Theorem 3.1.1. Cauchy-Goursat.
Let \(f\) be analytic on a simply connected domain \(D\text{.}\) Then
\begin{equation*}
\oint_\gamma f \, dz = 0
\end{equation*}
for any simple closed curve \(\gamma\) contained in \(D\text{.}\)
Section 3.1 Prerequisites
Subsection 3.1.1 Complex analysis
Let us begin with a very useful sort-of converse to the Cauchy-Goursat theorem
Theorem 3.1.2. Morera's theorem.
Let \(f\) be a continuous function defined on an open domain \(D\text{.}\) If
\begin{equation*}
\oint_\gamma f(\xi) \, d\xi = 0
\end{equation*}
for every simple closed path \(\gamma\) contained in \(D\text{,}\) then \(f\) is analytic on \(D\text{.}\)
Proof.
Assume without loss of generality that the domain \(D\) is connected (and so path-connected). Assume that \(f\) satisifies the hypotheses of the theorem. We will construct an anti-derivative for \(f\text{,}\) which will allow us to conclude that \(f\) is analytic.
Pick a basepoint \(z_0\) in \(D\text{,}\) and for any other \(z\) in \(D\text{,}\) let \(\gamma\) be a simple path from \(z_0\) to \(z\text{.}\) Let \(F\) be the function on \(D\) given by
\begin{equation*}
F(z) = \int_\gamma f(\xi) \, d\xi.
\end{equation*}
Note that this definition does not depend on the choice of path. Assume that \(\tau\) is another path from \(z_0\) to \(z\text{.}\) Then, by hypothesis,
\begin{align*}
0 \amp= \oint_{\tau\inv \gamma} f\\
\amp = \oint_{\tau\inv} f + \oint_\gamma f\\
\amp = -\oint_{\tau} f + \oint_\gamma f
\end{align*}
which implies that \(\oint_\tau f = \oint_\gamma f\text{.}\)
Note futher that a different choice of basepoint (say \(z_1\)) causes the resulting function to differ from \(F\) only by the constant \(C = \oint_C f\) where \(C\) is any simple path from \(z_1\) to \(z_0\text{.}\) Thus, choice of basepoint does not affect the derivative.
It remains to show that \(\frac{d}{dz} F(z) = f(z)\text{.}\) This boils down to a difference quotient argument. Consider \(F(z + h)\) for some small complex number \(h\text{.}\) Then
\begin{equation*}
F(z+h) - F(z) = \int_{z_0}^{z+h} f(\xi) \, d\xi - \int_{z_0}^z f(\xi) \, d\xi = \int_z^{z+h} f(\xi) \, d\xi.
\end{equation*}
(The notation is justified by the path independence of integrals of \(f\text{.}\)) Now,
\begin{align*}
\frac{1}{h} (F(z + h) - F(z)) - f(z) \amp= \frac{1}{h} (F(z + h) - F(z)) - \frac{1}{h} hf(z) \\
\amp = \frac{1}{h} (\int_z^{z + h} f(\xi) \, d\xi - h f(z))\\
\amp = \frac{1}{h} (\int_z^{z + h} f(\xi) \, d\xi - \int_z^{z+h} f(z) \, d\xi)\\
\amp= \frac{1}{h} \int_z^{z+h} f(\xi) - f(z) \, d\xi.
\end{align*}
Since \(f\) has path independent line integrals, choose the line segment between \(z\) and \(z + h\text{.}\) Since \(f\) is continuous,
\begin{equation*}
\abs{\frac{F(z + h) - F(z)}{h} - f(z)} \leq \frac{1}{h} \int_z^{z+h} \abs{f(\xi) - f(z)} \, d\xi \to 0 \text{ as } \abs{h} \to 0.
\end{equation*}
We conclude that \(F\) is differentiable at \(z\) and thus analytic at \(z\) with derivative \(f\text{,}\) which implies that \(f\) is analytic. Since this holds for all \(z\in D\text{,}\) we conclude that \(f\) is analytic on \(D\text{.}\)
Morera's theorem is often used with weaker (equivalent) hypotheses - that is, we need only check that
\begin{equation*}
\int_{\partial T} f\, dz = 0
\end{equation*}
around the boundary of any closed triangle \(T\) contained in \(D\text{.}\)
Morera's theorem is the standard method used to prove that a constructed function is holomorphic. The theorem is used so often and so widely that it may sometimes be invoked without reference (though not in our case).
One nice immediate result that we get from Morera is parallel to the (fundamental) result from real analysis that uniform limits of continuous functions are continuous.
Theorem 3.1.3.
The uniform limit of a sequence of analytic functions is analytic.
Proof.
Suppose that a sequence \(f_n\) of analytic functions converges uniformly to a continuous limit function \(f\) on an open disk. The Cauchy-Goursat theorem implies that, for all \(n\text{,}\)
\begin{equation*}
\oint_C f_n \, dz = 0
\end{equation*}
for any simple closed curve \(C\) contained in the disk. Using uniform converge to push the limit through the integral, we get
\begin{equation*}
\oint_C f \, dz = \oint_C \lim_n f_n \, dz = \lim_n \oint_C f_n \, dz = 0.
\end{equation*}
for every simple closed curve \(C\) in the disk, and so by Morera's theorem, \(f\) must be analytic on the disk.
This result extends to domains, as we can recreate this argument in a neighborhood of any \(z_0\) in \(D\text{.}\)
Since we'll be considering functions defined as series, it will be useful to combine the results above with a test for uniform convergence called the Weierstrass M-test.
Theorem 3.1.4. M-test.
Supposet that \(\{f_n\}\) is a sequence of functions on a common domain \(D\) and that there exists a sequence of non-negative numbers \(\{M_n\}\) satisfying
- \(\abs{f_n(z)} \lt M_n\) for all \(n \in \N\text{;}\)
- \(\displaystyle \sum_{n=1}^\infty M_n \lt \infty\text{.}\)
Then \(\sum_{n=1}^\infty f_n\) converges absolutely and uniformly on \(D\text{.}\)
Subsection 3.1.2 Measure theory
To be rigorous about convergence, we're going to need some extremely useful tools from analysis. The discussion here will mostly bracket out proofs, which are typically done in a graduate course in real analysis.
A measure is a function that assigns a numerical value to a set. The idea comes from an effort to generalize the notion of the length of an interval. A brief overview of the setup follows. We have a set \(X\) (for example, \(X=\R\)), and we want to define a function \(\mu: A \subset X \to [0,\infty]\) (note the inclusion of \(\infty\)). It turns out that we can't measure every subset of \(X\) (that darn axiom of choice). So we equip \(X\) with a sort of topology called a \(\sigma\)-algebra consisting of so called measurable sets.
Definition 3.1.5.
Given a set \(X\text{,}\) a \(\sigma\)-algebra on \(X\) is a non-empty collection \(\Sigma\) of subsets of \(X\) that is closed under complements, countable unions, and countable intersections.
Definition 3.1.6.
Let \(X\) be a set and \(\Sigma\) a \(\sigma\)-algebra on \(X\text{.}\) A set function \(\mu:\Sigma \to [0,\infty]\) is a measure if
- For all \(E \in \Sigma\text{,}\) \(\mu(E) \geq 0\text{.}\)
- \(\mu(\emptyset) = 0\text{.}\)
- For all countable collections \(\{E_j\} \subset \Sigma\) of pairwise disjoint sets,\begin{equation*} \mu(\bigcup_j E_j) = \sum_j \mu(E_j). \end{equation*}
One important example of a measure is Lebesgue measure, which directly generalizes the length of intervals. Lebesgue measure is usually defined on the Borel sets \(\mathcal B\text{,}\) which is the \(\sigma\)-algebra constructed by starting with the open intervals in \(\R\) and combining them in all ways to ensure that the conditions for a \(\sigma\)-algebra are met. Measures can be far stranger than Lebesgue measure. The Dirac measure at \(x_0\text{,}\) also called a point mass, is a measure that assigns a value of 1 to any set \(E\) containing \(x_0\) and \(0\) otherwise.
To work with measures, we'll also need a family of compatible functions.
Definition 3.1.7.
Suppose that \(X\) is a set and \(\Sigma\) is a \(\sigma\)-algebra on \(X\text{.}\) A function \(f: X \to \R\) is measurable if
\begin{equation*}
f\inv(B) \in \Sigma
\end{equation*}
for every Borel set \(B \in \mathcal B\text{.}\)
Measures allow a powerful generalization of integrals, replacing the familiar “\(dx\)”, which can be thought of as the length of a small interval, with \(d\mu\text{.}\) The essential approach is to approximate a function from below by step functions, which is called Lebesgue integration. That is, given a set \(X\text{,}\) a \(\sigma\)-algebra \(\Sigma\text{,}\) a measure \(\mu\) (together called a measure space (X, \(\Sigma, \mu)\)) and a function \(f:X \to [0,\infty]\text{,}\) any partition of \(X\) by subsets \(A_i\) of \(\Sigma\) leads to the integral-ish sum
\begin{equation*}
\sum_{i=1}^m \mu(A_i) \inf_{A_i} f.
\end{equation*}
You can view this as in the spirit of a Riemann sum - we're cutting \(X\) up into pieces (but that need not be intervals) and under-estimating \(f\) on those pieces. While this isn't strictly the area of a rectangle as in the Riemann case (as the \(A_i\) can be very non-interval), the idea is similar.
Definition 3.1.8.
When such a sum has a supremum over all \(\Sigma\)-partitions of \(X\text{,}\) we can define the integral of f with respect to \(\mu\) by
\begin{equation*}
\int f \, d\mu = \sup \{\sum_{i=1}^m \mu(A_i) \inf_{A_j} f: A_1, \ldots A_m \text{ is a } \Sigma-\text{partition of } X\}.
\end{equation*}
Lebesgue measure gives rise to integrals that directly generalize the Riemann integral - that is, if a function is Riemann integrable, then it is Lebesgue integrable and the integrals agree. (Note: unlike some authors, I do not admit improper Riemann integrals into the class of Riemann integrable functions.)
A more interesting example is to consider the Dirac measure \(\delta_a\) where \(a \in X\text{.}\) That is, \(\delta_a(E) = 1\) if \(a \in E\) and \(0\) otherwise. Notice that for any partition of \(X\text{,}\) \(a \in A_i\) for exactly one \(k\text{.}\) So
\begin{equation*}
\sum_{i=1}^m \delta_a(A_i) \inf_{A_i} f = 1 \inf_{A_k} f
\end{equation*}
and that the sup of these sums will be \(f(a)\text{!.}\) That is,
\begin{equation*}
\int_X f \, d\delta_a = f(a).
\end{equation*}
All this new machinery isn't much good if we don't get beefy new tools to work with. One of the big problems that Lebesgue integration is meant to solve is the failure in Riemann integration of being able to push limits inside integrals. In the Lebesgue setting, we have very powerful theorems describing when this can be done.
A sequence of functions \(f_k:X \to [0,\infty]\) is called increasing if for each \(x \in X\text{,}\) we have \(f_1(x) \leq f_2(x) \leq \ldots\text{.}\)
Theorem 3.1.9. Monotone convergence theorem.
Suppose that \((X, \Sigma, \mu)\) is a measure space and that \(0 \leq f_1 \leq f_2 \leq \ldots\) is an increasing sequence of measurable functions. Define \(f: X \to [0,\infty]\) by
\begin{equation*}
f(x) = \lim_{k \to \infty} f_k(x).
\end{equation*}
Then
\begin{equation*}
\lim_{k \to \infty} \int f_k \, d\mu = \int f \, d\mu.
\end{equation*}
An even beefier theorem states that we can push limits into integrals when the sequence of functions is dominated by a function with bounded integral. Note: I'm not covering the leap to complex measures here, but one can imagine that a real-valued measure is a sum of two positive measures, and that a complex measure is a sum of real measures in the obvious way.
Theorem 3.1.10. Dominated convergence theorem.
Suppose that \((X, \Sigma, \mu)\) is a measure space and that \(f_n\) is a sequence of complex-valued measurable functions. Suppose that \(f_n\) converges pointwise to \(f\) and that there exists a measurable function \(g\) such that
\begin{equation*}
\abs{f_n(x)} \leq g(x)
\end{equation*}
for all \(x\in X\) and all \(n\) and
\begin{equation*}
\int g \, d\mu \lt \infty.
\end{equation*}
Then \(f\) is integrable and
\begin{equation*}
\lim_{n\to \infty} \int f_n \, d\mu = \int f \, d\mu.
\end{equation*}
