§ 1. Integrals of Holomorphic Functions
21
μ(s) − μ(t) =
t
s
ν(x)dx
proving that μ is a primitive for ν is then obtained by passing to the limit.
Besides, it is obvious that in the space C
0 (I), the set C
1/2 (I; G) of μ ∈
C
1/2 (I) such that μ(I) ⊂ G is open in C
1/2 (I).
Let us now return to formula (2.13)
F (μ) =
μ
f (ζ)dζ =
1
0
f [μ(t)] μ
(t)dt
(3.4)
defining a function on C
1/2 (I; G). It may seem strange to differentiate it with
respect to μ, but as it it defined on the open subset C
1/2 (I; G) of the Banach
space C
1/2 (I), definition (1.1) which holds in R
2 can be imitated. Here, c
and h will be replaced by some μ ∈ C
1/2 (I; G) and some ν ∈ C
1/2 (I). So the
expression
F (μ + sν) =
f [μ(t) + sν(t)] [μ
(t) + sν
(t)] dt =
=
f [μ(t) + sν(t)] dμ(t) + s
f [μ(t) + sν(t)] dν(t)
must be differentiated with respect to s. For any ν, it is well-defined for
sufficiently small |s|. To differentiate under the
sign [Chapter V, § 2, Theorem 9 or § 9, formula (30.15)], it is sufficient to check that f [μ(t) + sν(t)]
is a continuous function of (s, t), which is obvious, and that its derivative
with respect s is a continuous function of (s, t); its existence is obvious – it
is f
[μ(t) + sν(t)] ν(t) – and so is its continuity since the functions f
, μ and
ν are continuous. Hence in telegraphic style,
f
(μ + sν)νdμ +
f (μ + sν)dν + s
f
(μ + sν)νdν =
=
f
(μ + sν)ν(dμ + sdν) +
f (μ + sν)ν
dt .
Integration by parts can justifiably be used to compute the last integral since
the functions f (μ + sν) and ν are of class C
1/2 . So it can also be written
f (μ + sν)ν
t=1
t=0
−
f
(μ + sν) (μ
+ sν
) νdt .
Since [μ
(t) + sν
(t)] dt = dμ(t) + sdν(t), the last integral can be written
f
(μ + sν)ν(dμ + sdν), canceling out the first term of the penultimate
formula. Hence finally,
d
ds
F (μ + sν) = f [μ(1) + sν(1)] ν(1) − f [μ(0) + sν(0)] ν(0) .
(3.5)
21
μ(s) − μ(t) =
t
s
ν(x)dx
proving that μ is a primitive for ν is then obtained by passing to the limit.
Besides, it is obvious that in the space C
0 (I), the set C
1/2 (I; G) of μ ∈
C
1/2 (I) such that μ(I) ⊂ G is open in C
1/2 (I).
Let us now return to formula (2.13)
F (μ) =
μ
f (ζ)dζ =
1
0
f [μ(t)] μ
(t)dt
(3.4)
defining a function on C
1/2 (I; G). It may seem strange to differentiate it with
respect to μ, but as it it defined on the open subset C
1/2 (I; G) of the Banach
space C
1/2 (I), definition (1.1) which holds in R
2 can be imitated. Here, c
and h will be replaced by some μ ∈ C
1/2 (I; G) and some ν ∈ C
1/2 (I). So the
expression
F (μ + sν) =
f [μ(t) + sν(t)] [μ
(t) + sν
(t)] dt =
=
f [μ(t) + sν(t)] dμ(t) + s
f [μ(t) + sν(t)] dν(t)
must be differentiated with respect to s. For any ν, it is well-defined for
sufficiently small |s|. To differentiate under the
sign [Chapter V, § 2, Theorem 9 or § 9, formula (30.15)], it is sufficient to check that f [μ(t) + sν(t)]
is a continuous function of (s, t), which is obvious, and that its derivative
with respect s is a continuous function of (s, t); its existence is obvious – it
is f
[μ(t) + sν(t)] ν(t) – and so is its continuity since the functions f
, μ and
ν are continuous. Hence in telegraphic style,
f
(μ + sν)νdμ +
f (μ + sν)dν + s
f
(μ + sν)νdν =
=
f
(μ + sν)ν(dμ + sdν) +
f (μ + sν)ν
dt .
Integration by parts can justifiably be used to compute the last integral since
the functions f (μ + sν) and ν are of class C
1/2 . So it can also be written
f (μ + sν)ν
t=1
t=0
−
f
(μ + sν) (μ
+ sν
) νdt .
Since [μ
(t) + sν
(t)] dt = dμ(t) + sdν(t), the last integral can be written
f
(μ + sν)ν(dμ + sdν), canceling out the first term of the penultimate
formula. Hence finally,
d
ds
F (μ + sν) = f [μ(1) + sν(1)] ν(1) − f [μ(0) + sν(0)] ν(0) .
(3.5)