Generalized Cauchy's Theorem

Question

Let $f$ analytic in $\mathbb{D}=\{|z|<1\}$ and continous on $\bar{\mathbb{D}}=\{|z|\leq1\}$. Then we have

$$\int_{|z|=1}f(z)dz=0$$ and

$$f(z_0)=\frac{1}{2\pi i }\int_{|z|=1}\frac{f(z)}{z-z_0}dz, z_0\in \mathbb{D}.$$

I was thinking to prove this using approximation by polynomials. I mean to use the fact the set of polynomials in $\bar{\mathbb{D}}$ is dense in the space of continuous functions in $\bar{\mathbb{D}}$ under the $\sup$ norm. But I am not sure if this is true. Any proof? Has it to do with that $f$ is uniformly continous on $\bar{\mathbb{D}}$?

Answer 1

For each $n\in\Bbb N\setminus\{1\}$,$$\int_{|z|=1-1/n}f(z)\,\mathrm dz=0\quad\text{and}\quad f(z_0)=\frac1{2\pi i}\int_{|z|=1-1/n}\frac{f(z)}{z-z_0}\,\mathrm dz\text{ if }|z_0|<1-\frac1n.$$Now, use the fact that$$\int_{|z|=1}f(z)\,\mathrm dz=\lim_{n\to\infty}\int_{|z|=1-1/n}f(z)\,\mathrm dz$$and that$$\int_{|z|=1}\frac{f(z)}{z-z_0}\,\mathrm dz=\lim_{n\to\infty}\int_{|z|=1-1/n}\frac{f(z)}{z-z_0}\,\mathrm dz.$$

Answer 2

Hint:

Fix an $r_0$ such that $|z_0|<r_0<1$. Then, by the usual Cauchy's theorem, for any $r_0\leq r<1$, \begin{align} f(z_0)=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z-z_0}\,dz \end{align} Now, use uniform continuity of the integrand on the annulus $r_0\leq |z|\leq 1$ to take limits $r\to 1^-$ on the RHS.