1. ## Complex Analysis

The Cauchy Integral formula states the following:

Suppose $U\in \mathbb{C}$ is an open set and $f$ is holomorphic on $U$. Let $z_0 \in U$ and $r>0$ such that $\overline{D}(z_0,r)\subset U$. Let $\gamma:[0,1]\to U$ be the $C^1$ curve $\gamma(t)=z_0+r\cos(2\pi t)+ir\sin(2\pi t)$. Then $f(z)=\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ for $z\in D(z_0,r)$

Now I have to show the Cauchy-integral formula is valid if we only assume that $F\in C^0(\overline{D})$, with $D$ some disc, and $F$ is holomorphic on $D$

I'm not sure if I understand what must be shown here exactly. Can someone clarify this a little?

2. Originally Posted by Dinkydoe
The Cauchy Integral formula states the following:

Suppose $U\in \mathbb{C}$ is an open set and $f$ is holomorphic on $U$. Let $z_0 \in U$ and $r>0$ such that $\overline{D}(z_0,r)\subset U$. Let $\gamma:[0,1]\to U$ be the $C^1$ curve $\gamma(t)=z_0+r\cos(2\pi t)+ir\sin(2\pi t)$. Then $f(z)=\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ for $z\in D(z_0,r)$

Now I have to show the Cauchy-integral formula is valid if we only assume that $F\in C^0(\overline{D})$, with $D$ some disc, and $F$ is holomorphic on $D$

I'm not sure if I understand what must be shown here exactly. Can someone clarify this a little?
My approach would be something like this:

Let $g_t(z)= \frac{1}{2\pi i} \int_{\gamma _t } \frac{f(\zeta )}{\zeta -z} d\zeta$ with $t\in (\frac{1}{2} ,1)$ and $\gamma _t= te^{2\pi is}$ (assuming $z_0=0$) argue that $\lim_{t\rightarrow 1} g_t(z)$ is the function you're given and since (by the usual Cauchy integral formula) $g_t(z)=f(z)$ for all $z\in \mathbb{D}_{\frac{1}{2} }$ by Riemann's theorem we would have that $g_1(z)=f(z)$ for all $z\in \mathbb{D}$

3. ok, I take it with $\mathbb{D}_{\frac{1}{2}}$ you mean the open ball with radius $r=1/2$ around $z_0=0$.

Thanks for your answer however what theorem do you refer to when you say : " Riemann's Theorem''