Uniform convergence in the proof of the Cauchy integral formula in Stein-Shakarchi

Question

The following is the proof of the Cauchy integral formula for derivatives of holomorphic functions in the Complex Analysis by Stein-Shakarchi:

Would anybody show me why one has uniform convergence in the integrand as $h\to 0$ (so that one can pass the limit inside the integral)?

---

[Updated:] As a particular case, consider $n=1$. Let $$ F(\zeta,h;z)=\frac{f(\zeta)}{(\zeta-z-h)(\zeta-z)}. $$ What I want to show is $$ \lim_{h\to 0}\int_CF(\zeta,h;z) d\zeta=\int_CF(\zeta,0;z)\ d\zeta. $$ So I need a uniform bound for $$ |F(\zeta,h;z)-F(\zeta,0;z)|, $$ which is the part I get stuck.

Answer 1

If $C$ is a closed simple curve and $z$ is in the inner region defined by $C$, then the distance $d(z,C)$ is positive because $C$ and $\{z\}$ are disjoint compact sets. Take $\delta:=\frac{1}{2}d(z,C)$. Consider the ball $B:=B_\delta(z)$ and observe that if $\zeta$ is in $C$ and $w$ in $B$ then $\frac{1}{|\zeta - w|}\leq \frac{1}{\delta}$. For any $h \in \mathbb{C}$ with $|h|<\delta$ the point $z+h$ is in $B$ and you have $$\left| \frac{h}{(\zeta-z-h)(\zeta-z)} \right| = \frac{|h|}{ |\zeta-(z+h)| |\zeta-z|} \leq \frac{|h|}{ \delta \, \cdot \, 2\delta}$$

for every $\zeta$ in $C$.

Note that the bound works with any $h$ such that $|h|<\delta$ and because of this taking limit $h\to 0$ is possible.

Answer 2

For each positive integer $k$, $g_k(\zeta):=(\zeta-z+k^{-1})^{-1}$ and $g(\zeta):=(\zeta-z)^{-1}$ in $E$ where $E$ a closed annulus containing $C$ but not containing $z$. Then $g_k\to g$ uniformly in $E$ as $k\to\infty$.

Proof: $$\exists r>0:|\zeta-z|>r\;\forall\;\zeta\in E\\ \implies |\zeta-z+k^{-1}|\ge||\zeta-z|-k^{-1}|\;\forall\;\zeta\in E$$

Assuming $k>r^{-1}$ leads to $$ |\zeta-z+k^{-1}|>r-k^{-1}>0\;\forall\;\zeta\in E\\ \implies \Big{|}{1\over \zeta-z+k^{-1}}-{1\over\zeta-z}\Big{|}={1\over k}{1\over|\zeta-z+k^{-1}|}{1\over|\zeta-z|}<{1\over kr(r-k^{-1})}={1\over r(rk-1)}\\ \implies \sup_{\zeta\in E}\Big{|}g_{k}(\zeta)-g(\zeta)\Big{|}<{1\over r(rk-1)} \implies \lim_{k\to\infty}\sup_{\zeta\in E}\Big{|}g_{k}(\zeta)-g(\zeta)\Big{|}=0 $$

$$\sum_{m=0}^{n-1}g_k(\zeta)^mg(\zeta)^{n-1-m}\to ng(\zeta)^{n-1}$$ uniformly in $E$.

Proof: We can use the fact that for bounded functions, $\phi_n\to\phi$ and $\theta_n\to\theta$ uniformly on the same domain $\implies \phi_n+\theta_n\to\phi+\theta,\; \phi_n\theta_n\to\phi\theta$ uniformly in that domain.

Answer 3

This is essentially Pipicito's answer. I would like to elaborate it more.

As a particular case, consider $n=1$. Fix $z$ inside $C$. Let $$ F(\zeta,h;z)=\frac{f(\zeta)}{(\zeta-z-h)(\zeta-z)}. $$ One wants to show $$ \lim_{h\to 0}\int_CF(\zeta,h;z) d\zeta=\int_CF(\zeta,0;z)\ d\zeta. $$ We calculate a uniform (in $\zeta\in C$) bound for $$ |F(\zeta,h;z)-F(\zeta,0;z)|. $$

Now $$ \left|F(\zeta,h;z)-F(\zeta,0;z)\right|=\left|\frac{f(\zeta)}{(\zeta-z)^2}\frac{h}{\zeta-z-h}\right| \leq M\cdot \left|\frac{h}{\zeta-z-h}\right| $$ where $M=\sup_{z\in C}|f(\zeta)/(\zeta-z)^2|$. Let $r>0$ be small enough so that $B(z,r)$is contained inside $C$. Let $$ d=\textrm{dist}(C,B(z,r)). $$ Note that $d>0$ (since it is the distance of two disjoint compact sets). Then for small enough $|h|$, for every $\zeta\in C$, $$ |\zeta-z-h|>d $$ and thus $$ \left|F(\zeta,h;z)-F(\zeta,0;z)\right|\leq \frac{M}{d} \cdot|h|. $$

For the general case, one can similarly give a uniform bound to $$ |A^{n-1}+A^{n-2}B+\cdots+B^{n-1}|. $$