Previously, we have seen that if has an antiderivative
on a domain
, then for any closed smooth contour
,
and vice versa. When does this hold?
This post will take heavy reference from this document.
Theorem 1 (Cauchy-Goursat Theorem). Let be open and
. If
is holomorphic on
, then for any bounded domain
with
,
Lemma 1. Theorem 1 holds whenever is a closed rectangle.
Proof. Firstly, since is holomorphic on
, it is continuous on
, so that there exists
such that
Suppose for a contradiction that equality holds. Then divide into
equal sub-rectangles. By the triangle inequality and the pigeonhole principle, at least one of these rectangles
with
must yield
Inductively obtain a decreasing sequence of rectangles such that
Each rectangle for each
has a center
. Since the sequence
is Cauchy, it converges to some
. Define the map
by
Fix . Since
is holomorphic at
, there exists
such that
Since , find
such that
. Since the map
has an anti-derivative ,
By construction, . By the ML-inequality,
Thus, the inequality yields a contradiction if
Therefore, set and take
to be sufficient large to produce the desired contradiction. Therefore,
Lemma 2. Suppose that is simply-connected and
is holomorphic on
. Then
has an anti-derivative
on
. Equivalently,
for any closed contour .
Proof. Fix . For any
, let
be two paths from
to
comprised only of finitely many horizontal and vertical paths. Then it is not hard to prove that there exist disjoint open rectangles
such that
and
Therefore,
Thus, the map defined by
where is any path from
to
comprised only of finitely many horizontal and vertical paths is well-defined. To establish the antiderivative property, work with the path
where each path
and
is (without loss of generality) parameterised by
Of course, we verify that
so that is indeed a path from
to
. Taking
sufficiently small and employing the continuity of
at
, the ML-inequality yields
establishing differentiability at . Since
is open and
is arbitrary,
is holomorphic on
with
.
We now handle the general case of Theorem 1.
Proof of Theorem 1. First suppose the special case that is simply connected. Then
is a closed contour, so that by Lemma 2,
For the general case, find a finite number of line segments with end-points in the smooth parts of
so that
is simply connected. Intuitively, we “chop”
up into a collection of simply connected “islands”. For each
, define
For sufficiently small ,
for some simply connected neighborhood
, so that
Then using arguments involving uniform continuity, the ML-inequality, and the definition of the contour integral, take to obtain
We took great pains to prove the Cauchy-Goursat theorem so that we can prove an even more incredible result—the Cauchy integral formula. This we will do the next time, as well as establish the many powerful properties of holomorphic functions that set it apart in the complex analytic setting from usual differentiability in the real-analytic setting.
—Joel Kindiak, 18 Aug 25, 1756H
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