16
VIII – Cauchy Theory
clearly the integral along this radius. Hence F (z +h)−F (z) =
f (z +th)hdt,
where, as usual, integration is over [0, 1]. It follows that
F (z + h) − F (z) − f (z)h =
[f (z + th) − f (z)] hdt;
f being continuous, |f (z + th) − f (z)| ≤ r for all t ∈ I provided |h| ≤ r
(uniform continuity on a compact set); so
F (z + h) = F (z) + f (z)h + o(h)
as h approaches 0, which proves the existence of F
(z) = f (z), qed.
(v) The case of a contractible domain. A naive attempt at constructing
a primitive without using theorem 1 would be to arbitrarily choose for every
z ∈ G a path μ z connecting a to z and to set F (z) = F (μ z ). Albeit strange,
this leads to the result provided μ z depends on z in not too. . . arbitrary a
manner. This, we will see, implies a drastic restriction on G. Formula (2)
used in the case of a star domain clearly falls within this framework, but is
based on an all too providential choice of μ z .
So let us assign to every z ∈ G a path
μ z : t ∈ [0, 1] −→ μ z (t)
connecting a to z = x + iy = (x, y) in G and set
H(z, t) = μ z (t).
So H(z, 0) = a, H(z, 1) = z for all z ∈ G. To show that the function
F (z) = F (μ z ) =
μz
f (ζ)dζ =
1
0
f [H(z, t)] DH(z, t).dt ,
(2.14)
where D = d/dt, is a primitive for f , it would suffice to show that, with
respect to x and y, it has partial derivatives D 1 F and D 2 F equal to f and if
respectively. For this, let us assume that differentiation is possible under the
sign without any difficulty– this would be miraculous if μ z was arbitrarily
chosen – and calculate as Euler or Cauchy would have done; the calculation
is similar to the one done for function (2) – only slightly harder. Using the
product and chain rules for differentiation and the relations DD 1 = D 1 D,
D 1 f = f
, the FT gives
12
12 In a notation such as DH(z, t).D1H(z, t), the point means that the operator D
is applied to H(z, t) and not to the product H(z, t)D1H(z, t).
VIII – Cauchy Theory
clearly the integral along this radius. Hence F (z +h)−F (z) =
f (z +th)hdt,
where, as usual, integration is over [0, 1]. It follows that
F (z + h) − F (z) − f (z)h =
[f (z + th) − f (z)] hdt;
f being continuous, |f (z + th) − f (z)| ≤ r for all t ∈ I provided |h| ≤ r
(uniform continuity on a compact set); so
F (z + h) = F (z) + f (z)h + o(h)
as h approaches 0, which proves the existence of F
(z) = f (z), qed.
(v) The case of a contractible domain. A naive attempt at constructing
a primitive without using theorem 1 would be to arbitrarily choose for every
z ∈ G a path μ z connecting a to z and to set F (z) = F (μ z ). Albeit strange,
this leads to the result provided μ z depends on z in not too. . . arbitrary a
manner. This, we will see, implies a drastic restriction on G. Formula (2)
used in the case of a star domain clearly falls within this framework, but is
based on an all too providential choice of μ z .
So let us assign to every z ∈ G a path
μ z : t ∈ [0, 1] −→ μ z (t)
connecting a to z = x + iy = (x, y) in G and set
H(z, t) = μ z (t).
So H(z, 0) = a, H(z, 1) = z for all z ∈ G. To show that the function
F (z) = F (μ z ) =
μz
f (ζ)dζ =
1
0
f [H(z, t)] DH(z, t).dt ,
(2.14)
where D = d/dt, is a primitive for f , it would suffice to show that, with
respect to x and y, it has partial derivatives D 1 F and D 2 F equal to f and if
respectively. For this, let us assume that differentiation is possible under the
sign without any difficulty– this would be miraculous if μ z was arbitrarily
chosen – and calculate as Euler or Cauchy would have done; the calculation
is similar to the one done for function (2) – only slightly harder. Using the
product and chain rules for differentiation and the relations DD 1 = D 1 D,
D 1 f = f
, the FT gives
12
12 In a notation such as DH(z, t).D1H(z, t), the point means that the operator D
is applied to H(z, t) and not to the product H(z, t)D1H(z, t).