Theorem 2.3. Let f(z) be a single S-valued holo-
morphic mapping on G. Then
(A)(Cauchy’s integral theorem).
Z
L
f(z)dz = 0,
where G =
Q
∞
j=1
G
j
;
(B)(Cauchy’s integral formula).
1
2πi
Z
L
f(t)(t − z)
−1
dz = f(z),
where z ∈ D
+
⊂ G, and (t − z)
−1
exists.
Definition 2.3. A subset X
0
of X is called a base-real
subspace of X and ¯u is called the conjugate element
of u if following conditions hold:
a) X is a vector space on C.
b) X
0
is a vector subspace of X on R.
c) For every u ∈ X, there exists an ¯u(∈ X) such that
u + ¯u ∈ X
0
and i(u − ¯u) ∈ X
0
hold satisfying a
unique decomposition u = ξ + ηi for ξ, η ∈ X
0
.
d) X
0
∩ iX
0
= {0}, where 0 is the zero element.
Theorem 2.4. If X is a complex vector space,
then there exists a base-real subspace X
0
such that
X = X
0
+ iX
0
, i.e. a complex vector space can be
represented by a direct sum of two spaces which are
generated by some real vector space.
Proof. For any x
0
(6= θ) ∈ X, let M
0
= Rx
0
=
{tx
0
: t ∈ R}. Then M
0
is a vector subspace of X on
the restricted number field R, and M
0
∩ iM
0
= {θ}.
Setting CM
0
= {zm
0
: z ∈ C, m
0
∈ M
0
}, we have
that CM
0
is a complex vector subspace of X. For
any x
1
∈ X\CM
0
we know that M
1
= M
0
+ Rx
1
is also a vector subspace on a restricted number field
R of X and M
1
∩ iM
1
= {θ}. By induction we
obtain a sequence {M
n
} of vector subspaces with
M
n
∩ iM
n
= {θ}. Assume that M is the family
of all vector subspaces on R, that M
0
= {M
0
∈ M :
M
0
∩ iM
0
= {θ}} (clearly, M
0
is nonempty), and
that {M
α
}
α∈J
is the family of totally ordered sub-
sets of M
0
, where J is an indexing set. Consequently,
X
M
= ∪
α∈J
M
α
is a vector subspace on R and the
supremum of {M
α
}
α∈J
. Further we have
X
M
∩ iX
M
=
S
α∈J
M
α
∩ i
S
α∈J
M
α
=
S
α∈J
M
α
∩ i
S
β∈J
M
β
=
S
α∈J
S
β∈J
(M
α
∩ iM
β
).
Since {M
α
}
α∈J
is a family of totally ordered
subsets with M
α
∩ iM
α
= {θ} for any α ∈ J
and M
α
∩ iM
β
= {θ} for any α, β ∈ J, we get
X
M
∩ iX
M
= {θ}. Now Zorn’s lemma yields that
M
0
has the maximum element X
0
.
Next we shall show that X
0
is the required
subspace. Firstly, CX
0
= X, here CX
0
is the
smallest complex subspace containing X
0
. In fact,
if CX
0
6= X, then there exists an x ∈ X\CX
0
. It
follows that X
0
0
= X
0
+ Rx is a vector subspace on
R containing X
0
with X
0
0
∩ iX
0
0
= {θ}. This is in
contradiction with the maximality of X
0
. Obviously,
CX
0
= X
0
+ iX
0
. Since X
0
∩ iX
0
= {θ},
X = X
0
+ iX
0
. If u = x + iy ∈ X (here x, y ∈ X
0
),
then ¯u = x − iy is the conjugate element of u, i.e.
X
0
is a base real subspace of X.
This proof is finished.
Suppose that e ∈ S is the idempotent element. The-
orem 5.3.2 (Hille and Phillips, 1957) may be extended
to the Fr´echet algebra S containing the Banach alge-
bra. Then using the result after extending we can get
exp (Log)z = z, exp(z + 2πie) = exp z,
where exp z =
P
∞
j=1
z
j
/j!. If z and e commute, then
exp(z + 2πine) = exp z, for n = 0, ±1, ±2, ....
It follows that
Log[exp z] = z +2πine = z
0
+i(z
00
+2πne), (2.2)
for any integer n and z
0
, z
00
+ 2πne ∈ S
0
, where
S = S
0
+ iS
0
, S
0
is a base-real Fr´echet algebra.
From (2.2) we can define the argument of exp z be-
ing z
00
+ 2πine. Because
exp z
0
z
00
= (exp z
0
1
z
00
1
, ..., exp z
0
j
z
00
j
, ...),
Logz = (log |z
1
|, ..., log |z
j
|, ...)
+i(Argz
1
, ..., Argz
j
, ...).
It follows that the argument Argz of z may be defined
by
(Argz
1
, ..., Argz
j
, ...). (2.3)
3 THE V H
∞
SPACES
Let |f (z)| = (|f
1
(z
1
)|, . . . , |f
j
(z
j
)|, . . .) be the vec-
tor modulus of f and kfk = (kf
1
k, . . . , kf
j
k,
. . .) the vector norm of f. For any a, b ∈ S, a ≤ (<
)b is a
j
≤ (<)b
j
for each j. Let C
+
j
= {z ∈ C
j
:
<z > 0} and S
+
=
Q
∞
j=1
C
+
j
. The set V H
p
(S
+
)
consists of all holomorphic mappings f : S
+
→ S
satisfying
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360