Understanding the proof of the Cauchy Integral Formula

Question

So the Cauchy integral formula goes like this,

Suppose $f$ is analytic on a domain G containing a simple closed contour $\gamma$, If $z_0$ is an interior point of $\gamma$, then $f(z_0)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-z_0}dz$.

The proof goes like this,

$\int_{\gamma}\frac{f(z)}{z-z_0}dz=\int_{C_r}\frac{f(z)}{z-z_0}dz$, where $C_r=\{|z-z_0|=r\}$, then $\int_{C_r}\frac{f(z)}{z-z_0}dz=\int_{C_r}\frac{f(z)+f(z_0)-f(z_0)}{z-z_0}dz=f(z_0)\int_{C_r}\frac{1}{z-z_0}dz+\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}dz=f(z_0)2\pi i+\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}dz$.

Now, here is where I don't understand. Define $g(z)=\begin{cases}\frac{f(z)-f(z_0)}{z-z_0}&z\neq z_0\\f'(z_0)&z=z_0\end{cases}$. Now let $z\rightarrow z_0\Rightarrow \exists M>0$ such that $|g(z)|\leq M\quad \forall z\in C_r\Rightarrow |f(z)|\leq M\Rightarrow |\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}dz|\leq M2\pi r$. Taking $r\rightarrow 0 \Rightarrow |\int_{C_r}\frac{f(z)}{z-z_0}dz|=0$. Thus proved.

So what I don't understand is how they deduced that there exists an M that is an upper bound for the function by taking $z$ to $z_0$, what theorem or definition would that come from? And how are they deducing that if |f| is bounded above from |g| being bounded above?

Answer 1

• If $|g(z)| < M$ for $|z-z_0| < R$ then for $r< R$ : $$\left|\int_{C_r} g(z)dz\right| = \left|\int_0^{2\pi} g(z_0+r e^{it}) r i e^{it}dt\right| < r 2\pi M$$

• And such an upper bounded exists because $g(z)$ is continuous (by definition of $f(z)$ is holomorphic at $z_0$) and $|z-z_0| \le R$ is compact

Answer 2

The statement "$|f(z)|\leq M$" occurring in the phrase

"Now let $z\rightarrow z_0$. Then there exists $M>0$ such that $|g(z)|\leq M\quad \forall z\in C_r\ $. [It follows that] $|f(z)|\leq M$. Therefore $|\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}dz|\leq M2\pi r\ $"

is wrong, and should not appear here. Instead write

"Now let $z\rightarrow z_0$. Then there exists $M>0$ such that $|g(z)|\leq M\quad \forall z\in C_r\ $. Therefore $$\left|\int_{C_r}\frac{f(z)-f(z_0)}{z-z_0}dz\right|=\left|\int_{C_r} g(z)dz\right|\leq M2\pi r\ ."$$ You then finish up by saying that $$\left|2\pi i f(z_0)-\int{f(z)\over z-z_0}\>dz\right|\leq 2\pi M r$$ for any $r>0$, hence the LHS has to be $=0$.