1. ## Complex Analysis

The Cauchy Integral formula states the following:

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

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

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

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

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

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

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