§ 2. Cauchy’s Integral Formulas
61
To finish the proof, consider a compact subset H of U and an upper bound
|f
(t, z)| ≤ p H (t) valid for t ∈ I and z ∈ H, with a positive μ-integrable
function p H . Then, for any compact set K ⊂ I,
I
f
(t, z)dμ(t) − g
K (z)
≤
I−K
p H (t)dμ(t)
for all z ∈ H, and since the right hand side approaches 0 as K “ approaches ” I,
it can be concluded that g
K (z), and hence also the partial derivatives of g K
with respect to x and y, converge uniformly, up to a factor i, to
f
(t, z)dμ(t)
on H. The function g = lim g K , therefore, has partial derivative with respect
to x and y obtained by passing to the limit over those of g K (Chap. III, § 4,
Theorem 19). They are continuous and like those of g K satisfy Cauchy’s condition, so that g is holomorphic,
40 with g
(z) = lim g
K (z) =
f
(t, z)dμ(t),
qed.
Second proof. Theorem 9 can also be proved by using Weierstrass’ theorem on uniform limits of holomorphic functions (Chap. VII, § 4, n
◦ 19, Theorem 17).
Let us first consider the case where I is compact. In what follows, suppose
that z remains in a compact subset H of U and set f t (z) = f (t, z). As I × H
is compact, f is uniformly continuous on it. In particular, for every r > 0
there exists r
> 0 such that
|s − t| < r
=⇒ ⇒f s − f t H < r .
(7.5)
Having said this, let us partition I into finitely many non-empty intervals I k
of length < r
, choose points t k ∈ I k and compare integral (1) to the Riemann
sum
f (t k , z)μ(I k ). Since, by (5), |f (t, z) − f (t k , z)| < r for all t ∈ I k and
all z ∈ H, it follows that
g(z) −
f (t k , z) μ (I k )
≤ ≤μr for all z ∈ H ,
where μ is the norm of μ. This means that the function g is a uniform
limit of holomorphic functions on every compact set H ⊂ U ; it is, therefore,
holomorphic. Moreover, since a limit of holomorphic functions can be differentiated term by term according to the same theorem, g
(p) (z) is the limit
of the expressions
f
(p) (t k , z)μ(I k ), which are just the Riemann sums with
respect to integral (2), and so the theorem follows when I is compact.
In the general case, replace I by a compact interval K ⊂ I and pass to
the limit as in the previous proof.
40 This is the “ useless ” result mentioned in Chap. III, § 5, at the end of n
◦ 22.
61
To finish the proof, consider a compact subset H of U and an upper bound
|f
(t, z)| ≤ p H (t) valid for t ∈ I and z ∈ H, with a positive μ-integrable
function p H . Then, for any compact set K ⊂ I,
I
f
(t, z)dμ(t) − g
K (z)
≤
I−K
p H (t)dμ(t)
for all z ∈ H, and since the right hand side approaches 0 as K “ approaches ” I,
it can be concluded that g
K (z), and hence also the partial derivatives of g K
with respect to x and y, converge uniformly, up to a factor i, to
f
(t, z)dμ(t)
on H. The function g = lim g K , therefore, has partial derivative with respect
to x and y obtained by passing to the limit over those of g K (Chap. III, § 4,
Theorem 19). They are continuous and like those of g K satisfy Cauchy’s condition, so that g is holomorphic,
40 with g
(z) = lim g
K (z) =
f
(t, z)dμ(t),
qed.
Second proof. Theorem 9 can also be proved by using Weierstrass’ theorem on uniform limits of holomorphic functions (Chap. VII, § 4, n
◦ 19, Theorem 17).
Let us first consider the case where I is compact. In what follows, suppose
that z remains in a compact subset H of U and set f t (z) = f (t, z). As I × H
is compact, f is uniformly continuous on it. In particular, for every r > 0
there exists r
> 0 such that
|s − t| < r
=⇒ ⇒f s − f t H < r .
(7.5)
Having said this, let us partition I into finitely many non-empty intervals I k
of length < r
, choose points t k ∈ I k and compare integral (1) to the Riemann
sum
f (t k , z)μ(I k ). Since, by (5), |f (t, z) − f (t k , z)| < r for all t ∈ I k and
all z ∈ H, it follows that
g(z) −
f (t k , z) μ (I k )
≤ ≤μr for all z ∈ H ,
where μ is the norm of μ. This means that the function g is a uniform
limit of holomorphic functions on every compact set H ⊂ U ; it is, therefore,
holomorphic. Moreover, since a limit of holomorphic functions can be differentiated term by term according to the same theorem, g
(p) (z) is the limit
of the expressions
f
(p) (t k , z)μ(I k ), which are just the Riemann sums with
respect to integral (2), and so the theorem follows when I is compact.
In the general case, replace I by a compact interval K ⊂ I and pass to
the limit as in the previous proof.
40 This is the “ useless ” result mentioned in Chap. III, § 5, at the end of n
◦ 22.