Compactness and Convergence in the Space Of Analytic Functions
A thesis
Submitted in partial fulfillment of the requirements for the award of the degree of
Master of Science In
Mathematics
By
Name: Basundhara Goswami Roll No: 409MA2065
(Under the Guidance of Dr.Suvendu Ranjan Pattanaik) Department Of Mathematics
National Institute Of Technology Rourkela, Orissa
2
National Institute Of Technology, Rourkela
DECLARATION
This is to certify that the thesis entitled “Compactness and Convergence in the Space Of Analytic Functions” submitted by Basundhara Goswami as one year project work for the requirement for the award of Master of Science in Mathematics, in the Department of
Mathematics, National Institute of Technology, Rourkela is a record of authentic work carried out by her under my supervision and guidance.
To the best of my knowledge, the matter embodied in this project work has not been submitted to any where for the award of any degree.
Date: Dr. Suvendu Ranjan Pattanaik Assistant Professor, Dept. Of Mathematics National Institute Of Technology Rourkela-769008
3
ACKNOWLEDGEMENTS
First of all I would like to thank my guide Dr.Suvendu Ranjan Pattanaik
Assistant Professor, Dept Of Mathematics, National Institute Of Technology, Rourkela , for his guidance, ever encouraging words of wisdom and for taking such an active interest in my project.
I am grateful to Prof .P. C. Panda, Director, National Institute of Technology Rourkela for providing facilities in the institute for carrying out research.
It is my great pleasure to acknowledge to Prof. G.K Panda, Head of the Department and all faculty and staff members of Department of Mathematics, National Institute of Technology, Rourkela for their help which made this research work a success.
I also wish to thank all of my friends for making my stay in this institute a memorable experience.
Last but not the least I would like to thank my parents and my family members for standing by me all along and for all the support and loving care given to me.
Basundhara Goswami 409MA2065
4
Table of Contents:
Abstract 5
Chapter- 1: 6 Space of continuous functions
Chapter- 2: 16 Spaces of analytic functions
Chapter-3: 20 Spaces of meromorphic functions
References 23
5
Abstract:
Here, our main aim is to study compactness and convergence in the space of continuous functions defined on a fixed region G subset of complex plane. First, we will define an appropriate metric in which we will study compactness and convergence. For defining compactness we will introduce the concept of normal set and then we will prove that normal closure is compact. Subsequently, we will prove a variant of a famous theorem i.e. Arzela-Ascoli theorem. Then we will divert our attention in studying compactness and convergence in the space of analytic functions defined on a fixed region G. The analytic functions are having an exceptional importance as this class is sufficiently large. It includes the majority of functions which are encountered in the principal problems of mathematics and applications to science and technology. Here, in our discussion we visualize these analytic functions as points in a metric space. Also, here we have proved Hurwitz and Montel theorem. In the last section of this dissertation we have studied the space of meromorphic functions defined in a fixed region G.
.
6
Chapter-1: Space of continuous functions
A mapping f :X Yis said to be continuous at a point ifGis an open subset ofY, then f1(G) should be open inX .In terms of (,)-criterion,let X (X,d)and ~)
, (Y d
Y be two metric spaces. Then a mapping f :X Yis said to be continuous at a pointx0Xif for every 0 there is a 0 such that~( ( ), ( )) satisfying ( , )
0
0 x X d x x
x f x f
d .
Suppose, G be an open set in C and (,d) is a complete metric space. Then C (G,) is the set all continuous functions from G to which will be never be empty as it contains the constant functions. If is C, then apart from constant functions it also contain all analytic function. And if it isC{}, then it contains meromorphic functions which will be our concern of latter discussion.
Proposition: (see [2]) If G is open in C, there is a sequence {Kn} of compact subsets of G such that
G =
1 n
Kn.
Moreover, the sets Kn can be chosen to satisfy the following conditions:
(a) Kn int Kn1
(b) K G and Kcompact implies K Kn for some n;
(c) Every component of CKn contains a component of CG Proof:
Consider
1} ) ,
( : { } :
{z z n z d z C G n
Kn for each positive n.
7
Now 1}] { : }
) ,
( : { } :
[{ z z n
G n C z d z n z
z .So ,Kn is bounded. Also, it is the
intersection of two closed sets and we know that every closed and bounded subsets of R2 is compact. So, it’s compact. We have drawn the following figure to explain this theorem properly.
Fig - 1
G is open inC,so CGis closed and so CGCG
.
We know xAd({x},A)0 . So, we can draw the following figure as follows.
Fig-2
8 From the above figure, we can say that,
1 n
Kn
G . Now we are required to prove the above mentioned (a), (b), (c).
(a) }
1 ) 1
, ( : { } 1 :
1 {
z z n z d z C G n
Kn ;
1} ) 1
, ( : { } 1 :
{
int 1
G n C z d z n
z z
Kn . This implies Kn int Kn1.
(b) Given, K G is compact where
1 n
Kn
G .We can also get
1
int
n
Kn
G . Hence,
1
int
n
Kn
K .Since Kis compact,so we can say{intKn}forms an open cover ofKwhich has also a finite subcover.
HereKn intKn1intKN KN. So, KKN for someN.
(c) C G isasubset of C Kn.The unbounded component ofC Knwill containand hence CGwill contain the same. The unbounded component of CKnalso
contains {z: z n}
.
Assume, D be a bounded component ofC Knwherez be a pointDsuch that 1;
) ,
(z C G n
d So, there exist a point 1.
with
in C G w z n
w This
implies C Kn
w n B
z 1)
;
( .
We have got zDCKnand C Kn w n
B 1)
;
( .So, D
w n B 1)
;
( .If D1is the component of CGthat contains w.It follows that D1 D.
Let
1 n
Kn
G
,
where each Knis compact andKn intKn1,
then define ) , (f g
n sup{d(f(z),g(z):zKn)}
9 for all functions f and ginC(G,). Also define
) , ( 1
) , ( 2
) 1 , (
1 f g
g g f
f
n n n
n
.
Since, we know
1 ,for n 1,..., )
, ( 1
) , (
g f
g f
n n
. So,
n
n n n
n f g
g
f
1 n
1 2
1 ) , ( 1
) , ( 2
1
.
But
n
n
1 2
1 is a geometric series, so it will converge.So, by comparison test
) , ( 1
) , ( 2
1
1 f g
g f
n n n
n
, it will also converge. Here, (f,g) is a metric.
Proposition: (see [2]) (C(G,),) is a metric space.
Proof:
(a) 1 ( , )
) , ( 2
) 1 , (
1 f g
g g f
f
n n n
n
0. (b) (f,g)0.This implies
) 0 , ( 1
) , ( 2
1
1
f g
g f
n n n
n
i .e. 0
) , ( 1
) , ( 2
1
g f
g f
n n n
for eachn.
It implies sup{d(f(z),g(z)):zKn}0.So d(f(z),g(z))0and hence f=g for allzKn. (c) (f,g)(g,f)
10
(d) Here, we need to prove the triangle inequality. Let0 . This implies and so
1
1 .
Now, n(f,g)sup{d(f(z),g(z)):zKn}which will be less than equal to }
)) ( ), ( ( { sup } )) ( ), ( ( {
sup d f z p z d p z g z . So, we can write
sup{ ( ( ), ( ))}
1
))}
( ), ( ( sup{
z g z f d
z g z f d
} )) ( ), ( ( { sup } )) ( ), ( ( { sup 1
} )) ( ), ( ( { sup } )) ( ), ( ( { sup
z g z p d z
p z f d
z g z p d z
p z f d
))}
( ), ( ( sup{
1
))}
( ), ( ( sup{
))}
( ), ( ( sup{
1
))}
( ), ( ( sup{
z g z p d
z g z p d z
p z f d
z p z f d
and hencefinally we get
(f,g)(f,p)(p,z). Lemma: (see [2])
Let the metric be defined as before. If 0is given then there is a 0 and a compact set G
K such that for f andgin C(G,),sup{d(f(z),g(z)):zK} will imply (f,g). Conversely, if 0 and a compact set K are given, there is an 0 such that for
), C(G, in and
g
f (f,g) ,implies sup{d(f(z),g(z)):zK} Proof
:
Choose 0 be fixed and coonsiderp be a positive integer such that
2 2 1
1
n
p n
. Assume Kp
K .
Take 0 such that 0t gives
2 1
t
t . Let f andg be functions in C(G,)which satisfy sup{d(f(z),g(z)):zK}. For 1n p , since Kn Kp K ,
so sup{d(f(z),g(z)):zKn} .That means, n(f,g), for 1n p.So, p
n 1 for 2 ) , ( 1
) ,
(
g f
g f
n
n . Therefore
11
) , ( 1
) , ( 2
) 1 , (
1 f g
g g f
f
n n n
n
.
This expression can also be written as
1 ( , )
) , ( 2
) 1 , (
1 f g
g g f
f
n n p n
n
) , ( 1
) , ( 2
1
1 f g
g f
n n n
p
n
and it is less than
p n
n
1 2
1 2
n
p n
1 2
1 and it is less than .Thus one part of the lemma is proved.
Converse part: Let Kand are given. Since
1 n
Kn
G =
1
int
n
Kn andK is compact, so, there exists an integer r1such thatK Kr. So,
} :
)) ( ), ( (
sup{d f z g z z Kr sup{d(f(z),g(z)):zK}.
That shows r(f,g)is greater than equal tosup{d(f(z),g(z)):zK}. Choose 0 appropriately so that 0 p2r implies
p p
1 . For any number r
t
t 2
1
implies
t t t t 1 1
1 i.e. t . Let(f,g) ,then it will automatically imply
} :
)) ( ), ( (
sup{d f z g z z K .
Proposition: (see [2])
(a) A set O(C(G,),) is open iff for each f in Othere is a compact set Kand 0 such thatO{g:d(f(z),g(z)),zK}.
(b) A sequence {fn}in (C(G,),)converges f iff {fn} converges to f uniformly on all compact subsets of G.
Proof: Let Ois open and let f O.Then for some 0 ,{g:(f,g)}O.So, by using the preceding lemma we can write [sup{d(f(z),g(z)):zK}]O. Hence the result follows.
Conversely let for each f in Othere is a compact set Kand a 0such that
12
O {g:d(f(z),g(z)),zK}.
Since for all zit happens so sup{g:d(f(z),g(z)),zK}O.
So, by preceding lemma there is an 0such that for f and ginC(G,), (f,g). So, Ois open.
(b) Let a sequence {fn}in (C(G,),)converges to f . Then for 0there exist N 0such that (fn, f) for all n N. Let Kbe a compact subset ofG. It will imply that there is a positive integer nsuch that K Kn.Now from(fn,f), we can get
) , ( 1
) , (
f f
f f
n n
n n
2n (say).
Then
) 1 , (fn f
n . Hence,
} 1 :
)) ( ), ( (
sup{d fn z f z z K . So, {fn}converges to f uniformly on all compact subsets ofG.
Conversely, let Kbe a arbitrary compact set. Let {fn} converges to f uniformly onK.Then 0
)) ( ), (
(fn z f z
as ntend to infinity and for all zin K. It will imply n(fn(z), f(z))0 as ntend to infinity .So, we can write sup{d(fn(z),f(z)),zK}tend to zero as ntend to infinity . Also we can say, for given 0, sup{d(fn(z), f(z)),zK}for all nN.Hence, there is as an such that (fn,f).
Proposition: (see [2]) C(G,) is a complete metric space.
Proof: Let {fm} be a Cauchy sequence in C(G,).So, (fm, fp) tend to zero as m,ptend to infinity. Then sup{d(fm(z),fp(z)):zKn}tend to zero. Let KGbe compact .ThenK Kn for somen and so sup{d(fm(z), fp(z)):zK}also tend to zero as m,ptend to infinity..Hence
}
{fm is a Cauchy sequence in K .For every 0, there is an integer nsuch that
} :
)) ( ), ( (
sup{d fm z fp z z K for m,pn.But {fm(z)}is a sequence in which is complete.
So, fm(z)converges to f(z)and f(z)also belongs to .Hence, (fm, f)tend to zero as
m .So for an 0there exist N 0such that sup{d(fm(z), f(z)):zK}for all N
m . So,fmconverges to f uniformly on all compact set.Hence f is continuous and so f belongs toC(G,) .
13
Proposition: (see [2]) A set C(G,)is normal iff its closure is compact.
Proof: First let C(G,)is normal .Then by definition of normality , each sequence in has a subsequence which converges to a function f in C(G,).Our claim is thatis compact.
Sinceis the set containing the setas well as all the limit points of .So, each converging function f of the subsequence of each sequence in is in and hence the result follows.
Now, let us assume is compact. Let {fn}is sequence in .So, it has a convergent subsequence { }
nk
f and suppose f f f
nk , .The set is a subset of C(G,)and since )
, (G
C is complete, so is also contained in C(G,).Hence f f , f C(G,)
nk ,i.e.
f is continuous and so it is normal.
Proposition: (see [2]) A set C(G,)is normal iff for every compact set K Gand 0 there are functions f1,f2,...,fnin such that for f in there is at least one k,1k n,with
} :
)) ( ), ( (
sup{d f z fk z z K .
Proof: Suppose C(G,)is normal.Let K Gbe compact and 0be given. is normal so, is compact and so it is totally bounded.
nk
fk
f f
1
} ) , ( : {
. Now(f,fk)it
will imply 2 ( )
) , ( 1
) ,
( say
f f
f
f n
k n
k
n
.So,we can say sup{d(f(z),fk(z)):zK} . For the converse, suppose has the above stated property.
Now fromsup{d(f(z), fk(z)):zK} we get (f,fk).So, it is totally bounded. So, is compact. Hence is normal.
Proposition: (see [2]) ( , )
1
d Xn
n
where dis defined by
) , ( 1
) , ( 2
1
1 n n n
n n n n
n d x y
y x d
is a metric space.If k {xnk}n1is in X= n
n
X
1
then k {xn}iff xnk xnfor each n.Also,if each )
,
(Xn dn is compact then X is compact.
Proposition: (see [2]) Suppose C(G,)is equicontinuous at each point of G; then is equicontinuous over each compact subset of G.
14
Proof: Fix a 0and let Kbe a compact subset of G.Using the equicontinuity concept, for any K
w there exist w 0such that wheneverww w then
) 2 ( )
(
w f w
f for all f in
i.e wB(w;w)forming an open cover of K.
B(w;w) G K
B(z;)
We know if Kis compact and it has a cover then,it has Lebesgue no,that means for each zK there exist 0such that B(z;) B(w;w).If zand zare in Kwith zz ,then there will be a wKsuch that zB(z;)B(w;w).So, we have got
) 2 ( )
(
w f z
f and
) 2 ( )
(
w f z
f .Finally we got f(z) f(z) and hence is equicontinuous overK. Arzela - Ascoli Theorem:(see [2]) A set C(G,)is normal iff the following two conditions are satisfied:
(a) for each zin G, {f(z): f }has compact closure in ; (b) is equicontinuous at each point of G.
Proof:First suppose thatC(G,)is normal. Let Az {f(z): f }.We need to show that Azis compact.Define a mapping T:C(G,) by the rule T(f) f(z) i.e
z w
15 Az
T() .Since is normal, so is compact .Let { fn} be a sequence in .Let fn f0i.e.
0 ) , (fn f0
as n .We are having
) , (fn f0
1 ( , )
) , ( 2
1
f0
f f f
n n
o n n n
and so, n(fn,f0)0 as n.
It means sup{d(fn(z), f0(z)):zK}0 as n and hence
K n
z z f z f
d( n( ), ( )): } 0 as
{ 0
So, Tis continuous .Hence Azis compact.
Secondly, we need to show is equicontinuous at each point of G.Let z0G.We are required to show that for every 0there exist 0such that zz0 f(z) f(z0) for all
f . That means whenzB(z0;), thend(f(z), f(z0)).
Let K B(z0;R)where R>0. Then K is compact. Using one of the previous proposition for 0 there are functions f1,f2,...,fnin such that for f in there is at least one k,1kn, with
} 3 :
)) ( ), ( (
sup{
K z z f z f
d k .
So we can write
)) 3 ( ), (
(
z f z f
d k .Also each fk is continuous, so there is a , 0 R such that whenever zz0 ,
)) 3 ( ), (
( 0
z f z f
d k k for 1k n.
)) ( ), ( ( )) ( ), ( ( )) ( ), ( ( )) ( ), (
(f z f z0 d f z f z d f z f z0 d f z0 f z0
d k k k k which is less than
3
3
3
i .e. less than .Hence proved. Note that for proof of converse part one may refer [1].
16
Chapter-2: Spaces of analytic functions
A function which is differentiable at each point of a domainD is called analytic in that domain.
Here, we are putting a metric on the set of all analytic functions on a fixed region and then our main aim is to discuss compactness and convergence with respect to this metric. Here, our discussion will be the analytic functions defined from Gto .Let it be denoted
byH(G,). We are considering hereH(G)as a subset of C(G,C)where H(G)is the collection of analytic functions onG.
Theorem: (see [2]) If {fn}is a sequence in H(G) and f belongs to C(G,C) such that fn f then f
is analytic and kn k
f
f for each integer k 1.
Proof: We will use the application of Morera’s Theorem for the first part of the proof the theorem. Let Gbe an open set in Cand Tis a tringle insideG. Since Tis compact and fn f . So, fn converges to f uniformly onT. Hence lim 0
T
nT
f
f So, f is analytic inG,since each fnis analytic.
To prove the second part, let DB(a;r)be a closed ball contained in G. Then there exist a number Rrsuch that B(a;R)G.Consider a circle which is za R.
Using Cauchy’s Integral formula, we will get,
z dw w
w f w f i z k
f z
fnk k
n k ( ) 1
) ( ) ( 2
) ! ( )
( for all zinD.
Then,
dw z
w
w f w k f
z f z
f k
n k
k
n
( ) 1
) ( ) ( 2
) ! ( )
( .
Take Mn sup{fn(w) f(w) :wa R}and we can write
1 1
1 {( ) ( )} ( )
)
(wz k wa za k Rr k .
17
So, 1
) ( ) ! ( )
(
k nk
k
n R r
R M z k
f z
f for za R. Since fn f . SolimMn 0. Hence, kn k
f
f
uniformly on B(a;R).For an arbitrary compact subset Kin Gand for 0rd(k,G) we are having somea1,a2,...,anin Ksuch that ( ; ).
1
nj
j r a B K
Since kn k
f
f uniformly on each )
; (a r
B j , hence the convergence is uniform onK. Corollary: (see [2]) H(G)is a complete metric space.
Proof: From the above theorem we got that H(G)is closed and it is the subset of C(G,C) which is complete. We know that closed subset of a complete set is complete. Hence, H(G)is complete.
Corollary: (see [2]) If fn :GCis analytic and
1
) (
n n z
f converges uniformly on compact sets to f(z) then
1
) ( )
(
n k n
k z f z
f .
Hurwitz’s Theorem: (see [2]) Let Gbe a region and suppose the sequence {fn} in H(G) converges to f . If f 0,B(a;r)Gand f(z)0for za R then there is an integer N such that for n N, f and fn have the same number of zeros in B(a;r).
Proof:
Given, f(z)0 for za R. Take inf{ f(z): za R}0. If fn f in za R, then this convergence will be uniform on za R. So, there exist an integer Nsuch that if
N
n and za R , then
) 2 ( )
(
z f z
f n .Also f(z) f(z) fn(z) and ( ) 2 f z
.So we
can write
) ( ) ( ) 2 (
) ( )
(z f z f z f z f z
f n n
.
Hence using Rouche’s theorem,we can say f and fnhave same number of zeros inB(a;r). Corollary: (see [2]) If {fn}H(G) converges to f in H(G)and each fnnever vanishes on G then either f 0or never vanishes.
18
Lemma: (see [2]) A set in H(G)is locally bounded iff for each compact set KGthere is a constant M such that f(z) M for all f in and zin K.
Proof:
Let Kbe a compact subset of Gand let there is a constant M such that f(z) M for all f in
and zin K. Since,Kis compact, so it is totally bounded and complete. And we know the definition of locally bounded is that a set in H(G)is locally bounded if for each point ain G there are constants Mandr 0such that for all f in, f(z) M , for za R. Since for every compact set it happens, so the set in H(G)is locally bounded.
Conversely, let a set in H(G)is locally bounded on a compact set K. Let f(z) M1, for
1 R
a
z ; f(z) M2, for za2 R and so on. Take supremum on all the upper bounds and let it be M.Then f(z) M .Proved.
Montel’s Theorem: (see [2]) A family in H(G)is normal iff is locally bounded.
Proof: Suppose in H(G)is normal and it is not locally bounded. Then for a compact set Kin Gwe can have
, }
: ) (
sup{fn z z K fn .We can also say sup{fn(z):zK,fn}n.Again since in H(G)is normal, so {fn}has a converging subsequence { }
nk
f such that f f
nk .So, we can also have the same expression for{ }
nk
f , i. e. fn z z K fn nk
k
k( ): , }
sup{ . Now nk
} :
) (
sup{f z z K
nk
and it is less than equal to
sup{f (z) f(z) :z K} sup{f(z):z K}.
nk
Letsup{f(z):zK}M. Then nk f z f z z K M
nk
sup{ ( ) ( ) : } . Whennk , then the expression will be Mwhich is a contradiction. Hence it is locally bounded.
Now let is locally bounded .We will use now the Ascoli- Arzela theorem to prove that it is normal. It easily satisfies the first condition of the theorem and so here we need to prove only the equicontinuous part. Let abe a fixed point in Gand letB(a,r)G. Also let there is a
0
M such that f(z) Mfor all zin B(a,r)and for all f in .Let
2 a r
z and f is in .Then using Cauchy’s Formula with (t)areit,0t 2,
19 z r a
z dw w a w
w z f
a
z dw w a w
z a w f
zdw w
w dw f
a w
w z f
f a f
4M
) )(
(
) ( 2
1
) )(
(
) )(
( 2
1
) ] ( )
[ ( 2 ) 1 ( ) (
By taking }
,4
min{2
M
r
r it follows that za gives f(z) f(a) for all f in. Corollary: (see [2]) A set in H(G)is compact iff it is closed and locally bounded.
Proof: Suppose inH(G)is compact. So it is closed and totally bounded. And we know if it is totally bounded then it will be locally bounded.
Converse part: Let in H(G)is closed and locally bounded. We are required to prove its compactness, i.e. only boundedness of. By using the previous lemma we are having if in
) (G
H is locally bounded, then for each compact setK in G, f(z) Mfor all f in and for all zin K.Since for each compact set in Git happens so we can conclude that is bounded on whole G.Hence proved.
20
Chapter-3: Spaces of meromorphic functions
We know the definition of a meromorphic function as, if Gis open and f is a function defined and analytic in Gexcept for poles , then f is a meromorphic function onG. The set of all meromorphic functions is denoted by M(G)and it is a subset of the space of continuous functionsC(G,C). Defining a metric onM(G), we will discuss here this metric space. The metric defined on C is
2 1 2 2 2
1 2 1 2
1
)]
1 )(
1 [(
) 2 , (
z z
z z z
z d
,
where z1,z2Cand
2 1 2) 1 ( ) 2 , (
z z
d
.We can check that
2 1 2
1
, 1 ) 1
,
(z z d z z
d and
) 1, ( ) 0 ,
(
d z z
d for z0.
Proposition: (see [2])
(a) If ais in Cand r0then there is a number 0such that B(a;)B(a;r).
(b) If 0is given and ais in Cthen there is a number r 0such that B(a;r) ).
; (a B
(c) If 0is given then there is a compact set KCsuch that C K B(;). (d) If a compact set KCis given, there is a number 0such that B(;) C K. Theorem: (see [2]) Let {fn}be a sequence in M(G)and suppose fn f in C(G,C).Then either f is meromorphic or f . If each fnis analytic then either f is analytic or f . Proof: Let we are having a point aGsuch that f(a)and letM f(a).From the above proposition we can get a number 0 such thatB(f(a);)B(f(a);M).But as we know
21 f
fn , so for an integern0,
)) 2 ( ), (
(
a f a f
d n for allnn0. We also have {f, f1, f2,...}is compact in C(G,C)and so,it is equicontinuous. So, there is an r 0 such that whenever
r a
z then
)) 2 ( ), (
(
z f z f
d n and hence d(fn(z), f(a)) for za rand fornn0. But choosein such a way that, fn(z) fn(z) f(a) f(a) 2Mfor all zin B(a,r)and
n0
n .Then ,
)) ( ), ( ( ) ( ) ) ( 4 1 (
2
2 f z f z d f z f z
M n n
forzin B(a,r) and nn0.
Asd(fn(z), f(z))0 uniformly for zinB(a,r), it implies that fn(z) f(z) 0uniformly for zinB(a,r). The sequence {fn} is bounded onB(a,r), so fnhas no poles and must be analytic nearzafornn0. Hence f is analytic in a disk abouta.
Again let there be a point ain Gwith f(a). Define g 1 by
) ( ) 1 1)(
( z g z
g if g(z)0or; 0
) 1)(
( z
g if g(z) and 1)( )
( z
g if g(z)0 wheregis a function in C(G,C).Hence )
, 1 (
C G C
g .Also, fn f in C(G,C)and so
f fn
1
1 in C(G,C). Since each function fn
1 is
meromorphic onG, so we can have a number r 0and an integer n0such that f 1 and
fn
1 are
analytic on B(a,r)for nn0 and fn
1 converges to f
1 uniformly onB(a,r). Using Hurwitz’s
Theorem we are having either 1 0
f or
f
1 has isolated zeros inB(a,r). So if f then 1 0 f and f must be meromorphic inB(a,r). Hence the result follows.
Consider each fnis analytic then fn
1 is having no zeros in B(a,r).Then from the corollary of
Hurwitz’s theorem that either 1 0
f or
f
1 never vanishes. But as f(a)we have that f 1 has at least one zero; thus f inB(a,r). Noting the first part of the proof we see that f or
f is analytic.
22
Corollary: (see [2]) M(G)is not complete but M(G){}is a complete metric space.
Proof: Let {fn(z)}{n}be a Cauchy sequence inM(G). When n ,it will converge to infinity inC(G,C). So, it will not be meromorphic and hence M(G)is not complete. But if we consider M(G){}then it will be a complete metric space
Corollary: (see [2]) H(G){}is closed in C(G,C)
Definition: If f is a meromorphic function on the region Gthen define (f):GRby 2
) ( 1
) ( ) 2
)(
(
z f
z z f
f
whenever zis not a pole of f , and
2 ) ( 1
) ( ) 2
)(
(
z f
z Lim f
a f
a
z
ifa is a pole of f .
Theorem: (see [2]) A family M(G)is normal in C(G,C)if and only if }
: ) ( { )
( f f
is locally bounded.
Proof-: One may refer [2].
23
REFFERENCE:
1-: Ahlfors, L. V., “Complex Analysis”, McGraw-Hill, Third Edition (1979).
2-: Conway, J. B., “Functions of One Complex Variable”, Springer , Second Edition (1980).
3- : Gamelin, T.W., “Complex Analysis”, Springer,(2001).
4-: Remmert, R., “Theory of Complex Functions”, Springer, (2002).
5-: Rudin, W., “ Function Theory in the Unit Ball of Cn”, Springer,(1980).
6-: L. Zalcman, “ Normal families: new perspective”, Bull. Amer. Math Soc. 35 (1998), pp.215- 230.
7-: P. Montel, “ Lecons sur les series de polynomes”, Gauthier-Villars (1910).
