I ntnat. J. Math. Math. Si.
(1978)97-112
Vol.
97
THE SPACE OF ENTIRE FUNCTIONS OF TWO
VARIABLES AS A METRIC SPACE
W. C. SISARCICK
Department of Mathematics
Marshall University
Huntington, West Virginia 25701
(Received October 17, 1977)
Section I.
Let F
Introduction.
2
If f(z,w) e
denote the space of entire functions of two variables
F2
f(z,w}=
a
m,n=0
mn z
m n
w
the series converging
absolutely for all (z,w) and uniformly in every
bicylinder centered at (0,0), [2]. Here, a metric is defined on
F 2 and three classes of linear functionals on F 2 are characterized.
We use the following notation.
W.C. SISARCICK
98
<am,n>m+n= k
(l)
a
m,n
<a
--a 0
--
lim
N
a as
m+n
,
a
The sequence
written
m, n
<am, n >m+n=k
is said to have limit
a, if and only if for any
lim a
m,n
e > 0, there exists an N
Lemma 1 2
0 ,k’ak+l, 0
m+n=0
m+ n
if m+n >
...a
0+a I, 0+a 0, l+a 2,0 +a I i+...
m+n=0
Definition i.i.
k ,0, ak_ 1,1
N(e)
>
lam,n -al
0 such that
N.
If f(z,w)
a
m, n=0
(z,w), the sequence
<am,nZ
m,n
z mwn
mn_
w
I’2
then for each
is such that
_+n=^
mu
N
Proof
Given (z,w), let S
exists, 0
lim S
N
N
N
la kzjwkl.
j+k=0 J’
lim S
N-,
(SN-SN_ l)
lim
N
N- 1
Hence given e > 0, there exists an M
aN_j,jzN-JwJ
0
an M
n E N and
<
for each j, 0
am,nzmwnl<
.
M(e) such that m+n > M
m n
lim am,n z w
m+ n
Lemma 1.3.
<
0.
mn
w =0
$ am, nZ
Since
}im S
N
lim
N-,
j +k=N
N
j, k
M(e) such that if N > M,
j E N.
Let m+n
-
N.
Then
Therefore given e > 0, there exists
am,nZmwn
<
Hence
A necessary and sufficient condition that
a
m, n=0 m,n
m
z wn
99
ENTIRE FUNCTIONS OF TWO VARIABLES
<lam,nll/m+n>m+n=l,
is an entire function is that for the sequence
am n ll/m+n
one has lim
m+n
Let
Proof.
a
m,n
m,n=0
0.
z mwn
F2
lam’nll/m+n"
m+n
lwl
and T
zl m
that lwl
If T > 0, choose (z,w) such that
).
T
Then choose p such
1
<(mk,nk)>
i such that <
k
(I/T
> p > I/T.
Then
mk+n0k>k= I
increases monotonically to
1
and
z
mk+nk
ask ’nk
m
k
w
Hence lim
m+ n
n
> i/p for all k.
k > I.
Hence
amk
This contradicts Lemma 1.2.
lam,nll/m+n
n
m n
z kw k
>
k
Therefore T
a
m, r=0
m,n
am n l/m+nm+n=l’
z mwn be a series such that for the
-am, n’l/m+n
|
one has lim
m+n
To
0.
show this series is an entire function, it sufficies to show
[2] the series converges for each (z,w).
Choose p such that
zl
=
Therefore
a
m+n=0
m, n
Consider (z,w) fixed.
w < p.
Let N be such that m+n
m
z wn [ <
Then for m+n > N,
< P and
am, nlllm+n < I/p.
CIz_l) m lwl) n,
m+n---N+l
p
p
> N
0.
0.
Conversely, let
sequence <
0 if
By definition of T, there exists a sequence
< 1 < T.
1
> I/T.
a
m n
zmwnl<
zmwn
.
am,n
Z;
[i) n
lzl)m
p
m+n=N+ 1 p
<
".
i00
W. C. SISARCICK
P
Let s
q
F. a
z mwn
m,n
n=0
F.
P,q
m=0
To show the series converges,
> 0, there exists an
it sufficies to show [I] that given
N
Is p,q -sm,n
N(e) such that
m
la
m+n= 0 m
F.
Since
such that N >
n
z wn
,
<
<
if P > m > N and q > n > N.
given e > 0, there exists an M
aj,kzJwk <
F.
max[M,l]
.
M()
Choose such an N.
j+k=N+l
P
=IF.
j=0
q
a
k=0
F.
J’ kZ
a
j+=N+
m
j wk
n
aj,k zJwkl
j=0 k--0
w
J’
k
<
a
m,n=0
b
m, n=0 m,n
m
z wn e
1.2
Proof.
The space
j+k=m+n
aj,kZ Jw
k
m,n
m m
z w
F2
and
define d(f,g)
sup[la0,0-b0,01, lam, n -bm,n ll/m+n:m+n
Theorem 1.5.
F.
.
Given f(z,w)
Definition 1.4.
g(z,w)
Sp,q-Sm,nl
For p > m > N and q > n > N,
N().
Then N
(2,d)
>
i].
is a metric space.
Given f, g as in Definition 1.4, the set
{lam,n -bm,n l/m+n: m+n i]
well defined.
and that d(f,g)
is a bounded set by Lemma 1 3, so d is
It is clear that d(f,g)
d(g,f).
Let h(z,w)
0 if and only if f
m
c
z wn
P2
m,n=0 m,n
g
sup[la0,0-c0, 0 am,n -bm,n i/m+n m+n > 1]
sup[I (a0,0-b0,0)+(b0,0-c0, 0)
(am, n -bm,n )+( bm,n- cm,n I/m+n :m+n
sup[ la0, 0-b 0,0 +Ib 0, 0-c 0,0
a
-b
m,n m,n ll/m+n+Ibm,n -c m,n ll/m+n
m+n > i] sup[ la0,0 -b0,0[’ am,n-bm,nll/m+n: m+n > i] + sup
Then d(f,h)
>
i]
I01
ENTIRE FUNCTIONS OF TWO VARIABLES
Ib0,0-c0,0 I’ Ibm,n
-C
d is a metric on V
2.
[
i/m+n
m,n
m+n
d(f,g) + d(g,h).
i]
>
Hence
The class of continuous linear functionals on F
Section 2.
2
A function F from F to
(complex plane)
2
is a linear functional if and only if for all f,g
1‘
Definition 2.1.
F(f)+F(g) and F(f)
Definition 2.2.
2.
,
,
F(f+g)
F(f).
2
A function F from F to
is said to be continuous
2
> 0 there exists a 6 > 0 such
if and only if for any
2
(f) -F (g)
that if g
F and d(f,g) < 6, then
at f e F
IF
Definition 2.3.
I<
A function F from I‘2 to
is said to be continuous
if and only if it is continuous at each f
The series
Lemma 2.4
<am, n >m+n=0
1‘2
b
converges for all sequences
a
m+n= 0 m,n m,n
such that lim
m+ n
<Ibm, n l’l/m+n>"m+n= 1
.
a
m, n
i/m+n_
0 if and only if
is a bounded sequence
<a
<bm..i/m+n>
m, n m+n= 0
n
m+n= 1 be a bounded sequence and
I/m+n
such that lim a
0. Choose M > 0 such that
m n
m+n
Proof Let
be
bm ,n .i/m+n
Jam ,n
I/m+n
Therefore
M if m+n >i and then N
1
2M
.
m+n=N+ 1
Then if m+n > N,
>
0 such that m+n > N
am,nbm,n
lam,nbm,n = m+n=N+ 1
1
2m+n
1
m+n
(2M)
m+n=
M
1
2m+n
W. C. SISARCICK
102
a
Hence the series
m+n= 0
b
m,n m,n
converges absolutely, hence it
converges.
<am, n >m+n=0
Conversely suppose for any sequence
lim
m+n
<
.am n i/m+n
1/m+n
Ibm, n
(mk,n k)
an
strictly
a
0, the series
m+n=0
___>m+n=]
b
converges.
m, n m, n
is not bounded, for each k
1
such that
Ibk,nk mk+n
increasing. Choose
mk+n k
k
Then
a
k
> k and
=
m,n
Z
If
+ there exists
<mk+nk>k=l
is
0 if (m,n)
nk
m+nlim
amk, nk
such that
1
mk+nk
k*lim
a
m+n=0
0.
But
Therefore
am,n bm,n
The series
> 1 for each k so
amk,nk bmk,nk
b
does not converge.
m,n m,n
is bounded.
only
1
<Ibm ,n ll/m+n>m+n=l
does not converge since the
0 terms are > 1 and there are an infinite number of
them
We now characterize the class of continous linear functionals
on 1-‘2
Theorem 2.5.
Let F be a function from F
2
to the complex plane.
2
Then F is a continous linear functional on F if and only if there
is a unique sequence <b
m
n>m+n=0
such that
<Ib0,01
is bounded and such that for all f(z,w)=
a
m+n=0
F(f)
a
b
m,n m,n
m+n= 0
m ,n
Ibm,nll/m+n>m+n=l
z mwn
F2
103
ENTIRE FUNCTIONS OF TWO VARIABLES
<Ib0,01 Ibm, n
such that b0,01
Let
Proof
M > 0 be
a
f(z,w)
m+n= 0
a
m+n= 0
m, n
a
bm,nll/m+n < M,
< M,
m n
2
z w e F
a
Then lim
2
a
m, n
z mwn
d(f,g) <
,
p2
We show
and d(f,g) < 6, then
Choose 6 > 0 such that 6M < 1 and
e.
.
2
<
C
Then if g(z,w)
m+n=0
IF(f-g)
IF(f)-F(g)
(a
m+n=0
g
0 sO
Hence we may define
F 2 be given.
there exists a 6 > 0 such that if g e
6 M
ll/m+n
1 and
to the complex plane by F (f)
m+n= 0
(i_-)
m+n
It is clear that F is a linear functional.
b
m,n m,n
IF(f)-F(g) [<
m n
m+n
Let e > 0 and f(z,w)=
6M +
be a bounded sequence,
>m+n=l
converges by Lemma 2.4.
b
m,n m,n
a function F from P
m+n=0
I/m+n
la0 0-C0 01M
6M +
m,n
m,n
m
z wn e F 2 and
-C
m,n )bm
nl
m+n
/
m+n=l
(6M)
lam,n -Cm,n IM
m+n
m+n= i
6M +
(6M)
m
n
n=l
m=l
6M +
(6M)
2
6M
(-’M+/u
< e
Conversely, let F be a continuous linear functional on F
Let
F(zmwn9
let
fN(z’w)
b
for all m+n
m ,n
N
> N]
0 as N
a
m+n=0
m,n
m
z wn
0.
Then
Given f(z,w)
d(fN,f)
sup[
so by the continuity of F, F(f
N)
2.
m n
z w
a
m+n=0 m,n
lam ,n
I/m+n :m+n
F(f) as N
.
W. C. SlSARCICK
104
But
a
F(fN
m+n=0
a
b
converges and F (f)
m+n=0 m,n m,n
Hence
Lemma 2.4, the sequence
<
N
Therefore lim
b
a
m,n m,n
N m+n=0
b
m,n m,n
Cm, n >
m+n=0
f(z,w)
Hence
m+n=0 m,n
Z+, F(zJwk)
<,b0,0,I/m+n>m+n=l
m
z wn
F2
F(f)
a
m+n= 0
Cj,k.
Let f,g e F 2
But
F(zJwk)
-f(z,w).
Then F
2
(f-g) (z,w)
f(z,w)g(z,w), (f)
becomes a commutative algebra with
In this section we characterize the continuous linear
Lemma 3.1.
Given e > 0 and (b,c) e
such that if f, g
Given e > 0 and (b,c)
e
x,
a
z mwn and g(z,w)
m,n=0 m,n
and d(f,g)< 6,
x,
there exists a 6 > 0
+
max[Ibl, Icl]
let R
(I_6-R)
2
< e.
Then if
z mwn are in
b
n
m, n=0 m,
If(b,c)-g(b,c)
(a
m+n=0
la0 0-b0 01
That is the
F 2 and d(f,g) < 6, then If(b,c)-g(b,c)
choose 6 > 0 such that 6R < 1 and 6+
f (z,w)
F2
Define
functionals on F 2 that preserve multiplication.
continuous scalar homomorphisms on F 2
Proof.
then for
C
m n m n
bj,k.
(complex field).
f(z,w) + g(z,w),
(f+g) (z,w)
a unit.
Suppose
The class of continuous scalar homomorphisms on
Section 3
(z,w)
is bounded.
and the sequence is unique.
bj,k
Cjk
By
b
a
m+n=0 m,n m,n
is a sequence such that for all
a
j, k e
F (f)
la -b IRm+n
m+n= 1 m, n m, n
m, n
-b
m, n
< 6 +
)bmc nl
(6R)
m+n= 1
m+n
p2
< e.
6 +
m
(6R)
n=l
m= 1
ENTIRE FUNCTIONS OF TWO VARIABLES
2
n
6R
(6R)
6 + (i_-) <
.
Then F is a continuous scalar homorphism on
x
there exists a unique (b,c)
a
m,n=0
m,n
p2
f(b,c).
0 continuous scalar homomorphism on I
a
m, n=0
each m and n, b
F(f)
a
m+n=0
b
m
z wn
F(zmwn)
m,n
m,n
m, n
m
n
by F(f)
F(f)
x,
(f-g) (b,c)
Section 4.
<
Then F is clearly a
.
=I f(b,c)-g(b,c)
such that
b
a
m+n=0 m,n m,n
,1"
Therefore
Then
<
IF(f)-F(g)
.
#
2
0 scalar homomorphism.
F 2 and d(f,g) < 6,
F(f-g)
Hence F is continuous.
The class of bounded linear functionals on F
Definition 4.1
For
define a function F from
> 0, let 6 > 0 be such that if f,g
f(b,c)-g(b,c)
m+n=0
f(bl,0,l).
1,0b0,1
f(b,c).
F2
F(z)mF(w)n
Conversely, given (b,c)
then
if and only if
"2.
Let F be a
for all f(z,w)
Given
0.
such that for all f(z,w)
By Theorem 2.5, there is a unique sequence <b
m
to
F
z mwn t
F(f)
Proof.
,
Let F be a function from F 2 to
Theorem 3.2.
105
Let F be a linear functional on F 2
2.
Then F
is said to be bounded if and only if there exists an M > 0 such
W. C. SlSARCICK
106
that for all f e i-
2
IF(f)
< Md(f,0).
function identically zero on
Here, 0 denotes the
x.
Let F be a linear functional on F
Lemma 4.2.
2
If F is bounded,
F is continuous but not conversely.
Let F be a bounded linear functional on F
Proof
Then if g e I‘2 and d(f ,g) < 6,
o
e/M+l
Choose 6
IF(fo-g)
Md(fo -g’)
<
F is continuous at f o
(m+l)d(fo ’g)
> 0
Let f o e F
Md(f,0)
IF(f)
F2
0 be such that for all f
be given and let M
2
2
IF(f o )-F(g)
< (M+I) 6 < e.
Therefore
Hence F is continuous.
For an example of a continuous linear functional that is not
bounded, let
n.
bm,n
Define a function F from
Then
F2
<Inll/m+n>m+n=l
to
is a bounded sequence.
a
by F
m,n=0
m
m,n
z wn
m+n=l
2
By Theorem 2.5, F is a continuous linear functional on F
is bounded, there exists an M
na
0 such that
m+n+l
sup
[la0,01, lam ,n
am ,n
a0,k
m+n
l/m+n
k,
am,n
I/m+n
nam,n
If F
M
m, n
m+n a I] for all <a m,n >m+n=0 such that
0 as m+n
0 if (m,n)
Let k e Z
(0,k).
+
k >
Then
max{M,2].
lam,nIl/m+n
since the sequence has only one non-zero term.
Let
0 as
But
107
ENTIRE FUNCTIONS OF TWO VARIABLES
na
m+n=l
k2
m,n
sup[
M
lam,nll/m’n
la0,01,
I/k <k.k I/k < k
M.k I/k < k,k
2,
:m+n
i]
>
Hence F is not
a contradiction.
bounded.
Let B denote the class of bounded linear
2
2
f e F
define (F+G)
For F,G
B,
functionals on F
Definition 4.3.
F(f)
,
IF[
-F(f),
(F) (f)
G(f),
inf[M
F2,
for all f
0
>
(f)
Md(f,0)].
F(f)
Theorem 4.4. With respect to Definition 4.3, B is a normed
linear space.
B.
Proof Let F
for some f
If not
O
e F
2
so choose e > 0 such that
IFII,
Definition of
Md(fo,0).
Therefore
F(f O )I >
IF(fo)
IFII
Hence
IFI Id(f,0)
IF(f)
d(f,0) for all F e
>
0 such that
0)
IF(f)
(I IFII
B,
2
on 12
For f
IIFI
d(f,0)+
F+G e B and
lel IF(f)
IF I"
F
,
+ )
> M
Then d(f
F
+ e
M, a contradiction.
for all f
F2
O
and
IFII
is the
F2
F+G and oF are clearly linear functionals
(F+G) (f)
IIGI d(f,0)=
IIF+GII = llFl +
lel IIFII
Md
+
smallest number to satisfy this inequality for all f e
For F, G
p2.
IFII d(f 0,0). Then d(fo,0)
IFII d(fo,0) + e d(fo,0). By
there exists an M
2
(f 0) for all f e F and
F(fO
FI
IF(f)
Then
IF(f) + IG(f)
IF(f)+G(f)l
(IIFII + llGll)d (f,0). Hence
l-F(f)
I811. Also l(e)(f)
d(f,0).
Hence
Suppose it is possible to have
=F
leFll
B and
<
lleFl
lel IFII.
<
<
II
Choose
0
W. C. SISARCICK
108
sFl
> 0 such that
IFil
d(f,O).
-
IF(f)
Therefore
contradiction.
Hence
s F I-
Fl
sl
+ e
IFII
)
(1
Then for all
d(f,O), a
F.
F
It is clear that
evaluated at the zero linear
2
0 for all
functional on F is 0 and if
0 then F(f)
f
F 2, hence F
m
F
Also the remaining properties required for B
0.
to be a normed linear space follow trivially.
linear space with respect to Definition 4.3.
2
Theorem 4.5. Let F be a function from F to
m,n=0
am ,n zmwn
B if
such that for
2
a0,0 a
F(f)
Then F
xx
and only if there exists unique (a,b,c,)
all f (z,w)
.
Hence B is a normed
+
al,0b
+
a0,1c.
Also,
Proof.
Let F
B.
Then F is continuou’s so there exist a unique
sequence <bm, n >n+n=0 such that F
and
m+n=0
a
b
m,n m,n
all sequences
Suppose b
(re,n)
lak, jll
k,j
IFI Isup[ la0
<am, n >m+n=O
such that
0
Ibk, j.
ak,j
Then
a
m+n=0
m,
nbm ,n
lam ,n ll/m+n:m+n 1 for
am ,n ll/m+n 0 as m+n ,.
>
- -
0 for some (k, j) with k+j > 2.
(k,j) and choose
MI
z mwn)
a
m,n=0 m,n
such that
lam,n ll/m+n
IF If
Choose
am,n
>
0 as m+n
,
0 if
ENTIRE FUNCTIONS OF TWO VARIABLES
Z
a
m+n=0
IFI .ak, jl k+3
lak. j bk, j
b
m,n m,n
109
Therefore
contradiction.
if k+j > 2.
Hence F
a
m,n=0
a
b
0 I 0,i"
IF(
Also
a
m, n=0
lal,01 Ibl,01
+
zmwn)
z w
)I
la0,11
a
<
a
(lbo,ol
IFII = lbO,Ol
Therefore
m n
m,n
+
l.b.,,ol
+
b
+
0,0 0,0
lbO,Zl.
(Ib 0 ,0 +
ie I
Ibo,ole,
f (z,w)
o
lbo,ol
b
e
+
b
1,0
i8 2
I ,0
Ib 1,0 le
i81
+ e -i82 z+e
lbZ Ol
’bO,11
+
b
-i@ 3
to
by F(f)
a
m,n=0
m,n
z mwn e
functional on
la0,11 Icl
2.
a0,0a
F2
+
Then
(lbo,ol
Conversely, given (a,b,c) e
p2
w.
el, 0 b
+
xx,
+
1""
o e
2
If b
Ib 0 1 le i83
0 1
al,0bl, 0
Ib0,11)d(f,0).
such that
O,O
choose
IF(f o
lbl,01
+
lbO,Zl)d(f O’ O)
define a function F from
+ a 0 I c, f(z,w)
By Theorem 2.5, F is a continuous linear
Since
IF(f)
lao,ol lal
(lal + Ibl + Icl)d(f,0),
+
lal,01 Ib
+
F e B.
Corollary 4.6. With respect to Definition 4.3, B is a Banach
space.
0
+
To show equality
Ibo,ll)d(f O, 0).
+
bk,j
0,0
lbl,ol
here, it sufficies to show there exists an f
F (f)
o
b
0,0
+
Hence
.
llO
to
<Fn>n
Let
Proof.
.
be a Cauchy sequence in B, F
lFn-Fmll
Hence each of <a >
n
anna, bnb
c c as n
n
F(
z mn
w .=
a
m,n=O
> 0
Given
+
/
N
m,n
Cm-Cn
Ic-c nl
n corresponding
Then for any e > 0, there exists an N
(an,bn,C n).
that m,n > N()
<
1
C. SISARCICK
.
ao, 0 a
< :.
That is
<bn> <Cn>
N(e) such
lan-aml+Ibn-bml
+
is a Cauchy sequence.
Let
Define a function F from I"2 to
+ a
0
b + a
there exists an N
c.
By Theorem 4.5
F
B
am-al + Ibm -b n
get la-an + [b-bnl
N(:) such that
< /2 if m,n > N ().
/2 if n > N().
0 I
by
Let m
to
> 0, there exists an
Hence given
N() such that if n > N(),
That is
If
IF n -FI
<
(al,bl,C I)
.
Therefore B is a Banach Space.
and
(a2,b2,c 2)
are in
x
,
and e
itis
clear that if addition and scalar multiplication are defined by
(al,bl,C I)
+
(a2,b2,c 2)
(a I, b I, c I), that
ll-ll s
on
defined by
xCx
xx
xx as follows: For
of xx such that F(f)
f (z,w)
a
m,n=0
m,n
z mwn e
F2
Also if
lal+LbJ+lcL, I-L,L s
into a Banach space.
from B to
element
is a vector space over
ll(a,b,c) ll
xx
making
.
(al+a 2, bl+b 2, Cl+C2), e(al,bl,C I)
F
a norm
Define a function Y
B, let (a,b,c) be the unique
a0,0 a
Let Y )
+ al,0 b +
a0,1 c
(a,b,c).
for all
The following
theorem is straightforward to prove so the proof is omitted
ENTIRE FUNCTIONS OF TWO VARIABLES
Theorem 5.4.7.
The spaces
B and
iii
are isometrically isomorphic Banach
spaces.
i.
Copson, E. T. Theor of Functions of a Complex Variable, Oxford Clarendon
Press, 1935.
2.
Funks, B. A.
Theor of Analytic Functions of Several Complex Variables,
American Mathematical Society, Providence, 1963.
ABSTRACr.
Three classes of linear functionals on the space of entire functions
of two variables are characterized.
ACKNO.
Several results are proved.
This work was supported by a 1976 Marshall University Sunmer
Research Grant.
KEV WORDS AND PHRASES. Space of
linemt
fna and
ee
funeX:ions, M space, continuouA
continuou scala homorphism.
AMS(MOS) SUBJECT CLASSIFICATIONS (1970). 30A66, 46E15.