Math 411: Honours Complex Variables - University of Alberta

Proof. There is no loss of generality supposing that N = 1.
Let x0,x ∈ D be such that every number between them is also in D. By the
Mean Value Theorem of single variable calculus, there exists, for each t ∈ [a,b], a real
number ξt between x0 and x such that
f(t,x)−f(t,x0)
x−x0
= ∂f
∂x (t,ξt).
Let ǫ > 0. By uniform continuity, there exists a δ > 0 such that
�
�
�
∂f ∂f
� (t,x1)−
∂x ∂x (t,x2)
�
�
�
ǫ
� <
b−a
for any t ∈ [a,b] and all x1,x2 between x0 and x with |x1 −x2|< δ.
Suppose that |x0 −x|< δ. Then we have:
� �
�
b
�
g(x)−g(x0) ∂f
� −
x−x0 a ∂x (t,x0)dt
�
�
�
� =
��
� b�
�
f(t,x)−f(t,x0)
� −
a x−x0
∂f
∂x (t,x0)
� �
�
dt�
�
� � b�
≤ �
f(t,x)−f(t,x0)
� −
a x−x0
∂f
∂x (t,x0)
�
�
�
�dt � � b�
= �
∂f ∂f
� (t,ξt)−
a ∂x ∂x (t,x0)
�
�
�
�dt � b
ǫ
≤
b−a dt, because |ξt −x0|< δ,
= ǫ.
This proves that g is differentiable at x0 with g ′ (x0) = � b
a
A similar (but easier) argument shows that g ′ is continuous.
a
∂f
∂x (t,x0)dt.
Lemma 5.4. Let D ⊂ C be open, and let f: [a,b]×D → C be continuous such that
: [a,b]×D → C exists and is continuous. Define
∂f
∂z
Then g is holomorphic with
for z ∈ D.
� b
g: D → R, z ↦→ f(t,z)dt.
g ′ � b
(z) =
a
a
∂f
∂z (t,z)dt
Proof. Apply Lemma 5.3 to Ref and Imf and note that the Cauchy Riemann differential
equations are satisfied.
37