C OMPOSITIO M ATHEMATICA
S. K. B OSE
D EVENDRA S HARMA
Integral functions of two complex variables
Compositio Mathematica, tome 15 (1962-1964), p. 210-226
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
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by
S. K. Bose and Devendra Sharma
1) 2) 1. Let 3)
be a function of the two
complex
variables zl and Z2’regular
forIZtl
rt, t = 1, 2.If rl
and r2 arearbitrarily large,
thent(Zl, Z2)
is an
integral
function of the twocomplex
variables. We know thatis the maximum modulus of
t(Zl Z2)
forIZtl
rt .In this paper we have defined maximum term and the ranks of the maximum term, and have extended the method of
systematic
determination of these as in the case of one variable
by
Newton’spolygon ((1),
p.28).
Also we have obtained relations betweenthese,
andinequalities involving
these and the maximum modulus.Further,
we have defined order and have obtained necessary and sufficient conditions for the function to be of finiteorder,
and alsothe same for functions of finite order and
type
T.2.
Let 4)
be a function of the two
complex
variables z, and z2 ,regular
inIz,
r, andIZ21
r2 .Writing
z, =r,. e’Oi
and Z2 =1’2ei02,
l} 1 regret to announce the sudden and untimely death of Devendra Sharma on 18th June, 1957.
2) We are thankful to the referee for the valuable criticism.
3) We have considered only two variables for simplicity. The results can easily
be extended to several variables.
4) If
(zo, zo)
is any given point, then by neighbourhood of this point we wouldmean a bicylinder
Iz,-el Tl,BZI-Z:1
rât, Tl > 0, TI > 0.and
This series is
convergent
for all values of01
and02
becauseI:1,ml=olaml,mllrlr:1
isuniformly convergent, by hypothesis.
We, therefore, multiply
both sidesby
cos(ml °1 + m2(2)
orsin (ml 01
+m2(2)
andintegrate
termby
term between the limitszero and 2n. We thus have
and
Multiplying
the secondby i
andadding,
weget
Hence
Again,
if weintegrate (2.1)
withrespect
to01
and0.
in the rangezero to 2n, we
get
From
(2.2)
and(2.3)
followsNow,
if U ispositive,
then theintegrand
isequal
to 2 U and if U isnegative
or zero, then theright
hand side is zero. Hence we haveTHEOREM 1.
If
thefunction f(zl’ Z2)
isregular for IZ11
ri andIZ21 |
r2 andil
A(r1’ r2)
is its m aximum realpart for IZ11
ri andIZ21
r2l thenf or
allpositive
valueso f
ml and m2, the numberla.,,,.,Iiî ir’a
is less than orequal
to thegreatest o f
the two numbers-2cxo,o
and 4A(ri, r2)-2ao,o
.COROLLARY.
If i(Zl z2)
is anintegral function,
thenf or
allpositive
valueso f
ml and m2’ andf or
all ri >rf ,
r2 >?2
Similar results can be obtained for minimum of U
(rI’
r2l°1’ (2).
Also for the maximum and minimum of
V(rl’
r2l°1’ (2).
THEOREM II.
If f(zi , Z2)
is anintegral f unction,
and ql and q2are two
tixed finite positive
numbers suchthat ) f(zi , z2)1
isalgebra- ically
less thanKrÎiri2 f or lzll =
ri andIZ21
= r2l where ri and r2are
arbitrary large
numbers and K is a constant, then/(zl, Z2)
is apolynomial of degree
notgreater
thanql+q2.
PROOF :
Since I/(zl’ z2)1
sKril11t for lzll
= riand IZ21
= T2’therefore,
from(2.4)
followsIf ml > q, or m2 > q2 , then the
right
hand side vanishes asr, or r2
respectively
tends toinfinity,
and soa,.,,.,
is zero forml > ql or m2 > q2.
THEOREM III.
I f f(Zi, Z2 )
is anintegral f unction o f
the twocomplex
variables z, and Z2’ thenf or
allRl
> rl> ri
andR2
> r2 >r2 ,
whereM(rl’ r2) = maxl/(zl’ z2)1, lor [zi[ ri and Iz?1
r2 .PROOF : We can write
(2.4)
asand
taking
r,R1
and r2R2’ it
follows thatCOROLLARY:
Il f(Zll Z2) is
zero at Zl = 0, Z2 = 0, thenf or RI
> rl ,R2
> r2 -3. Consider the moduli of the terms of the double series in the
expansion
of theintégral function 1 )
where
If we consider any column or row, then the sequence thus ob- tained tends to zero for all values
of r1
or T2. Hence for every valueof ri , keeping
m2and r2 fixed,
thereis, therefore,
one term ofthe sequence thus obtained which is
greater
than orequal
to allthe rest. This term will be the maximum term in that column and will be denoted
by ,u (m2;
rl,r2).
If there are more than one such term, then the term of thehighest
rank will beregarded
as themaximum term of this column and the rank will be denoted
by VI(M2; rl).
We nextgive
different values to m2 , i. e. consider different columns and suppose thegreatest
term occurs in thev2th column,
then the term,u (v2;
rl ,r2)
will be the maximum term withrespect
to columns and the rank of this term will be denotedby vl (rl ).
If there are more than one columncontaining
such term,then the term of the
highest
rank withregard
to column will beregarded
as maximum term, i.e. the column ofhighest
rankwhich contains the maximum term is
v2th. Similarly,
if we considermlth
row,keeping
ri, r2 and mlfixed,
then we shall have1 ) Suffix 1 will indicate row, for example, Mi, Pl, Vi etc. and suffix 2 will indicate
column, ml, P2’ 1’. etc.
u
(mi;
rl ,r2 )
as the maximum term in that row and the rank asv2 (m;
r2). Further,
for different valuesof ml,
suppose thegreatest
termoccurs in the
vlth
row, then the maximum term will be denotedby P("l; rl, r2)
and the rank of this term will be denotedby V2(r2)’
If there are more than one such term, then the same convention
as for columns is
adopted.
Hence we shall denote the maximumterm for
given
valuesof rl and r2 by ,u (r1, r2 )
and the rank of this term will be denotedby v(rl, r2).
For a
systematie study
offinding
the maximum term we shall extend the method of Valiron((1),
p.28)
for one variable.Let
log Cmi, m_ -gml,ma’
thenand
ml m.
Since
VCm m
andvc::-:
tend to zero as mland m2
respec-tively
tend toinfinity.
Taking OX, OY,
OZ as the axes ofcoordinates,
if weplot
thepoints Aml,m.
of coordinates(ml, m2’ gml,m.), then,
from(3.3),
it follows that we can construct a surface with
plane
faces andevery section of this surface
by planes parallel
to theXZ-plane
and
YZ-plane
form a Newton’spolygon, having
certain of thepoints Aml,m., lying
in this verticalplane.
Out of these some of them coincide with the vertices of thepolygon,
whilst the remain- der lie either on or above it. Let us denote this surfaceby S(f)
and call it Newton’s
polyhedron.
If VI
and v2 be
the rank of a maximum term as defined above andMi =A
vi orm2 =f=
V2,then,
iteasily
follows:Now,
let us consider thegeometrical interpretation
of thisinequality.
LetD 1’1,1’1
denote atangent plane, having
directioncosines
proportional
to-log
rl,-log
r2 and 1,passing through
the
point AVl,m2
orAml’v..
If we now draw theplane parallel
tothe
XZ-plane through AVl,ml’
then thosepoints Aml,m.
which liein the
plane
do not lie below the lineL1’l’
the line of intersection with theplane D,,,,, .
ofslope log
rl .Similarly,
if we consider theplane parallel
to theYZ-plane through Ami’’’.’
thosepoints Aml,ml
which lie in that
plane
do not lie below the lineL1’"
the line ofintersection with the
plane D" 1 " , 1
ofslope log
r2 . Thepoint Av l’ m .
or
Ami’ v.
is therefore apoint
of thepolyhedron S (f)
and theplane
D,1,rt
is atangent
to thispolyhedron.
Now,
we take a verticalplane parallel
toXZ-plane through
A v l’ m , .
then m2 isfixed,
i.e. in(3.2)
we areconsidering
a column.The intersection of the
plane
withthe
surface will include a New- ton’spolygon
and thetangent
lineL,.1
1 ofslope log rl
will passthrough AV1,m..
. Thepoint A,,,, 2is uniquely
determined in theplane,
whenlog
ri is notequal
to theslope
of one of the sides ofnl(!)’
and for such values of ri, there isonly
one term in the se-quence
(column
of(3.2 )
underconsidération) equal
to,u(m2; Tl’ r2 ).
When
log
rl isequal
to theslope
of a side ofni (f),
there are severalsuch terms and their number is
equal
to the number of thepoints
AVl,m.
which lie on this side of thispolygon.
When more than oneterm are
equal
to,u (m2;
rl,r2 ),
we shall take the term ofhighest
rank
amongst
them as the maximum term, withrespect
to m2 and rl .Thus having
obtained the maximum term or terms withrespect
to m2 and r1, we must draw a
plane parallel
toYZ-plane through
Aml’v.’
then ml isfixed,
i.e. in(3.2 )
we areconsidering mlth
row.Again
the intersection will include a Newton’spolygon
and thetangent
lineLr.
ofslope log
r2 will passthrough Aml’v..
Hereagain
thepoint Ami’
Va isuniquely
determined in theplane,
whenlog
r2 is notequal
to theslope
of one of the sides ofn2(/),
and forsuch values of r2 there is
only
one term in the sequence of terms in themlth
rowequal
to,u (ml;
ri ,r2 ).
Whenlog r2
isequal
to theslope
of a side ofn2(/),
there are several terms and their number isequal
to the number ofpoints A i,
". u which lie on the side of thepolygon.
As in the case of m2 and rl , we shall take the term ofhighest
rankamongst
them as the maximum term, withrespect
tomi and r2.
Finally
thegreater
of the two terms obtained above shall be denotedby ,u(TI’ r2).
We have
thus,
with thisconvention,
obtained one term as maxi-mum with
respect
to ml, m r and r2. ThusVI(m2; Tl’ 1:2)
or"l(TI’ r2 ),
y2fixed, 1’2(ml;
rl ,r2 )
orV2(rl, r2 ),
rifixed,
and"(Tl’ r2)
will be used to denote ranks of the maximum term of the double series.
"1(m2; Tl’ r2), 1’2(ml; Tl’ r2)
andv(rl, r2)
are unboundednon-decreasing
functions of r1 and r2 .Further, Vl(M2; rl, r2)
andV2(ml; r1, r2)
have left handdiscontinuity
wherever ri and r2respectively
passthrough
a value such thatlog
rl andlog r2 respectively equal
theslope
of one of the sides of thepolygons nl(/)
andn2 ( f ).
Hencev (r1, r2 )
has alsodiscontinuity
for suchvalues
of rl
and r2.4. Two functions
f(zi, Z2)
andg(z1 , Z2) having
the samepoly-
hedron will have the same maximum term and the rank.
Let us consider the function
where
Gm l’ m 1
is the Z-coordinate of thepoint,
whose X and Y coordinates are mi and m2 .The function
W(rl’ r2)
is a dominant function forf(zi, Z2)
and has the same maximum term. Also it is the
simplest
functioncorresponding
to thepolyhedron S(f).
The ratioand
of the coefficients in
W(rl, r2 ) corresponds
to the ratio ofaml,m.
and
am _1 m , am m
andam m _1’
andam m and am _1 m -1.
Weshall call these as rectified ratios. The
logarithm
ofR:. 1
isequal
tothe
slope
of the Newton’spolygon
obtainedby plotting
thepoints (ml , Gml,m.)
in aplane parallel
toXZ-plane
at a distance m2 , and is therefore anon-decreasing
function of mltending
toinfinity.
Similarly
thelogarithm
ofS’i
andRm l’ mare 1
arenon-decreasing
functions of ml , and ml and m2
tending
toinfinity.
Suppose
forsimplicity
thatGo,o
= 0. Then we haveand since
and
therefore,
We may also
put (4.2)
asand hence
Since
ju(rl, r2)
isgreater of u(vl;
r,,r2)
andIU(V2;
rl ,r2), therefore, log,u(rl, r2)
will begiven by
either(4.3)
or(4.4)
or both.We are now in a
position
to find a relation betweenIÀ(rl, r2 )
and
’V1(T1’ r2), V2(T1’ T2)
andv(rl, r2).
In the firstplace
Also it is obvious that
M(rl’ r2 )
does not exceed the value of the functionW(rl’ r2 ). Suppose
that Pl and P2 areintegers greater than vi = vl (r1 ) and v2
="2(r2)
and such that the rectified ratioRpi 1
> ri andS§§§
> r2.Then, for ql >
pl and q2 > P2’Hence
In order that the terms in the bracket may be
substantially equivalent,
we chooseand
which
implies
thatand
Hence we have the
following
result:5. We shall now define order. Here we shall restrict to
integral
functions of
finite
order.Let
and let then
and
f (zl, Z2)
is said to be anintegral
function offinite
order p.Further, pl(r.)
andp2(rl)
be defined asproximate
orders.Now,
for those functions whichsatisfy (5.1),
the relation(4.6)
appears in an
especially simple form,
and it may be writtenlog M (rl r2)
Hence
We may also define order as
We will now prove the
equivalence
of the two definitions.Let us start with the definition
(5.1).
From(4.3)
and(4.4)
andthe definitions of pl, , P2 and p, we have
or
where e =
max. (el, e2)
andkt >
1,k2 >
1, and sowhence,
in virtue of(4.6)
Next,
if we suppose thatthen
or
Hence
Thus,
from(5.3)
and(5.4),
theéquivalence
of the two definitions follows.6. We shall now come to the necessary and sufficient condition for an
integral
functionf(zl, Z2)
to be of finite order p. The result is as follows:THEOREM IV. The
integral function
is
o f finite
order,i f
andonly i f
is
finite ;
and then the order pof I(z!, z2)
isequal
to fl.PROOF: We first prove that
p>
,u. We may note that in case,u = oo the above statement is to be
interpreted
asmeaning
thatthe order is
infinite,
or elsef(zi , z2)
is not anintegral
function.We know that
(i) If À = 0, p k y,
since p is notnegative.
Let us suppose that0 e Jl oo.
Then,
from(6.1),
we havei.e.,
for an infinite sequence of values of ml and m2.
Also
(6.2)
may be written asLet rl =
(eml)l/p-e and r2
=(em2)1/p-e.
ThenSince
ju - e is independent of rl
and r2ltherefore,
Further,
e isarbitrary
and sop >
IÀ.(ii)
Next we prove thatp
IÀ. We note thatif g
= oo, the result is obvious. So we suppose that p oo. Let e > 0. Then from(6.1)
followsfor ml >
mO
and m2 >m2 ,
i.e.
or
Hence
Let
lhi
be thepart
of the double series in(6.4)
for whichml
(2rl)p+e m2 (2r2)p+e.
We estimateEl by taking
thelargest
value ofel le2 s.
Thensince the series in
(6.5 )
isconvergent and is independent of rl
and r2.Let
£2
contain the terms for which ml >(2rl)/I+e
and m2 >(2r )/I+e
and so inE
we have r1m-l/p+e 1
andr,m-ll#+e 1
and hence
Let
E3
be thepart
of the series for which ml(271)1’+8
andin2 > (2r2)P+8,
thenSince
and
therefore,
Let the
remaining part
of the double series in(6.4)
be denotedby E4
i.e. forml > (2rl)I’+8
and m2(2r2)1’+8,
thenFurther,
Hence, substituting
these values in(6.4),
weget
and hence
since e is
arbitrary
andindependent of rl
and r2.7.
Suppose
0 p co and let us defineand
The functions which
satisfy
the latterequality
are said to befunctions of
exponential type
T.If a = ezp and
using
theSterling’s
formula(7.1)
takes the formwhere Zl’ Z2 are any
(fixed) complex
numbers.If in this we
put
p = 1, thenWe now deduce the
following
result:THEOREM V.
Il
0 a 00, thefunction
is
o f
order p andtype
i,i f
andonly i f
a = eap.PROOF:
From (7.1),
we havefor e > 0 and ml’ m2
large.
We shall first prove that T
oc/ep.
Since we may add apolynom-
ial to
i(zl, z2 .)
withoutaffecting
itstype,
we may suppose that(7.3)
holds for almi
and m2 ,interpreting
itsright
hand side as 1for ml =
01 M2
= 0. ThenThe
general
term of theright
hand side does not exceed its maxi-mum.
Let
Then
for §
to bemaximum,
i.e.
Therefore,
Thus the maximum term is exp
((r£+r£)(ce+ E)fep),
attained formi
= ri (ce+ E) fe
and m2= r£ (ce+ E) fe.
LetEl
denote thepart
ofthe series for which
ml (ce+2E)r%
andm2 (cx+2e)r’.
ThenLet 12
denote thepart
of the series for whichrf mt/((X+2e)
and
r£ M2/(OC+28).
ThenLet
27g
denote thepart
of the series for whichml (rx+2e)ri
andm2 >
(ce+2E)r£.
Thensince
Therefore
Let the
remaining part
of the series be denotedby ~4’
i.e. for m1 >(oc+2e)rf
andm2 :S (oc+2e)rP. Then,
as in~3 ,
we obtainThus
Hence
since
t(Zl 1 Z2) ’S
ofexponential type
T.Next, to show that T >
oefep,
we haveagain
from(7.1),
for aninfinite sequence of values of ml and m2,
If zve
take rl and r2
such thatfor these values of ml and m2 , we have
for a sequence of values
of rl
and r2ltending
toinfinity.
HenceIt remains now to show that
f (zl , Z2)
is of order p at most ifa oo, and is of order p at least if ce > 0.
For
large
ml’ m2, we have from(7.1)
and hence
By (6.1),
the order off(z!, Z2)
is p at most.Similarly
if « > 0 the order ofI(z!, Z2)
is at least p.REFERENCE G. VALIRON,
[1] Theory of Integral Functions (1923).
(Oblatum 8-6-60).