# A version of Cauchy's integral formula

I'm trying to prove this version of Cauchy's integral formula that is little more general than what appears in Stein and Shakarchi "complex analysis":

[Let $\rho>0$ and assume that $f$ is continuous on the closed ball $\overline B_\rho(0)$ and holomorphic in the open ball $B_\rho(0)$. Let $w\in B_\rho(0).$ Prove $f(w)=\frac{1}{2\pi i}\int_{\partial B_\rho(0)} \frac{f(z)}{z-w}dz$ where the orientation is positive. Here, $B_\rho(0)$ is the open ball centered at 0 with radius $\rho$ and $\overline B_\rho(0)$ is the closure.]

What I'm referring to as Stein and Shakarchi's is as follows (p.45, slightly modified notation):

[Suppose $f$ is holomorphic in an open set that contains $\overline B_\rho(0)$. Then $f(w)=\frac{1}{2\pi i}\int_{\partial B_\rho(0)} \frac{f(z)}{z-w}dz$.]

What I'm thinking

If we fix $w$, then we can think of an open ball that contains $w$ that is contained in $B_\rho(0)$. Let's call it $B_r(0)$ with $r<\rho$. Applying the Stein and Shakarchi's, we get the equation $f(w)=\frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(z)}{z-w}dz$ hold for any point $w\in B_r(0).$ The remaining problem is that whether $\lim_{r\rightarrow \rho}\frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(z)}{z-w}dz=\frac{1}{2\pi i}\int_{\partial B_\rho(0)} \frac{f(z)}{z-w}dz$.

Am I on the right track? I would be grateful for any comment.

• Yes you are. Now take a parametrization of $\partial B_r(0)$ that approaches a parametrization of $\partial B_{\rho}(0)$ and use the uniform continuity of $f$ on $\bar B_{\rho}(0)$. – orangeskid Oct 1 '14 at 3:38
• Thank you very much, orangeskid. Am I right to simply say that the ∂Br(0) is γ(t)=rcost+irsint,0≤t≤2π, so it approaches the parametrization of $∂B_\rho(0) γ′(t)=ρcost+iρsint,0≤t≤2π$ as $r→ρ$, or do I need some argument for it? – stph Oct 1 '14 at 12:53
• No worries! Yes, for these parametrizations $|\gamma_r(t) - \gamma_{\rho}(t)| = (r - \rho)$ so $\gamma_{\rho}$ approaches $\gamma_r$ uniformly in $t$. – orangeskid Oct 1 '14 at 14:44
• Thank you very much. I will put up my answer sometime. – stph Oct 1 '14 at 23:23
• I posted an answer. I know it's not perfect, for instance, I don't know how to put in the argument about the parametrization approaching to another parametrization. Would you please help me refine it? – stph Oct 2 '14 at 0:12

Fix $w \in B_\rho (0)$. Choose $r>0$ such that $|w|<r<\rho$.

Then $\overline B_r(0)\subset B_\rho(0)$.

Apply Stein and Shakarchi's to obtain $f(w)=\frac{1}{2πi}\int_{∂B_r(0)}\frac{f(z)}{z−w}dz=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{f(re^{it})}{re^{it}-w}ire^{it}dt$.

Observe that $h(r,t)\equiv\frac{f(re^{it})}{re^{it}-w}ire^{it}$ is uniformly continuous on $\overline B_\rho(0)$. Now, take arbitrary sequence of points ${r_k}$ such that $r_k\rightarrow \rho$ as $k\rightarrow \infty$, put $g_k(t)\equiv h(r_k,t)$, and $g(t)\equiv h(\rho,t)$. Then $g_k\rightarrow g$ uniformly. Hence,

$\lim_{r\rightarrow\rho}\frac{1}{2πi}\int_{0}^{2\pi}\frac{f(re^{it})}{re^{it}-w}ire^{it}dt=\frac{1}{2πi}\int_{0}^{2\pi}\frac{f(\rho e^{it})}{\rho e^{it}-w}i\rho e^{it}dt=\frac{1}{2πi}\int_{∂B_\rho(0)}\frac{f(z)}{z−w}dz.$

Therefore, the formula

$f(w)=\frac{1}{2πi}\int_{∂B_\rho(0)}\frac{f(z)}{z−w}dz$

holds for arbitrary $w\in B_\rho (0)$.

• It wouldn't hurt to have a bound on the difference of the integrands: $$\left|\frac{r f(r e^{it})}{re^{it}-w} - \frac{\rho f(\rho e^{it})}{\rho e^{it}-w}\right|.$$ – Ted Shifrin Oct 2 '14 at 0:35
• I think the principled way is to look at the function $(r, \theta) \mapsto \frac{f(r e^{i\theta})}{r e^{i \theta} - w}$. This is a continuous funcion in $(r, \theta)$ so also uniformly continuous. That would prove the uniform convergence. – orangeskid Oct 2 '14 at 9:08
• Thank you very much for help. I still don't understand how I need a bound on the difference of the integrands, and also how I can really show uniform convergence. Do uniform continuity and compactness together imply uniform convergence? – stph Oct 6 '14 at 1:21