Dedicated to Ermanno Lanconelli.
1 Introduction
The present article studies holomorphic functions
f
:
Ω
→
X
and their application to linear partial differential equations with real analytic coefficients, where
Ω
⊂
C
n
(
n
≥
1
) is an open set and
X
is a complex topological vector space. The need for vector-valued holomorphic functions arises mainly in the theory of 1-parameter semigroups (cf. Arendt et al. [1]) and in analytic functional calculus (cf. Vasilescu [23]). For instance, families of holomorphic functions depending continuously on a parameter
t
∈
[
0
,
1
]
may be recast as holomorphic functions with values in the Banach space
X
=
C
(
[
0
,
1
]
,
C
)
of all
C
valued continuous functions on the unit interval
[
0
,
1
]
Cf. also Barletta and Dragomir [2,3], Barletta et al. [4], and Dragomir and Nishikawa [14].
Holomorphic functions of one complex variable, with values in a complex locally convex space
X
, were first studied by Grothendieck [17], who characterized the topological dual
O
(
Ω
,
X
)
∗
with
Ω
⊂
C
open. Properties of
X
-valued holomorphic functions on analytic spaces were studied by Bungart [8]. Bungart’s purpose in [8] is to extend Theorems A and B of Cartan (cf. [12]) to this more general class of holomorphic functions. Cartan’s Theorem B for vector-valued holomorphic functions was also proved by Bishop (cf. [6]) by a different methodological approach. To compensate for the lack of generality of the considerations in the present article, let
(
V
,
O
)
be a Stein space,
S
be a coherent analytic sheaf over
(
V
,
O
)
, and
H
(
V
,
S
)
be the space of cross-sections in
S
. Let
(
W
,
O
W
)
be a closed subvariety of
(
V
,
O
)
, and let
B
(
W
,
O
W
)
be the Banach space of bounded functions in
H
(
W
,
O
W
)
. The restriction map
H
(
V
,
O
)
→
H
(
W
,
O
W
)
admits a continuous linear right inverse
B
(
W
,
O
W
)
→
H
(
V
,
O
)
and the methods developed in [8] allow for a choice of this map
f
⟼
F
that can be represented by an integral formula:
(1.1)
F
(
z
)
=
∫
W
f
d
η
z
where
η
is a holomorphic mapping of
V
into the space of bounded measures on
W
. In particular, if
U
⊂
V
is a bounded domain, then there is a holomorphic mapping
η
from
U
into the space of measures on the distinguished boundary
∂
0
U
of
U
such that
(1.2)
f
(
z
)
=
∫
∂
0
U
f
d
η
z
,
z
∈
U
,
for any function
f
holomorphic on the closure
U
¯
of the domain. When
U
⊂
C
n
is a bounded domain with
C
∞
boundary, a choice of
η
with values in
X
=
C
∞
(
Λ
2
n
−
1
T
∗
(
∂
U
)
)
(organized as a Fréchet space) is available. Precisely (by Theorem 20.2 in [8], p. 343) there is a holomorphic map
α
:
U
→
X
such that
(1.3)
f
(
z
)
=
∫
∂
U
f
α
z
,
z
∈
U
,
for any
f
∈
O
(
U
¯
)
. A question raised in [8], i.e., whether
α
can be chosen to be a kernel, has been explored in [9], where the problem of building appropriate kernels that are holomorphic (anti-holomorphic) in the parameters involved (and not only real analytic, as is the case with Bochner-Martinelli formula, cf. [7] and [18]) is taken up within the line of thought started in [24] with the Cauchy-Weil formula. Cf. also [10] and [11]. It should be reminded that a related formula was discussed, together with its relationship to boundary kernels, in [5]. Both Cauchy-Weil and Bergman formulas hold solely for domains with particular boundary properties, while for existence theorems leading to representation formulas such as (1.3), holding for arbitrary smoothly bounded domains, one should rely on the findings in [8,15,16].
As a drawback of the approach in [8], the abstract existence theorems (Theorem 19.1 in [8], p. 340) underlying the representations (1.1)–(1.2) do not provide explicit representation formulas, as required by applications of sorts. In the present article, we aim to applications to the Cauchy problem:
(1.4)
P
(
x
,
D
)
u
=
f
in
Ω
,
(1.5)
∂
j
u
∂
x
n
j
x
n
=
0
=
φ
j
on
Ω
0
=
Ω
∩
{
x
n
=
0
}
,
0
≤
j
≤
m
−
1
,
P
(
x
,
D
)
≡
∑
∣
α
∣
≤
m
a
α
(
x
)
D
α
,
0
∈
Ω
⊂
R
n
,
a
α
∈
C
ω
(
Ω
)
,
f
∈
C
ω
(
Ω
,
X
)
,
φ
j
∈
C
ω
(
Ω
0
,
X
)
.
As well as in the case of scalar valued dependent variables, the Cauchy problems (1.4) and (1.5) may be transplanted to the holomorphic category and then restated as the integro-differential equation:
(1.6)
v
−
T
v
=
f
˜
,
where
T
:
O
(
U
,
X
)
→
O
(
U
,
X
)
is the linear operator
T
v
≡
−
∑
j
=
0
m
−
1
∑
β
∈
Z
+
n
−
1
∣
β
∣
≤
m
−
1
a
˜
β
j
(
z
)
z
n
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
(
D
z
′
β
v
)
(
z
′
,
t
z
n
)
d
t
.
The main ingredient needed in the treatment of (1.6) is a Cauchy integral formula involving a Bochner integral on the distinguished boundary of a polydisc. The less general (as compared to the representation formulas (1.1) and (1.2)) Cauchy integral formula is recovered to the case of vector-valued holomorphic functions by taking a different approach (with respect to that in [8]), i.e., by extending the arguments in [21] from functions of one complex variable (
n
=
1
) to the case of functions of several complex variables (
n
≥
2
). A key step is to show that weakly holomorphic functions with values in a complex Fréchet space are strongly holomorphic, and within to establish the strong continuity of weakly holomorphic functions (cf. Theorem 2.1 below). The proof makes use of a trick discovered in early works by Nicolescu (cf. [20]) and Ciorănescu (cf. [13]) itself relying on the fact that weak and strong boundedness in a locally convex space are equivalent. We would deceive the reader if we were not to mention that the solution to the Cauchy problems (1.4) and (1.5) does not rely solely on the Cauchy integral formula (2.5) but also on its more refined consequences such as Weierstrass’ theorem (that we recover to the case of uniformly convergent [on compact sets] sequences of vector-valued holomorphic functions in Section 3 [cf. Theorem 3.1 there]) and on the truncated Taylor formula with integral reminder (a vector-valued version of which is recovered, with a Bochner integral reminder, in Appendix B to the present article). To solve (1.6), we use an algorithm
A
(
T
,
F
)
learning from
v
0
∈
O
(
U
,
X
)
and producing a sequence
{
v
ν
}
ν
≥
1
that converges to the solution. A hard analysis ingredient in the proof of the (vector-valued analog to the) Cauchy-Kovalevskaja theorem, that is worth to mention, is the estimate (cf. [5.6] in Lemma 5.3) of the error committed by
A
(
T
,
F
)
at step
ν
, i.e.,
(1.7)
p
[
w
ν
(
z
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∣
z
n
∣
ν
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
,
p
∈
P
,
ν
∈
Z
+
,
z
∈
B
n
.
Here,
w
ν
≔
v
ν
+
1
−
v
ν
,
v
ν
+
1
=
T
v
ν
+
F
,
ν
≥
0
,
F
,
v
0
∈
O
(
U
,
X
)
,
B
n
¯
⊂
U
⊂
C
n
.
The frankly didactical proof of (1.7) provides the explicit calculation of the constants in (1.7) (as paralleled to the classical proof, where only existence of
M
(
p
)
>
0
and
C
(
m
)
>
0
is inferred):
M
(
p
)
=
sup
z
∈
B
n
¯
p
[
w
0
(
z
)
]
,
C
(
m
)
=
b
m
∑
j
=
1
m
m
j
,
a
=
max
0
≤
j
≤
m
−
1
max
β
∈
B
j
sup
z
∈
B
n
¯
∣
a
˜
β
,
j
(
z
)
∣
,
b
m
=
∣
B
0
∣
,
B
0
≔
{
β
∈
Z
+
n
−
1
:
∣
β
∣
≤
m
}
.
2 Vector-valued holomorphic functions of several complex variables
Let
Ω
⊂
C
n
be an open set (
n
≥
1
), and let
X
be a complex topological vector space. A function
f
:
Ω
→
X
is weakly holomorphic in
Ω
if
Λ
∘
f
∈
O
(
Ω
)
, i.e.,
Λ
∘
f
:
Ω
→
C
is a holomorphic function, for every
Λ
∈
X
∗
. Also
f
:
Ω
→
X
is strongly holomorphic if for any
a
∈
Ω
there is a neighborhood
a
∈
U
⊂
Ω
and a series
∑
∣
α
∣
≥
0
(
z
−
a
)
α
x
α
with
x
α
∈
X
such that
∑
∣
α
∣
=
0
∞
(
z
−
a
)
α
x
α
=
f
(
z
)
for any
z
∈
U
. Strongly holomorphic functions are weakly holomorphic, as well (because every
Λ
∈
X
∗
is linear and continuous). As to the converse, the following facts are classically known, though only for
n
=
1
(cf. Theorem 3.31 in [21], p. 82). Let
Ω
⊂
C
be an open set,
X
a locally convex space, and
f
:
Ω
→
X
a weakly holomorphic function. Then (i)
f
is strongly continuous. Let
Γ
(
Ω
)
denote the set of all closed rectifiable curves
Γ
=
{
γ
(
t
)
:
a
≤
t
≤
b
}
in
Ω
⊂
C
, and let us set
Ind
Γ
(
z
)
=
1
2
π
i
∫
Γ
d
ζ
ζ
−
z
,
z
∈
C
.
Let us assume that
co
¯
[
f
(
Γ
)
]
is a compact subset of
X
, for any
Γ
∈
Γ
(
Ω
)
. Then (ii)
(2.1)
∫
Γ
f
(
ζ
)
d
ζ
=
0
,
Γ
∈
Γ
(
Ω
)
,
Ind
Γ
(
z
)
=
0
,
z
∈
C
\
Ω
,
(2.2)
f
(
z
)
=
1
2
π
i
∫
Γ
(
ζ
−
z
)
−
1
f
(
ζ
)
d
ζ
,
z
∈
Ω
,
Ind
Γ
(
z
)
=
1
,
(2.3)
∫
Γ
1
f
(
ζ
)
d
ζ
=
∫
Γ
2
f
(
ζ
)
d
ζ
,
Γ
1
,
Γ
2
∈
Γ
(
Ω
)
,
Ind
Γ
1
(
z
)
=
Ind
Γ
2
(
z
)
,
z
∈
C
\
Ω
.
Let
X
be a complex Fréchet space. Then (iii)
f
is
C
-differentiable at each
z
0
∈
Ω
, i.e., the limit
lim
z
→
z
0
(
z
−
z
0
)
−
1
{
f
(
z
)
−
f
(
z
0
)
}
exists in the topology of
X
, for any
z
0
∈
Ω
. The formulas (2.1) and (2.2) are, respectively, Cauchy’s theorem and the Cauchy integral formula for
X
-valued holomorphic functions (cf. Theorem 1.5 and formula (1.59) in [22], pp. 42–49, for the scalar valued counterpart of (2.1) and (2.3)). Strong continuity of weakly holomorphic functions follows mainly from the Cauchy formula for scalar valued holomorphic functions (cf. [21], pp. 83–84).
We recall that a subset
E
⊂
X
is bounded if for any neighborhood
V
of
0
∈
X
there is a number
s
>
0
such that
E
⊂
t
V
for any
t
≥
s
. Also
E
is weakly bounded if
Λ
(
E
)
⊂
C
is bounded for any
Λ
∈
X
∗
. It should be observed that while basic notions make sense for an arbitrary topological vector space
X
, most basic results require more specialized values. The assumption that
X
is a locally convex space (in statement (i)) is exploited in two ways. First, any weakly bounded subset of a locally convex space is (strongly) bounded (cf. Theorem 3.18 in [21], p. 70). Second, for any locally convex space
X
the topological dual
X
∗
separates points on
X
(cf. Corollary to Theorem 3.4 in [21], pp. 59–60).
Let
Ω
⊂
C
n
be an open set (
n
≥
2
) and let
a
∈
Ω
. Let
ρ
=
(
ρ
1
,
…
,
ρ
n
)
be a polyradius (
ρ
j
>
0
) such that the polydisc
P
¯
(
a
,
2
ρ
)
=
{
z
∈
C
n
:
∣
z
j
−
a
j
∣
≤
2
ρ
j
,
1
≤
j
≤
n
}
is contained in
Ω
. Let
X
be a locally convex space. Let
f
:
Ω
→
X
be a weakly holomorphic function and let
Λ
∈
X
∗
. Let us set
a
(
j
)
=
(
a
1
,
…
,
a
j
)
∈
C
j
for any
1
≤
j
≤
n
. As
Λ
∘
f
:
Ω
→
C
is holomorphic, for every
z
∈
P
(
a
,
2
ρ
)
Λ
[
f
(
z
)
]
−
Λ
[
f
(
a
)
]
=
∑
j
=
1
n
{
Λ
[
f
(
a
(
j
−
1
)
,
z
j
,
…
,
z
n
)
]
−
Λ
[
f
(
a
(
j
)
,
z
j
+
1
,
…
,
z
n
)
]
}
=
(by the Cauchy formula)
=
1
2
π
i
∑
j
=
1
n
∫
∣
ζ
j
−
z
j
∣
=
2
ρ
j
Λ
[
f
(
a
(
j
−
1
)
,
ζ
j
,
z
j
+
1
,
…
,
z
n
)
]
ζ
j
−
z
j
d
ζ
j
−
∫
∣
ζ
j
−
z
j
∣
=
2
ρ
j
Λ
[
f
(
a
(
j
−
1
)
,
ζ
j
,
z
j
+
1
,
…
,
z
n
)
]
ζ
j
−
a
j
d
ζ
j
.
Let us set
M
(
Λ
)
=
sup
{
∣
Λ
[
f
(
ζ
)
]
∣
:
ζ
∈
P
¯
(
a
,
2
ρ
)
}
.
If
z
∈
P
¯
(
a
,
ρ
)
\
{
a
}
, then
Λ
[
f
(
z
)
]
−
Λ
[
f
(
a
)
]
≤
M
(
Λ
)
4
π
∑
j
=
1
n
∣
z
j
−
a
j
∣
ρ
j
∫
∣
ζ
j
−
z
j
∣
=
2
ρ
j
d
ζ
j
∣
ζ
j
−
a
j
≤
M
(
Λ
)
∣
z
−
a
∣
∑
j
=
1
n
1
ρ
j
(by
∣
ζ
j
−
z
j
∣
≥
∣
ζ
j
−
a
j
∣
−
∣
z
j
−
a
j
∣
=
2
ρ
j
−
∣
z
j
−
a
j
∣
≥
ρ
j
). Consequently, the set
(2.4)
{
∣
z
−
a
∣
−
1
[
f
(
z
)
−
f
(
a
)
]
:
z
∈
P
¯
(
a
,
ρ
)
\
{
a
}
}
is weakly bounded in
X
and hence strongly bounded, too. Let
V
⊂
X
be an open neighborhood of the origin
0
∈
X
. As the set (2.4) is bounded, there is
s
>
0
such that
f
(
z
)
−
f
(
a
)
∈
t
∣
z
−
a
∣
V
,
z
∈
P
¯
(
a
,
ρ
)
\
{
a
}
,
t
≥
s
.
We recall that a subset
B
⊂
X
of a topological vector space is balanced if for any
α
∈
C
with
∣
α
∣
<
1
one has
α
B
⊂
B
. As every topological vector space has a balanced local base of neighborhoods of the origin (cf. [21], p. 13), it follows that
f
is (strongly) continuous at the point
a
. We established the following:
Theorem 2.1
Let
X
be a locally convex space and let
Ω
⊂
C
n
be an open set (
n
≥
2
). Any weakly holomorphic function
f
:
Ω
→
X
is strongly continuous.
Let us get back to the one complex variable case (
n
=
1
) and the statements (i)–(iii) above. Once statement (i) was proved, one may conclude (by Theorem 3.27 in [21], p. 78) that the integrals
∫
Γ
f
(
ζ
)
d
ζ
and
∫
Γ
(
ζ
−
z
)
f
(
ζ
)
d
ζ
are well-defined elements of
co
¯
[
f
(
Γ
)
]
(as Bochner integrals, cf. Definition 3.26 in [21], p. 77, and Appendix A with this article). Then formulas (2.1) and (2.3) follow from the ordinary Cauchy formula and Cauchy’s theorem applied to the holomorphic functions
Λ
∘
f
for every
Λ
∈
X
∗
. Similarly, when
n
≥
2
, we may state
Theorem 2.2
Let
X
be a locally convex space and let
Ω
⊂
C
n
be an open set (
n
≥
2
). Let
a
∈
Ω
and let
ρ
=
(
ρ
1
,
…
,
ρ
n
)
be a polyradius such that
P
¯
(
a
,
ρ
)
⊂
Ω
. Let
f
:
Ω
→
X
be a weakly holomorphic function such that the closed convex hull of
f
[
∂
0
P
(
a
,
ρ
)
]
is a compact subset of
X
. Then for every
z
∈
P
(
a
,
ρ
)
,
(2.5)
f
(
z
)
=
1
(
2
π
i
)
n
∫
∂
0
P
(
a
,
ρ
)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
(
ζ
1
,
…
,
ζ
n
)
d
ζ
1
…
d
ζ
n
.
Here,
∂
0
P
(
a
,
ρ
)
=
∏
j
=
1
n
S
1
(
a
j
,
ρ
j
)
is the essential boundary of
P
(
a
,
ρ
)
and
S
1
(
z
0
,
r
)
=
{
z
∈
C
:
∣
z
−
z
0
∣
=
r
}
with
z
0
∈
C
and
r
>
0
.
Proof
By Theorem 2.1, the function
f
:
Ω
→
X
is continuous. Then (by Theorem 3.27 in [21]) the integral
∫
∂
0
P
(
a
,
ρ
)
f
(
ζ
)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
d
ζ
∈
co
¯
[
f
(
∂
0
P
(
a
,
ρ
)
)
]
is well defined, and
Λ
∫
∂
0
P
(
a
,
ρ
)
f
(
ζ
)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
d
ζ
=
∫
∂
0
P
(
a
,
ρ
)
Λ
[
f
(
ζ
)
]
∏
j
=
1
n
(
ζ
j
−
z
j
)
d
ζ
for any
Λ
∈
X
∗
. Then (2.5) follows from the Cauchy formula for
Λ
∘
f
, and the fact that
X
∗
separates points.
Let us go back once again to case
n
=
1
. Statement (iii) above required the further specialization of the values
X
, i.e., that
X
is a complex Fréchet space. This guarantees that
co
¯
[
F
(
Γ
)
]
is a compact set, where
F
(
z
)
=
z
−
2
f
(
z
)
,
Γ
=
{
ζ
∈
C
:
∣
ζ
∣
=
2
r
}
. Consequently, both Cauchy’s theorem (2.1) and the Cauchy integral formula (2.2) hold for the holomorphic function
F
(and the proof in [21], p. 84, does apply).□
Let
X
be a complex Fréchet space and
f
:
Ω
→
X
a weakly holomorphic function. For each
a
∈
C
n
, we set
Ω
j
,
a
=
{
z
∈
C
:
(
a
1
,
…
,
a
j
−
1
,
z
,
a
j
+
1
,
…
,
a
n
)
∈
Ω
}
,
f
j
,
a
(
z
)
=
(
a
1
,
…
,
a
j
−
1
,
z
,
a
j
+
1
,
…
,
a
n
)
,
z
∈
Ω
j
,
a
.
Each
f
j
,
a
is weakly holomorphic in
Ω
j
,
a
hence (by statement (iii) above)
f
j
,
a
is strongly holomorphic in
Ω
j
,
a
. It is a natural question whether, under these conditions, the function
f
:
Ω
→
X
is holomorphic. Equivalently, the question is whether Hartogs’ theorem holds for
X
-valued functions
f
with
f
j
,
a
holomorphic. We shall establish the following:
Theorem 2.3
Let
X
be a complex Fréchet space,
Ω
⊂
C
n
an open set (
n
≥
2
), and
f
:
Ω
→
X
a weakly holomorphic function. Then f is strongly holomorphic.
To establish Theorem 2.3, we follow the arguments in the proof of the classical Hartogs’ theorem (cf. [19], pp. 43–44). The proof is however easier because weakly holomorphic functions are already continuous, while Hartogs’ theorem assumes only separate analyticity. To prove Theorem 2.3, it suffices to show
Theorem 2.4
Let
X
be a complex Fréchet space, and let us consider the polydisc
Ω
=
{
z
∈
C
n
:
∣
z
j
∣
<
R
,
1
≤
j
≤
n
}
,
R
>
0
. Let
f
:
Ω
→
X
be a weakly holomorphic function. Then there is
0
<
r
<
R
, and there is a power series
∑
∣
α
∣
≥
0
z
α
x
α
with
x
α
∈
X
, converging uniformly on
P
(
0
,
r
)
and such that
f
(
z
)
=
∑
∣
α
∣
=
0
∞
z
α
x
α
for any
z
∈
P
(
0
,
r
)
. Here,
r
=
(
r
,
…
,
r
)
.
Proof
Let
D
r
(
0
)
=
{
x
∈
C
:
∣
z
∣
<
r
}
. As
f
:
Ω
→
X
is continuous (cf. Theorem 2.1), it is Bochner integrable on the product of circles
T
ρ
=
∏
j
=
1
n
{
ζ
j
∈
C
:
∣
ζ
j
∣
=
ρ
}
,
0
<
ρ
<
R
,
i.e.,
f
∈
L
1
(
T
ρ
,
X
,
d
ζ
)
. Let
z
′
=
(
z
1
,
…
,
z
n
−
1
)
such that
∣
z
j
∣
<
R
for any
1
≤
j
≤
n
−
1
. Note that
D
ρ
(
0
)
⊂
Ω
n
,
z
′
. Also, as argued earlier,
f
n
,
z
′
is holomorphic in
Ω
n
,
z
′
, and in particular in
D
ρ
(
0
)
. Hence, (by statement (ii))
f
(
z
′
,
z
n
)
=
1
2
π
i
∫
∣
ζ
n
∣
=
ρ
(
ζ
n
−
z
n
)
−
1
f
(
ζ
′
,
z
n
)
d
ζ
n
,
∣
z
n
∣
<
ρ
.
For fixed
z
1
,
…
,
z
n
−
2
with
∣
z
j
∣
<
R
,
1
≤
j
≤
n
−
2
, and fixed
ζ
n
∈
D
ρ
(
0
)
, the function
f
(
z
1
,
…
,
z
n
−
1
,
ζ
n
)
is holomorphic in
∣
z
n
−
1
∣
<
ρ
, and hence, we may repeat the aforementioned procedure. In the end, for any
∣
z
j
∣
<
ρ
,
1
≤
j
≤
n
, one has
(2.6)
(
2
π
i
)
n
f
(
z
1
,
…
,
z
n
)
=
∫
∣
ζ
n
∣
=
ρ
d
ζ
n
∫
∣
ζ
n
−
1
∣
=
ρ
d
ζ
n
−
1
…
∫
∣
ζ
1
∣
=
ρ
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
(
ζ
1
,
…
,
ζ
n
)
d
ζ
1
.
Let
P
¯
(
0
,
r
)
=
{
z
∈
C
n
:
∣
z
j
∣
≤
r
,
1
≤
j
≤
n
}
, where
r
=
(
r
,
…
,
r
)
and
0
<
r
<
ρ
. Let
z
∈
P
¯
(
0
,
r
)
and
ζ
∈
T
ρ
. Then
(2.7)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
=
∑
∣
α
∣
=
0
∞
z
α
ζ
α
+
1
,
where
α
+
1
=
(
α
1
+
1
,
…
,
α
n
+
1
)
and the series in the right-hand of (2.7) converges uniformly for
ζ
∈
T
ρ
and
z
∈
P
¯
(
0
,
r
)
. Let
f
α
(
z
,
ζ
)
=
(
z
α
/
ζ
α
+
1
)
f
(
ζ
)
.□
Lemma 2.5
For any
z
∈
P
¯
(
0
,
r
)
, the series
∑
∣
α
∣
≥
0
f
α
(
z
,
ζ
)
is convergent in the topology of
X
, uniformly with respect to
ζ
∈
T
ρ
.
Proof
Let
B
be a balanced local base (of neighborhoods of the origin in
X
). As every
Λ
∘
f
(
Λ
∈
X
∗
) is continuous, the set
f
(
T
ρ
)
is weakly bounded and then bounded in
X
. Hence, for any
V
∈
B
, there is
s
>
0
such that
f
(
T
ρ
)
⊂
t
V
for any
t
≥
s
. Let us set
s
ν
(
z
,
ζ
)
=
∑
∣
α
∣
=
0
ν
z
α
ζ
α
+
1
,
S
ν
(
z
,
ζ
)
=
∑
∣
α
∣
=
0
ν
f
α
(
z
,
ζ
)
.
As
∑
∣
α
∣
≥
0
z
α
/
ζ
α
+
1
is convergent (uniformly for
ζ
∈
T
ρ
and
z
∈
P
¯
(
0
,
r
)
), there is
N
≥
1
such that
∣
R
ν
μ
(
z
,
ζ
)
∣
<
1
s
,
ν
≥
μ
>
N
,
ζ
∈
T
ρ
,
∣
z
j
∣
≤
r
,
where
R
ν
μ
=
s
ν
−
s
μ
. Next, as
V
is balanced
R
ν
μ
(
z
,
ζ
)
f
(
ζ
)
∈
R
ν
μ
(
z
,
ζ
)
f
(
T
ρ
)
⊂
R
ν
μ
(
z
,
ζ
)
s
V
⊂
V
,
ν
≥
μ
>
N
,
ζ
∈
T
ρ
,
z
∈
P
¯
(
0
,
r
)
.
Hence, for each
z
∈
P
¯
(
0
,
r
)
, the sequence
{
S
ν
(
z
,
ζ
)
}
ν
≥
0
is Cauchy in
X
uniformly for
ζ
∈
T
ρ
.□
By Lemma 5, we may integrate
∑
∣
α
∣
≥
0
f
α
(
z
,
ζ
)
term by term (integration with respect to
ζ
∈
T
ρ
) and obtain (by (2.6)):
f
(
z
)
=
∑
∣
α
∣
=
0
∞
z
α
x
α
,
z
∈
P
¯
(
0
,
r
)
,
0
<
r
≤
ρ
,
x
α
=
1
(
2
π
i
)
α
∫
∣
ζ
n
∣
=
ρ
d
ζ
n
∫
∣
ζ
n
−
1
=
ρ
d
ζ
n
−
1
…
∫
ζ
1
∣
=
ρ
(
1
/
ζ
α
+
1
f
(
ζ
)
d
ζ
1
∈
X
.
To end the proof of Theorem 2.4, we need to establish the following:
Lemma 2.6
The series
∑
∣
α
∣
≥
0
z
α
x
α
converges in
X
uniformly for
z
∈
P
¯
(
0
,
r
)
.
Proof
Let
P
be a separating family of seminorms, determining the topology of
X
as a locally convex space. Let us set
V
(
p
,
k
)
=
x
∈
X
:
p
(
x
)
<
1
k
,
p
∈
P
,
k
∈
Z
,
k
≥
1
.
The collection
B
of all finite intersections of the sets
V
(
p
,
k
)
is a convex balanced local base of
X
. Let
p
∈
P
and
k
∈
Z
,
k
≥
1
. As
f
(
T
ρ
)
is bounded in
X
, there is
s
>
0
such that
f
(
T
ρ
)
⊂
s
V
(
p
,
k
)
, i.e.,
p
[
f
(
ζ
)
]
<
s
⁄
k
for any
ζ
∈
T
ρ
. Then (by Theorem A1 in Appendix A to the present article),
p
(
x
α
)
≤
1
(
2
π
)
n
∫
∣
ζ
n
∣
=
ρ
d
ζ
n
∫
∣
ζ
n
−
1
∣
=
ρ
d
ζ
n
−
1
…
∫
∣
ζ
1
∣
=
ρ
p
[
f
(
ζ
)
]
∣
ζ
α
+
1
∣
d
ζ
1
≤
s
k
(
2
π
)
n
∫
T
ρ
d
ζ
ζ
α
+
1
=
s
k
ρ
−
∣
α
∣
,
and hence,
p
∑
∣
α
∣
=
μ
+
1
ν
z
α
x
α
≤
∑
∣
α
∣
=
μ
+
1
ν
∣
z
α
∣
p
(
x
α
)
≤
s
k
∑
∣
α
∣
=
μ
+
1
ν
∏
j
=
1
n
∣
z
j
∣
α
j
ρ
−
∣
α
∣
≤
s
k
∑
∣
α
∣
=
μ
+
1
ν
r
ρ
∣
α
∣
<
1
k
for some
N
≥
1
and any
ν
>
μ
≥
N
.□
3 Weierstrass theorem
Let
X
be a complex Fréchet space,
Ω
⊂
C
n
an open set, and
{
f
ν
}
ν
≥
0
⊂
O
(
Ω
,
X
)
a sequence of holomorphic functions. We establish the following vector-valued analog to Weierstrass’ theorem.
Theorem 3.1
If
{
f
ν
}
ν
≥
1
converges uniformly on any compact subset of
Ω
, then the pointwise limit
f
=
lim
ν
→
∞
f
ν
is holomorphic in
Ω
, i.e.,
f
∈
O
(
Ω
,
X
)
. Moreover, for every
α
∈
Z
+
n
, the sequence
{
D
α
f
ν
}
ν
≥
0
converges to
D
α
f
uniformly on compact subsets of
Ω
.
Let
a
∈
Ω
and let
r
=
(
r
1
,
…
,
r
n
)
be a polyradius such that
P
(
a
,
r
)
¯
⊂
Ω
. We need
Lemma 3.2
f
∈
C
(
P
(
a
,
r
)
¯
,
X
)
.
Proof
Any strongly holomorphic function is weakly holomorphic, hence (by Theorem 2.1) each
f
ν
is strongly continuous, and in particular,
f
ν
:
P
(
a
,
r
)
¯
→
X
is strongly continuous. We claim that
f
ν
:
P
(
a
,
r
)
¯
→
X
is also strongly bounded. Indeed, as any
Λ
∈
X
∗
is continuous, the (scalar valued) function
Λ
∘
f
ν
:
P
(
a
,
r
)
¯
→
C
is continuous on the compact set
P
(
a
,
r
)
¯
, and hence, the set
{
Λ
[
f
ν
(
z
)
]
:
z
∈
P
(
a
,
r
)
¯
}
is bounded, which is to say that the set
{
f
ν
(
z
)
:
z
∈
P
(
a
,
r
)
¯
}
is weakly bounded. Any weakly bounded subset of a locally convex space is strongly bounded, so the claim is proved. As
X
is a Fréchet space, its topology is compatible with some invariant metric. So
f
ν
:
P
(
a
,
r
)
¯
→
X
,
ν
≥
0
, is a sequence of continuous and bounded mappings of metric spaces, converging uniformly to
f
∣
P
(
a
,
r
)
¯
. Consequently,
f
∣
P
(
a
,
r
)
¯
is continuous.□
By the Cauchy integral formula,
(3.1)
2
π
i
f
ν
(
z
)
=
∫
∣
ζ
n
∣
=
r
n
d
ζ
n
∫
∣
ζ
n
−
1
∣
=
r
n
−
1
d
ζ
n
−
1
…
∫
∣
ζ
1
∣
=
r
1
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
ν
(
ζ
1
,
…
ζ
n
)
d
ζ
1
for any
z
∈
P
(
a
,
r
)
.
Lemma 3.3
For any
z
∈
P
(
a
,
r
)
,
lim
ν
→
∞
∫
∂
0
P
(
a
,
r
)
F
ν
(
ζ
)
d
ζ
=
∫
∂
0
P
(
a
,
r
)
F
(
ζ
)
d
ζ
,
where
F
ν
(
ζ
)
=
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
ν
(
ζ
)
,
F
(
ζ
)
=
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
(
ζ
)
.
Proof
If
X
and
Y
are metric spaces, let
C
(
X
,
Y
)
denote the metric space of all continuous and bounded functions
F
:
X
→
Y
with the distance function
d
∞
(
F
,
G
)
=
sup
x
∈
X
d
Y
(
F
(
x
)
,
G
(
x
)
)
. The uniform convergence
f
ν
→
f
on compact subsets of
Ω
yields convergence of
F
ν
to
F
in
C
(
∂
0
P
(
a
,
r
)
,
X
)
, i.e.,
d
∞
(
F
ν
,
F
)
→
0
as
ν
→
∞
. Let
P
=
{
p
m
:
m
≥
1
}
be a countable separating family of seminorms on
X
determining the topology of
X
, and let us set
d
X
(
x
,
y
)
=
∑
m
=
1
∞
1
2
m
p
m
(
x
−
y
)
1
+
p
m
(
x
−
y
)
,
x
,
y
∈
X
,
so that
d
X
is an invariant metric on
X
compatible to the topology of
X
. For any
ε
>
0
, there is
ν
ε
∈
Z
+
such that for any
ν
≥
ν
ε
ε
>
d
∞
(
F
ν
,
F
)
=
sup
z
∈
∂
0
P
(
a
,
r
)
d
X
(
F
ν
(
z
)
,
F
(
z
)
)
,
i.e., for every
z
∈
∂
0
P
(
a
,
r
)
,
ε
>
d
X
(
F
ν
(
z
)
,
F
(
z
)
)
=
∑
m
=
1
∞
1
2
m
p
m
(
F
ν
(
z
)
,
F
(
z
)
)
1
+
p
m
(
F
ν
(
z
)
,
F
(
z
)
)
=
sup
{
S
m
(
ν
,
z
)
:
m
≥
1
}
,
where
S
m
(
ν
,
z
)
≔
∑
k
=
1
m
2
−
k
p
k
(
F
ν
(
z
)
,
F
(
z
)
)
[
1
+
p
k
(
F
ν
(
z
)
,
F
(
z
)
)
]
−
1
, that is, for any
ν
≥
ν
ε
, any
z
∈
∂
0
P
(
a
,
r
)
, and any
m
∈
N
p
m
(
F
ν
(
z
)
−
F
(
z
)
)
<
2
m
ε
1
−
2
m
ε
yielding
p
m
∫
∂
0
P
(
a
,
r
)
F
ν
(
ζ
)
d
ζ
−
∫
∂
0
P
(
a
,
r
)
F
(
ζ
)
d
ζ
≤
∫
∂
0
P
(
a
,
r
)
p
m
[
F
ν
(
ζ
)
−
F
(
ζ
)
]
d
ζ
≤
2
m
ε
1
−
2
m
ε
∣
∂
0
P
(
a
,
r
)
∣
.
Here,
∣
A
∣
is the Lebesgue measure of
A
⊂
R
2
n
. Let
N
∈
N
and let
ε
(
m
,
N
)
≔
2
−
(
m
+
1
)
1
+
N
∣
∂
0
P
(
a
,
r
)
∣
,
ν
(
m
,
N
)
≔
ν
ε
(
m
,
N
)
,
so that
∫
∂
0
P
(
a
,
r
)
F
ν
(
ζ
)
d
ζ
−
∫
∂
0
P
(
a
,
r
)
F
(
ζ
)
d
ζ
∈
V
(
p
m
,
N
)
for any
m
,
N
∈
N
and any
ν
≥
ν
(
m
,
N
)
. However, as well known,
{
V
(
p
,
N
)
:
p
∈
P
,
N
∈
N
}
is not a local base for the topology of
X
. Instead, the family
B
of all finite intersections
B
=
V
(
p
m
1
,
N
1
)
∩
…
∩
V
(
p
m
k
,
N
k
)
is a local base. Let
ν
(
B
)
≔
max
{
ν
(
m
i
,
N
i
)
:
1
≤
i
≤
k
}
. Then
∫
∂
0
P
(
a
,
r
)
F
ν
(
ζ
)
d
ζ
−
∫
∂
0
P
(
a
,
r
)
F
(
ζ
)
d
ζ
∈
B
for any
ν
≥
ν
(
B
)
.□
Let us pass to the limit with
ν
→
∞
in (3.1). Then (by Lemma 3.3)
(3.2)
2
π
i
f
(
z
)
=
∫
∣
ζ
n
∣
=
r
n
d
ζ
n
∫
∣
ζ
n
−
1
∣
=
r
n
−
1
d
ζ
n
−
1
…
∫
∣
ζ
1
∣
=
r
1
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
f
(
ζ
1
,
…
ζ
n
)
d
ζ
1
,
and hence,
f
∈
O
(
P
(
a
,
r
)
,
X
)
. By (3.2), together with Lemma 8.6 [23], p. 29,
f
is holomorphic in each variable separately, and then (by Hartogs’ theorem together with our Theorem 2.3)
f
∈
O
(
P
(
a
,
r
)
,
X
)
for arbitrary
a
∈
Ω
, and hence,
f
∈
O
(
Ω
,
X
)
, thus proving the first statement in Theorem 3.1.
Let now
K
⊂
⊂
Ω
be a compact subset. There is a compact subset
A
⊂
Ω
such that
A
∘
⊃
K
. As
f
ν
→
f
for
ν
→
∞
uniformly on
A
, for any
p
∈
P
, and any
N
∈
N
, there is
ν
(
p
,
N
)
≥
0
such that for any
ν
≥
ν
(
p
,
N
)
, and any
z
∈
A
:
f
ν
(
z
)
−
f
(
z
)
∈
V
(
p
,
N
)
or
p
(
f
ν
(
z
)
−
f
(
z
)
)
<
1
⁄
N
, which yields
∥
f
ν
−
f
∥
p
,
A
<
1
N
,
ν
≥
ν
(
p
,
N
)
.
Here, for every
F
∈
C
(
Ω
,
X
)
, we adopted the notation:
∥
F
∥
p
,
K
=
sup
z
∈
K
p
(
F
(
z
)
)
.
We also write briefly
‖
⋅
‖
m
,
K
=
‖
⋅
‖
p
m
,
K
[so that
{
‖
⋅
‖
n
,
K
:
m
∈
N
,
K
⊂
⊂
Ω
}
is a family of seminorms on
C
(
Ω
,
X
)
organizing it as a complex Fréchet space]. Let
a
∈
K
and let
r
=
(
r
1
,
…
,
r
n
)
be a polyradius such that
P
(
a
,
r
)
¯
⊂
A
∘
. Then
f
ν
,
f
∈
O
(
P
(
a
,
r
)
,
X
)
∩
C
(
P
(
a
,
r
)
¯
,
X
)
.
Lemma 3.4
For every
F
∈
O
(
P
(
a
,
r
)
,
X
)
∩
C
(
P
(
a
,
r
)
¯
,
X
)
,
(3.3)
F
(
z
)
=
∑
α
∈
Z
+
(
z
−
a
)
α
α
!
(
D
α
F
)
(
a
)
,
z
∈
P
(
a
,
r
)
.
Moreover, for every
α
∈
Z
+
n
, one has
D
α
F
∈
O
(
P
(
a
,
r
)
,
X
)
and
(3.4)
p
m
[
D
α
F
)
(
a
)
]
≤
α
!
r
α
∥
F
∥
m
,
K
,
where
K
=
P
(
a
,
r
)
¯
.
Statement (3.3) for
n
=
1
is Lemma 8.6 in [23], p. 29 (an immediate consequence of the Cauchy formula for
X
valued holomorphic functions of one complex variable), while statement (3.4) for
n
=
1
is formula (8.3) in [23], p. 29. Similarly (3.3) and (3.4) follow from the Cauchy integral formula (2.5) in Theorem 2.2. The proof is left as an exercise to the reader.
By Lemma 3.4, one has
D
α
f
ν
,
D
α
f
∈
O
(
P
(
a
,
r
)
,
X
)
and
p
m
[
(
D
α
f
ν
)
(
a
)
−
(
D
α
f
)
(
a
)
]
≤
α
!
r
α
∥
f
ν
−
f
∥
m
,
P
(
a
,
r
)
¯
≤
α
!
r
α
∥
f
ν
−
f
∥
m
,
A
<
1
N
for any
ν
≥
ν
(
p
m
,
N
)
.
4 Cauchy problems versus integro-differential equations
Let
X
be a complex topological vector space. Let
Ω
⊂
R
n
be an open set with
0
∈
Ω
, and let
P
(
x
,
D
)
u
≡
∑
∣
α
∣
≤
m
a
α
(
x
)
D
α
u
be a PDO of order
m
, with coefficients
a
α
:
Ω
→
C
. Let
f
:
Ω
→
X
and
φ
j
:
Ω
0
→
X
,
0
≤
j
≤
m
−
1
, where
Ω
0
=
Ω
∩
{
x
n
=
0
}
. The Cauchy problem is to find a neighborhood of the origin
U
⊂
Ω
and a solution
u
:
U
→
X
to
P
(
x
,
D
)
u
=
f
in
U
such that
(4.1)
∂
j
u
∂
x
n
j
x
n
=
0
=
φ
j
on
U
0
=
U
∩
{
x
n
=
0
}
,
0
≤
j
≤
m
−
1
.
We seek to establish a vector-valued analog to the classical Cauchy-Kovalevskaja theorem, i.e.,
Theorem 4.1
Let
X
be a complex Fréchet space,
Ω
⊂
R
n
an open neighborhood of the origin, and let
f
∈
C
ω
(
Ω
,
X
)
and
φ
j
∈
C
ω
(
Ω
0
,
X
)
,
0
≤
j
≤
m
−
1
, be given functions. If
a
α
∈
C
ω
(
Ω
)
and
(4.2)
a
(
0
,
…
,
0
,
m
)
(
0
)
≠
0
,
then there is a neighborhood of the origin
0
∈
U
⊂
Ω
and a unique solution
u
∈
C
ω
(
Ω
,
X
)
to the Cauchy problem (4.1) for
P
(
x
,
D
)
u
=
f
.
As well as in the classical scalar valued case, it suffices to prove Theorem 4.1 for
φ
j
≡
0
,
0
≤
j
≤
m
−
1
, and
a
(
0
,
…
,
0
,
m
)
≡
1
. More important, the Cauchy problem (4.1) may be replaced with a formally similar problem in the holomorphic category. Indeed let
u
∈
C
ω
(
U
,
X
)
be a solution to
(4.3)
P
(
x
,
D
)
u
=
f
in
U
,
∂
j
u
∂
x
n
j
=
0
on
U
0
,
0
≤
j
≤
m
−
1
.
There is an open set
Ω
˜
⊂
C
n
such that
Ω
˜
R
=
Ω
, and there are holomorphic extensions to
Ω
˜
of the real analytic functions
a
α
and
f
, i.e., functions
a
˜
α
∈
O
(
Ω
˜
)
and
f
˜
∈
O
(
Ω
˜
,
X
)
such that
a
˜
α
∣
Ω
=
a
α
and
f
˜
∣
Ω
=
f
. Similarly, there is a connected open neighborhood of the origin
U
˜
⊂
Ω
˜
and a holomorphic function
u
˜
∈
O
(
U
˜
,
X
)
such that
U
˜
R
=
U
and
u
˜
∣
U
=
u
. Throughout, if
A
⊂
C
n
is a set, then
A
R
=
{
(
z
1
,
…
,
z
n
)
∈
A
:
z
j
∈
R
,
1
≤
j
≤
n
}
. Then
(4.4)
P
(
z
,
D
z
)
u
˜
=
f
˜
in
U
˜
,
∂
j
u
˜
∂
z
n
j
=
0
on
U
˜
0
,
0
≤
j
≤
m
−
1
,
where
U
˜
0
=
U
˜
∩
{
z
n
=
0
}
and
P
(
z
,
D
z
)
≡
∑
∣
α
∣
≤
m
a
˜
α
(
z
)
D
z
α
,
D
z
α
≡
∂
∣
α
∣
∂
z
1
α
1
…
∂
z
n
α
n
.
Viceversa, if
u
˜
∈
O
(
U
˜
,
X
)
is a solution to (4.4) for some open set
0
∈
U
˜
⊂
Ω
, then
u
=
u
˜
∣
U
with
U
≔
U
˜
R
is a solution to (4.3), where
a
α
=
a
˜
α
∣
U
and
f
=
f
˜
∣
U
.
Let
u
˜
∈
O
(
U
˜
,
X
)
be a solution to (4.4) with
a
˜
(
0
,
…
,
m
)
≡
1
. We may assume w.l.o.g. that the set
U
˜
is balanced in the
z
n
variable, i.e.,
(
z
′
,
t
z
n
)
∈
U
˜
for any
z
∈
U
˜
and any
∣
t
∣
≤
1
. Here,
z
′
=
(
z
1
,
…
,
z
n
−
1
)
so that
z
=
(
z
′
,
z
n
)
. Let us consider the
C
∞
function
ψ
:
[
−
1
,
1
]
→
X
given by
ψ
(
t
)
=
u
˜
(
z
′
,
t
z
n
)
,
∣
t
∣
≤
1
.
As
(
∂
j
u
˜
/
∂
z
n
j
)
(
z
′
,
0
)
=
0
for any
0
≤
j
≤
m
−
1
, the truncated Taylor development of
ψ
(with the rest in Bochner integral form, cf. Appendix B to this article) is expressed as follows:
ψ
(
t
)
=
∑
j
=
0
m
−
1
t
j
j
!
d
j
ψ
d
t
j
(
0
)
+
R
m
−
1
(
t
;
0
)
=
R
m
−
1
(
t
;
0
)
,
R
m
−
1
(
t
;
0
)
=
1
(
m
−
1
)
!
∫
0
t
(
t
−
τ
)
m
−
1
d
m
ψ
d
t
m
(
τ
)
d
τ
,
∣
t
∣
≤
1
.
Moreover,
ψ
(
1
)
=
u
˜
(
z
)
and
ψ
(
1
)
=
R
m
−
1
(
1
;
0
)
imply
(4.5)
u
˜
(
z
)
=
z
n
m
(
m
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
1
v
(
z
′
,
t
z
n
)
d
t
,
where we have set for simplicity
v
∈
O
(
U
˜
,
X
)
,
v
(
z
)
=
∂
m
u
˜
∂
z
n
m
(
z
)
,
z
∈
U
˜
.
It will be useful to observe that the functions
J
s
:
U
˜
→
X
,
s
≥
0
,
J
s
(
z
)
=
∫
0
1
(
1
−
t
)
m
−
1
v
(
z
′
,
t
z
n
)
,
z
∈
U
˜
,
satisfy the following recurrence identities
Lemma 4.2
For any
s
≥
1
,
(4.6)
z
n
∂
J
s
∂
z
n
(
z
)
=
s
J
s
−
1
(
z
)
−
(
1
+
s
)
J
s
(
z
)
.
Proof
For any fixed
z
∈
U
˜
, let us consider the function
φ
:
[
0
,
1
]
→
X
given by
φ
(
t
)
=
v
(
z
′
,
t
z
n
)
. Then
J
s
(
z
)
=
∫
0
1
(
1
−
t
)
s
φ
(
t
)
d
t
,
d
φ
d
t
(
t
)
=
z
n
∂
v
∂
z
n
(
z
′
,
t
z
n
)
,
so that
∂
J
s
∂
z
n
(
z
)
=
1
z
n
∫
0
1
t
(
1
−
t
)
s
d
φ
d
t
(
t
)
d
t
.
Finally, integration by parts [together with the elementary identity
d
d
t
{
t
(
1
−
t
s
)
}
=
(
1
−
t
)
s
−
1
−
(
1
+
s
)
t
(
1
−
t
)
s
−
1
] yields (4.6).□
Formula (4.5) reads
u
˜
(
z
)
=
z
n
m
(
m
−
1
)
!
J
m
−
1
(
z
)
,
z
∈
U
˜
,
hence (by Lemma 4.2)
∂
u
˜
∂
z
n
(
z
)
=
z
n
m
−
1
(
m
−
2
)
!
J
m
−
2
(
z
)
.
The iterative calculation of partials leads to
∂
j
u
˜
∂
z
n
j
(
z
)
=
z
n
m
−
j
(
m
−
j
−
1
)
!
J
m
−
j
−
1
(
z
)
,
0
≤
j
≤
m
−
1
,
or
(4.7)
∂
j
u
˜
∂
z
n
j
(
z
)
=
z
n
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
v
(
z
′
,
t
z
n
)
d
t
for any
0
≤
j
≤
m
−
1
. Next, let us consider the linear operator
T
:
O
(
U
˜
,
X
)
→
O
(
U
˜
,
X
)
,
T
g
≡
−
∑
j
=
0
m
−
1
∑
∣
β
∣
≤
m
−
j
a
˜
β
j
(
z
)
z
n
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
(
D
z
′
β
g
)
(
z
′
,
t
z
n
)
d
t
for any
g
∈
O
(
U
˜
,
X
)
, where
D
z
′
β
≡
∂
∣
β
∣
∂
z
1
β
1
…
∂
z
n
−
1
β
n
−
1
,
β
∈
Z
+
n
−
1
.
Then
f
˜
=
P
(
z
,
D
z
)
u
˜
=
∑
∣
α
∣
≤
m
a
˜
α
(
z
)
∂
∣
α
∣
u
˜
∂
z
1
α
1
…
∂
z
n
α
n
=
(by
a
˜
(
0
,
…
,
0
,
m
)
≡
1
)
=
∂
m
u
˜
∂
z
n
m
+
∑
j
=
0
m
−
1
∑
∣
β
∣
≤
m
−
j
a
˜
β
j
(
z
)
∂
∣
β
∣
+
j
u
˜
∂
z
1
β
1
…
∂
z
n
−
1
β
n
−
1
∂
z
n
j
=
v
+
∑
j
=
0
m
−
1
∑
∣
β
∣
≤
m
−
j
a
˜
β
j
(
z
)
∂
∣
β
∣
∂
z
1
β
1
…
∂
z
n
−
1
β
n
−
1
∂
j
u
˜
∂
z
n
j
=
(by (4.7))
=
v
+
∑
j
=
0
m
−
1
∑
∣
β
∣
≤
m
−
j
a
˜
β
j
(
z
)
z
n
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
(
D
z
′
β
v
)
(
z
′
,
t
z
n
)
d
t
=
v
−
T
v
.
Therefore, for every solution
v
∈
O
(
U
˜
,
X
)
to the equation
(4.8)
v
−
T
v
=
f
˜
,
the function
u
˜
defined by (4.5) is a solution to the Cauchy problem (4.4) with
a
˜
(
0
,
…
,
0
,
m
)
≡
1
.
5 Cauchy-Kovalevskaja theorem
We need the following:
Lemma 5.1
Let
m
∈
N
be a positive integer, and let
C
>
0
and
γ
∈
(
0
,
1
)
be constants. There is
ε
∈
(
0
,
1
)
such that
(5.1)
C
∣
z
n
∣
(
1
−
max
1
≤
j
≤
n
∣
z
j
∣
)
−
m
≤
γ
for any
z
∈
B
ε
(
0
)
¯
.
Proof
Note that
z
∈
B
ε
(
z
)
¯
⇒
max
1
≤
j
≤
n
∣
z
j
∣
<
ε
⇒
1
−
max
1
≤
j
≤
n
∣
z
j
∣
≥
1
−
ε
⇒
C
∣
z
n
∣
(
1
−
max
1
≤
j
≤
n
∣
z
j
∣
)
−
m
≤
C
ε
(
1
−
ε
)
−
m
.
Let us consider the monotonously increasing function
ϕ
:
(
0
,
1
)
→
(
0
,
+
∞
)
given by
ϕ
(
ε
)
=
C
ε
(
1
−
ε
)
−
m
. In particular
ϕ
is bijective, hence for any
γ
∈
(
0
,
1
)
, there is a unique
ε
γ
∈
(
0
,
1
)
such that
ϕ
(
ε
γ
)
=
γ
. Therefore, for every
z
∈
B
ε
(
0
)
¯
,
□
C
∣
z
n
∣
(
1
−
max
1
≤
j
≤
n
∣
z
j
∣
)
−
m
≤
C
ε
γ
(
1
−
ε
γ
)
−
m
=
ϕ
(
ε
γ
)
=
γ
.
Let
P
be a separating family of seminorms on
X
determining its topology as a locally convex space. By Theorem 1.37 in [21], pp. 27–28
Every
p
∈
P
is continuous,
A set
E
⊂
X
is bounded
⇔
every
p
∈
P
is bounded on
E
.
Lemma 5.2
Let
p
∈
P
and let
f
∈
O
(
B
n
,
X
)
such that
(5.2)
p
(
f
(
z
)
)
≤
B
(
1
−
max
1
≤
j
≤
n
∣
z
j
∣
)
b
for some constants
b
>
0
and
B
>
0
, and every
z
∈
B
n
. Then for every multi-index
α
∈
Z
+
n
and every point
z
∈
B
n
,
(5.3)
p
[
(
D
z
α
f
)
(
z
)
]
≤
B
e
∣
α
∣
(
b
+
1
)
…
(
b
+
∣
α
∣
)
(
1
−
max
1
≤
j
≤
n
∣
z
j
∣
)
b
+
∣
α
∣
.
Proof
It suffices to prove (5.3) for
∣
α
∣
=
1
and apply the resulting inequality to the derivatives of
f
. Also, we may assume w.l.o.g. that
α
=
(
0
,
…
,
0
,
1
)
∈
Z
+
n
. For every
z
∈
B
n
, there is a polyradius
ρ
=
(
ρ
1
,
…
,
ρ
n
)
such that
∑
j
=
1
n
ρ
j
2
<
1
,
z
∈
P
(
0
,
ρ
)
,
P
(
0
,
ρ
)
¯
⊂
B
n
.
Let us set
ε
j
≔
ρ
j
−
∣
z
j
∣
so that
ε
≔
(
ε
1
,
…
,
ε
n
)
lies on the cube
I
n
, where
I
=
(
0
,
1
)
. Therefore,
P
(
z
,
ε
)
⊂
P
(
0
,
ρ
)
. Next, let us set
m
z
≔
max
1
≤
j
≤
n
∣
z
j
∣
,
ε
≔
ε
n
=
1
−
m
z
1
+
b
,
ρ
=
ρ
n
=
∣
z
n
∣
+
1
−
m
z
1
+
b
,
so that
∑
j
=
1
n
−
1
ρ
j
2
=
∑
j
=
1
n
−
1
(
∣
z
j
∣
+
ε
j
)
2
<
1
−
∣
z
n
∣
+
1
−
m
z
1
+
b
2
,
and moreover,
z
∈
D
ρ
(
0
)
and
D
ε
(
z
n
)
¯
⊂
D
ρ
(
0
)
¯
⊂
D
1
(
0
)
. Here,
D
r
(
ζ
0
)
=
{
ζ
∈
C
:
∣
ζ
−
ζ
0
∣
<
r
}
. Let us consider the planar domain
D
ρ
,
ε
=
{
ζ
∈
C
:
∣
ζ
∣
<
ρ
,
∣
ζ
−
z
n
∣
>
ε
}
.
Then
D
ρ
,
ε
¯
⊂
D
1
(
0
)
and
∂
D
ρ
,
ε
=
{
ζ
∈
C
:
∣
ζ
∣
=
ρ
}
∪
{
ζ
∈
C
:
∣
ζ
−
z
n
∣
=
ε
}
=
∂
D
ρ
(
0
)
∪
∂
D
ε
(
z
n
)
⊂
∂
D
1
(
0
)
.
Let us consider the function
f
z
′
:
D
ρ
(
0
)
¯
→
X
given by
f
(
ζ
)
=
f
(
z
′
,
ζ
)
. Then
f
z
′
∈
C
(
D
ρ
(
0
)
¯
,
X
)
∩
O
(
D
ρ
(
0
)
,
X
)
hence (by the Cauchy formula)
f
(
z
′
,
ζ
)
=
1
2
π
i
∫
∂
D
ρ
(
0
)
1
ζ
−
z
n
f
(
z
′
,
ζ
)
d
ζ
,
and then
(5.4)
∂
f
∂
z
n
(
z
)
=
1
2
π
i
∫
∂
D
ρ
(
0
)
1
(
ζ
−
z
n
)
2
f
(
z
′
,
ζ
)
d
ζ
.
Let us set
ω
=
1
(
ζ
−
z
n
)
2
f
(
z
′
,
ζ
)
d
ζ
.
Then
ω
∈
C
∞
(
T
∗
(
M
)
⊗
X
)
, i.e.,
ω
is a
X
valued differential 1-form on
M
=
D
1
(
0
)
\
{
z
n
}
. As
ζ
↦
1
(
ζ
−
z
n
)
2
f
(
z
′
,
ζ
)
is holomorphic, the differential 1-form
ω
is closed, i.e.,
d
ω
=
0
. Consequently (by the Stokes theorem),
0
=
∫
D
ρ
,
ε
d
ω
=
∫
∂
D
ρ
,
ε
ω
⇒
∫
D
1
(
ρ
)
ω
=
∫
D
ε
(
z
n
)
ω
and (5.4) becomes
∂
f
∂
z
n
(
z
)
=
1
2
π
i
∫
D
ε
(
z
n
)
1
(
ζ
−
z
n
)
2
f
(
ζ
′
,
z
n
)
d
ζ
or (with
ζ
=
z
n
+
ε
e
i
θ
)
∂
f
∂
z
n
(
z
)
=
1
2
π
ε
∫
0
2
π
e
−
i
θ
f
(
z
′
,
z
n
+
ε
e
i
θ
)
d
θ
.
Let us set
z
ε
(
θ
)
=
(
z
′
,
z
n
+
ε
e
i
θ
)
for the sake of brevity. Then
(5.5)
m
z
ε
(
θ
)
≤
m
z
+
ε
and (by Theorem A1 in Appendix A, the assumption (5.2) in the current lemma, and (5.5))
p
∂
f
∂
z
n
(
z
)
≤
1
2
π
ε
∫
0
2
π
p
[
f
(
z
ε
(
θ
)
)
]
d
θ
≤
1
2
π
ε
∫
0
2
π
B
d
θ
[
1
−
m
z
ε
(
θ
)
]
b
≤
B
ε
(
1
−
m
z
−
ε
)
b
.
On the other hand,
(
1
−
m
z
−
ε
)
b
=
1
−
m
z
−
1
−
m
z
1
+
b
b
=
(
1
−
m
z
)
b
1
−
1
1
+
b
b
≥
(
1
−
m
z
)
b
1
−
1
b
b
≥
1
e
(
1
−
m
z
)
b
,
hence,
□
p
∂
f
∂
z
n
(
z
)
≤
B
e
(
1
+
b
)
(
1
−
m
z
)
b
+
1
.
Let
U
⊂
C
n
be an open set such that
B
n
¯
⊂
U
, and let
F
,
v
0
∈
O
(
U
,
X
)
. Next, let
v
ν
,
w
ν
∈
O
(
U
,
X
)
,
ν
≥
1
, be defined recurrently by
v
ν
+
1
=
T
v
ν
+
F
,
w
ν
=
v
ν
+
1
−
v
ν
,
ν
≥
0
.
Lemma 5.3
For any seminorm
p
∈
P
, nonnegative integer
ν
∈
Z
+
, and point
z
∈
B
n
(5.6)
p
[
w
ν
(
z
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∣
z
n
∣
ν
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
−
m
ν
,
M
(
p
)
=
sup
z
∈
B
n
¯
p
[
w
0
(
z
)
]
,
C
(
m
)
=
b
m
∑
j
=
1
m
m
j
,
a
=
max
0
≤
j
≤
m
−
1
max
β
∈
B
j
sup
z
∈
B
n
¯
∣
a
˜
β
,
j
(
z
)
∣
,
b
m
=
max
0
≤
j
≤
m
−
1
∣
B
j
∣
=
∣
B
0
∣
,
B
j
≔
{
β
∈
Z
+
n
−
1
:
∣
β
∣
≤
m
−
j
}
.
Proof
The proof of Lemma 5.3 is by induction over
ν
∈
Z
+
. Let
P
(
p
,
ν
,
z
)
be the predicate in (5.6). Let
p
∈
P
. Then
p
[
w
0
(
z
)
]
≤
M
(
p
)
is equivalent to
P
(
p
,
0
,
z
)
, so that
P
(
p
,
0
,
z
)
is true for any
z
∈
B
n
. Let
ν
∈
Z
+
,
ν
≥
1
, such that
P
(
p
,
ν
,
z
)
is true for any
p
∈
P
and any
z
∈
B
n
. Note that, independently from the induction hypothesis
w
ν
+
1
=
v
ν
+
2
−
v
ν
+
1
=
T
v
ν
+
1
+
F
−
(
T
v
ν
+
F
)
=
T
(
v
ν
+
1
−
v
ν
)
=
T
w
ν
,
hence for every
z
∈
B
n
(by the very definition of
T
)
w
ν
+
1
(
z
)
=
−
∑
j
=
0
m
−
1
∑
β
∈
Z
+
n
−
1
∣
β
∣
≤
m
−
j
a
˜
β
,
j
(
z
)
z
n
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
(
D
z
′
β
w
ν
)
(
z
′
,
t
z
n
)
d
t
.
Consequently,
(5.7)
p
[
w
ν
+
1
(
z
)
]
≤
∑
j
=
0
m
−
1
∑
β
∈
Z
+
n
−
1
∣
β
∣
≤
m
−
j
∣
a
˜
β
,
j
(
z
)
∣
∣
z
n
∣
m
−
j
(
m
−
j
−
1
)
!
∫
0
1
(
1
−
t
)
m
−
j
−
1
p
[
(
D
z
′
β
w
ν
)
(
z
′
,
t
z
n
)
]
d
t
.
For any
z
∈
B
n
and any
0
<
t
<
1
,
∥
(
z
′
,
t
z
n
)
∥
2
=
‖
z
′
‖
2
+
t
2
∣
z
n
∣
2
<
‖
z
′
‖
2
+
∣
z
n
∣
2
=
‖
z
‖
2
<
1
,
hence
φ
(
z
′
)
≔
w
ν
(
z
′
,
t
z
n
)
is well defined and (by the induction hypothesis)
P
(
p
,
ν
,
(
z
′
,
t
z
n
)
)
is true, i.e.,
(5.8)
p
[
φ
(
z
′
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
t
ν
∣
z
n
∣
ν
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
is true. The function
D
z
′
β
w
ν
is holomorphic in a neighborhood
U
of
B
n
¯
, so in particular
D
z
′
β
w
ν
∈
O
(
B
n
,
X
)
for every
β
∈
Z
+
n
−
1
. At this point, it should be observed that (5.8) is equivalent to the hypothesis (5.2) in Lemma 5.2, provided the data in Lemma 5.2 is replaced as follows:
B
n
f
B
b
⟼
B
n
−
1
φ
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
t
ν
∣
z
n
∣
ν
m
ν
.
Therefore, we may apply Lemma 5.2, so that to conclude that (5.3) holds with the new data, i.e.,
(5.9)
p
[
(
D
z
′
β
w
ν
)
(
z
′
,
t
z
n
)
]
=
p
[
(
D
β
φ
)
(
z
′
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
t
ν
∣
z
n
∣
ν
e
∣
β
∣
(
m
ν
+
1
)
…
(
m
ν
+
∣
β
∣
)
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
+
∣
β
∣
.
Thus, (by (5.7) and (5.9))
p
[
w
ν
+
1
(
z
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∑
j
=
0
m
−
1
∑
β
∈
Z
+
n
−
1
∣
β
∣
≤
m
−
j
∣
a
˜
β
,
j
(
z
)
∣
×
∣
z
n
∣
m
−
j
+
ν
e
∣
β
∣
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
+
∣
β
∣
(
m
ν
+
1
)
…
(
m
ν
+
∣
β
∣
)
(
m
−
j
−
1
)
!
∫
0
1
t
ν
(
1
−
t
)
m
−
j
−
1
d
t
=
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∑
j
=
0
m
−
1
∑
β
∈
Z
+
n
−
1
∣
β
∣
≤
m
−
j
∣
a
˜
β
,
j
(
z
)
∣
×
∣
z
n
∣
m
−
j
+
ν
e
∣
β
∣
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
+
∣
β
∣
(
m
ν
+
1
)
…
(
m
ν
+
∣
β
∣
)
(
m
−
j
−
1
)
!
B
(
ν
+
1
,
m
−
j
)
,
where
B
(
r
,
s
)
=
∫
0
1
t
r
−
1
(
1
−
t
)
s
−
1
d
t
is the Euler function of the first kind, hence
B
(
ν
+
1
,
m
−
j
)
=
Γ
(
ν
+
1
)
Γ
(
m
−
j
)
Γ
(
ν
+
m
−
j
+
1
)
=
ν
!
(
m
−
j
−
1
)
!
(
ν
+
m
−
j
)
!
.
Also
∣
z
n
∣
ν
+
m
−
j
≤
∣
z
n
∣
ν
+
1
,
e
∣
β
∣
≤
e
m
,
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
+
∣
β
∣
≥
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
m
ν
+
m
,
so that
(5.10)
p
[
w
ν
+
1
(
z
)
]
≤
a
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∣
z
n
∣
ν
+
1
e
m
ξ
ν
(
1
−
max
1
≤
k
≤
m
−
1
∣
z
k
∣
)
m
(
ν
+
1
)
,
ξ
ν
≔
∑
j
=
0
m
−
1
∑
β
∈
B
j
ν
!
(
m
ν
+
1
)
…
(
m
ν
+
∣
β
∣
)
(
ν
+
m
−
j
)
!
,
ν
≥
1
.
We claim that the sequence
{
ξ
ν
}
ν
≥
1
⊂
(
0
,
+
∞
)
is bounded by the constant
C
(
m
)
. Indeed
0
<
ξ
ν
≤
∑
j
=
0
m
−
1
∣
B
j
∣
ν
!
(
m
ν
+
1
)
…
(
m
ν
+
m
−
j
)
(
ν
+
m
−
j
)
!
≤
b
m
ν
!
∑
j
=
0
m
−
1
(
m
ν
+
1
)
…
(
m
ν
+
m
−
j
)
(
ν
+
m
−
j
)
!
=
b
m
m
ν
+
1
ν
+
1
+
m
ν
+
1
ν
+
1
m
ν
+
2
ν
+
2
+
…
+
m
ν
+
1
ν
+
1
m
ν
+
2
ν
+
2
…
m
ν
+
m
ν
+
m
.
Let us consider the monotonously increasing functions
f
j
:
[
1
,
+
∞
)
→
R
given by
f
j
(
x
)
=
(
m
x
+
j
)
⁄
(
x
+
j
)
for
x
≥
1
and
1
≤
j
≤
m
, and let us note that
lim
x
→
+
∞
f
j
(
x
)
=
m
. Finally,
□
0
<
ξ
ν
≤
b
m
∑
j
=
1
m
m
j
=
C
(
m
)
.
Then (5.10) implies
P
(
p
,
ν
+
1
,
z
)
, and we are done.
Let us recall that Lemma 5.1 builds a number
0
<
ε
<
1
starting with the data
(
m
,
C
,
γ
)
, where
m
∈
N
,
C
>
0
, and
0
<
γ
<
1
. An inspection of the proof of Lemma 5.1 shows that its output
ε
depends on the constant
C
. Let us apply Lemma 5.1 with the input data
(
m
,
a
e
m
C
(
m
)
,
γ
)
, where
m
is the order of the given PDO and
0
<
γ
<
1
is arbitrary. By Lemma 5.1, there is
ε
=
ε
γ
(
m
)
∈
(
0
,
1
)
such that
(5.11)
a
e
m
C
(
m
)
∣
z
n
∣
(
1
−
max
1
≤
k
≤
n
∣
z
k
∣
)
m
≤
γ
for any
z
∈
B
ε
(
0
)
¯
. By Lemma 5.3, for any nonnegative integer
ν
∈
Z
+
and any
z
∈
B
n
p
[
w
ν
(
z
)
]
≤
M
(
p
)
[
a
e
m
C
(
m
)
]
ν
∣
z
n
∣
ν
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
−
m
ν
,
hence (by
(
1
−
max
1
≤
k
≤
n
−
1
∣
z
k
∣
)
−
1
≤
(
1
−
max
1
≤
k
≤
n
∣
z
k
∣
)
−
1
together with (5.11))
p
[
w
ν
(
z
)
]
≤
M
(
p
)
γ
ν
for any
z
∈
B
ε
(
0
)
¯
. As
∑
ν
=
0
∞
γ
ν
<
∞
the series
∑
ν
≥
0
w
ν
converges uniformly on
B
ε
(
0
)
¯
(cf. also Appendix C to this article). Let us set
w
≔
∑
ν
=
0
∞
w
ν
,
s
ν
≔
∑
κ
=
0
ν
w
κ
,
so that
s
ν
=
∑
κ
=
0
ν
(
v
κ
+
1
−
v
κ
)
=
v
ν
+
1
−
v
0
. Thus, for any
p
∈
P
and any
N
≥
1
, there is
ν
(
p
,
N
)
∈
Z
+
such that for any
ν
≥
ν
(
p
,
N
)
and any
z
∈
B
ε
(
0
)
¯
,
v
ν
+
1
(
z
)
−
(
v
0
+
w
)
(
z
)
=
s
ν
(
z
)
−
w
(
z
)
∈
V
(
p
,
N
)
,
i.e., the sequence
{
v
ν
}
ν
≥
0
converges to
v
≔
v
0
+
w
uniformly on
B
ε
(
0
)
¯
. In particular,
v
∈
C
(
B
ε
(
0
)
¯
,
X
)
. Let
ε
≔
(
ε
,
…
,
ε
)
∈
Z
+
n
. Then
P
(
0
,
ε
)
¯
⊂
U
. Therefore,
v
ν
∈
O
(
P
(
0
,
ε
)
,
X
)
∩
C
(
P
(
0
,
ε
)
¯
,
X
)
,
ν
∈
Z
+
,
hence (by the Cauchy integral formula),
v
ν
(
z
)
=
1
(
2
π
i
)
n
∫
∂
0
P
(
0
,
ε
)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
v
ν
(
ζ
1
,
…
,
ζ
n
)
d
ζ
1
…
d
ζ
n
for any
z
∈
P
(
0
,
ε
)
. Passing to the limit with
ν
→
∞
yields
(5.12)
v
(
z
)
=
1
(
2
π
i
)
n
∫
∂
0
P
(
0
,
ε
)
∏
j
=
1
n
(
ζ
j
−
z
j
)
−
1
v
(
ζ
1
,
…
,
ζ
n
)
d
ζ
1
…
d
ζ
n
for any
z
∈
B
ε
(
0
)
¯
. On the other hand, the right-hand side of (5.12) is a holomorphic function in
B
ε
(
0
)
⊂
P
(
0
,
ε
)
so that
v
∈
O
(
B
ε
(
0
)
)
. By (the vector-valued analog to) Weierstrass’ theorem (i.e., Theorem 3.1) for every
α
∈
Z
+
n
, the sequence
{
D
α
v
ν
}
ν
≥
0
converges to
D
α
v
∈
O
(
B
ε
(
0
)
,
X
)
, uniformly on any compact subset of
B
ε
(
0
)
. Hence, (by the very definition of
T
),
lim
ν
→
∞
T
v
ν
=
T
v
on any compact subset of of
B
ε
(
0
)
. Let
δ
>
0
such that
B
δ
(
0
)
¯
⊂
B
ε
(
0
)
. Then
v
=
lim
ν
→
∞
v
ν
+
1
=
lim
ν
→
∞
(
T
v
ν
+
F
)
=
T
v
+
F
on
B
δ
(
0
)
, that is
v
∈
O
(
B
δ
(
0
)
,
X
)
is the unique solution to the integro-differential equation (4.8) on
B
δ
(
0
)
.
Acknowledgements
The article was written while the Authors were members of G.N.S.A.G.A. of INdAM, Rome, Italy.
Appendix A Vector-valued integration
Let
X
be a topological vector space such that
X
∗
separates points on
X
. Let
Q
be a compact Hausdorff space and let
μ
be a Borel probability measure on
Q
. If
f
:
Q
→
X
is a continuous function and
co
¯
[
f
(
Q
)
]
is compact in
X
then there is a unique
y
∈
co
¯
[
f
(
Q
)
]
such that
Λ
y
=
∫
Q
(
Λ
∘
f
)
d
μ
,
Λ
∈
X
∗
.
∫
Q
f
d
μ
≔
y
is the Bochner integral of
f
.
Theorem A1
Let
X
be a Fréchet space, and let
P
be a separating family of semi-norms on
X
determining the topology of
X
as a locally convex space. Let
μ
be a positive Borel measure on the compact Hausdorff space Q. Then for any continuous function
f
:
Q
→
X
,
p
∫
Q
f
d
μ
≤
∫
Q
(
p
∘
f
)
d
μ
,
p
∈
P
.
Theorem A1 is proved in [20] only when
X
is a Banach space (cf. Theorem 3.29 in [20], p. 81). We provide a proof holding for every Fréchet space
X
. We make use of the following corollary to Hahn-Banach theorem (cf. Theorem 3.3 in [20], pp. 58–59).
Corollary A2
Let
X
be a locally convex space and let
x
0
∈
X
. Let
P
be a separating family of seminorms determining the topology of
X
. For every
p
∈
P
there is
Λ
∈
X
∗
such that
Λ
x
0
=
p
(
x
0
)
and
∣
Λ
x
∣
≤
p
(
x
)
for any
x
∈
X
.
Once again, the proof in [20], p. 59, is confined to the case where
X
is a normed vector space.
Proof of Corollary A2
A function
Λ
:
X
→
C
as in thesis of Corollary A2 may be built as follows. If
x
0
=
0
, we set
Λ
=
0
. If
x
0
≠
0
let
Y
be the 1-dimensional space spanned by
x
0
, i.e.,
Y
=
C
x
0
. For a fixed seminorm
p
∈
P
, let us consider
f
:
Y
→
C
,
f
(
λ
x
0
)
=
λ
p
(
x
0
)
,
λ
∈
C
.
One checks easily that
f
:
Y
→
C
is linear. Moreover, for every
x
=
λ
x
0
∈
Y
,
∣
f
(
x
)
∣
=
∣
λ
p
(
x
0
)
∣
=
∣
λ
∣
p
(
x
0
)
=
p
(
λ
x
0
)
=
p
(
x
)
.
At this point, one may apply the Hahn-Banach theorem to the data
Y
=
C
x
0
and
f
:
Y
→
C
and
p
to produce a linear functional
Λ
:
X
→
C
extending
f
such that
∣
Λ
x
∣
≤
p
(
x
)
for any
x
∈
X
. As
p
:
X
→
R
is continuous, it follows that
Λ
is continuous at the origin, and hence on
X
. Finally,
x
0
∈
Y
and hence
Λ
(
x
0
)
=
f
(
x
0
)
=
p
(
x
0
)
.
Let us apply Corollary A2 for
x
0
=
y
=
∫
Q
f
d
μ
and
p
∈
P
. We may then consider a linear and continuous functional
Λ
∈
X
∗
such that
Λ
(
y
)
=
p
(
y
)
and
∣
Λ
x
∣
≤
p
(
x
)
for any
x
∈
X
. In particular, for
x
∈
f
(
Q
)
,
∣
Λ
f
(
s
)
∣
≤
p
(
f
(
s
)
)
,
s
∈
Q
.
Finally,
□
p
∫
Q
f
d
μ
=
p
(
y
)
=
Λ
(
y
)
=
∫
Q
(
Λ
∘
f
)
d
μ
=
∫
Q
(
Λ
∘
f
)
d
μ
≤
∫
Q
∣
Λ
∘
f
∣
d
μ
≤
∫
Q
(
p
∘
f
)
d
μ
.
B Taylor’s formula with Bochner integral reminder
Let
X
be a locally convex space over
R
, on which
X
∗
separates points, and let
f
∈
C
n
+
2
(
U
,
X
)
where
U
⊂
R
is an open set. Let
t
0
∈
U
and let us set
(A1)
P
n
(
t
;
f
,
t
0
)
=
∑
k
=
0
n
1
k
!
(
t
−
t
0
)
k
f
(
k
)
(
t
0
)
,
(A2)
R
n
(
t
;
f
,
t
0
)
=
f
(
t
)
−
P
n
(
t
;
f
,
t
0
)
.
Lemma A3
The reminder is of order
o
(
∣
t
−
t
0
∣
n
)
, i.e.,
lim
t
→
t
0
1
(
t
−
t
0
)
n
R
n
(
t
;
f
,
t
0
)
=
0
.
The reminder admits the integral representation formula
R
n
(
t
;
f
,
t
0
)
=
1
n
!
∫
t
0
t
(
t
−
s
)
n
f
(
n
+
1
)
(
s
)
d
s
.
Proof
(i) Linear and continuous functionals
Λ
∈
X
∗
commute with derivatives of any order. Hence,
(A3)
Λ
[
P
n
(
t
;
f
,
t
0
)
]
=
P
n
(
t
;
Λ
(
f
)
,
t
0
)
,
where
Λ
(
f
)
≔
Λ
∘
f
. As
Λ
(
f
)
is a scalar valued function of class
C
n
+
1
, for any
t
∈
U
, there is
α
=
α
(
t
,
t
0
,
Λ
(
f
)
)
∈
[
0
,
1
]
such that
(A4)
R
n
(
t
;
Λ
(
f
)
,
t
0
)
=
(
t
−
t
0
)
n
+
1
(
n
+
1
)
!
d
n
+
1
Λ
(
f
)
d
t
n
+
1
(
ξ
)
,
where
ξ
=
(
1
−
α
)
t
0
+
α
t
. Consequently, as
Λ
(
f
)
is of class
C
n
+
2
the representation formula (A4) implies
(A5)
lim
t
→
t
0
R
n
(
t
;
Λ
(
f
)
,
t
0
)
(
t
−
t
0
)
n
+
1
=
1
(
n
+
1
)
!
d
n
+
1
Λ
(
f
)
d
t
n
+
1
(
t
0
)
.
On the other hand (by (A3)),
R
n
(
t
;
Λ
(
f
)
,
t
0
)
(
t
−
t
0
)
n
+
1
=
1
(
t
−
t
0
)
n
+
1
[
Λ
(
f
(
t
)
)
−
P
n
(
t
;
Λ
(
f
)
,
t
0
)
]
=
Λ
1
(
t
−
t
0
)
n
+
1
R
n
(
t
;
f
,
t
0
)
,
hence (by (A5)) for any sequence
{
t
ν
}
ν
≥
1
⊂
U
with
lim
ν
→
∞
t
ν
=
t
0
the limit
lim
ν
→
∞
Λ
1
(
t
ν
−
t
0
)
n
+
1
R
n
(
t
ν
;
f
,
t
0
)
exists and is finite. Consequently,
Λ
1
(
t
ν
−
t
0
)
n
+
1
R
n
(
t
ν
;
f
,
t
0
)
:
ν
≥
1
⊂
R
is a bounded set for every
Λ
∈
X
∗
, i.e., the set
E
=
1
(
t
ν
−
t
0
)
n
+
1
R
n
(
t
ν
;
f
,
t
0
)
:
ν
≥
1
⊂
X
is weakly bounded. From now on, we assume that
X
is a locally convex space. By Theorem 3.18 in [20], p. 70, every weakly bounded set in
X
is also strongly bounded. Hence, for any neighborhood
V
of
0
∈
X
, there is
s
V
>
0
such that
E
⊂
s
V
for any
s
>
s
V
, i.e.,
1
(
t
ν
−
t
0
)
n
R
n
(
t
ν
;
f
,
t
0
)
∈
(
t
ν
−
t
0
)
s
V
for any
ν
≥
1
. Consequently, the sequence
1
(
t
ν
−
t
0
)
n
R
n
(
t
ν
;
f
,
t
0
)
ν
≥
1
is strongly convergent to 0 as
ν
→
∞
.
(ii) Let
X
be a topological vector space on which
X
∗
separates points. By a classical representation formula
R
n
(
t
;
Λ
(
f
)
,
t
0
)
=
1
n
!
∫
t
0
t
(
t
−
s
)
n
d
n
+
1
(
Λ
∘
f
)
d
t
n
+
1
(
s
)
d
s
for any
t
∈
U
. Then
Λ
R
n
(
t
;
f
,
t
0
)
−
1
n
!
∫
t
0
t
(
t
−
s
)
n
f
(
n
+
1
)
(
s
)
d
s
=
0
for any
Λ
∈
X
∗
, where
∫
t
0
t
(
t
−
s
)
n
f
(
n
+
1
)
(
s
)
d
s
is a Bochner integral. As
X
∗
separates points, we may conclude that
(A6)
R
n
(
t
;
f
,
t
0
)
=
1
n
!
∫
t
0
t
(
t
−
s
)
n
f
(
n
+
1
)
(
s
)
d
s
for any
t
∈
U
.□
Lemma A4
Let
X
be a locally convex space on which
X
∗
separates points. Let
U
⊂
R
be an open neighborhood of
t
0
∈
R
and let
F
∈
C
n
+
1
(
U
,
X
)
such that
F
(
t
0
)
=
F
′
(
t
0
)
=
…
=
F
(
n
)
(
t
0
)
=
F
(
n
+
1
)
(
t
0
)
=
0
. Then
lim
t
→
t
0
1
(
t
−
t
0
)
n
F
(
t
)
=
0
.
Proof
Λ
(
F
)
∈
C
n
+
1
(
U
,
R
)
for every
Λ
∈
X
′
and (by applying repeatedly the classical l’Hôspital theorem),
lim
t
→
t
0
Λ
(
F
(
t
)
)
(
t
−
t
0
)
n
+
1
=
lim
t
→
t
0
d
d
t
[
Λ
(
F
(
t
)
)
]
(
n
+
1
)
(
t
−
t
0
)
n
=
…
=
lim
t
→
t
0
d
n
d
t
n
[
Λ
(
F
(
t
)
)
]
(
n
+
1
)
!
(
t
−
t
0
)
=
lim
t
→
t
0
d
n
+
1
d
t
n
+
1
[
Λ
(
F
(
t
)
)
]
(
n
+
1
)
!
=
1
(
n
+
1
)
!
Λ
[
F
(
n
+
1
)
(
t
0
)
]
=
0
,
hence
lim
ν
→
∞
Λ
(
F
(
t
ν
)
)
(
t
ν
−
t
0
)
n
+
1
=
0
for any sequence
{
t
ν
}
ν
≥
1
⊂
U
such that
lim
ν
→
∞
t
ν
=
t
0
. Consequently, the set
1
(
t
ν
−
t
0
)
n
+
1
F
(
t
ν
)
:
ν
≥
1
⊂
X
is weakly bounded, and then strongly bounded, in
X
. Then for any neighborhood
V
⊂
X
of 0, there is
s
>
0
such that
1
(
t
ν
−
t
0
)
n
F
(
t
ν
)
∈
(
t
−
t
0
)
s
V
,
ν
≥
1
,
yielding
lim
ν
→
∞
1
(
t
ν
−
t
0
)
n
F
(
t
ν
)
=
0
strongly in
X
.□
Let
X
be a Fréchet space. Let
A
⊂
R
n
be an open set and let
f
∈
C
k
(
A
,
X
)
and
x
0
∈
A
. There is
R
>
0
such that
B
¯
R
(
x
0
)
⊂
A
. Let
w
∈
R
n
such that
‖
w
‖
=
1
, and let us consider the function
F
:
(
−
R
,
R
)
→
X
,
F
(
t
)
=
f
(
x
0
+
t
w
)
,
∣
t
∣
<
R
.
Then
F
∈
C
k
(
(
−
R
,
R
)
,
X
)
and then by Taylor’s formula with a reminder for
X
valued functions of one real variable, cf. (A1) and (A2)]
(A7)
F
(
t
)
=
∑
j
=
0
k
t
j
j
!
F
(
j
)
(
0
)
+
R
k
(
t
;
F
,
0
)
.
On the other hand,
(A8)
F
(
h
)
(
t
)
=
h
!
∑
∣
α
∣
=
h
w
α
α
!
(
D
α
f
)
(
x
0
+
t
w
)
,
and by choosing
w
=
1
‖
x
−
x
0
‖
(
x
−
x
0
)
,
t
=
‖
x
−
x
0
‖
,
x
∈
B
R
(
x
0
)
,
the formula (A7) becomes
f
(
x
)
=
F
(
‖
x
−
x
0
‖
)
=
∑
j
=
0
k
‖
x
−
x
0
‖
j
j
!
F
(
j
)
(
0
)
+
R
k
(
‖
x
−
x
0
‖
;
F
,
0
)
=
[for (A8) with
t
=
0
]
=
∑
j
=
0
k
‖
x
−
x
0
‖
j
∑
∣
α
∣
=
j
1
α
!
(
x
−
x
0
)
α
‖
x
−
x
0
‖
∣
α
∣
(
D
α
f
)
(
x
0
)
+
r
k
(
x
;
f
,
x
0
)
,
where we have set
r
k
(
x
;
f
,
x
0
)
=
R
k
(
‖
x
−
x
0
‖
;
F
,
0
)
,
F
(
t
)
=
f
x
0
+
t
‖
x
−
x
0
‖
(
x
−
x
0
)
,
∣
t
∣
<
R
.
The Taylor formula we seek for is
(A9)
f
(
x
)
=
∑
∣
α
∣
≤
k
(
x
−
x
0
)
α
α
!
(
D
α
f
)
(
x
0
)
+
r
k
(
x
;
f
,
x
0
)
.
Next let us assume that
f
∈
C
k
+
2
(
A
,
X
)
so that
F
∈
C
k
+
2
(
(
−
R
,
R
)
,
X
)
, and hence,
R
k
(
t
;
F
,
0
)
=
1
k
!
∫
0
t
(
t
−
s
)
k
F
(
k
+
1
)
(
s
)
d
s
,
where from (by (A8) for
h
=
k
+
1
)
r
k
(
x
;
f
,
x
0
)
=
1
k
!
∫
0
‖
x
−
x
0
‖
(
‖
x
−
x
0
‖
−
s
)
k
(
k
+
1
)
!
∑
∣
α
∣
=
k
+
1
(
x
−
x
0
)
α
α
!
‖
x
−
x
0
‖
∣
α
∣
(
D
α
f
)
x
0
+
s
‖
x
−
x
0
‖
(
x
−
x
0
)
d
s
or (by a change of variable
r
=
‖
x
−
x
0
‖
−
s
)
(A10)
r
k
(
x
;
f
,
x
0
)
=
k
+
1
‖
x
−
x
0
‖
k
+
1
∑
∣
α
∣
=
k
+
1
(
x
−
x
0
)
α
∫
0
‖
x
−
x
0
‖
r
k
(
D
α
f
)
x
−
r
‖
x
−
x
0
‖
(
x
−
x
0
)
d
r
.
Let
X
be a Fréchet space. Let
P
be a separating family of seminorms defining the topology of
X
as a local convex space.
Lemma A5
Let
A
⊂
R
n
be an open set and let
f
∈
C
k
+
2
(
A
,
X
)
. Let
x
0
∈
A
and
R
>
0
such that
B
R
(
x
0
)
⊂
A
. Then
(A11)
p
[
r
k
(
x
;
f
,
x
0
)
]
≤
M
k
+
1
∥
x
−
x
0
∥
k
+
1
×
max
∣
α
∣
=
k
+
1
sup
0
≤
τ
≤
1
p
[
(
D
α
f
)
(
(
1
−
τ
)
x
+
τ
x
0
)
]
for any
p
∈
P
and any
x
∈
B
R
(
x
0
)
.
Here,
M
ℓ
is the cardinality of the set
{
α
∈
Z
+
n
:
∣
α
∣
=
ℓ
}
.
Proof of Lemma A5
p
[
r
k
(
x
;
f
,
x
0
)
]
≤
k
+
1
‖
x
−
x
0
‖
k
+
1
∑
∣
α
∣
=
k
+
1
∏
j
=
1
n
∣
x
j
−
x
j
0
∣
α
j
∫
0
‖
x
−
x
0
‖
r
k
p
(
D
α
f
)
x
−
r
‖
x
−
x
0
‖
(
x
−
x
0
)
d
r
≤
(as
∏
j
=
1
n
∣
x
j
−
x
j
0
∣
α
j
≤
‖
x
−
x
0
‖
∣
α
∣
)
≤
(
k
+
1
)
∫
0
‖
x
−
x
0
‖
r
k
d
r
×
∑
∣
α
∣
=
k
+
1
sup
0
≤
r
≤
‖
x
−
x
0
‖
p
(
D
α
f
)
x
−
r
‖
x
−
x
0
‖
(
x
−
x
0
)
≤
M
k
+
1
∥
x
−
x
0
∥
k
+
1
max
∣
α
∣
=
k
+
1
sup
0
≤
r
≤
‖
x
−
x
0
‖
p
(
D
α
f
)
x
−
r
‖
x
−
x
0
‖
(
x
−
x
0
)
.
Setting
τ
=
r
⁄
‖
x
−
x
0
‖
∈
[
0
,
1
]
, one may conclude that
□
p
[
r
k
(
x
;
f
,
x
0
)
]
≤
M
k
+
1
∥
x
−
x
0
∥
k
+
1
max
∣
α
∣
=
k
+
1
sup
0
≤
τ
≤
1
p
[
(
D
α
f
)
(
(
1
−
τ
)
x
+
τ
x
0
)
]
.
C Series in Fréchet spaces
Let
X
be a topological vector space and let
{
x
ν
}
ν
≥
0
⊂
X
.
Lemma A6
If
∑
ν
≥
0
x
ν
is convergent, then
x
ν
→
0
in
X
as
ν
→
∞
.
Proof
Let
W
be a neighborhood of the origin in
X
. As the map
(
x
,
y
)
↦
x
−
y
is continuous, there is a neighborhood of the origin
V
⊂
X
such that
V
−
V
⊂
W
. Let
S
μ
=
∑
ν
=
0
μ
x
ν
and
S
=
lim
μ
→
∞
S
μ
. There is
N
=
N
(
V
)
≥
1
such that
S
ν
−
S
∈
V
for any
ν
≥
N
. Hence,
x
ν
=
(
S
ν
−
S
)
−
(
S
ν
−
1
−
S
)
∈
W
for any
ν
≥
N
.□
Let
X
be a Fréchet space and
{
x
ν
}
ν
≥
0
⊂
X
. Let
P
be a family of seminorms defining the topology of
X
as a locally convex space.
Lemma A7
If
∑
ν
=
0
∞
p
(
x
ν
)
<
∞
for every
p
∈
P
, then the series
∑
ν
≥
0
x
ν
is convergent in
X
.
Proof
Let
σ
μ
(
p
)
=
∑
ν
=
0
μ
p
(
x
ν
)
and
S
μ
=
∑
ν
=
0
μ
x
ν
. Let
p
∈
P
and
k
∈
N
. As
{
σ
ν
(
p
)
}
ν
≥
0
⊂
R
is a Cauchy sequence, there is
N
=
N
(
p
,
k
)
≥
1
such that
∣
σ
μ
(
p
)
−
σ
ν
(
p
)
∣
<
1
⁄
k
for any
μ
>
ν
≥
N
. Then
p
(
S
μ
−
S
ν
)
≤
∣
σ
μ
(
p
)
−
σ
ν
(
p
)
∣
<
1
k
,
μ
>
ν
≥
N
,
that is
{
S
ν
}
ν
≥
0
⊂
X
is a Cauchy sequence in
X
. Yet the topology of
X
(as a Fréchet space) is compatible to a complete invariant metric, and hence,
{
S
ν
}
ν
≥
0
is convergent in
X
.□
Let
X
be a complex Fréchet space and
{
x
ν
}
ν
≥
0
⊂
X
.
Lemma A8
If there is
z
0
∈
C
\
{
0
}
such that
∑
ν
≥
0
z
0
ν
x
ν
is convergent, then
∑
ν
≥
0
z
ν
x
ν
is convergent, for any
z
∈
D
∣
z
0
∣
(
0
)
. Also
∑
ν
≥
0
z
ν
x
ν
is uniformly convergent for
z
∈
D
r
(
0
)
for any
0
<
r
<
∣
z
0
∣
.
Proof
As
z
0
ν
x
ν
→
0
in
X
as
ν
→
∞
, for any
p
∈
P
and any
k
∈
N
, there is
N
=
N
(
p
,
k
)
≥
1
such that
∣
z
0
∣
ν
p
(
x
ν
)
<
1
⁄
k
for any
ν
≥
N
. If
z
∈
D
∣
z
0
∣
(
0
)
then
p
(
z
ν
x
ν
)
<
q
ν
⁄
k
, where
q
=
∣
z
⁄
z
0
∣
so that
0
≤
q
<
1
. Therefore,
∑
ν
=
0
∞
p
(
z
ν
x
ν
)
<
∞
so that
∑
ν
≥
0
z
n
x
ν
is convergent in
X
. Finally, for each
0
<
r
<
∣
z
0
∣
, one has
□
sup
∣
z
∣
<
r
p
(
z
ν
x
ν
)
≤
1
k
r
∣
z
0
∣
ν
.
The convergence radius of
∑
ν
≥
0
z
ν
x
ν
is
R
=
sup
∣
z
0
∣
:
∑
ν
≥
0
z
0
ν
x
ν
is convergent in
X
.
For each
p
∈
P
, we set
ℓ
(
p
)
=
limsup
ν
→
∞
p
(
x
ν
)
1
⁄
ν
. If
0
<
ℓ
(
p
)
<
∞
let
z
∈
D
1
⁄
ℓ
(
p
)
(
0
)
so that
∣
z
∣
ℓ
(
p
)
<
1
. Then
∣
z
∣
[
ℓ
(
p
)
+
ε
]
<
1
for some
ε
>
0
. As
ℓ
(
p
)
+
ε
>
ℓ
(
p
)
, there is
N
≥
1
such that
ℓ
(
p
)
+
ε
>
p
(
x
ν
)
1
⁄
ν
for any
ν
≥
N
. Thus,
p
(
z
ν
x
ν
)
=
∣
z
∣
ν
p
(
x
ν
)
<
(
∣
z
∣
[
ℓ
(
p
)
+
ε
]
)
ν
so that
∑
ν
=
0
∞
p
(
z
ν
x
ν
)
<
∞
.
Proposition A9
If
0
<
ℓ
(
p
)
<
a
for some
a
>
0
and any
p
∈
P
, and we set
R
=
inf
1
ℓ
(
p
)
:
p
∈
P
then
R
>
0
and the series
∑
ν
≥
0
z
ν
x
ν
is convergent (respectively divergent) for any
z
∈
D
R
(
0
)
(respectively for any
z
∈
C
\
D
R
(
0
)
¯
).
If
0
<
ℓ
(
p
)
<
∞
for any
p
∈
P
yet, there is a sequence
{
p
j
}
j
≥
1
⊂
P
such that
lim
j
→
∞
ℓ
(
p
j
)
=
∞
or
ℓ
(
p
)
=
∞
for some
p
∈
P
, then
∑
ν
≥
0
z
ν
x
ν
is divergent for any
z
∈
C
\
{
0
}
.
If
ℓ
(
p
)
=
0
for some
p
∈
P
, then let us set
P
0
=
{
p
∈
P
:
ℓ
(
p
)
=
0
}
,
R
=
inf
1
ℓ
(
p
)
:
p
∈
P
\
P
0
.
If
sup
{
ℓ
(
p
)
:
p
∈
P
\
P
0
}
<
∞
, then
R
>
0
and the series
∑
ν
≥
0
z
ν
x
ν
is convergent for any
z
∈
D
R
(
0
)
, while if
sup
{
ℓ
(
p
)
:
p
∈
P
\
P
0
}
=
∞
, then
R
=
0
and
∑
ν
≥
0
z
ν
x
ν
is divergent for any
z
∈
C
\
{
0
}
.
Proof
(1) If
R
=
0
, then for any
ε
>
0
there is
p
ε
∈
P
such that
ℓ
(
p
ε
)
>
1
⁄
ε
. Hence,
ℓ
(
p
ε
)
→
∞
as
ε
→
0
+
, a contradiction. Hence,
R
>
0
. Let
z
∈
D
R
(
0
)
. Then
∑
ν
=
0
∞
p
(
z
ν
x
ν
)
<
∞
for any
p
∈
P
hence,
∑
ν
≥
0
z
ν
x
ν
is convergent. If in turn
∣
z
∣
>
R
, then
∣
z
∣
ℓ
(
p
0
)
>
1
for some
p
0
∈
P
. Hence, there is
δ
>
0
such that
∣
z
∣
[
ℓ
(
p
0
)
−
δ
]
>
1
. On the other hand,
ℓ
(
p
0
)
−
δ
<
ℓ
(
p
0
)
hence, for any
n
≥
1
, there is
ν
(
n
)
≥
n
such that
ℓ
(
p
0
)
−
δ
<
p
0
(
x
ν
(
n
)
)
1
⁄
ν
(
n
)
.
Consequently, there is a sequence
{
ν
j
}
j
≥
1
⊂
N
such that
1
≤
ν
1
<
ν
2
<
…
↑
∞
,
ℓ
(
p
0
)
−
δ
<
p
0
(
x
ν
j
)
1
⁄
ν
j
,
j
≥
1
.
Finally,
p
0
(
z
ν
j
x
ν
j
)
=
∣
z
∣
ν
j
p
0
(
x
ν
j
)
>
∣
z
∣
ν
j
[
ℓ
(
p
0
)
−
δ
]
ν
j
>
1
,
j
≥
1
.
Yet
p
0
is continuous, so
∑
ν
≥
0
z
ν
x
ν
is divergent.
(2) Let
ℓ
(
p
j
)
→
∞
as
j
→
∞
. Then, for any
A
>
0
, there is
j
(
A
)
≥
1
such that for any
j
≥
j
(
A
)
and any
n
≥
1
there is
ν
≥
n
satisfying
p
j
(
x
ν
)
1
⁄
ν
>
A
. Given
z
∈
C
\
{
0
}
, one may pick
A
≔
1
⁄
∣
z
∣
and choose
j
0
≥
j
(
A
)
. Consequently, there is a sequence
1
≤
ν
1
<
ν
2
<
…
↑
∞
such that
p
j
0
(
z
ν
k
x
ν
k
)
>
1
,
k
≥
1
,
so that (by the continuity of the seminorm
p
j
0
) the series
∑
ν
≥
0
z
ν
x
ν
is divergent. A similar argument may be provided when
ℓ
(
p
)
=
∞
for some
p
∈
P
.
(3) Let
p
∈
P
0
. Then for any
ε
>
0
, there is
n
ε
≥
1
such that
p
(
x
ν
)
1
⁄
ν
<
ε
,
ν
≥
ν
ε
.
Let
z
∈
C
\
{
0
}
and let us choose
0
<
ε
<
1
⁄
∣
z
∣
. Then
p
(
z
ν
x
ν
)
<
(
ε
∣
z
∣
)
ν
,
ν
≥
n
ε
,
so that
∑
ν
=
0
∞
p
(
z
ν
x
ν
)
<
∞
. If
P
0
=
P
then
∑
ν
≥
0
z
ν
x
ν
is convergent for any
z
∈
C
. If
P
\
P
0
≠
∅
, then let
R
=
inf
1
ℓ
(
p
)
:
p
∈
P
\
P
0
.
The remainder of the proof is similar that of part (1) (when
sup
p
∈
P
\
P
0
ℓ
(
p
)
<
∞
) and part (2) (when
sup
p
∈
P
\
P
0
ℓ
(
p
)
=
∞
).□
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and
Francesco Esposito
