I am trying to show:
If $f: B(0,R) \to \mathbb C$ is analytic then the Taylor series of $f$ at $0$ converges uniformly to $f$ in $B(0,r)$ for all $r\in (0,R)$
But I got stuck with my proof. Please can somebody help me? Here is what I have so far:
From here I have
$$ f(z) = \sum_{n=0}^\infty z^n {1\over 2 \pi i}\int_C {f(w) \over w^{n+1}}dw$$
Therefore the Taylor series of $f$ at $0$ is
$T_0(z) = \sum_{n=0}^\infty z^n c_n$
where $c_n = {1\over 2 \pi i}\int_C {f(w) \over w^{n+1}}dw$.
My idea is to apply the Weierstrass $M$-test here for $f_n(x) = z^n c_n$. It holds that $|z^n| < r^n$ and $|c_n|\le {M \over r^n}$ and therefore $|f_n(z)|\le M$. The problem is how to show
$\sum_{n=0}^\infty M < \infty$? It is obviously not true (consider $R=1$). Where is my mistake?
Best Answer
We have to show, that $\sum_{n=0}^\infty z^n c_n$ is unformly convergent for $|z|<r$. The trick is, that $C\subset B(0,R)$ can be any smooth closed curve rounding the origin. So take $\varrho$ such that $r<\varrho<R$ and let $C:=\{z\,|\,|z|=\varrho\}=\{\varrho e^{it}\,|\,0\leq t\leq 2\pi\}$. Then $$ |c_n| =\left|{1\over 2 \pi i}\int_C {f(w) \over w^{n+1}}dw\right|\leq \frac{1}{2\pi}\int_C\frac{|f(w)|}{\varrho^{n+1}}|dw|\leq \max_{w\in C}|f(w)|\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{\varrho^{n+1}}\varrho\,dt=\max_{w\in C}|f(w)|\frac{1}{\varrho^n}. $$ Thus it follows $$ \left|\sum_{n=0}^\infty z^n c_n\right|\leq\sum_{n=0}^{\infty} \max_{w\in C}|f(w)|\frac{r^n}{\varrho^n}=\max_{w\in C}|f(w)|\sum_{n=0}^{\infty} \left(\frac{r}{\varrho}\right)^n=\max_{w\in C}|f(w)|\frac{1}{1-(r/\varrho)}. $$ (Here we used that $0<(r/\varrho)<1$.)