Cauchy’s theorem for triangles

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complex.pdf page41

Theorem 7.2. Suppose that $U ⊆ C$ is an open subset and let $T ⊆ U$ be a triangle whose interior is entirely contained in $U$. Then if $f : U → C$ is holomorphic we have$$\int_{T} f(z)\mathrm{\ d}z=0$$ Proof. The proof proceeds using a version of the "divide and conquer" strategy one uses to prove the Bolzano-Weierstrass theorem. Suppose for the sake of contradiction that $\int_T f(z)\mathrm{\ d}z \neq 0$, and let $I=\left|\int_T f(z)\mathrm{\ d}z\right|>0$. We build a sequence of smaller and smaller triangles $T^n$ around which the integral of $f$ is not too small, as follows: Let $T^0=T$, and suppose that we have constructed $T^i$ for $0 \leq i< k$. Then take the triangle $T^{k-1}$ and join the midpoints of the edges to form four smaller triangles, which we will denote $S_i(1 \leq i \leq 4)$.

Figure 2

Then we have $\int_{T^{k-1}} f(z)\mathrm{\ d}z=\sum_{i=1}^4 \int_{S_i} f(z)\mathrm{\ d}z$, since the integrals around the interior edges cancel (see Figure 2). In particular, we must have \[ I_k=\left|\int_{T^{k-1}} f(z)\mathrm{\ d}z\right| \leq \sum_{i=1}^4\left|\int_{S_i} f(z)\mathrm{\ d}z\right| \] so that for some $i$ we must have $\left|\int_{S_i} f(z)\mathrm{\ d}z\right| \geq I_{k-1} / 4$. Set $T^k$ to be this triangle $S_i$.
Then by induction we see that $\ell\left(T^k\right)=2^{-k} \ell(T)$ while $I_k \geq4^{-k} I$
Now let $\mathcal{T}$ be the solid triangle with boundary $T$ and similarly let $\mathcal{T}^k$ be the solid triangle with boundary $T^k$. Then we see that $\operatorname{diam}\left(\mathcal{T}^k\right)=2^{-k} \operatorname{diam}(\mathcal{T}) \rightarrow 0$, and the sets $\mathcal{T}^k$ are clearly nested. It follows from Lemma 205 that there is a unique point $z_0$ which lies in every $\mathcal{T}^k$. Now by assumption $f$ is holomorphic at $z_0$, so we have \[ f(z)=f\left(z_0\right)+f^{\prime}\left(z_0\right)\left(z-z_0\right)+\left(z-z_0\right) \psi(z), \] where $\psi(z) \rightarrow 0=\psi\left(z_0\right)$ as $z \rightarrow z_0$. Note that $\psi$ is continuous and hence integrable on all of $U$. Now since the linear function $z \mapsto f^{\prime}\left(z_0\right) z+f\left(z_0\right)$ clearly has a primitive, it follows from Theorem 5.19 \[ \int_{T^k} f(z)\mathrm{\ d}z=\int_{T^k}\left(z-z_0\right) \psi(z)\mathrm{\ d}z \] Now since $z_0$ lies in $\mathcal{T}^k$ and $z$ is on the boundary $T^k$ of $\mathcal{T}^k$, we see that $\left|z-z_0\right| \leq \operatorname{diam}\left(\mathcal{T}^k\right)=2^{-k} \operatorname{diam}(T)$. Thus if we set $\eta_k=\sup _{z \in T^k}|\psi(z)|$, it follows by the estimation lemma that\begin{aligned} I_k=\left|\int_{T^k}\left(z-z_0\right) \psi(z)\mathrm{\ d}z\right| & \leq \eta_k \cdot \operatorname{diam}\left(T^k\right) \cdot\ell\left(T^k\right) \\ &=4^{-k} \eta_k \cdot \operatorname{diam}(T) \cdot \ell(T) . \end{aligned}But since $\psi(z) \rightarrow 0$ as $z \rightarrow z_0$, it follows $\eta_k \rightarrow 0$ as $k \rightarrow \infty$, and hence $4^k I_k \rightarrow 0$ as $k \rightarrow \infty$. On the other hand, by construction we have $4^k I_k \geq$ $I>0$, thus we have a contradiction as required.

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lectures14-Earl.pdf page68
Lemma 205 Let $M$ be a compact metric space and $C_n$ a decreasing sequence of closed nonempty subsets. Then $\bigcap C_n \neq \varnothing$.

Proof Let $U_n=M \backslash C_n$ so that the $U_n$ form an increasing sequence of open subsets. Suppose that the intersection $\bigcap C_n$ is empty and then, by De Morgan's Law
\[
\bigcup U_n=\bigcup\left(M \backslash C_n\right)=M \backslash \bigcap C_n=M,
\]
shows that the $U_n$ form an open cover for $M$. By compactness there are $n_1<n_2<\cdots<n_k$ such that
\[
U_{n_k}=U_{n_1} \cup \cdots \cup U_{n_k}=M
\]
but this leads to the conclusion that $C_{n_k}=\varnothing$, a contradiction.