Proving $g'(z)=\frac{1}{2\pi{}i}\int_{C}\frac{g(u)du}{(u-z)^2}$ for $g(z)$ holomorphic in and on contour $C$, and $z$ in $C$'s interior

Question

Prove that if $g(z)$ is holomorphic everywhere inside and on a simple closed contour $C$, taken in a positive sense, and $z$ is any point interior to $C$, then $$g'(z)=\dfrac{1}{2\pi{}i}\int_{C}\dfrac{g(u)du}{(u-z)^2}$$

So far, I know this: If you let $f$ be holomorphic everywhere inside and on a simple closed contour $C$ taken in the positive sense, and if $z_0$ is any point interior to $C$, then

$$f(z_0)=\frac{1}{2\pi{}i}\int_{C}\frac{f(z)dz}{z-z_0}$$

However, I do not know how to attempt this proof using the knowledge I currently have.

Answer 1

For another approach (without power series), we use the definition of derivative. You want to compute

$f'(z_0)= \lim_{z \to z_0} \frac{1}{z - z_0} \left [ \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w - z} \: dw - \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w - z_0} \: dw \right ].$

The right hand side of this simplifies to

$\lim_{z\to z_0} \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z)(w - z_0)} \: dw.$

Now the issue is the interchange of the integral with the limit, for if this is valid, then you get

$\lim_{z\to z_0} \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z)(w - z_0)} \: dw=\frac{1}{2\pi i} \int_{\gamma} \lim_{z\to z_0}\frac{f(w)}{(w - z)(w - z_0)} \: dw=\int_{\gamma}\frac{f(w)}{(w - z_0)^2} \: dw$, as desired.

All that remains therefore is to justify the limit/integral switch: a fancy way would be to use the dominated convergence theorem, which applies since $f$ is bounded on $\gamma$ and there is a number $\delta$ such that $d(z_0,\gamma)=\delta>0$ and we may also assume that $d(z,\gamma)>\delta$ for all $z$ close enough to $z_0.$

But you can also argue directly: Let $M\max_{w\in \gamma} {|f(w)}|,\ \delta$ as in the above remark and $\ell(\gamma)$ the length of $\gamma.$ Then,

$\begin{align}\left |\int_{\gamma} \frac{f(w)}{(w - z)(w - z_0)} \: dw-\int_{\gamma}\frac{f(w)}{(w - z_0)^2} \: dw\right |\le \frac{M}{\delta}\cdot \left |\int_{\gamma} \frac{1}{(w - z)} \: dw-\int_{\gamma}\frac{1}{(w - z_0)} \: dw\right |\le \frac{M}{\delta}\cdot \int_{\gamma}\left|\frac{z-z_0}{(w-z)(w-z_0)}\right|dw\le \frac{M}{\delta^3}\int_{\gamma}|z-z_0|dw\le \frac{M\cdot \ell((\gamma)}{\delta^3}|z-z_0|\end{align}$

To finish, let $z\to z_0.$

For the $kth$ derivative, this calculation goes through virtually unchanged.

Answer 2

To prove the general case you can take limits of difference quotients and use induction (or expand the Cauchy integral in a power series as I do below). In this case, since you only want the formula for the first derivative, you can simply pass the derivative through the integral; namely, $$f^\prime(z) = \frac{d}{dz}\left(\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta - z} \ d\zeta \right) = \frac{1}{2\pi i}\int_C \frac{d}{dz}\left(\frac{f(\zeta)}{\zeta - z}\right) \ d\zeta.$$

Edit: The final step is to compute $$\frac{d}{dz}\left(\frac{f(\zeta)}{\zeta-z}\right) = f(\zeta)\frac{d}{dz}\left(\frac{1}{\zeta-z}\right) = \frac{f(\zeta)}{(\zeta-z)^2}.$$

Edit 2: I will actually not do the inductive proof, but rather a proof based on expanding a Cauchy integral as a power series. If you want to see the inductive proof, there are plenty of websites with said proof.

First, note that we can write $$f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta - z} \ d\zeta,$$

Fix a point $z_0$ off $\gamma$, set $R = \mathrm{dist}(z_0,\gamma)$ and let $z$ be a point satisfying $\lvert z - z_0 \rvert < R$. Notice that $$\frac{1}{\zeta-z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0}\left(\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}\right) = \sum_{n=0}^\infty \frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}},$$ where the power series converges locally uniformly wrt $\zeta$ in $\bigg\lvert\frac{z-z_0}{\zeta-z_0}\bigg\rvert<1$ and so converges uniformly on $\gamma$. Since $f$ is bounded on $\gamma$, the series $$\frac{1}{2\pi i}\sum_{n=0}^\infty \frac{f(\zeta)(z-z_0)^n}{(\zeta-z_0)^{n+1}}$$ converges uniformly on $\gamma$ to $\frac{f(\zeta)}{\zeta - z}$. Therefore, we can integrate the power series term-by-term over $\gamma$: $$f(z) = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} \ d\zeta\right) (z-z_0)^n.$$ Therefore, the terms $\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} \ d\zeta$ are the coefficients in the power series for $f$ centered about $z=z_0$ and therefore $$\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} \ d\zeta = \frac{f^{(n)}(z_0)}{n!}.$$

Edit 3: This is a bit of a tangent, so feel free to skip it if you aren't in the mood to indulge me. One reason why I like (and chose to present) this proof is that, in addition to proving the Cauchy integral formula (for derivatives), it actually can (very clearly) tell us something about Cauchy integrals in general. Let $\gamma:[a,b] \to \mathbb{C}$ be a piecewise-$C^1$ closed curve and let $\phi: G \supset \gamma \to \mathbb{C}$ (slightly abusing notation) be continuous on $\gamma$. Define $$F(z) = \int_\gamma \frac{\phi(\zeta)}{\zeta - z} \ d\zeta.$$ Then, the (essentially) same proof as above shows that $F$ is analytic on $\mathbb{C}\setminus\gamma$. So, we can start with a function which is merely continuous, define its Cauchy integral and get back an analytic function. It also, of course, tells us that all holomorphic functions are analytic.