Possible proof of Cauchy’s Integral Formula for derivatives – completion and verification

Question

First, let me state Cauchy's Integral formula:

Let $U$ be an open region in the complex plane and $D = \{z : |z-z_0| \leq R\}$ a disk in $U$. If $f : U \to \mathbb C$ is holomorphic and $\gamma$ is the boundary of $D$, then for all points $a$ inside the disk
$$ f(a) = \frac 1{2πi}\oint_\gamma \frac{f(z)}{z-a}\ dz$$

and Cauchy's Integral Formula for the derivatives:

$$f^{(n)}(a) = \frac{n!}{2πi}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\ dz$$

The majority of books prove the second formula by differentiation of the first formula and then proceeding by induction. However, it seems like Wikipedia wants to suggest a different proof that I couldn't find anywhere else, but they only give hints:

Since ${\displaystyle 1/(z-a)}$ can be expanded as a power series in the variable ${\displaystyle a}$:
${\displaystyle {\frac {1}{z-a}}={\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}}$
— it follows that holomorphic functions are analytic, i.e. they can be expanded as convergent power series. In particular f is actually infinitely differentiable

So I took the time to try to understand what they are actually suggesting. What follows is my attempt to turn Wikipedia's hint into a proof, however, it's incomplete because I don't know how to justify the red-marked interchange of integral and series:

Let $a$ be a point inside $D$ and $w$ a point in the largest open disk around $a$ contained in $D$. Then, $|w-a| < |z-a|$ for all $z$ on $\gamma$, so $\left|\frac{w-a}{z-a}\right|<1$ and then
$$\sum_{n=0}^\infty \frac{(w-a)^n}{(z-a)^{n+1}} = \frac{1}{z-a} \frac{1}{1-\frac{w-a}{z-a}} = \frac 1{(z-a)-(w-a)} = \frac 1{z-w} $$

Applying Cauchy's Integral formula:
$$\begin{align} f(w) &= \frac 1{2πi}\oint_\gamma \frac{f(z)}{z-w}\ dz \\ &= \frac 1{2πi}\oint_\gamma \sum_{n=0}^\infty f(z)\frac{(w-a)^n}{(z-a)^{n+1}}\ dz \\ &\color{red}{=} \sum_{n=0}^\infty \frac 1{2πi}\oint_\gamma f(z)\frac{(w-a)^n}{(z-a)^{n+1}}\ dz \\ &= \sum_{n=0}^\infty \left(\frac 1{2πi}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\ dz\right)(w-a)^n \end{align}$$
but this is a power series in $w$ around $a$. And we know that power series are infinitely-differentiable in their radius of convergence with
$$ f(w) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(w-a)^n $$ therefore $$ f^{(n)}(a) = \frac{n!}{2πi}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\ dz $$

Is the red-marked interchange valid? Apart from the interchange, is everything alright?

Answer 1

I felt like elaborating more so here's a more complete answer. Verify the following for yourself:

• Use the Weierstrass $M$-test to show the following: Consider the functions $f_n(\zeta)=\zeta^n$, defined for $|\zeta|<1$. Then, for any $0\leq r_0<1$, the series $\sum_{n=0}^{\infty}f_n(\zeta)$ converges absolutely and uniformly (to $\frac{1}{1-\zeta}$). As long as we have uniform convergence, we can easily interchange limits with integrals, series (which is itself a limit) with integrals.

I think this power series argument is very slick because it completely bypasses the differentiation under the integral sign issue, and it directly proves holomorphic implies analytic, and furthermore that the power series expansion is valid on the whole disc of holomorphicity (if you differentiate under the integral sign, all you manage to show is that $f$ is infinitely complex differentiable, but the existence of a power series expansion still needs to be proved. I just checked and this whole argument takes about 2-3 pages in Ahlfors's book, but merely under 1 page in Cartan's book).

On the other hand, one has to be slightly careful about the radii of the various discs; this is one of those "open set chasing games". Regarding this, I have one issue with your proposed proof:

Let $a$ be a point inside $D$ and $w$ a point in the largest open disk around $a$ contained in $D$. Then, $|w−a|<|z−a|$ for all $z$ on $\gamma$,

I'm not sure that this is good enough. Sure, this way, for each $w$ we can expand as a series $\frac{1}{z-w}=\sum_{n=0}^{\infty}\frac{(w-a)^n}{(z-a)^{n+1}}$, but I doubt the convergence is uniform (because the Weierstrass M-test only guarantees uniform convergence on compact subsets of the unit disc), so we need to ensure $\sup_{z,w}\left|\frac{(w-a)^n}{(z-a)^{n+1}}\right|<1$, and based on the current way you've phrased it I doubt this is true. So, as written, your series and integral interchange is unjustified.

Here's how I'd phrase the argument (essentially what Cartan does, except he WLOG assumes $z_0=0$) to prove holomorphic implies analytic (and hence easily "read off" the formula for the derivatives).

We will show that for each $0\leq r<R$, there is some power series centered at $z_0$ which converges absolutely and uniformly to $f$ in the disk $\{|w-z_0|\leq r\}$. But now by the uniqueness of power series expansions about a point, the coefficients of this series actually do not depend on $r$. Since $0\leq r<R$ is arbitrary, it follows that the power series has radius of convergence $\geq R$, and has sum equal to $f(w)$; this will complete the proof.

Now, let $0\leq r<r_0<R$ be given (the $r_0$ is the "extra buffer" we're giving ourselves). Let $\gamma_0$ denote the positively oriented circle of radius $r_0$ centered at $z_0$. Now, for any $|w-z_0|\leq r$ and $z$ lying on $\gamma_0$, we have $\left|\frac{w-z_0}{z-z_0}\right|\leq \frac{r}{r_0}<1$. This provides us with a uniform estimate so that we can use the Weierstrass M-test below to deduce uniform convergence and thus interchange the series and integral. So, for any $|w-z_0|\leq r$, we have that $w$ lies interior to $\gamma_0$ so we can use Cauchy's formula to get \begin{align} f(w)&=\frac{1}{2\pi i}\int_{\gamma_0}\frac{f(z)}{z-w}\,dz\\ &=\frac{1}{2\pi i}\int_{\gamma_{0}}\frac{f(z)}{z-z_0}\cdot\frac{1}{1-\frac{w-z_0}{z-z_0}}\,dz\\ &=\frac{1}{2\pi i}\int_{\gamma_{0}}\sum_{n=0}^{\infty}\frac{f(z)}{(z-z_0)^{n+1}}(w-z_0)^n\,dz\\ &=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma_{0}}\frac{f(z)}{(z-z_0)^{n+1}}\,dz\right)(w-z_0)^n. \end{align} Thus, we have found an appropriate power series expansion, and all the steps are justified (perhaps up to some typo), so the proof is complete.

Some remarks I should make: when I made the series expansion above, because we ensured \begin{align} \sup\limits_{\substack{|w-z_0|\leq r\\ |z-z_0|=r_0}}\left|\frac{w-z_0}{z-z_0}\right| &\leq \frac{r}{r_0}<1, \end{align} the geometric series converges absolutely and uniformly with respect to both $w$ and $z$. So, after integrating with respect to $z$, the convergence of the series to $f(w)$ is still uniform with respect to $w$; so everything really is as nice as it can be. Lastly, in hindsight, we can invoke Cauchy's theorem once again to apply a homotopy to the path $\gamma_0$ without changing any of the integral values. Thus, we do not have to assume our path is a circle of a given radius and having center $z_0$.