Cauchy's Integral Theorem for Rectangles

# Cauchy's Integral Theorem for Rectangles

We are now ready to prove a very important (baby version) of Cauchy's Integral Theorem which we will look more into later; called Cauchy's Integral Theorem for Rectangles.

Theorem 1: Let $A \subseteq \mathbb{R}$ be open and let $f : A \to \mathbb{C}$. If $R \subset A$ is a rectangle that is contained in $A$ and whose interior is contained in $A$ and if $f$ is analytic on $A$ then $\displaystyle{\int_R f(z) \: dz = 0}$. |

*The notation "$\displaystyle{\overline{R}}$" used below denotes the closure of $R$. In this proof, the closure of the rectangle $R$ is the rectangle along with its interior, i.e., a closed rectangular box.*

**Proof:**Let $R \subset A$ be a rectangle (with a given orientation) that is contained in $A$ and whose interior is contained in $A$.

- Take $R$ and subdivide it into four congruent sub-rectangles $R1$, $R2$, $R3$, and $R4$ whose exterior edges constitute $R$ and its orientation.

- Furthermore we have that $\displaystyle{\int_R f(z) \: dz = \int_{R1} f(z) \: dz + \int_{R2} f(z) \: dz + \int_{R3} f(z) \: dz + \int_{R4} f(z) \: dz}$ because the integral of $f$ along the inner edges of $R1$, $R2$, $R3$, and $R4$ cancel each other since each inner edge is traversed once in a positive orientation and traversed once in a negative orientation. So by the triangle inequality:

\begin{align} \biggr \lvert \int_R f(z) \: dz \biggr \rvert \leq \sum_{k=1}^{4} \biggr \lvert \int_{Rk} f(z) \: dz \biggr \rvert \end{align}

- In particular, there exists some $i \in \{ 1, 2, 3, 4 \}$ for which:

\begin{align} \biggr \lvert \int_{Ri} f(z) \: dz \biggr \rvert \geq \frac{1}{4} \biggr \lvert \int_R f(z) \: dz \biggr \rvert \end{align}

- Let $R_1 = Ri$. We repeat this process of subdivision on $R_1$ and so forth to get that:

\begin{align} \biggr \lvert \int_{R_n} f(z) \: dz \biggr \rvert \geq \frac{1}{4^n} \biggr \lvert \int_R f(z) \: dz \biggr \rvert \\ \end{align}

- So for each $n \in \mathbb{N}$ we have that:

\begin{align} \quad \biggr \lvert \int_R f(z) \: dz \biggr \rvert \leq 4^n \biggr \lvert \int_{R_n} f(z) \: dz \biggr \rvert \quad (*) \end{align}

- Furthermore, $\overline{R_n} \subset \overline{R_{n-1}} \subset ... \subset \overline{R_1} \subset \overline{R}$ for each $n \in \mathbb{N}$ is a sequence of nested compact boxes that shrink to some point $z_0 \in A$ as $n \to \infty$. Since $f$ is analytic at $z_0$ we must have that $f'(z_0)$ exists, that is $\displaystyle{f'(z_0) = \lim_{n \to \infty} \frac{f(z) - f(z_0)}{z - z_0}}$ exists. So for all $\epsilon_1 > 0$ there exists a $\delta > 0$ such that if $\mid z - z_0 \mid < \delta$ then:

\begin{align} \quad \biggr \lvert \frac{f(z) - f(z_0)}{z - z_0} - f'(z_0) \biggr \rvert < \epsilon_1 \end{align}

- For a given $\epsilon_1 > 0$ we can then choose $n$ sufficiently large so that $\displaystyle{\frac{1}{2^n} \mathrm{diag} (R) < \delta}$ and such that:

\begin{align} \quad \mid f(z) - f(z_0) - f'(z_0)(z - z_0) \mid \leq \frac{\epsilon_1}{2^n} \mathrm{diag} (R) \end{align}

- Now note that $f'(z_0)$ is a constant (and a continuous function on $\overline{R_n}$) and $f'(z_0)(z - z_0)$ is a continuous function on $\overline{R_n}$ where $\overline{R_n}$ is an open and connected set. Hence $\displaystyle{\int_{R_n} f'(z_0) \: dz = 0}$ and $\displaystyle{\int_{R_n} f'(z)(z - z_0) \: dz = 0}$. So we rewrite $\displaystyle{\int_{R_n} f(z) \: dz}$ as follows:

\begin{align} \quad \biggr \lvert \int_{R_n} f(z) \: dz \biggr \rvert & = \biggr \lvert \int_{R_n} f(z) \: dz - \int_{R_n} f(z_0) \: dz - \int_{R_n} f'(z_0)(z - z_0) \: dz \biggr \rvert \\ &= \biggr \lvert \int_{R_n} [f(z) - f(z_0) - f'(z_0)(z - z_0)] \: dz \biggr \rvert \\ & \leq \int_{R_n} \mid f(z) - f(z_0) - f'(z_0)(z - z_0) \mid \mid dz \mid \\ & \leq \int_{R_n} \frac{\epsilon_1}{2^n} \mathrm{diag} (R) \mid dz \mid \\ & \leq \frac{\epsilon_1}{2^n} \mathrm{diag} (R) \int_{R_n} \mid dz \mid \\ \end{align}

- But note that $z = x + yi$ so $dz = x'(t) + y'(t)i$ and $\mid dz \mid = \sqrt{(x'(t))^2 + (y'(t))^2} \: dt$. Noting that $l(R_n) = \frac{1}{2^n} l(R)$ we have that:

\begin{align} \quad \biggr \lvert \int_{R_n} f(z) \: dz \biggr \rvert \leq \frac{\epsilon_1}{2^n} \mathrm{diag} (R) \int_{R_n} \sqrt{(x'(t))^2 + (y'(t))^2} \: dt = \frac{\epsilon_1}{2^n} \mathrm{diag} (R) l(R_n) = \frac{\epsilon_1}{2^n} \mathrm{diag} (R) \frac{l(R)}{2^n} = \frac{\epsilon_1}{4^n} \mathrm{diag} (R) l(R) \quad (**) \end{align}

- Using $(*)$ and $(**)$ together yields:

\begin{align} \quad \biggr \lvert \int_R f(z) \: dz \biggr \rvert \leq 4^n \cdot \frac{\epsilon_1}{4^n} \mathrm{diag} (R) l(R) = \epsilon_1 \mathrm{diag} (R) l(R) \end{align}

- Note that $\epsilon_1$ is arbitrary and $\mathrm{diag} (R)$ and $l(R)$ are fixed. So for all $\epsilon > 0$ we have that $\displaystyle{\biggr \lvert \int_R f(z) \: dz \biggr \rvert < \epsilon}$. Thus:

\begin{align} \quad \int_R f(z) \: dz = 0 \quad \blacksquare \end{align}