,t I}xt = lim sup {m!° I Bm I}m. m- e0
1_00
so that we have
Hence this formula also holds for S = 1
(70)
1
lim sup {A,!° I B,, I}ae = lim sup {m!'' I Bm IF-. m-oo
l-r o0
Moreover, it follows from (68), with A replaced by At, that for I = 0, 1, ... we have the inequalities 8) (71)
1 Mat I o,
ZIAOBI, I-
)1_+
AnBn+a, 1
I AnBn t I n n At 1-1+0 n! 0'' n=1 J 1 A. I /n + At11- I +n 00 Q IAOB1,l-I B,, I T !n n=1 n
ZIA
R2t I-
IA n
IAOB,,I-IB,,1 2, -
=1 (1 +
(because of
=IBa,l Iaol-
At)
-i-o
-1 - 77> 0 and 0-77>0, see L
(64))
(because of (62))
00
if we put L = Y, I An 1, which series, because of (63), converges. n=1
8) All the series occurring in (71) are convergent since the series in the last
but one line of (71) converges because of (63). 141
Consequently we have lim sup {A,!" I Mat 1-+00
-
L
1
lim sup {A,!" I Bat I}ar lim
I ao I - - 1
1-+00
1-,.0o
-i L
z lim sup A,!" I Ba, I C ao - -
a,
(1 + 1
1
= lim sup {A,!" ( Bat I}zt = lim sup {m!" I B,,, lion
mz-.o
(because of (70)).
Besides, since the sequence {I Ma, 1} (1 = 1, ...) is a sub-sequence of the sequence {I Ma I} (A = 1, ...) we have the formula 1
lim sup {A!" I Ma I }
1
Z lim sup {A,!" I Ma= I).
From this, from the last but one formula and from (67) it follows that formula (69) is correct. This means that the formula 1
(72)
1
1
lim sup A ! 0 (A r I Ma a-.a0
, 1 A! a I
a = 00
holds. Moreover, from theorem 9 we already know that the function h(x) is an integral function of the same kind at most as the function y(x). So, according to definitions 8 and 5, it does not exceed the maxi-
mum type of the order a. Then for x = 0 lemma 8, applied with I (x) = h(x) and a = a, tells us that for every positive S 1
(73)
lim A! a+' 2! 4-900
1
1
Ma
1
1 a = 0.
2!o
It now follows from (73) and (72) that the function h(x) has the
property P 2 with e = a, so that it is an integral function of the order a. Besides it follows from (72) that for the function h(x) formula (P, 10) with n = A, p = a and
att=A!"Ma
li A!Q
holds. Consequently the function h(x) is of the maximum type of the order a. According to definition 7 it is therefore of the same kind as the function y(x). 142
Herewith theorem 14 is proved for the case in which as 0 0. Now we assume ao = 0. In this case we may furnish the proof in literally the same way as in the similar case of the proof in theorem 12, provided that in the latter proof we replace the expression "case a)"
by "case c)" and the expression "not exceeding the normal type, less
than (ar) o , of the order 1 " by "the order of which is less than 1 ". This completes the proof of theorem 14. Finally we consider the case where the function y(x) is an integral function of the order zero. For this case we have Theorem 15. A. C o n d i t i o n s: The function y(x) is a transcendental integral function of the maximum type of the order zero. The differential operator
0 F(D) is defined by F(D) _ Y,
a are not
n=0 all equal to zero. A s s e r t i o n s: 1. 1/ the numbers a have the property that there exists a finite real number v, depending neither on n nor on z, such that the expression
H(z) _
(74)
"z" n. a
n=0
defines an integral function the order of which is less. than 1, then the integral function h(x) = F(D) - y(x) is of the same kind as the function Y(x)
2. The function h(x) is a transcendental function as well. B. C o n d i t i o n s: The function y(x) is a polynomial of degree
0 q > 0. The differential operator F(D) is defined by F(D) = T, a, D".
Assertions :
n=0 1.
If of the numbers a0, ... , a,_1 at least one
differs from zero, then the integral function h (x) = F(D) -> y(x) is of the same kind as the function y(x). 2. The function h(x) mentioned in the preceding assertion is a polynomial of degree q - r, where r is the suffix of the first number in the sequence a0, . . ., a,_1 that differs from zero. 3. 1/ all the numbers a0, ... , a,_1 are equal to zero, then the function h(x) is of a lower kind than the function y(x). 4. The function h(x) mentioned in assertion* 3 is identically equal to a constant. C.
C o n d i t i o n s:
The function y(x) is identically equal to a
constant. The differential operator F(D) is defined by F(D)
_
00
n=0 143
A s s e r t i on: The integral function h(x) = F(D) -* y(x) is of the same kind as the function y(x). Proof.
We shall begin by proving the assertions in A. From
theorem 10 it follows that in this case h(x) is an integral function. First we assume ao 0. Let the integral function y(x) have the 00 b,,,xm. As the function y(x) is a power series expansion y(x) _ 0
transcendental function (of the order zero) it is not a polynomial so that in this power series expansion infinitely many of the coefficients bin differ from zero. The function y(x) has the property P 1 with e = 0, so that for ever YPositive e formula (P, 4), with n = m, e = 0, b = 1 and an = b9n. holds. So we have for every positive e 1
lim {m!` I
(75)
0.
If the order of the integral function H(z) defined by the expression (74), is equal toe (0 S e < 1) then the function H(z) has the property P 1, so that for every positive number S formula (P, 4), with an replaced
by
ii
(76)
holds. Therefore the formula a n
lim lyt!e+a n .oo `
n =0
holds for every positive 6 and so also for b = 1 - e. If now in (76) we replace 6 by 1 - N we find 1
(77)
lira {n!'-' I an 11-It = 0. n-,.oo
Now we define the numbers An by n!1_q
(78)
an = An
(n = 0, 1, ...).
Then (79)
AO = ao a 0.
From (77) it follows that the numbers An just defined satisfy the relation (80)
lim I A n I n-oo
144
= 0.
In formula (75), which holds for every positive e, we now put e equal to a number ,u, satisfying (81) It > max (l, v), and of a fixed choice.
With the help of the number it thus chosen we then define the numbers Bm in the following way
(m = 0, 1, ...).
m!"bm = Bm
(82)
it follows
From this definition and from formula (75) with e that the numbers Bm satisfy the relation 1
lim I Bm I n+ = 0.
(83)
M-00
In chapter II we have seen (formula (II, 6)) that h(x) = ado
{
470
an bn+x (n { 2) ! }.
Because of (78) and (81) it follows from this formula that Xx
00
h(x) _ I00- {nI n!`-'An(n + a ,)!-"Bn+,t (n + A)!} x I 00 A.B. y., (n x=o
-It -w
n=0
n.
\\
A)' -"}
n
Now putting
un + A 11-" Nx = 0 AnBn+x n!"-q n ) n-0
(84)
(A = 0. 1. ...).
it follows from the last but one formula that
h(x) _
(85)
N,t
We then put (86)
n r" -
= atn (A = 0, 1, ... ; n = 0, 1, ... ),
( n
by which (84), after interchanging the right- and left-hand members, becomes 00
(87)
I axnBn+a = Nz
n=0
(A = 0, 1, ...).
In which we have because of (86) and (79) (88)
a,,0 = A0 = a0 0 0
(A.= 0, 1, ...). 145 10
Now we divide both members of formula (87) by aao and we put (89)
aan axo
N.,
= an
(A = 0, 1, ... ; n = 0, 1, ... ),
Nx
aa0
by which (87) becomes 00
(90)
n=0
anBn+N.,*
(A=0, 1, ...).
ao=1
(A= 0,1,...).
So here (91)
Now for the quantities aa*n occurring in (90) we have for A = 0,1, ... 9) 00
(92)
n=1 anI2=1+an12
n=0
+
I J
+ -1T
sl+ =
a0
J2 ni
1
}0
J ao 12 ri
J An J2 '1
(because of (89))
J aan 12
nt2(/, -,,)
(n -
A)2_2/
I
J An 12 T12 J2
a
(because of (88) and (86))
(because of (81))
n
K2,
if we put
1+ Iao121>
K2
(K>0).
From (80) and (81) it follows that the number K is finite. Moreover,
K does not depend on A and for A = 0, 1.... we have as we have just seen in (92) 00
(93)
n=0
an
J2
K2.
Now if the function h(x) were a polynomial it would follow from (85), if g (g z 0) is the degree of this polynomial, that (94)
Na=Nx =0
for Azg+1.
9) All the series occurring in (92) are convergent because the series oc-
curring in the last but one line of (92) converges because of (80) and (81). 146
Then we consider the following system, connected with (90), (95)
7,
a system of infinitely many linear equations with $,+x, $9+2 as unknowns. Since this system is homogeneous and (91) and (93) are
satisfied, we may apply to this system lemma 20 with M = K. According to this lemma system (95) has a solution having the property that I
(96)
lim sup
1
< -K
and this solution is the only solution with the property (96).
However, the system (95) has the trivial solution n = 0 (n = g + 1, ... ). Hence this trivial solution is the only solution of system
(95) that has the property (96). But from (90) with N.,* = 0 for A Z g + I (because of (94)) it follows that also S n = Bn (n = g + 1, .. ) is a solution of the system (95) and because of (83) this solution n = Bn
(n = g + 1, ...) has the property (96) as well. This means that the
solutions n = 0 (n=g+ 1, ...) and n = B (n=g+ 1, ...) of the system (95) are identical, so that we find B,, = 0 (n = g + 1, ...). Then it follows from (82) that also bn = 0 (n = g + 1, ...). But then there are not infinitely many of the coefficients b,n of the power series
expansion of the function y(x) different from zero. Here we meet a contradiction with the assumption that infinitely many of the coefdiffer from zero. Thus we conclude that the function h(x) ficients is not a polynomial. Hence it is a transcendental function. Besides, theorem 10 tells us that the function h(x) is an integral function of the same kind at most as the function y(x). Then, according
to definitions 8 and 6, the function h(x) is an integral function not exceeding the maximum type of the order zero. Since it is not identically equal to a constant (it is not even a rational integral function
of a positive degree as we have just proved), according to what we know from the Preparatory Chapter, it is an integral function of the maximum type of the order zero. According to definition 7 it is therefore of the same kind as the function y(x). We now assume a0 = 0. In this case we may prove in literally the same way as in the similar case in the proof of theorem 12, provided that in this theorem we replace the expression "case a)" by "case d)" and the passage "G2(z)
= n=0 Y
an+1v zn
n! 147
defines an integral function not exceeding the normal type, less than 1
(ar) o , of the order I" by "H1(z)
= n=0
_an
. zn n!
defines an integral function the order of which is less than 1 ". This completes the proof of assertions I and 2 of case A. Now we are going to prove the assertions of case B. Here the function
y(x) is a polynomial of degree q > 0 so that it is an integral function
of the maximum type of the order zero. If of the numbers of the sequence a 0 .
. .
. . a0_1 at least one differs from zero and if ar is the first
number in this sequence that is not equal to zero, then 00
00
h(x) = F(D) - y(x) = (2; a,Dn) - Y(x) = ay(t)(x) + Y, any(n)(x), n=r+1
n=0
from which appears that the function h(x) is a polynomial of degree
q - r z 1. Hence the function h(x) is also an integral function of the maximum type of the order zero, so that, according to definition 7, it is of the same kind as the function y(x). If none of the numbers a0, . . ., av_1 differs from zero, then 00
h(x) =
any(n)(x),
n=q
so that in this case the function h(x) is identically equal to a constant. Then, according to definition 9, it is of a lower kind than the function Y(x)
Herewith the assertions of case B are proved 10). The assertion of case C is obvious. Hence theorem 15 is proved conclusively.
10) The assertions of case B may also be proved with the help of lemma 19. 148
CHAPTER IV
ON THE DIFFERENTIAL EQUATION F(D) --> y(x) = h(x)
If the function y(x) is a given integral function not exceeding the minimum type of the order I and if the differential operator F(D) co
a,,Dn, where the numbers a.R satisfy certain is defined by F(D) _ n=0 conditions, then, as we have proved in chapter III, the function h(x) defined by F(D) a y(x) = h(x) (1) .
is always an integral function of the same kind as the function y(x), except in one case. If the function y(x) is namely a polynomial, then it may occur that the function h(x) is of a lower kind than the function y(x).
In this chapter we shall treat the "converse" case. Indeed, we now suppose that the function h(x) instead of the function y(x) is given. Of this function h(x) we assume that it is an integral function not exceeding the minimum type of the order 1. Moreover we assume
that the numbers an satisfy certain conditions that resemble very much the conditions they satisfied in chapter III in the corresponding cases concerning the function y(x). We shall now prove that in case as 0 (1), considered as a differential equation for the function y(x), has one and only one solution y(x) = fi(x) that is an integral function of the same kind as the function h(x). Moreover we shall prove that this differential equation has no solution that is an integral function of a lower kind than the function h(x). If ao = 0- but at least one of
the numbers an 0, then the differential equation has more than one solution that is an integral function of the same kind as the function h(x). In case h(x) is not identically equal to a constant the number of such solutions is oo where P denotes the number of 0 begin-coefficients of the power series 7, anDn which are equal to zero. n=0
The case where h(x) is identically equal to a constant 1) will be treated 1) Here zero is included. In this case the differential equation is a homogeneous one. 149
in theorem 19 where also other results are stated for the case where the function h(x) is an integral function of the order zero. In theorem 16 we consider the case where the integral function h(x) is of the normal type r of the order a (0 < a < 1). In theorem 17 the case is treated where the integral function h(x) is of the minimum type of the order a (0 < a S 1). Theorem 18 deals with the case where the integral function h(x) is of the maximum type of the order a (0 < a < 1). Finally, in theorem 19 the case where the integral function h(x) is of the order zero comes up for discussion. Before proceeding to treat the first case we give a lemma. Lemma 21. If the function 4p(x) is an integral function of finite order, then the indefinite integral [(x)dx
is an integral function of the same order as the function qp(x).
Let the function T(x) have the power series expansion 71 cnx". If the order of this function is equal to P (Lo Z.0),
Proof.
00
na0
then this function has the property P 1. So for every positive number
6 we have 1
1
1im {n! ¢+d I C, J)n = 0,
(2)
n-.oo
while, if e > 0, we have for every positive 6, less than 1
1
lim sup {n!¢-a I C. IF = oo.
(3)
n-+oo
Now f92(x)dx =
I+
c
nmO n + 1
xn F 2
=
Ynx",
n=0
where I' denotes a finite (complex) constant of integration and Yo = T, Yn
__
c"-1
n
(n= 1, 2, ...).
For n Z 1 we have for every positive number 6 {n! e+°
1
1
y, I}= Ini¢+a
1 cn-1 I l n
n [{(n - 1)!¢+d
150
1 Cn-: In ,(n¢ +d '1)n,
from which, in connection with (2) it follows that for every positive 6 1
1
lim {n! a+e
(4)
I
Y. 1} n = 0.
n-soo
In case e = 0 it follows from this formula that fq,(x)dx has the property P 3 with A =
and an = y,,. Hence it is an integral function
of the order zero and therefore it is of the same order as the function 1F(x).
In case e > 0 we may similarly prove by using formula (3) that for every positive 6, less than 1
1
lim sup {n!e-e yn }n = 00.
(5)
n- oo
Then it follows from (4) and (5) that the function f T(x)dx has the property P 2. Hence it is an integral function of the order a so that it is of the same order as the function q)(x).
We now proceed to treat the case where the function h(x) is an integral function of the normal type r of the order a (0 < a < 1). Theorem i6. C o n d i t i o n s: The function h(x) is an integral function of the normal type r of the order a (0 < a < 1). The differential 00
operator F(D) is defined by F(D) =
where the numbers an n=o are not all equal to zero. A s s e r t i o n s: 1. If the numbers an have the property that the expression
G(z) = 7
(6)
an
n=U n!
1
zn
defines an integral function not exceeding the normal type, less than 1
(ar) - Q , of the order 1, then if ao 0 0 the differential equation (7)
F(D) -* y(x) = h(x)
has one and only one solution y(x) = rl(x) that is an integral function of the same kind at most as the function h(x). 2. The function n(x) mentioned in assertion i is an integral function of the same kind as the function h(x). 3.
1/ the numbers a have the property mentioned in assertion r
and if, moreover, ao - ... = a9-1 = 0, a9 0 0 (p z 1), then the set of all the solutions of the differential equation (7) that are integral 151
functions of the same kind at most as the function h(x), consists of oon functions. The difference of each pair of these functions is a polynomial
of a degree not exceeding p - I. 4. The solutions mentioned in assertion 3 are all of the same kind as the function h(x). Proof. First we shall prove assertion 1. As ao 0 0 we are allowed 1 we may divide both members to assume as = 1, because in case ao of equation (7) by a0.
Since the numbers a have the same property as in theorem 12, formula (III, 16) of the proof of theorem 12 now also holds. With the help of the number y defined in the proof of theorem 12 and which therefore satisfies (III, 18), we again define the numbers A,,
(n = 0, 1, ...) by means of (III, 19). So these numbers A have the property (III, 21). We have AO = 1 because we assumed ao = 1. Now let the function h(x) have the power series expansion h(x) _ 57,00dAxA.
(8)
A=O
From the datum that this function h(x) is an integral function of the normal type r of the order a (0 < a < 1) it follows that the numbers dx have the property I
(9)
1
1
lim sup {.1!Q I dA }x= (ar);,
because of formula (P, 9), with o = a, a,, = dx and y = r. As in the proof of theorem 12 we now choose a positive number e such that inequality (III, 22) holds, which is possible because of (III, 18). Then we define the number r1(e) as in (III, 24) so that from (III, 22) and (III, 24) it follows that yr1(e) < 1.
(10)
Moreover we define the number r1(0) by (III, 25), i.e. by r1(0) _ (ar)
(11}
Now the numbers D2 are defined as follows (12)
A!
(2 = 0, 1, ...).
dx = (r1(e))2DA
Therefore these numbers D2 depend on the choice of e. Then it follows from (12), (9) and (11) that 1
(13)
lim sup DD A =
tc0 ri(e)
152
.
We are going to prove now that the differential equation (7) has one and only one solution y(x) = 27(x) that is an integral function of the same kind at most as the function h(x). According to definitions 8 and 3 such a function n(x) is an integral function not exceeding the
normal type a of the order a (0 < a < 1). If it has the power series expansion n(X) _
(14)
Nmxm,
nx=0
then it follows from lemma 5, with I (x) replaced by q(x), A by r and a by a, and with x = 0, that the numbers 1 m satisfy the relation 1
(15)
1
lim sup {M! a
i
(Or) 0.
I } nl.
?11_
Conversely, if the numbers #,,, occurring in the right-hand member of formula (14), satisfy the relation (15), then it follows from lemma 9 (formula (17)) with n replaced by m, a by a, A by r and f(x) by i(x), that ?I(x) is an integral function not exceeding the normal type r of the order a. Then, according to definitions 3 and 8, it is of the same
kind at most as the function h(x). Hence condition (15) is necessary and sufficient for the expression ,I(x) in (14) to define an integral function of the same kind at most as the function h(x).
In connection with our definition of the notion "solution of a differential equation of the form (7)" we now have to prove that we can determine the numbers (3,,, uniquely, such that a) they satisfy (15) ; /3) the differential operator F(D) _ is applicable to the function 77(x) defined by (14);
n=o
y) the function (16)
S)
(17)
k(x) = F(D) - n(x) is an integral function; the relation k(x) - h(x) is satisfied.
Since the numbers a have the property mentioned in assertion I and since, (15) being satisfied, the function q(x) defined by (14) is an integral function not exceeding the normal type r of the order a (a > 0), according to theorem the differential operator F(D) is applicable to the function fi(x). 1
So condition fl) is fulfilled as soon as (15) is satisfied. 153
As the function j(x) is an integral function not exceeding the normal type r of the order a (0 < a < 1) it follows from the assertion
mentioned in the remark following the proof of theorem 7, with y(x) replaced by n(x), that the function k(x) is an integral function. Therefore condition y) mentioned above is fulfilled as soon as formula (15) holds.
So it remains to be proved that we can choose the numbers am in one and only one way such that (15) and (17) are both satisfied. (Here k(x) is deduced from (14) by means of (16), while h(x) is the given function occurring in (8)). As we have seen in chapter II (formula (II, 6) with h(x) = k(x) and bn+z = Nn+z) the function k(x) has the power series expansion k(x) _
(18)
{
z=o A! n=o
anon+z {n + A) !j.
Therefore we have to prove that we can determine the numbers 1 m (m = 0, 1, ...) with the property (15) in one and only one way such that they satisfy the following system of infinitely many linear equations (19)
n=O
an #n+A (n + A)! = dz
(A = 0, 1, ...),
which system arises from (17), (18) and (8) by equating the coefficients of equal powers of x.
Making use of (III, 19) and (12) we see that we may write system (19) in the form 2; y A,, n!0
{A = 0, 1, ...}.
Nn+a (n + A)! = A!a
Obviously the latter system is equivalent to the system of equations 00 1 /n + A 1 YnA- n+z (n + 2)!1 n } n
1
a
_ (i1(e))zDz
{A = 0, 1, ... )
and this system in turn is equivalent to the system 1
(20)
00
Q
n- {YZ1{E))"A++ Nn
{t({E))
1
A)'
A)
n+ Instead of the unknowns fl,. (m = 0, 1, ...) we now introduce l0
new unknowns ,um (m = 0, 1, ...) by way of the substitution 1
/21)
((
M! m
154
=fm
(m = 0, 1, ... ).
This substitution transforms (20) into
+ (22)
n=o
(Yrl(E))"An
n
)
1- 1
(A=0, 1, ...).
fln+x = D2
Now in the equation with number A (A z 0) of this system the coefficient of ,ul is equal to 1, while for A z 0, because of 0 < a < 1, we have (Ii J
5 1 -+
_
+n
IA
n=1 °°
S (Yrl(e))" I
A.
1
1+
2
2 n=1
I
J {
1
A)1
-a
2
1
(Yil(E))2tt I An
12.
00
Obviously the series 5 in the last member of this formula does n=1
not depend on A and because of (10) and (III, 21) it converges. In connection with the formula -Cl
(23)
\ 2) > zl(0)'
which holds because of (III, 24) and (11), this is the cause that to the number s already introduced there exists a non-negative integer A such,
that for every integer A z A the inequality 2
1+
I 2
(1+A)I
2
I (Yr1(E))2n I A. 12 <
holds.
From the last but one formula it now follows that for A Z A 2
1_
1+
(yrl(E))n I A. I
I
2
<
Cn n
ti\2) i.
al(0)
Moreover, because of the inequality 2E
zl
) < r1(E),
which follows from (III, 24), and because of (13), we have 1
lim sup I D;[ I x = Ti(0) 2-,oo
< tI(0 E 1
r1(E)
Ti C
2 155
Therefore to the system of equations i
00
(24)
1 (Yzi E ())"A n
n=o
(n+A \ n )I
n +a =
1,...)
D .A a
(this is system (22) without the first A equations if A > 0) we may e
Zl
apply lemma 20 with dh = Da, M =
- and
tip
ch,
This lemma tells us that the system (24) has a solution {,u, } (m = A, A + 1, ...) satisfying lim suP I ftm I nL <
(25)
nt-.oo
zl (0)
--T'(
2
Moreover it tells us that this solution is the only solution of (24). that has the property (25). Let this solution be /Im = ,u,* (m = A,
In case A > 0 the numbers um (m = 0, ..., A - 1) remain to be determined. To that end we write the first A equations of the system (22) as follows I
(26)
1L=o
(y 1(E))"A" +\
/l
n
Da - n=n-7. (Yri(E))"A. (
Q
it +
A) Yn+a
n
(A = 0,
..., A-1)
00
Now we shall prove that the series
in the right-hand members
of the A equations in (26) converge if in them we replace g.+,, by ,un+a (27)
(n + A = A, A + 1,
... ).
rn + 2)1 n
a
Because of 44;t
1 - 1 < 0 we have v
(Yr,(e))n
I An I I jun+a 1
Now it follows from (25) with um = ,u* that there exists a positive quantity K, not dependent on m, with the property
*I
156
0 [Tit2)1.
(m=A,A±1,...).
From this and from (27) it follows that n
SKI
+ n
}
1/in
I
(because of (23)).
K I A. I (Y=1(e))"
(Ytl(e))'
2")]
Now from this inequality, from (III, 21) and from (10) it follows at once that the series in the right-hand members of the A equations in (26) converge if in them we replace ,u.+a by un+z Then from these A equations, beginning with the (A - 1)th, we can compute PA- 1, ... , ,uo successively and uniquely since in the Ath equation (A = A - 1, ... , 0) the coefficient of ,ux is equal to 1. The numbers IuA_ 1, ... , ,u0 thus
._p* respectively. determined we call So we have proved that the system of equations (22) has one and only one solution {,um} (m = 0, 1, ...) that satisfies (25). This solution is fem=fUm (m=0, 1, ...). Because of (21) the solution {,u*,} just indicated gives rise to a solution {#m} (m = 0, 1, ...) of the system (19), viz. (28)
m(Zl(1))m..
Nm =
(m = 0, 1, ...).
mta
On account of (25), with ,um = p.*,, this solution has the property 1
(29)
lim sup {m! ° M-.00
I
fln, I}- < zl(0)il(E)
,
Zl \ 2
and this solution is the only solution of the system (19) that has the property (29) since the solution {,u,*} (m = 0, 1, ...) is the only solution of the system (22) that satisfies (25) with um = ,u,* (m = 0, 1, ...). Now if we take a number e1 with 0 < el < e we find quite analogously that the system (19) has one and only one solution that has the property lim sup {m!° I /9m I}m <1(0)zx(el) m-oo
Ti
el \
\
2
Since, because of (III, 24), the right-hand member of this formula is less than the right-hand member of (29), the solution that we find 157
corresponding to the number s, is identical with the only solution of the system (19) that has the property (29). Hence this solution does not depend on the choice of the number e, provided that E satisfies (III, 22). And, since, because of (II1, 24), the limit for E - 0 of the
right-hand member of formula (29) exists and equals z,(0), this solution has the property i i lim sup {m!R I flm 1121-1 5 r,(0). "I-00 1
Moreover, since z,(0) is equal to (oz) o (see (11)), the system (19) has one and only one solution that satisfies (15). Hence assertion I is proved.
We now proceed to proving assertion 2. We already know that the function 77(x) is an integral function not exceeding the normal type r of the order a and hence (according to definitions 3 and 8) it is of the
same kind at most as the function h(x). Now if the function 77(x) should not be of the same kind as the function h(x), then it would be of a lower kind than the function h(x). Then it would follow from definitions 9 and I that in this case it would be either of the normal type t (0 < t < r) of the order a or it would not exceed the minimum type of the order a. In the latter case, according to definitions 1 and 3, the function 77(x) certainly does not exceed the normal type u of the order a, where we may choose u arbitrarily in the interval 0 < u < r. Hence in both cases there exists a number v with 0 < v < r such that the function 77(x) is an integral function not exceeding the normal type v of the order a. From the datum concerning the numbers a and from the _ i
_i
formula (ar) Q < (av) Q , which holds because of 0 < v < r and 0 < a < 1, it then follows that the numbers a,, have the property mentioned in the assertion, occurring in the remark that follows the proof of theorem 7, if in this remark we replace y(x) by i7(x) and r by v. Then this assertion tells us that the function k(x) = F(D) -* 77(x) is an integral function not exceeding the normal type v of the order a. According to definitions 3 and 9, the function k(x) is then of a lower kind than the function h(x). Since this contra-
dicts formula (17) we have proved that the assumption that the function 77(x) would not be of the same kind as the function h(x) leads to a contradiction. Hence assertion 2 is proved. Now we prove assertions 3 and 4. Since the numbers a,, have the property mentioned in assertion 1 and since this property is the same as that mentioned in assertion 2 case a) of theorem 11, the numbers according to assertion 1 of lemma 19, have the same property, i.e. 158
that the expression an
G1(z) =
9 zn
n=0 yyla
defines an integral function not exceeding the normal type, less than i
of the order 1. Now we put F(D) = F*(D)DP, so that
(ar)
F*(D)
(30)
Since a,
0,
n=0
an+,Dn
it follows from assertion
I
that the differential
equation
F*(D) - y(x) = h(x)
has one and only one solution that is an integral function of the same kind at most as the function h(x), while assertion 2 tells us that this solution is of the same kind as the function h(x). We write this solution as the pth derivative of a function fi(x), i.e. as (31)
dxP
The function (31) is therefore determined uniquely and we have
F*(D)_. j d ,fi(x)} = h(x).
(32)
The left-hand member of this equation is equal to 00
00
an+v dxn
an+, (n+,)(x) = F(D)
n-0
{x),
so that formula (32) may be written in the form F(D) -> fi(x) = h(x).
This means that the functions $(x), obtained by integrating the function (31) p times indefinitely, are solutions of the differential equation (7).
Now it follows from lemma 21 that each of the functions is an integral function of the same order as the function
$(x) dD
dxD
fi(x), so that the functions $(x) are all of finite order.
Moreover, the function h(x) is not a polynomial since its order a is 159
dD
positive. Then the function
P E(x) is neither a polynomial, since
dxP
it is an integral function of the same kind as the function h(x) (see definition 7). Hence it is clear that none of the functions . (x) is a polynomial. So these functions fi(x) are all transcendental integral functions of finite order. According to lemma 18 (assertion 3y)) with M(z) = E(x), together dD
with the subjoined remark, the function dx fi(x) is of the same kind as each of the functions fi(x). This means that all the functions fi(x) D
are of the same kind as the function
dxP
fi(x). Therefore they are of
-
the same kind as the function h(x). Thus we see that the differential equation (7) has solutions that are of the same kind as the function h(x). These solutions arise from the function (31) by integrating the latter function p times indefinitely. Hence the solutions stated form a set of ooP functions. Clearly each pair of this set has a difference that is a polynomial of a degree not exceeding p - 1. Moreover, besides the solutions belonging to the set of ooP functions
mentioned, the differential equation (7) has no other solution that is an integral function of the same kind at most as the function h(x).
In fact, if this differential equation should have another solution that was an integral function of the same kind at most as the function dv
h(x), e.g. the function y(x) = t9(x), then the function z(x) = dP 8(x) would satisfy the differential equation
F*(D) - z(x) = h(x).
(33)
D
However, the function -- z9(x) is an integral function of the same dxP
kind at most as the function 8(x) as may be seen from lemma 18 (assertion 4), with M(z) = $(x), and the subjoined remark. Hence this dD
function dxP 8(x) is of the same kind at most as the function h(x) (see D
definition 8). The function dx P$(x), however, is the only solution of the differential equation (33) that is an integral function of the same
kind at most as the function h(x) as we have proved. This means that dxP 160
8(x)
dxP e(x).
Therefore the function 19(x) does indeed belong to the set of oon solutions mentioned above. Hence assertions 3 and 4 are proved. This proves theorem 16 conclusively.
We now suppose that the function h(x) is an integral function of the minimum type of the order a, where 0 < a S 1. For this case we have Theorem 17. Conditions: The function h(x) is an integral function of the minimum type of the order a (0 < a S 1). The differential 00
operator F(D) is defined by F(D) = Y, a,,Dn, where the numbers a n=o are not all equal to zero. A s s e r t i o n s: If the numbers an have the property that the expression an
G(z) _
(34)
Z,,
n=0 rylo
defines an integral function not exceeding the normal type of the order 1, then, if ao 0, the differential equation F(D) --)- y(x) = h(x)
(35)
has one and only one solution y(x) = rt(x) that is an integral function of the same kind at most as the function h(x). 2. The function rt(x) mentioned in assertion r is an integral function of the same kind as the function h(x). 3.
If the numbers a,, have the property mentioned in assertion r 0 (p Z 1) then the moreover, ao = ... = a9_1 = 0, a,,
and if,
set of all the solutions of the differential equation (35) that are integral functions of the same kind at most as the function h(x), consists of 009 functions. The difference of each pair of these functions is a polynomial
of a degree not exceeding p - 1. 4. The solutions mentioned in assertion 3 are all of the same kind as the function h(x).
Proof. We shall first prove assertion 1. As as 0 we are allowed to assume ao = 1, because in case ao 0 1 we may divide both members of equation (35) by ao.
Since the numbers an have the same property as in theorem 13, formula (III, 39) of the proof of theorem 13 now also holds. With the help of the number I defined in the proof of theorem 13 and which therefore satisfies (III, 42), we again define the numbers A,, 161 II
(n = 0, 1, ...) by means of (III, 43). So these numbers A,, have the property (III, 45). Consequently A0 = I because we assumed ao = 1. Let the function h(x) have the power series expansion 00
h(x) =
(36)
dx;t. A-0
From the datum that this function h(x) is an integral function of the minimum type of the order a (a > 0) it follows that the numbers dA have the property i
r
(37)
lim {.Z! Q d, J) T = 0, A-00
because of formula (P, 7) with P = a and a = dA. Now we define the numbers Dx by i
A i d x = DA
(38)
(2 = 0, 1, ... ).
Because of (37) we therefore have i
urn D [1 = 0.
(39)
t-.oo
Now we are going to prove that the differential equation (35) has one and only one solution y(x) = 22(x) that is an integral function of the same kind at most as the function h(x). According to definitions 8 and I such a function n(x) is an integral function not exceeding the
minimum type of the order a (0 < a S 1). If it has the power series expansion 00
(40)
#mx-,
77(x) _
m=0
it follows from lemma 7 (formula (13)) with /(x) replaced by q(x) and a by a, and with x = 0, that the numbers /,,, satisfy the relation 1
(41)
1
lim {in!0 /',,, J}»+ = 0. I
na- 00
Conversely, if the numbers #m in the right-hand member of formula (40) satisfy the relation (41) it follows from lemma 9 (formula (19)), with n replaced by in, a by a and I (x) by rl(x), that q(x) is an integral function not exceeding the minimum type of the order a. Then, according to definitions I and 8, the function q(x) is of the same kind at most as the function h(x). Hence condition (41) is necessary and sufficient for the expression 27(x) in (40) to define an integral function of the same kind at most
as the function h(x). 162
In a similar way as in the proof of assertion I of theorem 16 we may now conclude 2) that assertion 1 is proved as soon as we have shown that we can determine the numbers #m (m = 0, 1, ...) in one and only one way such that they have the property (41) and that they satisfy the following system of infinitely many linear equations 00
(A = 0, 1, ... ).
a.Nn+a (n + A) ! = d1
(42)
If we make use of (III, 43) and (38) we may write this system in the form CO Q Di I
A!
I nAnn!a 1
n=0
(A=0,1,...)
P. +z(n+A)!=
A!a
and hence in the form 00
(43)
n+ A 1- a
1
Pa
CnAn Nn+a (n + A) !
n=O
(A = 0, 1, ...).
Now we choose a number a exceeding C (so a > 0) and we multiply both members of the equation with number A of the system (43) with
a'. Then we find the following system (44)
0o
Y
n=0
a
)'Ananpn I
x(n+A)!a
(
n+A l=axDa (A=0,1,...). n )
Instead of the unknowns Nm (m = 0, 1, ...) we now introduce new unknowns vm (m = 0, 1, ...) by the substitution ,q amflmmla = vm
(45)
(m = 0, 1, ...).
Then the system (44) becomes 0o
2: (a")n An ( n-0
(46)
yy + A 1 n }
0
vn+a =
aA
2
(A = 0, 1, ...).
In the equation with number A (A Z 0) of the system (46) the coefficient of vx is equal to 1, while for A Z 0, because of 0 < or 5 1, we have the inequality
I+
A}1-alas
{(--)
n1A01(nn
n-1
a)
11z2 .
00
The series
n=1
in the right-hand member of this inequality does not
depend on A and because of (III, 45) it converges since we chose 2) Here we make use of the remark that follows the proof of theorem 8. 163
a > C. If its sum is equal to K2 -
(K
I
1)
it follows from the
last inequality 1
a 2
(47)
1+
{t a )m A I I
5 K2
}1
Cn n
(A = 0, 1, ... ).
Moreover the numbers DA satisfy (39) so that I
1
limIaDzIa =0
)n A \ / 1
a a Then this lemma tells us that the system (46) has a solution {v,,,} Chk
(m = 0, 1, ...) satisfying the relation I
1
(48)
lira sup Ivmm < -K m-'.oo
and besides this lemma gives that this solution is the only solution of the system (46) that has the property (48). Let this solution be
vm=vm (m= 0,1,...). This means that the system (44) and so also the system (42) has a solution {#m} (m = 0, 1, ...) with the property 1
1
(49)
lim sup {M! ° I Nm 1)m <
I
,
aK which formula follows from (48) with v,m = v* and from (45) with m +oo
This solution is the only solution of the system (42) that has the property (49) since the solution {v*,} (m = 0, 1, ...) is the vm =
only solution of the system (46) that satisfies (48) with vm = v,*m.
If now we take a number a1 > a we may prove analogously that the system (42) has one and only one solution that satisfies 1
lim sup {m! m-.oo
1
I}"t <
I
a1K1
(K1 --;- 1).
From this it follows that the solution we find corresponding to the number a1 is identical with the only solution of the system (42) that has the property (49). Hence this solution does not depend on the choice of the number a, provided that a satisfies the inequality
a > . Since for each a we have K1 Z 1 and therefore lim -1 = 0, this solution has the property (41). 0+00 aK1 164
Hence the system (42) has one and only one solution satisfying (41). Thus assertion I is proved. We now prove assertion 2. We already know that the function rl(x) is an integral function not exceeding the minimum type of the order a. Hence it is of the same kind at most as the function h(x). Now if the function n(x) should not be of the same kind as the function h(x), then it would be of a lower kind than the function h(x). Then it
follows from definition 9 that the order of the function rt(x) is less than a. If this order is equal to ti, then there exists a number d satisfying
e < A < a. Hence the function rt(x) does not exceed the maximum type of the order A (see definition 5).
From theorem 2 it follows that the differential operator F(D) is applicable to all integral functions of the maximum type of the order A.
Hence the numbers an have the property mentioned in theorem 3, with a replaced by A. Therefore they also have the property men-
tioned in theorem 9. Then the remark that follows the proof of theorem 9, in which we replace y(x) by rj(x), tells us that the function k(x) = F(D) --- TI(x) is an integral function not exceeding the maxi-
mum type of the order A. Hence, according to definitions 5 and 9, this function k(x) is of a lower kind than the function h(x). This contradicts the fact that the function n(x) is a solution of (35) so that F(D) -- n(x) = h(x). This proves assertion 2. The proofs of the assertions 3 and 4 run mutatis mutandis quite analogous to the proofs of the similar assertions of theorem 16. Therefore they will be omitted.
Next we assume that the function h(x) is an integral function of the maximum type of the order a, where 0 < a < 1. Then we have Theorem i8. C o n d i t i o n s: The function h(x) is an integral function of the maximum type of the order a (0 < a < 1). The differential 00
operator F(D) is defined by F(D) _ I anDn, where the numbers an
n=0 are not all equal to zero. A s s e r t i o n s: 1. If the numbers an have the property that the
expression (50)
G(z)
00
an
n=0 n1a
zn
defines an integral function the order of which is less than 1, then, if ao 0, the differential equation (51) F(D) -* y(x) = h(x) 165
has one and only one solution y(x) - rt(x) that is an integral function of the same kind at most as the function h(x). 2. The function rt(x) mentioned in assertion i is an integral function of the same kind as the function h(x).
If the numbers a have the property mentioned in assertion z and if, moreover, ao = ... = a,_1 = 0, aD 0 (p Z 1), then the set 3.
of all the solutions of the differential equation (51) which are integral functions of the same kind at most as the function h(x), consists of oov functions. The difference of each pair of these functions is a polynomial
of a degree not exceeding p - 1. 4. The solutions mentioned in assertion 3 are all of the same kind as the
function h(x). Proof. We shall first prove assertion 1. As ao 0 we are allowed to assume ao = 1, because in case ao 1 we may divide both members of the equation (51) by a0. Let the order of the integral function G(z) defined by (50) be equal
to P < 1. This function has the property P 1, with a replaced by an 1
, so that for every positive S we have
n!a 1
_ n = 0.
a
n !B+e n-.oo
Now we choose a number 0 such that it satisfies (53)
0<0<min
(I-1, -1 -1), 0
if0<<1,
0 <0 <-1, v
if p=0.
and 1
(54)
Then if follows from (52) that for the number 0 thus chosen the formula (55)
lim n!1+, n-.oo
a1
1
n=0
n!Q
holds. For if o satisfies the inequality 0 < P < 1, we have I + 0 < Consequently there exists a positive number 6 such that 1
In case e = 0 we may simply take 6 = 166
o-
S
=1 + 0.
With the help of the number 0 we now define the numbers An in the following way 1-1+6 (56)
n!
an = An
(n = 0, 1, ...).
Then (57)
AO = ao = 1,
while from (56) and (55) it follows that 1
(58)
limIAnIn=O.
n-.oo
Let the function h(x) have the power series expansion h(x) _
daxa.
00
a=o
From the datum that this function h(x) is an integral function of the order a (v > 0), it follows that formula (P, 4), with n = A, P = a, 6 = E and an = da, for every positive e holds, i.e. that we have for every
positive e the formula 1
1
(59)
lim {A!+a J da J} T = 0.
Now it follows from (53) and (54) respectively, that we may choose
the number s such that it satisfies the relation 1
1
Then it follows from this and from (59) that also (60)
lim{).!Q-°Jd,,J}z =0.
Now we define the numbers Da by (61)
1 -e
A!a
da=Da
(A= 0,1,...).
From this and from (60) it then follows that the numbers DA satisfy the relation 1
(62)
limJD2
= 0.
Now we shall prove that the differential equation (51) has one and only one solution y(x) = 77(x) that is an integral function of the same kind at most as the function la(x). According to definitions 8 and 5 167
such a function fi(x) is an integral function not exceeding the maximum
type of the order a (0 < a < 1). If it has the power series expansion (63)
#mxmr
n(X)
M=O
it follows from lemma 8, with /(x) replaced by fi(x) and a by a, and
with x = 0, that the formula I
1
lim {m!Q
(64)
0
holds for every positive number e. Conversely, if the numbers Nm in the right-hand member of (63) for every positive s satisfy the relation (64), then it follows from lemma 11, where we replace n by m, a by a, 6 by e, a by Nm and /(x) by q(x), that i(x) is an integral function the order of which does not exceed a.
Hence, according to definition 5, it does not exceed the maximum type of the order or. Then it follows from definition 8 that the function 7(x) is of the same kind at most as the function h(x).
Hence, a necessary and sufficient condition for the expression n(x) in (63) to define an integral function of the same kind at most as
the function h(x), is, that the numbers
Nm
have the property that
formula (64) holds for every positive e.
In a similar manner as in the proof of assertion of theorem 16 we may draw the conclusion 3) that we have proved assertion I if we have shown that we can determine the numbers N m (m = 0, 1, ...) in one and only one way such that they satisfy the system of infinitely many linear equations 1
1
00
(A=0,1,...)
2 I a,,..8.+, (n + A) ! = dA
(65)
and moreover have the property that formula (64) holds for every positive E.
Making use of (56) and (61) we may write the system (65) in the form m
A,,n'!
_(I
+e)
0-
a
h'n+x(n+A)!=A!
T
and hence in the form 1
(66)
n=O
(n + ) ! °
+ 1 ©n
(
n
>
°DA (A=0,1,...)
+0
= DA (A = 0, 1, ... ).
3) Now we make use of the remark that follows the proof of theorem 9. 168
Instead of the unknowns /gym (m = 0, 1, ...) we now introduce the new unknowns co,,, (m = 0, 1, ...) by the substitution e
(m = 0, 1, .... ). Owing to this substitution the system (66) is transformed into Pmnt ! ° - = u>n7
(67)
c0o
L7 An
(68)
+0
1
TL=o
)
tit
Wn+x = D2
(A = 0, 1, ...).
Now in the equation with number A (A -- 0) in the latter system the coefficient of cue is equal to 1, while for A Z 0, because of 1
I
-1 -!
n
+ 0 < 0 (as follows from (53) and (54) respectively), we have
0 11 AnI(n+)')I \ n /
a 1012
I).
K2
I
n=1
1
Obviously the series in the middle member of this formula does not depend on A and, because of (58), it converges. Therefore the number K is finite. Besides it follows from formula (62) that the numbers Dx satisfy the inequality 1
I
limsun D 1a
Consequently, to the system (68) we may apply lemma 20, with
dh=D2, M=K and
c,, = A.
+
/9t + 92
A/1
1
Then this lemma tells us that the system (68) has a solution
(m = 0, 1, ...) that satisfies 1
(69)
lim sup j w
m
<
{W m}
1
A
Moreover this lemma tells us that this solution is the only solution
of the system (68) that has the property (69). Let this solution be Wm=COm (m=0, 1, ...). Then it follows from (67) that system (66) and therefore also system
(65) has a solution {1m} (m = 0, 1, ...) with the property (70)
1
lim sup {m! a M-00
1
0 1
f4,,, j}
which formula follows from (69) with
co,n
<
1
K co7* and from (67) with 169
w,,,, = w,*n. Moreover, this solution is the only solution of the system (65) that has the property (70) since the solution {co} ,*n(m = 0, 1, ... ) is the only solution of the system (68) that satisfies (69) with co',' = w,*m.
(m = 0, 1, ...) so that
Let this solution be
1
(71)
1
B
lim sup {in! °
/3 n I }'" <
1
K
.
Now we assert that from formula (71) we can deduce that for every positive E the formula 1
1
(72)
1im {,n ! Q
*j}
== 0
t
M-M
holds. In fact, in the first place it follows from formula (71) that formula
(72) is correct for every positive e satisfying e > B. Now if formula (72) should not hold for every positive e, then there would exist a 0, such that number f.1, satisfying 0 < el 1
(73)
-e
lim sup {m!
1
I
> 0.
p,* j}
We now choose the number 01 such that it not only satisfies (53) and (54) respectively, 0 being replaced by 0, but also 0 < 01 < e1. Repeating the proof from formula (54) to formula (71) inclusive, always replacing B by ©1, we find that the system (65) has one and only one
solution {iI,,,} (m = 0, 1, ...) with the property 1
1
<
-0
1im sup {m!
rt-.
K
(K1 2 1).
1
Since 0l < 0 this solution is identical with the solution
we
found corresponding to the number 0. Therefore we have 1
lim sup {9n! ° -0,
1
P* j}"' <
1
K.
However, this formula contradicts formula (73), because of O1 < e1. Hence for the solution {9;*,t} formula (72) holds for every positive e. Thus we have shown that the system (65) has one and only one solution
that has the property that formula (63) holds for every positive e. This proves assertion 1. We shall now prove assertion 2. We have already seen that the function 77(x) is an integral function not exceeding the maximum type
of the order a, and so an integral function of the same kind at most as the function h(x). If the function 71(x) should not be of the same kind as the function h(x), then it would be of a lower kind than the 170
function lz(x). Then it follows from definitions 9 and 1 that the function
,)(x) would be an integral function either of the normal type 7(0 < r < oo) of the order a, or not exceeding the minimum type of the order a. In both cases there exists a number t with 0 < t < oo, such that the function rt(x) is an integral function not exceeding the normal type t of the order a. It now follows from theorem 3 that the differential operator F(D) is applicable to all the integral functions of the normal type t of the
order a. Hence the numbers an have the property mentioned in theorem 1, where we replace s by t. Therefore they also have the property stated in the assertion occurring in the remark following the proof of theorem 7. This assertion, with y(x) replaced by,t(x), tells us
that the function k(x) = F(D) ->. ,t(x) is an integral function not exceeding the normal type t of the order a. According to definition 9,
this function is therefore of a lower kind than the function h(x). This, however, contradicts the fact that the function ,fi(x) is a solution of (51) so that F(D) -* n(x) = h(x). Our assumption, that the function n(x) should not be of the same kind as the function h(x) leads therefore to a contradiction. This proves assertion 2. The proofs of assertions 3 and 4 run mutatis mutandis quite analogous to the corresponding assertions of theorem 16. Therefore they will be omitted.
Finally we deal with the case where the function h(x) is an integral function of the order zero. As we know from the Preparatory Chapter, then the function h(x) is either a transcendental integral function of the order zero, or a rational integral function the degree of which is
positive or zero. For this case we have Theorem ig. Conditions: The function h(x) is an integral function of the order zero. The differential operator F(D) is defined by 00
F(D) _ I anDn where the numbers an are not all equal to zero. n=o
A s s e r t i o n s: A. It the function h(x) is a transcendental integral function of the order zero, hence of the maximum type of the order zero, then the following holds: 1. If the numbers an have the property that there exists a finite real
number v, neither depending on n nor on z, such that the expression H(z)
-a
n_o n.v zn
defines an integral function the order of which is less than 1, then, if 171
ao T 0, the differential equation
F(D) -* y(x) = h(x)
(74)
has one and only one solution y(x) = n(x) that is an integral function of the same kind at most as the function h(x). 2. The function it(x) mentioned in assertion z is an integral function of the same kind as the function h(x). It is a transcendental integral function of the order zero.
If the numbers a have the property mentioned in assertion i
3.
and if, moreover, ao = ... = aD_1 = 0, a9:0 (p L- 1), then the set of all the solutions of the differential equation (74) that are integral functions of the same kind at most as the function h(x), consists of oop functions. The difference of each pair of these functions is a polynomial
of a degree not exceeding p - 1. The solutions mentioned in assertion 3 are all of the same kind as the
4.
function h(x) and they are all transcendental integral functions of the order zero.
I/ the function h(x) is a rational integral function of degree l
B. (1
1.
0), then the following holds : If ao 0, then the differential equation
(75)
F(D) -> y(x) = h(x)
has one and only one solution y(x) = rt(x) that is a polynomial. This polynomial is also of degree 1. 2. 1/ ao 0 and if, moreover, the numbers a have the property mentioned in assertion i of part A, then the differential equation (75) has no other solution that is an integral function of the order zero except the function rt(x). Then there exists therefore no solution in particular that is a transcendental integral function of the order zero. 3.
If ao = ... = a9_1 = 0, a9 0 0 (p Z 1), then the set of all the
solutions of the differential equation (75) that are polynomials, consists of oo' functions. Each of these functions is a polynomial of the exact degree l + p, if h(x) 0- 0, and of a degree not exceeding p - 1, if h(x) = 0.
The difference of each pair of these polynomials is a polynomial of a degree not exceeding p - 1. If h(x) is not identically equal to a constant, then all these oop functions are of the same kind as the function h(x). If h(x) is identically equal to a constant = 0, then all these oo9 functions are of a higher kind than the function h(x). If h(x) = 0 then the set of cot' functions mentioned contains all the constants. These constants are all of the same kind as the function h(x). 172
All the further functions belonging to this set of oov functions are therefore
not constant; hence they are of a higher kind than the function h(x). 4.
If as = ... = aq_1 = 0, a,
0 (f Z 1) and if, moreover, the
numbers a,, have the property mentioned in assertion i of part A, then the differential equation (75) has no other solution that is an integral function of the order zero except for the oo' solutions mentioned in assertion 3. Then there exists therefore no solution in particular that is a transcendental integral function of the order zero. Proof. First we prove assertion I of case A. Since ao 0 we are allowed to assume ao = 1, because in case ao 1 we may divide both members of equation (74) by ao. Since the numbers a have the same property as in case A of theorem 15, now formula (III, 77) of the proof of theorem 15 holds too. The
numbers A. (n = 0, 1, ...) we again define by means of formula (III, 78). So these numbers A have the property (III, 80). We now have A. = I since we have assumed ao = 1. Let the function h(x) have the power series expansion W
h(x) = Y, dAx2.
From the datum that this function h(x) is a transcendental integral function of the order zero, it follows that infinitely many of the coefficients dA differ from zero. Besides, for every positive number A
formula (P, 4), with n = .t, o = 0,
A and a = dA, holds, i.e.
we have for every positive number A i lim {A!4 dx IF, = 0.
A-W
We now choose the number A equal to a positive number B that satisfies the inequality (76)
B > max (1, v).
Consequently 1
(77)
lim {R!n I
dA
I}x = 0.
A-.W
Now we define the numbers DA by (78)
2!$ dA = DA
(A = 0, 1, ...).
From this definition and from (77) it follows that the numbers D. 173
have the property that the formula 1
lim I Dx I ;C = 0
(79)
x--. 00
holds.
We are now going to prove that the differential equation (74) has one and only one solution y(x) = fi(x) that is an integral function of the same kind at most as the function h(x). According to definitions 8 and 6 such a function 1j(x) is an integral function of the order zero.
If it has the power series expansion W
(80)
#,,,X-,
27(x) ,92=0
then, because the function 21(x) has the property P I with for every positive e formula (P, 4), with n = na, e = 0,
0,
= e and
a,,= Nm, holds. Therefore we have for every positive e 1
1im {m!` I fl Nm I}-n
(81)
0.
9Y!-2 00
Conversely, if the numbers Nm in the right-hand member of formula (80) for every positive e satisfy relation (81), then n(x) has the property
P 3. Hence n(x) is then an integral function of the order zero. According to definition 8 it is of the same kind at most as the function h(x).
Therefore a necessary and sufficient condition for the expression (80) to define an integral function of the same kind at most as the function h(x), is, that the numbers Nm have the property that formula (81) holds for every positive e. In a similar manner as in the proof of assertion 1 of theorem 16 we may now conclude that assertion 1 is proved if we have shown that
we can determine the numbers P. (nt = 0, 1, ...) in one and only one way such that they satisfy the following system of infinitely many linear equations
-
(82)
a! n=0
a9#..+A(n + A) ! = da
(,? = 0, 1, ... )
and moreover have the property that for every positive e formula (81) holds.
Making use of (III, 78) and (78) we may write the system (82) in the form I
00
2; n! v-' AJLJf+A (n + 2)! = 1.! .-0 174
A!-F DA
(A = 0, 1, ...)
and hence in the form 00
(83)
1
Ann!,-BNn-k-a
n-0
(nn + 1.)!B
n + ]")1 - B
l
n
= Da
(A = 0, 1, ...).
Instead of the unknowns #,,, (3n = 0, 1, ...) we now introduce the new unknowns z,,, (m = 0, 1, ...) by means of the substitution (3n = 0, 1, ...).
,,,.332!B = :rqa
(84)
This substitution transforms the system (83) into
In +
(85)
n
n=o n!
A11-B
l
Zn+a = D2
(A = 0, 1, ... ).
In the equation with number A (A 0) of the latter system the coefficient of z, is equal to ; because of (76) we have for A Z 0 1
1+
Bnl n=1
{I7t.
A)1_B
i+AnI2=K2 (KZ 1) n=1
('2n
00
The series
I A 12 does not depend on A, and, because of (III, 80),
it converges so that the number K is finite. Besides, it follows from formula (79) that the numbers Dx satisfy the relation I
1
Jim sup IDA 1A < K
.
Hence to the system (85) we may apply lemma 20 with d,, = DA,
M=K and Ch
B-q Cn n
All
-B
7l .
Then this lemma tells us that the system (85) has a solution n,,, (m = 0, 1, ...) that satisfies the relation 1
(86)
1
lim sup I z,,, i tm < K m- 00
Moreover, this lemma tells us that this solution is the only solution of the system (85) that has the property (86). Let this solution be nm = z*
(n2=0,1,...).
Consequently, the system (83) and therefore also the system (82) has a solution (m = 0, 1, ...) that has the property (87)
lim sup {m!B I Pm I} "+ < 1 M-.00
,
which formula follows from (86) with a,,, = z* and (84) with 72,n = zm. 175
This solution is the only one of the system (82) that has the property (87) since the solution {z,*n} is the only solution of the system (85)
that satisfies (86) with a,,, = n . Let this solution be Nm = (m = 0, 1, ... ). Hence we have i
1
Jim sup {rn!n I #* I}. <
(88)
igm
K
m-0o
We now assert that from formula (88) we can derive that for every
positives we have lim {m!° I * i} ra- = C.
(89)
M-00
In the first place it follows from formula (88) that for every e satisfying 0 < e < B formula (89) holds. If formula (89) should not hold for every positive e, then there would exist a number el that satis-
fies el Z B and that has the property that lira sup (M!, I ,* I } m > 0.
(90)
We then choose the number Bl such that it not only satisfies formula (76), with B replaced by B1, but also Bl > el. Repeating the proof from formula (76) as far as formula (88) inclusive, always replacing B by B1, we find that the system (82) has one and only one solution {fl,,,}, such that
/
lim sup {nt!nl I P. I}""
<j K Because BI > B this solution is identical with the solution {#.*) we found corresponding to the number B. Hence we have M-00
1
lim sup {m !1?1 I fl n J} m < K .
M-
This contradicts formula (90), however, because Bl > e . Thus we
conclude that for the solution
formula (89) holds for every
positive e.
This proves that the system (82) has one and only one solution that has the property that for every positive e formula (81) holds. Hence assertion I is proved. We proceed to prove assertion 2. Let v(x) be a polynomial of degree
r (r Z 0). Then 00
(91)
r
F(D) - v(x) = Y, anv(n)(x) = L.r anv(n)(x) n=0 n=0
Because ao 176
0 it follows from this that the function F(D) -
z,(x)
is a polynomial (of degree r). Then the function 77(x) _
Pnx'" m=0
cannot be a polynomial because F(D) -> rl(x) = h(x) and the function h(x) is a transcendental integral function (of the order zero). So the function -n(x) like the function h(x) is a transcendental integral function of the order zero. Therefore they are both of the maximum type of
the order zero so that, according to definition 7, the function 'j(x) is of the same kind as the function h(x). Hence assertion 2 is correct. The proofs of assertion 3 and the first part of assertion 4 run mutatis in'u andis quite analogous to those of assertions 3 and 4 of theorem 16. Therefore they will be omitted. That the second part of assertion 4 is true we may prove in the same way as we proved assertion 2. This settles case A. We shall now prove assertion I of case B. In the proof of assertion 2 of case A we have seen that the function F(D) --> v(x) is a poly-
nomial of degree r if the latter is the case with the function v(x). Now, if v(x) is a solution of the differential equation
F(D) - v(x) = h(x), where h(x) is a polynomial of degree l (l -,-- 0), then the polynomial v(x) is also of degree l and we have t
(92)
2,
h(x)
n=o
Besides, by the latter relation the polynomial v(x) is determined uniquely. For, if we equal the coefficients of x', xi-1, ... in the left-
hand and right-hand members of (92) we obtain for the l + I coefficients of the polynomial v(x) exactly l + 1 linear equations from which we can determine these coefficients unambiguously because ao 0. Hence there exists one and only one polynomial v(x) that satisfies (75) and this polynomial is of the same degree as the polynomial h(x). This proves assertion 1. That assertion 2 is true we may prove as follows. If, besides the polynomial solution y(x) = 77(x), the differential equation (75) should have a transcendental integral function of the order zero, y(x) = w(x) for a solution, then the transcendental integral function w(x) - rj(x) of the order zero would satisfy the homogeneous differential equation F(D) -)- y(x) = 0. Then let k(x) be an arbitrary transcendental integral function of the order zero. According to assertion I of part A the differential equation
F(D) - y(x) = k(x) then has one and only one solution y(x) = fi(x) that is an integral (93)
177 12
function of the same kind at most as the function k(x). The function y(x) _ fi(x) + w(x) - ?I(x) is an integral function of the order zero, hence
of the same kind at most as the function k(x). Moreover it is also a solution of the differential equation (93). Hence we should have $(x) + w(x) - i7(x) = E(x), so that w(x) = q(x). Here we have a contradiction.
We now prove assertion 3. In the proof of assertion 2 of part A we have already seen that formula (91) holds, if v(x) is a polynomial of degree r (r -a 0). Then we have V(X)
F(D) -->
r
=I a,,v(n)(x) n=p
r-p
_ n=0 an+,v(n+P)(x),
if '>r.
F(D) -> v(x) = 0, Putting (94)
v(p)(x) ` w(x),
it follows from the two preceding formulae that we have r-p F(D) - v(x) = an+pw(n)(x) = F*(D) - w(x), (95)
if
{
F(D) -- v(x) = 0
= F*(D) -a w(x),
if b
r,
if P > r,
00
where F*(D) =Ian+pDn n=0
If the polynomial v(x) is a solution of the differential equation (75),
then it follows from (95) that the polynomial w(x), connected with v(x) by means of the formula (94), satisfies the differential equation (96) F*(D) - y(x) = h(x)It now follows from assertion I that the latter equation has one and only one solution that is a polynomial and, moreover, that this polynomial is of degree 1. This polynomial, which, as a consequence, is identical with w(x), is = 0 in case h(x) _ 0 and it is 0 0 in case h(x) = 0. This means that all polynomials that are solutions of the differential equation (75) can be found by solving the equation (94). As is well-known, every solution of the differential equation (94) is a polynomial and that a polynomial of degree I + p, if h(x) $ 0 (for then w(x) $ 0) and of degree ; p - I if h(x) _ 0 (for then w(x) - 0)_ In both cases the coefficients of x0, ... , xp-1 may be chosen arbitrarily; the remaining coefficients (if there are any) are completely determined.
Hence there are oov polynomial solutions and the difference of each pair of these polynomials is a polynomial the degree of which does not
exceed 5 - 1. 178
If the function h(x) is not identically equal to a constant, then Z > 0, so that h(x) is an integral function of the maximum type of the order zero. As this is also the case with all polynomials v(x), (in fact, the
degree of each of them is equal to l + P > 0) all the oov functions v(x) are of the same kind as the function h(x). 0, then all polynomials If h(x) is identically equal to a constant v(x) are therefore of degree p > 0 so that they are all of a higher kind than the function h(x). If h(x) = 0, then only those of the oov polynomial solutions are of the same kind as the function h(x) that are identically equal to a constant. All remaining polynomials are not identically equal to a constant
and hence they are of a higher kind than the function h(x). Finally we prove assertion 4. From assertion 3 it follows that we have proved assertion 4 as soon as we have shown that the differential equation (75) has no solution that is a transcendental integral function of the order zero. If the differential equation (75) should have such a solution y(x) _ fi(x), then from F(D)
si(x) =
an+vS(n+P)(x)
n=p
n=0
n=0
it would follow that 00
an+v (n+D1(x) = h(x). n=0
If we put cv)(x) = 0(x)
then 0(x) is also a transcendental integral function of the order zero, and we have 00
n=0
that is F*(D)
0(x) = h(x).
Hence the function 0(x) satisfies the differential equation (96). Then it follows from assertion 2 that the function 0(x) is a polynomial. Therefore it would not be a transcendental function. Here we meet a contradiction. This proves theorem la conclusively.
179
CHAPTER V
CHAPTERS III AND IV CONTINUED
In this chapter we first consider the case which we left open in chapter III, viz. the case where the function y(x) is an integral function of the normal type of the order 1. If this integral function y(x) is of the
normal type r of the order 1 and if the differential operator F(D) is 00
defined by F(D) =
anDn, then we have already seen in chapter I n=0
(theorem 1 with a = 1) that this operator is applicable to the function y(x) if the numbers -an have the property that the expression F(z)
=
anzn n=0
defines a function that is analytic for I z s a. We now put again h(x) = F(D) - y(x). Then it follows from the definition of applicability (definition 11) that h(x) is a function defined for every finite value of x in the complex
x-plane. According to definition 7 this function h(x) is an integral function with the property that it is of the same kind at most as the function y(x).
In theorem 20 we shall investigate under what conditions the function h(x) is of the same kind as the function y(x) and under what conditions the function h(x) is not of the same kind as the function y(x). Since the function h(x) is of the same kind at most as the function y(x), it is, according to definitions 8, 7 and 9, of a lower kind than the function y(x), if it is not of the same kind as this function. Then in this chapter we shall consider the differential equation F(D) -± y(x) = h(x),
assuming that the function h(x) is a given integral function of the
normal type of the order 1. This case has not been treated in chapter IV. In theorem 21 we shall examine whether the differential 180
equation mentioned has a solution that is of the same or of a lower kind than the function h(x), if the latter is an integral function of the normal
type r of the order 1. A.o. it will appear that there exists no solution that is an integral function of a lower kind than the function h(x). In both investigations we distinguish between the case where the function F(z) has no zeros that lie on or within the circle I z I = r and the case where this function does have zeros that lie on or within the circle I z I = r. In the latter case these zeros play an important role. Then in theorem 22 we occupy ourselves with the case of the homogeneous differential equation
F(D) -- y(x) = 0. Next we recapitulate in theorem 23 a part of the results of theorems 12, 13, 14, 15 and 20 in order to prove theorem 24 which is a general-
ization of a theorem of J. M. Whittaker. For the formulation of the latter we refer to the Introduction 1). Finally, we give in theorem 25 a survey of a part of the results of theorems 16, 17, 18, 19 and 21. A very special case of theorem 25 is theorem 26. The latter theorem is a form improved by us of a theorem which we have formulated in the Introduction 2) and which has been proved by J. M. Whittaker.
We draw the reader's attention to the fact that in this chapter we do not explain the notation of some formulae, as e.g. formula (17). In such a case the Introduction is each time tacitly referred to. We now first prove some lemmas. Lemma 22. If the function I (x) is an integral function not exceeding the normal type r of the order 1, then to every bounded closed region M of the complex x-plane and to every positive a there exists a constant L such that on M uniformly in x the inequality I f(n)(x) I < L(r + e)n
(n = 0, 1, ...)
holds. This constant L is not dependent on n. Proof. Let the point xa belong to M. From the datum concerning the function I (x) and from lemma 5, where we replace A by r and a by 1,
it follows that to an arbitrary but fixed choice of the positive number
e there exists a number K, not dependent on h, such that the 1) Theorem H. 2) Theorem I. 181
inequality I fu,)(xo)
(1)
I S K(r +
(k = 0, 1, ...)
e)h
holds.
Now let the positive number r be chosen so large that all the points of M lie within the circle with xo as the centre and of which r is the length of the radius.
If x denotes an arbitrary point of M, then for n = 0, 1, .. , we
have
(n)(x) = f(n) (x0 + (x - x0)) = f(n+kl(x0) (x
x0)k
k
I
From this and from (1) it then follows that for n = 0, 1, ... we have f(n)(x) I <
00
K (r + s)n+k
kL o
z'
= K(r + eW n er(r+e).
k!
Putting Ker(r+e) = L, we see that lemma 22 is true. Lemma 23. C o n d i t i o n s: The function I (x) is an integral function not exceeding the normal type r of the order 1. 00
anzn and G(z) = f, chzh are both analytic
The functions F(z) _
for IzI5r.
h=o
"=o
The differential operators F(D) and G(D) are defined by 00
00
F(D) _ Y, anDn and G(D) _ n=0
Assertion :
chDh.
h=0
For every finite value of x we have
G(D) - {F(D) - I (x)} = {G(D)F(D)} -+ I (x).
(2)
Proof. According to theorem 1, with y(x) = f (x) and a= 1, the differential operator F(D) is applicable to the function I(x) and the numbers an have the property stated in the assertion in the remark that follows the proof of theorem 7, with a = 1. Then this assertion tells us that the function k(x) = F(D) f (x) is an integral function not exceeding the normal type z of the order 1. Then it follows from theorem 1, with y(x) replaced by k(x) and F(D) by G(D), that the differential operator G(D) is applicable to the function k(x). In connection with definition 11 it follows from this that for every finite value of x the left-hand member of formula (2) is significant. Since the functions F(z) and G(z) are analytic for I z I S x, the radii of convergence of their power series expansions exceed r. From this 182
it follows, in connection with formula (B) of p. 55, that there exists a positive number e, such that
(n=0,1,...)
(3) (4)
I
Ch I S K2 (z +
(h = 0, 1, ...).
2E)-1
where K1 and K2 are not dependent on n and h respectively 3).
If M denotes an arbitrarily chosen bounded closed region of the complex x-plane, then it follows from lemma 22 that to the number s already mentioned there exists a constant L not dependent on n, such that on M uniformly in x the inequality
(n=0, 1, ...)
I f(n)(x) I
(5)
00
holds. From this and from (3) we see that on M the series I an f (n)(x) n=0
converges uniformly. Because of 00
k(x) _
an f(n)(x),
n=0
it follows 4) that for h = 0, 1, ... and for every point of M we have Dh -> k(x) = Dh -* {2, an f(n)(x)} = n=0
00
an f(n+h)(x).
n=0
From this formula it follows that for every point x, belonging to M the formula 00
(6)
00
G(D) - {F(D) - I (x)) = I c, {2; a,,/(n) (x)}(h) h=0
n=0
00
00
ch{2, a(n+h)(x)} h=O
n=0
holds. Since, because of (4), (3) and (5), we have 00
00
1 Ch I Y, I an/(n+h)(x) I < K2(r + 2E)-h I K,(z + 2e)-'L(z + E)n+h n=0
n-0
K ZK1L
r+
+ z+
2E) n e o
E
On'
}
the repeated series in the right-hand member of (6) appears to converge 3) If e.g. the radius of convergence of the power series expansion of the function F(z) is equal to R, then R > T. Then we choose the number e such 1
that R > t + 2E. From (B) it then follows that lim sup I an I n < (z + 2s) and from this follows (3). 1) Cf. Titchmarsh [1] p. 95.
183
absolutely. Then, according to a well-known theorem of the theory of double series a), the right-hand member of (6) is equal to the sum by diagonals, i.e. it is equal to
(1
Chain-h)f'-'(x).
m=0 h=0
Then it follows from this and from (6) that for every point x belonging to M we have 0o
m
G(D) - {F(D) -* I(x)) _ I (Y. chain-,)I( m)(x).
(7)
n&=0 h=0
Moreover, 00
00
00
t
G(z)F(z) = (I chzh) (Y. anzn) = I (I Chain-h)zm, to=0 h=0 n-0 h=0
so that m (1 Chain-h)D'
00
G(D)F(D) =
m=0 h=0
.
From this we see that the right-hand member of (7) may be written in the form {G(D)F(D)} - /(x).
And since this holds for every point x of M and M may be chosen arbitrarily we see that lemma 23 holds. Lemma 24. Conditions: The function w(x) is an integral function of the normal type 0 of the order 1. m denotes a positive integer. A,, denotes, for ,u = 1, ... , m, a (complex) number with the property A,, I < 0, while in case m Z 2 the numbers AI, ..., Am are distinct. A s s e r t i o n: For every choice of the m polynomials p,(x)
(u = 1, ..., m) the function m
w(x) + I N=1
is an integral function of the same kind as the function w(x).
Proof. We choose a system of polynomials p,,(x) (,u = 1, ..., m). A,, I < 0 for y z = 1, ... , m, there exists a positive number e with the property that Because of I
IA,,I <0-e
(,u= 1, ...,m).
b) See e.g. Ferrar [I], theorem 58 together with the corollary following it. 184
Then obviously there exists a positive number K, not dependent on x, nor on u, such that for every finite value of x we have I p,,(x)eA,,x I < Ke(8-E)I x I. Consequently
m p,,(x)e.-x I < mKe(B-)I x
In connection with definition A it follows from this that the order m
of the integral function qq(x) _
=1
p,,(x)e2Nx does not exceed
1.
If
this order is equal to 1, it follows from definition B that the function p(x) is either of the minimum type of the order 1, or of the normal
type t of the order I where t S 0 - E < 0. Then according to definition 9 the function T(x) is of a lower kind than the function w(x), so that the function w(x) + q(x) is of the same kind as the function w(x) 6).
Lemma 25. C o n d i t i o n s: The function l(x) is an integral function of the normal type of the order 1. m denotes a positive integer. A,, denotes, for It = 1, .. ., m, a (complex) number with the property
A,, (S $, while in case m z 2 the numbers R1, ..., Am are distinct. v,, denotes, for It = 1, . . ., m, a positive integer. Assertion: Every solution of the differential equation m
(8)
11
(D - A,,)`N - v(x) = l(x)
µ=1
is an integral function of the same kind as the function l(x) 7). Proof. First we consider the case where m = 1 and v1 -- 1. As the integral function 1(x), the power series expansion of which we write in the form (9)
I(X) _
-" xn,
n=o n!
6) This follows easily from definitions 9, A and B. 7) It can be proved that the assertion of lemma 25 also holds if the function l(x) is an integral function of finite order that is of the maximum type of the order 1 or of the order g > 1 and the inequality I .1,, k S is omitted. The assertion of lemma 25 does not hold any longer,if one of the numbers R
has the property j Am I > as may be seen from the example (D - 2) -> v(x) = ex. Of this differential equation the totality of all solutions is given by
v(x) = ce2x - ex. If c
0, the function v(x) is of the normal type 2 of the
order I and hence, according to definition 9, of a higher kind than the function ex. 185
is of the normal type
of the order 1, it follows from formula (P, 9),
an
with an = -- , P = I and y
that for the numbers an the formula
n!
limsup IanJn= n-.oo
holds. From this it follows, in connection with formula (B) of p. 55, that the radius of convergence of the power series 00
(10)
n=o
anxn
is equal to
We now know from the Preparatory Chapter that every solution of the differential equation (8) is an integral function. Consequently every solution v(x) of the differential equation (8), with m = 1 and vt = 1, is an integral function. All the derivatives v(n)(0) (n = 0, 1, ...) are therefore significant. We write 00
v(x) = 7 n=o
v(n)(0)
xn n!
and for the numbers v(n)(0) we deduce a recurrent relation.
From the differential equation (8), with m = I and v1 = 1, it follows that Dn-1(D - A1) -* v(x) = Dn-1 -) l(x)
(n = 1, 2, ..).
If in this formula we take x = 0, we find v(n)(0)
-
A1v(n-1)(0) = l(n-1)(0)
(n = 1, 2, ...).
From this, in connection with (9), we may deduce the recurrent relation required by applying the principle of mathematical induction.
This relation is (12)
Al-1a0
v(n)(0)
+ ... + ).1an-2 + an-1 (11 = 0, 1, ...). = Aiv(0) + This relation also holds for At = 0; for 2t = 0 and n = 0 the righthand member is to be equalled to v(0). With the help of these numbers v(n)(0) we now form the power series 00
(13)
1 v(n)(0) xn. n =0
On account of formula (12) we may write this power series in the form (14) 186
n-0
(!1v(0) +
Al-la0
+ ... +
Alan-2
+ an-1)x"
We can now easily prove that the latter power series defines an analytic function. In fact, we have 00
(15)
1 0"iv(0) + Al-1ao + ... + Alan-2 + an-1)xn
n-0
00
00
(v(0) +
n-1
an-Ixn) (` ;.lxn), n=0
where for 71 = 0 the second factor in the right-hand member is to be put equal to 1, while for 7.1 = 0 in the left-hand member A° is to be put equal to 1. The series occurring in the first factor of the product in the right-
hand member of this formula converges for I x I < f , since the radius of convergence of the power series (10) is equal to
Obviously the
.
series in the second factor of the right-hand member of formula (15)
converges for
Z
'
Al I
I x I < iAl 18). Now we have assumed I Al I S , so
Consequently the radius of convergence of the power series
expansion of the product in the right-hand member of (15) is not less
than
.
However, this power series expansion occurs in the left-hand
member of (15). The latter power series expansion is therefore the
expansion of a function that is analytic for I x < We now assert that on the circle
Ix
1
.
+
there lies a singular
point of this function. In fact, this follows from the fact that the circle Ix
is the circle of convergence of the power series 00
v(0) +
an-1xn n=1
Hence of the function defined by this series there lies at least one singular point on the circle I x I =
This singularity cannot be re-
moved by the function represented by the power series which occurs in the second factor in the right-hand member of formula (15). 8) For Al = 0, the series stated is identically equal to its first term, i.e. to 1. Then it converges for every value of x. 187
On the circle
Ix
there lies therefore a singular point of the
function defined by the power series (14) and hence by (13) so that
the radius of convergence of the latter power series is equal to From this it follows, in connection with formula (B) of p. 55, that the
numbers v(")(0) (n = 0, 1, ...) have the property that 1
Jim sup I v(")(0) I n =
(16)
"yam
With the help of this formula it is now easy to see that the function v (x) in (11) has the property P 2, with A = 1. Hence it is of the order 1. v(n)(0)
From formula (16) and formula (P, 9), with an _
n!
I
and
y =, it then follows that the function v(x) is an integral function of the normal type of the order 1. This proves lemma 2 for the case m = v, = 1. We now assume that m = v, = 1 does not hold. Then we write the differential equation (8) in the form 9) ,7L
(D - A,) - {(D - R,)°1-1 fJ (D -
(17)
v(x)) = l(x)
µ=2
and we put
(D -
(18)
71L
(D -
v(x) = w(x).
.-=2
From this and from (17) it then follows that (D - A,) -± w(x) = l(x). To the latter differential equation we apply lemma 25 with m = 1,
v, = 1. Hence every solution w(x) of this differential equation is an integral function of the same kind as the function l(x). With this the problem has been reduced to the similar problem with respect to the differential equation (18) 10) the order of which is I less
than the order of the differential equation (8). In a finite number, 9>L
v,,, of steps we thus can prove the correctness of the assertion for
viz. u=1
the case where m = v, = I does not hold. This proves lemma 25 conclusively. 9) In case m = I, v, > 1 the product is empty and therefore its value is, as usual, is equal to 1. 10) In (18) we may have v, - 1 = 0. Then we put (D - A,)91-1 equal to 1. 188
Lemma 26. Conditions: The function w(x) is an integral function of the normal type 0 of the order 1. in denotes a positive integer. A denotes, for y = 1, . . ., m, a (complex) number with the property
A I S 0, while in case m z 2 the numbers 7.1, ..., Am are distinct. v,, denotes, for u = 1, . . ., m, a positive integer. It is not possible to choose the m polynomials p,,(x) (,u = 1, ..., m) of a degree not exceeding v,, - i respectively, such that the integral function m
w(x) - 7i7, p,,(x)ex,,x H=1
is of a lower kind than the function w(x) 11).
Assertion: The integral function q(x) defined by (19)
w(x)
q(x) = 1 1 (D P=1
is of the same kind as the function w(x). Proof. In connection with formula (P, 16) of the Preparatory Chapter we write formula (19) in the form m
(20)
q(x) = p=2 I (D - Ap)'i`
{(D - A1)'1-; w(x)}.
Then we first prove that the function (21)
r(x) = (D - A1)" - w(x)
is of the same kind as the function w(x).
In case Al = 0 this assertion follows at once from assertion 3y of lemma 18 together with the subjoined remark, both with p = v1. We now assume Al 0 0. For this case we shall prove the assertion
indirectly. For a moment we therefore suppose that the function r(x) is not of the same kind as the function w(x). From lemma 7, with F(D) = (D - A1)", y(x) = w(x) and h(x) = r(x) 12), it follows that the function r(x) is of the same kind at most as the function w(x). 11) From lemma 24 it follows that this condition is satisfied automatically, ?.,, I < 0 for fe = 1, ..., m. If at least one of the numbers I A,, is equal to 0 then this condition is significant as may be seen from the example we shall give to assertion 4 of theorem 20. This example is to be found in remark 1 that follows theorem 20. if
I
I
12) Obviously the condition of theorem 7 is fulfilled, because F(D) is a polynomial in D. 189
If the function r(x) is not of the same kind as the function w(x) then it follows from definitions 8, 7 and 9, that it is of a lower kind than the function w(x). Then, according to definition 9, for the function r(x) the following three cases are possible:
A. It is of the normal type t of the order 1, where I A, I S t < 0. This case is only possible, if Al I < 0. I
B. It is of the normal type t of the order 1, where 0 < t < Al C. It does not exceed the minimum type of the order 1. In case A we apply lemma 25 to the differential equation (D - Al)" v(x) = r(x), with m = 1 and l(x) replaced by r(x), E by I. Hence every solution of this differential equation is an integral function of the same kind as the function r(x). This differential equation is also satisfied by the function w(x). Therefore the function w(x) is of the same kind as the function r(x). However, we assumed that the function r(x) was not of the same kind as the function w(x). Consequently we meet a contradiction; hence in case A our assumption that the function r(x) is not of the same kind as the function w(x) appears to be incorrect. If the function r(x) comes under case B or case C then it follows from definitions 1 and 3 that there exists a number,u with the property 0 < It < I Al I such that the function r(x) is an integral function not
exceeding the normal type u of the order 1. Then we consider the function O(z) = 1z
This function is analytic for I z I < I Al I and therefore certainly for I z < u. We now expand the function O(z) into a power series of ascending powers of z. Let this power series expansion be O(z) =
00
cnzn
n=0
(I Z < I Al I). Then, according to theorem
1,
the
differential operator 00
O(D) =
c0Dn
n=0
is applicable to the function r(x) and, moreover, the numbers cn have the property mentioned in the assertion occurring in the remark
that follows the proof of theorem 7, if in this remark we replace y(x) by r(x), a by 1, i by u, a,, by cn, F(D) by O(D). This remark then
tells us that the function (22) 190
v(x) _ O(D) -- r(x)
is an integral function not exceeding the normal type it of the order 1. Because of it < Al S 0, according to definition 9, this function v(x) is of a lower kind than the function w(x) which is of the normal type 0 of the order 1. Then it follows from (22) that
(D -
v(x) = (D - Al)"1-->
r(x)}
and from this and from lemma 23, with I (x) = r(x), F(D) = O(D) and G(D) = (D - Al)"1, we see that (D - Al)"1- v(x) = {(D - A1)"1q(D)} -> r(x) = I -i- r(x) = r(x).
This means that the function v(x) is a solution of the differential equation (23)
(D - Al)"1- n(x) = r(x).
The totality of all solutions of this differential equation may be written in the form rl(x) = p(x)e"' + v(x),
where p(x) is a polynomial in x of degree vl - I with undetermined finite coefficients. It now follows from (21) that the function w(x) is also a solution of
the differential equation (23). Consequently the coefficients of the polynomial p(x) can be chosen such that there arises a polynomial p1(x) with the property that w(x) =
p1(x)e*lx + v(x),
where, as we have seen, the function v(x) is of a lower kind than the function w(x). However, this contradicts the datum concerning the function w(x). The assumption that the function r(x) is not of the same kind as the function w(x) is therefore neither correct in cases B and C. Hence the function r(x) is of the same kind as the function w(x). Now it follows from (20) and (21) that we may write nt
q(x)=fJ(D-r(x). Repeating for this formula the reasoning applied to formula (20) and continuing this process, we may prove in a finite number, viz. m, of steps that the function q(x) is of the same kind as the function w(x). This proves lemma 26. 191
We now proceed to the investigation mentioned at the beginning of this chapter and we prove Theorem 20. Conditions: The function y(x) is an integral function of the normal type r of the order 1. The differential operator 00
anDn, where the numbers an are not all
F(D) is defined by F(D) =
e,
n=O
equal to zero and have the property that F(z) _
is analytic for I z j 5 r.
n-0
anzn is a function that
0 for every value of z satisfying 1. If F(z) r, then the function h(x) = F(D) - y(x)
A s s e r t i o n s: zI
(24)
is an integral function of the same kind as the function y(x). 2. Let the function F(z) in the closed region I z 15 r have m (m Z 1) distinct zeros A. (,u = I, ... , m) with multiplicity v,, respectively. Moreover, let it be possible 13) to choose the polynomials P,, (x) (,u = 1, . . ., m) of
a degree not exceeding v,, - 1 respectively, such that the function ,n
(25)
u(x) = Y(x) +µ=1I
pµ(x)e'µx
is of a lower kind than the function y(x). Then the function h(x) defined by (24) is an integral function of a lower kind than the function y(x). 3.
Let all conditions mentioned in assertion 2 be satisfied. If we
choose the polynomials p,,(x) such that the function u(x) in (25) is equal to a function u1(x) that has the property that for every other choice of the polynomials p,,(x) the function u(x) given by (25) is of the same kind at least as the function ul(x), then the function h(x) defined by (24) is an integral function of the same kind as the function ul(x), except in one case. This exceptional case is that for which at the same time the following holds :
a) one of the numbers A,,, e.g. Al, is equal to zero,
9) the function u1(x) is a polynomial of degree r, where 0 < r S v1. In this exceptional case the function h(x) is of a lower kind than the function u1(x). Then it is identically equal to a constant. 4. Let the function F(z) have the property mentioned in assertion 2. Let it then, in contradistinction with 2, not be possible to choose the polynomials p,,(x) mentioned in 2 such that the function u(x) is of a lower kind than the function y(x). Then the function h(x) defined by (24)
is of the same kind as the function y(x). 13) In connection with lemma 24 this implies that at least one of the numbers l,, is equal to r. j
192
REMARK 1.
a) Let us mention an example for the case of assertions
2 and 3. y(x) = x - xex -+ e2x; F(z) = (z - 1)2(z - 2)(z - 3).
Here r=2, m=2, Al = 1, A2=2, v1=2, v2 = 1. So p1(x) is a polynomial of a degree not exceeding 1, p2(x) is identically equal to a
constant. Hence the function u(x) in (25) takes the form (26)
u(x) = x - xex + e2x + (ax + b)ez + Ce2x
where a, b and c are constants. If we take a = 0, b = 1, c = - 1, 1, then the function u(x) becomes so that p1(x) - 1, p2(x) u(x) = x - (x - l)ex. It is of the normal type 1 of the order 1 and, according to definition 9, of a lower kind than the function y(x).
If we take a = 1, b = 0, c = - 1, so that p1(x) = x, p2(x) = - 1, then u(x) = x
Therefore it is now of the maximum type of the order zero. Moreover, from (26) we see that it is not possible to choose the constants a, b, c and with them the polynomials p1(x) and p2(x) such that the function
(26) is of a lower kind than the function u(x) = x just found. Hence we have in this special case u1(x) = x. In this example the function u1(x) is therefore determined uniquely. P) An example for the case of assertions 2 and 3, where the function
u1(x) is not determined uniquely is:
y(x) = x - xex + e2z; F(z) _ (z - 1)(z - 2)(z - 3).
We now have therefore r = 2, m = 2, Al = 1, A2 = 2, v1 = v2 = 1. So p1(x) and p2(x) are both identically equal to a constant. In this case the function u(x) occurring in (25) has the form u(x) = x - xex + e2x + bex + ce2x,
where b and c are constants. If we now take c = - 1, then the function
u(x) takes the form u(x) = x + (b - x)ex.
Obviously for every choice of the constant b this function is of the normal type 1 of the order 1. Hence we have u1(x) = x - xez, but it is just as well allowed to take u1(x) = x + (b1 - x)ez, where b1 denotes an arbitrary, but fixed, constant. 193 13
y) An example for the case of assertion 4 is: Y(x) = e2z + e2ix; F(z) = z - 2.
Now T = 2, m = 1, Al = 2, v1 = 1. So p1(x) is identically equal to a constant c. The function u(x) in (25) has therefore the form u(x) = e2z + e2ix + ce2x.
For every choice of the constant c this function is of the normal type 2 of the order 1 and hence, according to definition 7, of the same kind as the function y(x).
Now we shall show that in the case of assertion 2 there always exists a function u1(x) with the property mentioned in assertion 3. To that end we start from the function u(x) in (25). According to assertion 2 this function is of a lower kind than the function y(x). Then it follows from definitions 9 and I that for the function u(x) the following two cases are possible: REMARK 2.
a) it is an integral function of the normal type t of the order I with
0
it is an integral function not exceeding the minimum type of the order 1. In connection with lemma 24 both in case a) and in case b) at least one of the numbers I A , , (u = 1, ... , m) is equal to r. We first consider case b). There are two cases possible, viz.: b1) none of the numbers A,, is equal to zero; b2) one of the numbers A,, is equal to zero (so m z 2 and since the b)
I
numbers A,, are distinct, one of the numbers A,, at most is equal to zero).
In case b1) it is obvious that for every choice of the polynomials (u = 1, ..., m), where they are not all identically equal to zero, the function M.
(27)
u(x) +
q,,(x)ez--x
µ=1
is of a higher kind than the function u(x). In fact, if v is the largest of the absolute values of the numbers A,, occurring in the exponents N,.
of the terms in the sum
µ-1
that are not identically equal to zero, then
with the help of definitions A and B it is easy to prove that the whole expression (27) is of the normal type v of the order 1. Hence, according to definition 9, it is of a higher kind than the function u(x). This means that in this case the function u1(x) is identical with u(x). In case b2) we assume A,,, = 0. Then for every choice of the poly194
nomials q,,(x) (,u = 1, ..., m), where the polynomials q,,(x) (u = 1, ...,
m - 1) are not all identically equal to zero, the expression m-1
u(x) +
p=1
g,,(x)e
+ gm(x)
is of a higher kind than the function u(x), because for every choice of the function u(x) + qm(x) does not exceed the minimum type of the order 1. Therefore if an expression of the form (27) shall not be of a higher kind than the function u(x), then in this case it is necessary
that q,,(x) - 0 (,u = 1, ..., m - 1). We now consider the expression u(x) + q,,,(x),
(28)
where qm(x) is a polynomial of a degree not exceeding v,,, - 1. Then the following two cases are possible: b21) the function u(x) is not a polynomial of a degree not exceeding
V. - 1; b22) the function u(x) is a polynomial of a degree not exceeding
V. - 1. In case bL1) it is clear that for every choice of the polynomial
the degree of which does not exceed v, - 1, the expression (28) is of the same kind as the function u(x). Hence we now have u1(x) = u(x).
In case b22) the function (28) is of a lower kind than the function 2 and we take qm(x) = - u(x). Then u1(x) - 0. However, if v,,, = 1 then the functions u(x) and q,,,(x) are both identically equal to a constant. So this is also the case with the expression (28). Hence, for every choice of the polynomial qm(x) of a degree not exceeding v. - 1, this expression is now of the same kind as the function u(x) so that now u1(x) = u(x). This means that we have settled case b). We now consider case a). Here are two possibilities, viz.: a1) all numbers A,, exceed t, so u(x), if vm
t
(µ=1,...,m);
a2) not all numbers A I exceed t. Now let
St (29)
t
i
for,u=l,...,x
(x<m) for y=%+ 1, ...,m.
In case a,) we have u1(x) = u(x). We may prove this in a similar way as in case b1). In case a2) for every choice of the polynomials q,,(x) (,u= x + I , ... , m)
that are not all identically equal to zero and for every choice of the 195
polynomials q (x) (,u = 1, ... , x) the function
u(x) +
m
7"
µ=1
µ=x+1
q,,(x)e'#
is of a higher kind than the function u(x). This follows from (29)
and from the fact that the function u(x) +
N
q,,(x)eaµx does not
µ=1
exceed the normal type t of the order 1. Hence, if the expression (27)
shall not be of a higher kind than the function u(x) it is necessary 0 (,u = x + 1, ..., m). Now we distinguish that we have the following two cases :
a21) none of the numbers ?.,, (,u = 1, ..., x) is equal to t; a22) at least one of the numbers I A,, Cu = 1, ... , x) is equal to t. In case a21) it follows from lemma 24 that for every choice of the polynomials q,,(x) (a = 1, ..., x) the function I
I
x
u(x) +
gq(x)eaML
µ=1
is of the same kind as the function u(x). Hence we now have u1(x)=u(x).
In case a22) we assume that we have
t=1A,I? ... Z IAxl. Let I
Al I =
... =
I
where w1 S x,
A(,,, I,
and I A. 1
I > I A, , I
co,
for
< x, then in the expression x gN(x)eaµ'
v(x) = u(x) + µ=1
for every choice of the polynomials q,,(x) of a degree not exceeding v,, - 1 respectively, the sum gµ(x)e -` µ=rut -1-1
is of a lower kind than the function u(x). We now consider the function v,(x) = u(x) + I gµ(x)eas', v=1
where q,,(x) (,u = 1, ... , co,) are polynomials of a degree not exceeding
v,, - I respectively. Then the following two cases are possible: 196
a221) there do not exist polynomials q,,(x) (,u = 1, ... , w1) of a
degree not exceeding v - I respectively, such that the
function v,(x) is of a lower kind than the function u(x) ; a222) there exists a system of polynomials q,,(x) =q, (x) (,u=1, ... , wt) of a degree not exceeding v,, - 1 respectively, such that the function vl(x) is of a lower kind than the function u(x). In case a221) for every choice of the polynomials q,,(x) (u = 1, ... , co,)
of a degree not exceeding v,, - I respectively, the function v,(x) is of the same kind as the function u(x). Then the function v(x) is always of the same kind too as the function u(x). Hence we may take u,(x)=u(x). In case a222) the function w,(x) = 21(x) +
q*(x)e''px
p-1
is of a lower kind than the function u(x). For co, < x we make an arbitrary, but fixed, choice of the polynomials q* (x) (,u = w, + 1, ... , x)
such that the degree does not exceed v - 1 respectively. Then we consider the function (30)
w(x) = u(x) +
=1 q*(x)e'"x
which therefore is of a lower kind than the function u(x). With respect to the function u(x) there now presents itself a similar
case with regard to the function w(x) as to the function u(x) with respect to the function y(x) in the beginning of this remark, with this difference, however, that the number of numbers A,,, occurring in the exponents of the expression (30), is equal to x and hence is less than the corresponding number in (25) which is equal to m. Now for the function w(x) we may consider two cases which are similar to cases a) and b) respectively for the function u(x).
Proceeding in this way after a finite number of steps we finally reach either a case which is similar to one of the cases b), a,), a2,), a221)
already investigated, or an expression of the form r(x) +
x s(x)e,
is a polynomial of a degree not exceeding v - 1, and which expression comes under case a222). Hence then there exists a polynomial s,*(x) of a degree not exceeding vx - 1, such that the function where
r,(x) = r(x) + Sx (x)e
is of a lower kind than the function r(x).
If then A. 0 0, then the function r(x) is of the normal type J .l 197
of the order 1. If A. = 0, then the function r(x) is a polynomial of a positive degree. In fact, if it were a constant, then it would not be possible for the function r1(x) to be of a lower kind than the function r(x).
There does not exist a polynomial sx*(x) now of a degree not exceeding v,, - I such that the function r2(x) = r(x) + sx *(x)e)x
is of a lower kind than the function rl(x). In order to prove this we write r2(x) = ri(x) + {sx *(x) - sX (x))e--,
0 0 the function
so where sx * (x) - s,* (x) 0 0. In case {s,**(x) - s,*(x))e
of the order 1 and hence of a higher kind is of the normal type I than the function r1(x). Then the function r2(x) is not of a lower kind than the function r1(x). In case k = 0 r(x) is a polynomial of a positive degree. Now the function r1(x) is identically equal to a conAl
I
stant. Then the function r2(x) cannot be of a lower kind than the function r1(x). Hence we have u1(x) = rl(x). Therefore in all cases after a finite number of steps we find a function 141(x).
Proof of theorem 20. From theorem 1, with a = 1, it follows that h(x) is a function that is defined for every finite value of x and from theorem 7 it appears that the function h(x) is an integral function of
the same kind at most as the function y(x). Hence, according to definitions 8 and 3, it is an integral function not exceeding the normal type r of the order 1. We shall now prove assertion 1. Since the function F(z) is analytic
and = 0 for I z I S r, the function O(z) =
(31)
1
F(z)
is analytic for I z S r. Let this function O(z) have the power series
c"z". Then it follows from theorem 1, with 00 F(D) replaced by O(D) _ c"Dn and y(x) by h(x), that the difexpansion O(z) _
00
n=0
n=0
ferential operator #(D) is applicable to the function h(x). According to the definition of applicability (definition 11), the expression (32) 198
#(D) - h(x)
is significant for every finite value of x. Then it follows from (24) that O(D) - h(x) _ ¢(D) -; {F(D) -> y(x)}. Applying lemma 23, with I (x) = y(x) and G(D) = O(D), we find that the right-hand member of this formula is equal to {¢(D)F(D)} -> y(x) = I
y(x) = y(x),
so that we have (33)
¢(D) - h(x) = y(x)-
If the integral function h(x) should not be of the same kind as the function y(x), then from definitions 8, 7 and 9 it would follow that it is of a lower kind than the function y(x) since it is of the same kind at most as the function y(x). Then from definitions 9 and I we deduce that for the function h(x) there exist the following two possibilities: a) it is of the normal type of the order 1, where 0 < < r; b) it does not exceed the minimum type of the order 1.
For a function h(x) belonging to one of these two classes there always exists a number t with 0 < t < r, such that the function h(x) does not exceed the normal type t of the order 1 (definition 3). Then
the assertion mentioned in the remark that follows the proof of theorem 7, with y(x) replaced by h(x), a by 1, r by t and F(D) by ¢(D), tells us that the function O(D) - h(x) does not exceed the normal type t of the order 1. However, this contradicts formula (33), since the function y(x) is of the normal type r (r > t) of the order 1. Hence, our
assumption that the function h(x) is not of the same kind as the function y(x) leads to a contradiction. Therefore it is of the same kind as the function y(x) so that assertion I is proved. Now we shall prove assertion 2. We now assume that we have chosen the polynomials (u = 1, ... , m) such that the function u(x) in (25) is of a lower kind than the function y(x). Then from defi-
nitions 9 and 3 it follows that the function u(x) does certainly not exceed the normal type r of the order 1, so that, according to theorem I with y(x) replaced by u(x), the differential operator F(D) is applicable to the function u(x).
Besides, it follows from definition B that the integral function p,,(x)eAHx, where It may take each of the values 1, ... , m, is of the normal type I A I of the order 1, provided that 2, 0. There, according to the datum, I A,, 15 r. In case 1,, = 0 the integral function p,,(x)e'px is a polynomial and therefore of the order zero. According to definition 3, in both cases the function p,,(x)e'µx is an integral function not ex199
ceeding the normal type r of the order 1. According to theorem 1 the differential operator F(D) is therefore also applicable to the functions p,,(x)eµx (,u = 1, ..., m). This means that from h(x) = F(D) -> y(x) = F(D) -> {u(x) -
p,,(x)el
}
µ=1
it follows that 'n
u(x) - I {F(D) -
h(x) = F(D)
p,,(x)ezpy}.
µ=1
Putting nt
F(z) = V(z) JJ (z µ=1
we have rn
F(D) = p(D) fl (D µ=1
so that h(x) = F(D)
m
u(x) -
nt
{y (D) fT (D µ=1
p,,(x)expx}.
µ=1 nt
Applying lemma 23 with G(D) - p(D), F(D) = fT (D - ,,)Yµ µ=1
and f(x) = p,(x)elµ', we find n
nt
h(x) = F(D) - u(x) -
{yo(D) - (fT (D µ=1
p,,(x)ez )}
µ=1
m
= F(D) -* u(x) -
V(D) - 014) µ=1
Hence (34)
h(x) = F(D) -* u(x).
As the function u(x) is of a lower kind than the function y(x), according to definitions 9 and 1, it comes under one of the cases a), b) mentioned in the proof of assertion 1. Then we can easily prove that
there exists a number t with 0 < t < r, such that the function h(x) does not exceed the normal type t of the order 1. According to definition 9 the function h(x) is therefore of a lower kind than the function Y(x)
Next we prove assertion 3. Now we assume that we have chosen the polynomials p,,(x) (u = 1, ..., m) such that the function u(x) in (25) is equal to a function u1(x) that has the property that it is of a 14) The function
is a solution of the differential equation
(D 200
- 7(x) = 0-
lower kind than the function y(x), while for every other choice of the the function u(x) in (25) is of the same kind at least polynomials
as the function u1(x). Then we now have h(x) = F(D) - ul(x), (35) as we may prove quite similarly to the proof of formula (34). Since the function ul(x) is of a lower kind than the function y(x), according to definition 9 it falls under one of the following cases: a) b)
it is of the normal type of the order 1, where 0 < < i; it is of the normal type a of the order a, where 0 < a < 1 and
0
d) e)
/)
it is of the minimum type of the order a, where 0 < a 5 1; it is of the maximum type of the order a, where 0 < a < 1; it is of the maximum type of the order zero; it is identically equal to a constant.
In case a) we distinguish the following two sub-cases: a,) the function F(z) does not take the value zero for ; z J S
a,r) the function F(z) takes the value zero at at least one point z with In case a,) we apply assertion 1, with y(x) replaced by u1(x) and r by . Therefore the function h(x) is now of the same kind as the function u1(x).
In case all) we assume that the zeros of the function F(z) of which the absolute value does not exceed , are Au (u = 1, ... , n, so where n 5 m because < z). We now put F(z) = x(z) 17 (z -
(36)
µ=1
Then the function x(z) is analytic and follows that
0 for I z S
.
From (36) it
F(D) = x(D) 17 (D - A,)"a µ=1
and from this and from (35) we see that
h(x) _ {x(D) j7 (D 'U=1
u1(x).
Hence, by applying lemma 23 with G(D) = x(D), f (x) = u1(x) n and F(D) = 17 (D - Au)'A, we get the formula N-1
n
(37)
h(x) = X(D) - {11 (D µ=1
-
u1(x)). 201
Putting n
(38)
rf (D - A,J) ,- - u1(x) = k(x),
'U=1
it follows from (37) that the formula (39)
h(x) = x(D) - k(x)
holds.
To (38) we now apply lemma 26 with m replaced by n, 0 by
,
w(x) by u1(x) and q(x) by k(x). This lemma tells us that the function k(x) is of the same kind as the function u1(x).
Next, at formula (39) we make use of the fact that, as we have 0 for I z I 5 . And since the seen, the function x(z) is analytic and function k(x) is of the same kind as the function u1(x) and therefore, according to definition 7, of the normal type $ of the order 1, to (39) we may apply assertion 1. This assertion then tells us that the function h(x) is of the same kind as the function k(x). Hence it is of the same
kind as the function u1(x). This proves the assertion for case ail). In cases b), c) and d) it is again easy to see 15) that the conditions of theorems 12, 13 and 14 respectively are fulfilled, if in them we replace y(x) by u1(x). Hence in these cases the function h(x) is of the same kind as the function u1(x).
In case e) we apply theorem 15 with y(x) replaced by u1(x). If the function u1(x) is a transcendental integral function of the order zero, then the function h(x) is of the same kind as the function u1(x). If the function u1(x) is a polynomial of degree q > 0, then the function h(x) is also of the same kind as the function u1(x), unless the condition of assertion 3 of part B of theorem 15 is satisfied. In fact, in the latter case z = 0 is a zero with multiplicity q of the function F(z). Then
one of the numbers A,,, e.g. AI, is equal to zero and v1 z q. Then, according to assertion 3 of part B of theorem 15, the function h(x) is of a lower kind than the function u1(x). According to assertion 4 of part B of theorem 15 the function h(x) is now identically equal to a constant. In case f) the function u1(x) is identically equal to a constant and hence the function h(x) also. According to definition 7 the function h(x)
15) In fact, according to theorem 1, the differential operator F(D) is applicable in each of these cases to all integral functions that are of the same kind as the function u1(x). This means that the numbers a satisfy the conditions of theorems 1, 2 and 3 respectively (and also those of theorem 4 in case e)). Hence, the conditions of theorems 12, 13 and 14 respectively (and also those of theorem 15 in case e)) are fulfilled. 202
is now also of the same kind as the function u1(x). This proves assertion
3 completely. The proof of assertion 4 runs quite analogous to the proof of case arr) which we considered when we proved assertion 3. In that proof we only have to replace u1(x) by y(x), by z and n by in. This proves theorem 20 conclusively.
Now we treat the "converse" problem. Theorem 21. Conditions: The function h(x) is an integral function of the normal type r of the order 1. The differential operator F(D) 00
a,LD", where the numbers a are not all equal 00 n-o anz" is a function that is to zero and have the property that F(z) _ n=o analytic for I z 15 T. A s s e r t i o n s: 1. If the function F(z) does not take the value zero for I z I < r, then the differential equation
is defined by F(D) _
F(D) -± y(x) = h(x)
(40)
has one and only one solution y(x) = i7(x) that is an integral function of the same kind at most as the function h(x). 2. The function q(x) mentioned in assertion i is an integral function of the same kind as the function h(x). 3. If for I z I < a the function F(z) has in (m 1) distinct zeros 7/L
A,,
(,u = 1, ... , m) with multiplicity v,, respectively and if p = X v,,, Y=1
then the differential equation (40) has oon solutions that are integral functions of the same kind at most as the function h(x). 4.
The solutions mentioned in assertion 3 are all of the same kind
as the function h(x).
REMARK. A differential equation of the type considered here can indeed have solutions of a higher kind than the function h(x).
An example of such an equation is
l
(2n + 1) i
D2n+1' - y(x) = ex.
As may easily be seen, y(x) = ex + e2x is a solution of this equation
and this function is of the normal type 2 of the order 1. Hence it is of a higher kind than the function h(x) = ex. 203
First we shall prove assertion 1. Since the function F(z)
Proof.
is analytic and
0 for J z
r, the function O(z) =
1
F(z)
is analytic
r. Let this function O(z) have the power series expansion
for I z 00
O(z) _
Then we define the differential operator O(D) by
n=o 00
¢(D) =
c,,Dn. According to theorem 1, with y(x) replaced by h(x) 9:=0
and F(D) by /(D), this differential operator O(D) is applicable to the function h(x) and according to theorem 7, likewise with y(x) replaced by
h(x) and F(D) by ¢(D), the function O(D) > h(x)
(41)
is an integral function of the same kind at most as the function h(x). Now we assume that the differential equation (40) has a solution y(x) = 77(x) that is an integral function not exceeding the normal type r of the order 1. Then it follows from what we have just said with respect to the function (41) and from (40) with y(x) = 77(x), that we have, by applying lemma 23, with G(D) = O(D) and /(x) _ 77(x), O(D) -- h(x) _ O(D) - {F(D) > 77(x)} = {O(D). F(D)} -- i(x) _ 71(x).
Hence the function j(x) is determined uniquely and we see at the same time that it is an integral function of the same kind at most as the function h(x). This function 21(x) is indeed a solution of the differential equation (40). In fact, we have
F(D) - fi(x) = F(D) - {O(D) - h(x)} and applying lemma 23, with G(D) replaced by F(D), F(D) by O(D) and I (x) by h(x), we see that F(D) -* {o(D) > h(x)) = {F(D)O(D)} --* h(x) = I ->. h(x) = h(x),
so that
F(D) - n(x) = h(x). Assertion is therefore proved. We shall now prove assertion 2. According to definitions 8 and 1, 1
for the function n(x) determined just now there are three cases possible, viz. the two cases a) and b) indicated in the proof of theorem 20 and the third case, where the function 71(x) is of the normal type r of the
order 1. In cases a) and b) there exists a number t with 0 < t < r, such that the function F(D) > i(x) does not exceed the normal type t of the order 1. According to definition 9 it is in these cases of a lower 204
kind than the function h(x). Since, however, F(D) - r7(x) = h(x) it appears that cases a) and b) cannot occur. Hence the third case is the only possible one and, according to definition 7, in this case the function i(x) is of the same kind as the function h(x). This proves assertion 2. Next we prove assertions 3 and 4. We write the function F(z) in the form In
(42)
F(z) = y,(z) fl (z µ=1
Then the function ,(z) is analytic and
0 for I z
T. If this function
V(z) has the power series expansion V(z) =oof dnzn, then we define the n=o
differential operator V(D) by ,V(D) _
00
d,,Dn. Now we assume that
n==o
the differential equation (40) has a solution y(x) = E(x) that is an integral function of the same kind at most as the function la(x). Then, according to definitions 8 and 3, this integral function E(x) does not
exceed the normal type r of the order 1. From lemma 23, with G(D) = p(D), F(D) _ fl (D - A,,)"N, I (x) _ $(x), it then follows that µ=1
we may write (40), with y(x) = $(x), in the form 77L
(43)
(D - AH)µ > $(x)} = h(x).
zv(D) > { µ=1
Putting M
(D - Aµ)"µ > fi(x) = w(x),
(44) µ=1
it follows from the assertion in the remark that follows theorem 7, that the function w(x) is also an integral function not exceeding the normal type r of the order 1. From (43) and (44) we see that this function w(x) is a solution of the differential equation (45)
'p(D) > w(x) = h(x).
According to assertion I the latter differential equation has one and only one solution w(x) = ip(x) of the same kind at most as the function h(x) and this function qq(x) is an integral function of the same kind as the function h(x), as follows from assertion 2. This means, that,
if the function (z) is an integral function of the same kind at most as the function h(x) and if, moreover, it satisfies the differential equation (40), it is also a solution of the differential equation (44) the right-hand member of which is a function which is uniquely 205
determined and, according to definition 7, of the normal type r of the order 1. Every solution of the latter differential equation, however, is an integral function of the normal type r of the order 1, as follows from lemma 25, with l(x) replaced by w(x) and by T. So we see that,
if the differential equation (40) has a solution that is an integral function not exceeding the normal type r of the order 1, this solution occurs among the solutions of the differential equation (44), which
solutions are all of the normal type r of the order I and which, as follows from the Preparatory Chapter, can be written in the form T u(x).
fi(x) = 2,
µ=1
In it u(x) is a solution of the equation (44), while the functions
are polynomials of degree v,, - I respectively, with undetermined m
finite coefficients. Hence the number of these coefficients is p =
v,,.
µ=1
Consequently the number of solutions of equation (44) is oov and these are all of the same kind as the function h(x) (see definition 7). Finally we shall show that all these solutions of the differential equation (44) are also solutions of the differential equation (40). In fact, if fi(x) is such a solution of (44) then, because of lemma 23, (44) and (45), we have F(D) -* fi(x) = tp(D) -- {
(D - ,,)'
fi(x)} = V(D) - w(x) = h(x).
This proves theorem 21. REMARK. Both in theorems 16, 17, 18 and 19 of chapter IV and in theorem 21 just proved we speak of oov solutions. In all these theorems the definition of the notion "oov functions" is applicable (definition C). Nevertheless it is important to point out a difference
in the manner in which the p linear independent functions gpl(x), ..., ppv(x), with the help of which the oov solutions are formed, arise in theorems 16, 17, 18 and 19 on the one side and in theorem 21 on the other side. In theorems 16-19 there occur oov solutions if for the generating
power series 71 a,,zn of the differential operator F(D) = n=0
00
n=0
anDn
we have a0 = ... = a,_1 = 0, a -A 0 (p Z 1). This means that then we may write 00
F(z) = z" 2; a,,,, z". n=0
206
If we wished so, we might therefore say that p is the multiplicity with which F(z) is equal to zero at the point z = 0 16). Now the p linear independent functions ry1(x), ... , T ,,(x) of definition C are the functions x0, .. ., x9-1. So all are powers of x. 1) solutions, if the generating In theorem 21 there occur ooz, (p 00
a function that is analytic
power series I
n0
i, of which the for I z S z and that has zeros in the closed region I z 1, ... , m) are these zeros sum of the multiplicities is equal to p. If A,, (i 111
v,,. In this case the
with multiplicity v respectively, then p P=1
linear independent functions T ,(x), ... , (p,(x), with the help of which the oo' solutions are formed, are the functions e'Ax,
... ,
(Ft = 1, ... , M).
x"v-1 e-µ2
Hence they are exponential functions now, if d,, 0 0. Only in case in = 1, R1 = 0, v1 = p they coincide with the p linear independent functions, with the help of which in theorems 16-19 the cot' solutions are formed.
We now proceed to treat the homogeneous differential equation. For this equation the following theorem holds: Theorem 22. In the homogeneous differential equation
F(D) - y(x) = 0
(46)
00
the differential operator F(D) is defined by F(D) _
a are not all equal to zero.
n=°
Assertions: A.
In case ao - 0 the following holds: 1.
If there exist a number a with 0< a< I and a number r with 0 < x <_ oo, such that the expression
(47)
a i--
G(z) = Y n=0 n!
Z.
16) This pronouncement has only a formal value since, strictly speaking, it only comes in for consideration if F(z) is analytic at z = 0. However, this is only the case in theorem 17 for all differential operators F(D) considered and then only if a = 1. In theorem 17 with 0 < a < I and in theorems 16, 18 and 19 differential operators are also admitted of which the generating power series converges for z = 0 only. Then there is hardly any sense in speaking of point z =0 as a zero of F(z). 207
defines an integral function not exceeding the normal type, of the order 1, then the differential equation (46) has no solution that is an integral function not exceeding the
less than (ar)
2.
a,
normal type r of the order a and, moreover, that is not = 0. 1/ there exists a number a with 0 < a < 1, such that the expression (47) defines an integral function of an order less than 1,
then the differential equation (46) has no solution that is an integral function not exceeding the maximum type of the order Cr
and, moreover, that is not ; 0. 3. If there exists a number a with 0 < a 5 1, such that the expression
B.
(47) defines an integral function not exceeding the normal type of the order 1, then the differential equation (46) has no solution that is an integral function not exceeding the minimum type of the order a and, moreover, that is not 0. 1) the following holds. In case ao = ... = a9_1 = 0, ap 0 (p 1. The differential equation (46) has oop solutions that are polynomials. All these polynomials are of a degree not exceeding p - 1 17). 2. If there exist two numbers a and r with the properties mentioned
in assertion A i, then the differential equation (46) has oop solutions that are integral functions not exceeding the normal type r of the order a. All these solutions are polynomials; they are the same as those mentioned in assertion B i. 3.
If there exists a number a with the properties mentioned in assertion A 2, then the differential equation (46) has oop solutions that are integral functions not exceeding the maximum type of the order a. All these solutions are polynomials; they are the same as those mentioned in assertion B i.
4.
If there exists a number a with the properties mentioned in assertion A 3, then the differential equation (46) has oop solutions
that are integral functions not exceeding the minimum type of the order a. All these solutions are polynomials; they are the same as those mentioned iu assertion B i. Co
C.
1.
If the expression F(z) _
anzn defines a function that is n=0
analytic for I z I a r and that has no zero for I z I S r, then the
differential equation (46) has no solution that is an integral function not exceeding the normal type r of the order 1 and more-
over, that is not = 0. ") See also the remark which precedes this theorem. 208
If the expression F(z) _ I00 anzn defines a function that is
2.
0
analytic for I z I S t and that in the closed region I z S i has m (m Z 1) distinct zeros A,, (y = 1, ... , m) with multiplicity v respectively, then the differential equation (46) has oon solutions that are integral functions not exceeding the normal type z of the nt
order 1. There p = Proof.
v,, 17).
µ=1
First we prove assertion A 1. Let h(x) be an integral function
of the normal type r (0 < t < oo) of the order a (0 < a < 1). According to assertion (48)
I
of theorem 16 the differential equation F(D) -* y(x) = h(x)
then has one and only one solution y(x) = n(x) that is an integral function not exceeding the normal type r of the order a (see also definition 3). Now if the equation (46) should have a solution y(x) = fi(x)
that is an integral function not exceeding the normal type r of the order a and that is not 0, then the function n(x) + fi(x) would be an integral function not exceeding the normal type r of the order a and this function is not identically equal to the function 71(x). Now it follows from theorem I that the differential operator F(D) is applicable to the function rt(x) + fi(x). Then, like the function rl(x), the function rl(x) + fi(x) would be a solution of the differential equation (48). Hence this equation would have two different solutions that are both integral functions not exceeding the normal type r of the order a. As this is not possible assertion A I is proved.
For the proofs of assertions A 2 and A 3 we may similarly refer to
of theorems 18 and 17 respectively. assertion Assertion B 1 is contained in assertion 3 of part B of theorem 19. We now prove assertion B 2. As in the proofs of assertions 3 and 4 of 1
theorem 16 we put F(D) = F*(D)D9, 00
where F*(D) _
no
an+,Dn.
Because aD
0 the differential equation
F*(D) - y(x) = 0
has, according to assertion A 1, no other solution that is an integral
function not exceeding the normal type z of the order a than the trivial solution y(x) = 0. This means that the solutions required are those functions y(x) which we obtain by integrating p times indefinitely
the function y(x) - 0 18). Hence the number of these solutions is 18) See a related reasoning in the proof of assertions 3 and 4 of theorem 16. 209 14
oon; they are all polynomials of which the degree does not exceed p - 1. Hence they are the same as those mentioned in assertion B 1. For the proofs of assertions B 3 and B 4 we may similarly refer to assertions A 2 and A 3.
The proof of assertion C 1 we can give analogously to that of assertion A 1 by using assertion I of theorem 21. In order to prove assertion C 2 we write F(z) as in formula (42). Then it follows from assertion C 1 that the differential equation (45),
with h(x) = 0, has one and only one solution that is an integral function not exceeding the normal type r of the order 1, viz. w(x) - 0.
With the help of the differential equation (44) with w(x) - 0 we may now easily complete the proof. This proves theorem 22. REMARK 1.
The different cases considered in the assertions of
part A, give us an opportunity to refer in some of them to the theorems 16-18 of chapter IV. We need not consider separately a case where we might make use of the assertions of part A of theorem 19. In fact, if the conditions mentioned in assertion A 1 of theorem 19 are satisfied, then there exists a. finite real number v, independent both of n and of z, such that the expression H(z) mentioned there, defines an integral function of an order less than 1. In case v > 1 1
we take - = v so that then 0 < a < 1 and we may apply assertion Cr
A 2 of theorem 22. In case v 5 I the expression (49)
K(z)
_
a ni zn
°°
nta
n=O
defines for every a with 0 < or < I an integral function of an order less than 1. Hence now also assertion A 2 of theorem 22 may be applied.
In fact, if the order of the function H(z) is equal to e, then it follows from property P 1, that for every positive 6 we have Ian l Jim in! e + a !r
1
n= 0
n
Then also for every positive 6 1
lim n!e+a n +o0
+ 1 _y a
_a I
-
n! a
210
I
1
n
If we now choose 6 = I - e we find lira n !
--
+1-Ia
U I
n = 0.
Because of 0 < a < I and v -z- I we have 1 +
1 - v > 1. It now
follows from property P 8 that the expression (49) defines an integral
/
function the order of which does not exceed ( 1 + \ the order is certainly less than 1.
1-v
.
Hence
REMARK 2. From theorems 16 and 22 we easily deduce that in the case with ao = ... = av_1 = 0, an 0 (p Z 1) the set of the oov
solutions mentioned in assertion 3 of theorem 16 may be represented by the sum of a solution y(x) = T(x) of the differential equation F(D) -* y(x) = h(x), (50)
that is an integral function not exceeding the normal type r of the order a (0 < a < 1), and the set of the oov solutions of the homogeneous differential equation
F(D) - y(x) = 0, mentioned in assertion B 2 of theorem 22. (51)
In fact, if we represent the set of all the solutions of (50) that are integral functions not exceeding the normal type r of the order a, by
y(x) = 90) + V(x), then the function ip(x) satisfies the differential equation (51). Hence the function ?p(x) represents the set of all the solutions of the latter differential equation that are integral functions not exceeding the normal type r of the order a. It may easily be constructed with the help of assertion B 2 of theorem 22.
A similar result holds for the solutions mentioned in assertion 3 of theorem 17, in assertion 3 of theorem 18 and in assertion 3 of theorem 21.
Moreover, a similar result holds for the solutions mentioned in assertion 3 of part A of theorem 19 and for those mentioned in assertion 4 of part B of theorem 19, as may easily be seen in connection with
assertion B 3 of theorem 22 and remark I which precedes this remark. Also a similar result holds for the solutions mentioned in assertion 3
of part B of theorem 19 for the case where the function h(x) 0 0 as may be proved with the help of assertion B I of theorem 22. We shall now show that our theorems 12, 13, 14, 15 and 20 enable us to prove a generalization of a theorem of J. M. Whittaker. In the 211
Introduction we have formulated Whittaker's theorem as theorem H. As we have pointed out there, our generalization is a consequence of
the fact that we replace Whittaker's "asymptotic period" of an integral function of finite order y(x) in case this function y(x) does not exceed the normal type of the order 1 by our number (o to which we attach a wider meaning than Whittaker to his asymptotic period. Our generalization of theorem H we formulate in theorem 24. To prove theorem 24 we need a part of the results of our theorems 12, 13, 14, 15 and 20. For convenience' sake we recapitulate these results needed in theorem 23. In the Introduction we have already said that in a sense theorem 23 is a far-going generalization of our theorem 24 and hence of Whittaker's theorem H. Now we formulate theorem 23.
Theorem 23. Conditions: 1. Let the function y(x) be either
a) an integral function of the normal type r of the order ; or b) an integral function of the minimum type of the order a (0 < a S 1) ; or 1
c) an integral function of the normal type r of the order a (0 < z
0
a transcendental integral function o the maximum type of the order zero; or f) a polynomial the degree of which is p sitive; or
e)
g) identically equal to a constant. 00
The differential operator F(D) is defined by F(D) _ I a,,Dn,
2.
n=o
where in the cases a) -f) mentioned above the numbers a are not all equal to zero and have the following property respectively 19): 00
In case a) that the expression F(z) _ I that is analytic for I z 15 r. In case b) that the expression
n=o
a G(z) _
(52)
a function
zn
n= o niv
defines an integral function not exceeding the normal type of the order 1.
In case c) that the expression (52) defines an integral function 19) In case f) the numbers an are only required to be not all equal to zero. In case g) on the numbers an no condition is imposed. 212
I
not exceeding the normal type, less than (ir) o , of the order 1. In case d) that the expression (52) defines an integral function of an order less than 1. In case e) that there exists a finite real number v, not dependent on n nor on z, such that the expression an H(z)= -z" n=0 n. 0o
defines an integral function of an order less than 1.
Assertions: 1.
The expression
h(x) = F(D) - y(x) defines an integral function that in the following cases is of the same kind as the function y(x): A. In case a) mentioned above, if one of the following two conditions a and P is satisfied : a. The function F(z) is zA 0 for every value of z that satisfies
IzI5T.
(i.
The function F(z) has m (m Z 1) distinct zeros A. (p. 1, ... , m) in the closed region I z I -< r with multiplicity vN respectively; there do not exist polynomials pN(x) (,u = 1, ... , m) the degree
of which does not exceed v, function
-I
respectively, such that the
In
pN(x)eaNz
u(x) = y(x) + N=1
is of a lower kind than the function y(x). B. In case b) mentioned above. C. In case c) mentioned above. D. In case d) mentioned above. E. In case e) mentioned above. F. In case f) mentioned above, if the following holds :
The function y(x) is a polynomial of degree q > 0, while, moreover, of the numbers a0, ... , a4_1 at least one differs from zero.
G. In case g) mentioned above. 2. The expression
h(x) = F(D) - y(x) defines an integral function that is of a lower kind than the function y(x) in the following cases:
T. In case a) mentioned above, if the following holds: In the closed region I z M S z the function F(z) has m (m Z 1) distinct zeros 213
IF (,u = 1, . . ., m) with multiplicity vE, respectively; there exist polynomials p,,(x) (,u = 1, ... , m) the degree of which does not exceed vµ - 1 respectively, such that the function Y/L
u(x) = y(x) -}-
p0(x)e,,x
µ=1
is of a lower kind than the function y(x) 20). IL In case /) mentioned above, if the following holds : The function
y(x) is a polynomial of degree q > 0, while, moreover, a0 = .. .
=a._1=0. follows in cases A, B, C, Proof. The correctness of assertion D, E, F and G from the theorems 20, 13, 12, 14, 15, 15 and 15 respectively. The correctness of assertion 2 follows in cases I and II from the theorems 20 and 15 respectively. 1
Now our generalization of Whittaker's theorem reads as follows: Theorem 24. Conditions: The function y(x) is an function not exceeding the normal type of the order 1. 0) for which the function B is the set of numbers w (w k(x) = (e0'D - 1) _ y(x) (53)
integral
is of a lower kind than the function y(x). A s s e r t i o n s: 1. If the integral function y(x) is of the normal type r of the order I and if there exists a number j9 with I P I = r and with the property that two constants cl and c2 may be found such that the function v(x) = y(x) + cle'x + c2e-dx
is of a lower kind than the function y(x) 21), then set B is formed by the
points k
(k = ± 1, ± 2,
... ). 1/ such a number fi does not exist, then
set B is null. 2. If the integral function y(x) does not exceed the minimum type of the order 1, but if it is not a polynomial of degree 1, then set B is null. 3. 1/ the integral function y (x) is a polynomial of degree 1, then set B is formed by all complex numbers co 0 0. 20) In connection with lemma 24 this implies that for at least one of the numbers .t we have I A. I = r. 21) Of such a number l; the argument is modulo n determined uniquely as will follow from the proof of assertion 1. 214
Proof. We first prove assertion 1. Let the integral function y(x) be of the normal type z of the order 1. It then comes therefore under case a) of theorem 23.
0
If a differential operator F(D) is defined by F(D) _
a"D", n=0
where the numbers a" have the property for case a) mentioned in condition 2 of theorem 23 and if the function F(z) has at least one zero z with I z I S z, then it follows from assertions I A and 2 I of theorem 23 that the function h(x) = F(D) -± y(x)
is of a lower kind than the function y(x), if and only if there exist polynomials P(x) the degree of which does not exceed v, - I, with the
property that the function 24X) = AX) +
eaµX
Pµ (x) µ=1
is of a lower kind than the function y(x). In it A. are those zeros, with multiplicity v,, respectively, of the function F(z) of which the absolute value does not exceed T. In our special case the differential operator F(D) is defined by F(D) = e',) - I (w 0) so that wn
(54)
a0 = 0, an = n (n = 1, 2, ... ). i
Obviously these numbers an have the property mentioned in condition 2 of theorem 23 for case a). The zeros of the function
F(z) = e" - 1 are all simple ones and they lie on a straight line through the point z = 0. Let this line cut the circle I z = z in the points z = and (55)
z=-j9.
From the above it follows that the function k(x) in (53) is of a lower
kind than the function y(x), if and only if there exist m constants c,, (,u = 1, ... , m) such that the function m
(56)
u(x) = y(x) +
µ=1
cµ ezµl
is of a lower kind than the function y(x). There A,*, (,u = 1, ... , m) are those (simple) zeros of the function (55) of which the absolute value does not exceed rr.
We shall now prove that, as a consequence, the function k(x) is of a lower kind than the function y(x), if and only if the points z = # 215
and z = - fi, already mentioned, are zeros of the function (55) and moreover, there exist two constants cl and c2 such that the function v(x) = Y(x) + cle'X + c2e-px
(57)
is of a lower kind than the function y(x). In fact,
if I A* I < a
(p = 1, ..., m), then it follows from lemma 24, with w(x) = y(x), Aµ = A* and 0 = z, that the function u(x) in (56) is of the same kind as
the function y(x). However, if z = # and z = - # are zeros of the
function (55), then they occur under the numbers X* (,u = 1, ..., m). Now I i4 I = z and since the points z = X' all lie on the linesegment joining the points z = fi and z = - 9, for all the remaining numbers R* we have the inequality I R* I < z. Hence the sum E c,,eP formed with the latter numbers A* is of a lower kind than the function y(x). Consequently these zeros ,f*, for which therefore A* < z, play no part here 22). Hence if there does not exist a number fi with r and with
the property that two constants belong to it, c1 and c2, such that the function (57) is of a lower kind than the function y(x), then set B is null. If, on the contrary, there does exist a number fl with f /1 I = z
and with the property that two constants belong to it, c, and c2, such that the function (57) is of a lower kind than the function y(x),
then the number
ni
is an element of B. 2ni
For, z = fi is a zero of the function e a Z - 1. Moreover, each of the numbers k2 Z (k = N
1, ± 2, ...) is an element of B, because z = k2xi
is also a zero of the function e a - 1. We now assert that B contains no number y (y
not belong to the sequence of numbers k
0) which does
(k = + 1, ± 2, ...). Z
In fact, if B did contain such a number y, then the function k(x) = (e" - 1) - y(x)
would be of a lower kind than the function y(x). Then to this number y
there would exist a number S with 16 = z and with the property that there exist two constants dl and d2, such that the function w(x) = y(x) + dl?-' + d2e-ex is of a lower kind than the function y(x). From this and from the 22) This follows easily from the definitions 9, A and B. 216
fact that the function v(x) in (57) is of a lower kind than the function y(x), it follows that the function (58)
v(x) - w(x) = cleez +
c2e-sx - dleex -
d2e-ax
is of a lower kind than the function y(x). Consequently 6 is equal to # or to - fl. In fact, if this should not be the case then we should use the fact that at least one of the numbers cl and c2 is different from zero as follows from (57) and from the assumption that the function 0. Then we put v(x) is of a lower kind than the function y(x). Let cl re-iargd (r > 0) by which the exponent fix becomes positive, viz. x= equal to fire-i°'g" = (fl ( r, while the real parts of the remaining three exponents in the right-hand member of (58) are less than I fi Jr. From this it now follows easily, in connection with definitions A and B, by letting r --* oo, that the right-hand member of (58) is an integral function of the normal type. I fi I = r of the order and hence of the same kind as the function y(x). Thus we meet a contradiction. 1
Therefore we have 6 = (3, or 6 = - P. Then z = (3 is a zero of the function e''Z - 1, so that y does occur in the sequence of numbers k
27tZ
(k1,+2,...).
Thus we have proved that B is formed by the numbers k (k = ± 1, ± 2, . .). This proves assertion 1.
2zi
.
16
We shall now prove assertions 2 and 3. In these cases the function y(x) does not exceed the minimum type of the order 1. Then, according to definition 1, for this function there are the possibilities b), c), d), e), /) and g) mentioned in condition I of theorem 23. In each of these cases
the numbers (54) have the property which for the case in question is mentioned in condition 2 of theorem 23 23) In cases b), c), d), e) and g) it follows from assertions I B, I C, I D, I E and I G of theorem 23 respectively, that the function k(x) in (53) is of the same kind as the function y(x). Since this holds for every finite value of co (w 0 0), in each of these cases B is null.
In case /) there are two possibilities. Because ao = 0, a1 0 (see (55)) it follows from assertion I F of theorem 23 that the function k(x) is of the same kind as the function y(x) if y(x) is not a polynomial of degree 1. So in this case B is null. If the function y(x) is a polynomial of degree 1, then, according to assertion 2 II of theorem 23, the function k(x) is of a lower kind than the function y(x) and it has this
property for every finite value of co (co 0 0). In this case set B is 23) This may easily be proved. Cf. footnote 1s) 217
formed by all (complex) numbers u, 0. Therefore assertions 2 and 3 are proved. This proves theorem 24 conclusively. REMARK. If the case occurs for which assertion I holds and if co is an element of set B, then from theorem 24 it follows only that the function k(x) is of a lower kind than the function y(x). Theorem 24 does not answer the question "of which kind" the function k(x) then exactly is. The answer to this question is given by theorem 20, and that in this sense, that theorem 20 mentions a function ul(x) which
has the property that the function y(x) is of the same kind as the function u1(x).
In conclusion to this work we want to show that our theorems 16, 17, 18, 19 and 21 provide us with a means to prove an improved form of theorem I; the latter we formulated in the Introduction and it originated with Whittaker. In proving this improved form we make
use of a part of the results of our theorems 16, 17, 18, 19 and 21. Of the results needed we give a recapitulation in the form of theorem 25.
As we already have pointed out in the Introduction theorem 25 is a
generalization of theorem 26 and therefore also of Whittaker's theorem I.
1.
Theorem 25. Conditions: The function h(x) is either
a) an integral function of the normal type r of the order 1; or
b) an integral function of the minimum type of the order a c) an integral function
of
the
normal type r of the order a
(0 < r
f) identically equal to a constant. The differential operator F(D) is defined by F(D) _
Co
n=0
anDn, where
the numbers a are not all equal to zero and in cases a) -e) mentioned above have the following properties respectively z'):
In case a) that the expression F(z) = that is analytic for I z I
r.
n-o
anzn defines a function
24) In case f) the numbers an arc only required not to be all equal to zero. 218
In case b) that the expression °°
G(z) = 1
an
i n=onto
(59)
x"
defines an integral function not exceeding the normal type of the order 1.
In case c) that the expression (59) defines an integral function not i
exceeding the normal type, less than (a r)
a,
Of the order 1.
In case d) that the expression (59) defines an integral function the order of which is less than 1. In case e) that there exists a finite real number v, not dependent on n nor on z, such that the expression an H(z) n=0 =°° -n." zn
defines an integral function the order of which is less than 1.
Assertions: 1.
The differential equation
F(D) - y(x) = h(x)
(60)
has one and only one solution y(x) = rt(x) that is an integral function of the same kind at most as the function h(x) in the following cases: 1.
In case a) mentioned above if the function F(z) for
IzI
r
does not take the value zero. 11.
In each of the cases b) - f) mentioned above, provided that ao ZA 0.
2. In each of the cases a)-/) the function 77 (x) mentioned in assertion z is of the same kind as the function h(x).
3. The differential equation (6o) has oop solutions y(x) = rt(x) that are integral functions of the same kind at most as the function h(x) in the following cases: 1. In case a) mentioned above, if for I z I r the function F(z) has m (m -a 1) distinct zeros A,, (,et = 1, ... , m) with multiplicity nt
v respectively. There p =
v,,.
µ=1
II. In each of the cases b)-e) mentioned above, if ao =
a,,-i=0, a,, :0 (pz I)-
... _
In case f) mentioned above if the function h(x) is identically equal to zero and, moreover, ao = 0. Then p = 1. 4. In each of the cases mentioned in assertion 3 the functions q(x) 111.
mentioned there are all of the same kind as the function h(x).
5. In case f) mentioned above the differential equation (6o) has no 219
solution that is an integral function of the same kind at most as the function h(x), if h(x) is identically equal to a constant 0 0 and, moreover, ao = 0. Proof. The correctness of the assertions 1 -5 follows easily from theorems 16, 17, 18, 19 and 21.
Our improved form of theorem I mentioned in the Introduction, in so far as it concerns the case where the integral function f(x) does not exceed the normal type of the order 1, reads as follows: Theorem 26. C o it d i t i o n s: The function h(x) is an integral function not exceeding the normal type of the order 1.
w denotes a (complex) numbers 0.
Assertions:
The difference equation
1.
y(x + co) - y(x) = h(x)
(61)
has a solution y(x) = rt(x) that is an integral function of the same kind at most as the function h(x) and that, apart from an arbitrary additive constant, is uniquely determined, in the following cases
a)-E): a) The function h(x) is of the normal type r of the order 1 and w f3)
satisfies the inequality r I co I < 2n. The function h(x) is of the minimum type of the order a (0 < a S 1).
y) The function h(x) is of the normal type r of the order
(0
or
6) The f unction h(x) is of the maximum type of the order or (0S a< 1). E) The function h(x) is identically equal to zero.
In each of the cases a)-e) mentioned in assertion i the function
2.
rt(x) mentioned there is of the same kind as the function h(x). 3. The difference equation (6z) has no solution that is of the same kind at most as the function h(x), if the function h(x) is identically equal to a constant 0. Proof.
Since here only integral functions y(x) are in question
it follows from 00 wi Ax + w) - y(x) = n=0 n. 0 y1n,(x) - y(x) _ n=1 n.
y(n)(x)
that equation (61) may be written in the form of the differential equation (62) 220
F(D) - y(x) = h(x),
where con
(63)
F(D)
Dn.
From the latter formula it follows that we have o,
u,n
F(z)=2;-- zn=e°'Z-1, n=1 nI
so that the generating power series of the differential operator F(D)
defines an integral function. Hence the function F(z) is certainly analytic for I z 15 r. As is well-known the zeros of the function F(z) are
z=kz
(k=0,±1,+2,...)
CO
and these zeros are all simple ones. In the case on which assertion 1 a) bears we have assumed r I to I < 2ii.
From this it follows that z = 0 is the only zero of the function F(z) that lies on or inside the circle I z I = T. If we apply assertion 3 I of theorem 25, with m = 1, A, = 0, v1 = 1, then we see that ih = 1, so that the number of solutions of equation (62) that are of the same kind at most as the function h(x), is equal to ool. Then the remark that follows the proof of theorem 21 tells us that the function 991(x) mentioned there, with the help of which we may form these ool solutions, is equal to x°, i.e. equal to 1. Hence we may represent the set of these ool solutions by y(x) = 77(x) + c. Here the function q(x) is an integral function of the same kind at most as the function h(x) and it satisfies the differential equation (62) and hence the difference equation (61). Therefore assertion la is proved. In case a) assertion 2 follows from assertion 4 of theorem 25. In the cases of assertions 1 fl, 1 y, 16 and 1 e and in the cases fl, y and 6
of assertion 2 we make use of the fact that z = 0 is a simple zero of the
function F(z), so that we may refer 25) to assertions 3 II and 3 III,
with a° = 0, P = 1, and to assertion 4 of theorem 25. Finally the correctness of assertion 3 follows from assertion 5 of theorem 25. This proves theorem 26. 26) It is easy to prove that in all these cases the numbers an mentioned in condition 2 of theorem 25, which, in case of the operator F(D) considered here may immediately be read from (63), have the property mentioned in this condition 2. Cf. footnote 15). 221
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