Complex analysis: Using Taylor expansion to show $|c_n| ≤ \frac{1}{r^n}\sup_{z∈C_r(0)}|f(z)|$ Consider the function $f$ is defined through the power series
$$f(z) := c_0 + \sum_{n=1}^\infty c_nz^n$$
and assume that the series on the right has a radius of convergence $R > 0$. Show that
$$|c_n| ≤ \frac{1}{r^n}\sup_{z∈C_r(0)}|f(z)|$$
for any r ∈ (0, R), where $C_r(0) := \{z ∈ \mathbb{C} | |z| = r \}$. (Hint: what is the Taylor expansion
for f?)
My attempt thus far:
Finding the Taylor expansion:
General formula $$T_{m-1}(z)=\sum_{n=0}^{m-1} \frac{f^n(z_0)}{n!}(z-z_0)^n$$
$$R_m(z)=\frac{f^m(\xi_z)}{m!}(z-z_0)^m$$
$$f(z)=T_{m-1}(z)+R_m(z)$$
for m=4 and $f(z)=\sum_{n=0}^\infty c_nz^n$
$$f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2}(z-z_0)^2+\frac{f^3(z_0)}{6}(z-z_0)^3+\frac{f^4(\xi_z)}{24}(z-z_0)^4$$
$$f(z)=\sum_{n=0}^\infty c_nz_0^n+(\sum_{n=1}^\infty nc_nz_0^{n-1})(z-z_0)+\frac{\sum_{n=2}^\infty n(n-1)c_nz_0^{n-2}}{2}(z-z_0)^2+\frac{\sum_{n=3}^\infty n(n-1)(n-2)c_nz-0^{n-3}}{6}(z-z_0)^3+\frac{\sum_{n=4}^\infty n(n-1)(n-2)(n-3)c_n\xi_{z}^{n-4}}{24}(z-z_0)^4$$
I now have no clue where to go from there.
Other thoughts:
$$\lim_{n\to\infty}\sup|c_n|^{\frac{1}{n}}≤\frac{1}{r}\Rightarrow\lim_{n\to\infty}\sup|c_n|≤\frac{1}{r^n}$$
\begin{align}|f(z)|=&\sum_{n=0}^\infty |c_nz^n|\\
&=\sum_{n=0}^\infty |c_n||z|^n\\
&=\frac{1}{1-|z|}\sum_{n=0}^\infty |c_n|\\
&=\frac{1}{1-r}\sum_{n=0}^\infty |c_n|\\
\end{align}
Again I have no idea what to do from here.
 A: Did you try using Cauchy's integral formula? Since $f$ is analytic you have $$f^{(n)}(0) = \frac {n!}{2\pi i} \int_{C_r(0)} \frac{f(z)}{(z-0)^{n+1}} \, dz$$
so that
$$|f^{(n)}(0)| \le \frac{n!}{2\pi} \cdot \max_{|z| = r} |f(z)| \cdot \frac{2\pi r}{r^{n+1}} = \frac{n!}{r^n} \max_{|z| = r} |f(z)|.$$
Now consider the Taylor series. What is $f^{(n)}(0)$ equal to?
A: Fix $0<r<R$ and consider
$$f(re^{i\theta})=\sum_{n=0}^\infty c_nr^ne^{in\theta}$$
Since the series converges uniformly on $[-\pi,\pi]$,
$$c_nr^n=\frac{1}{2\pi}\int_{-\pi}^{+\pi}f(re^{i\theta})e^{-in\theta}d\theta$$
Therefore,
$$|c_n|r^n\leq\sup_{z\in C_r(0)}|f(z)|$$
for every $0<r<R$.
A: By Cauchy's integral formula, the coefficients of the Taylor expansion of a function around $ 0 $ are given by
$$ c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z^{n+1}}\, dz $$
(C is any closed curve in the neighborhood of $ a $ where $ f $ is analytic) where
$$ f(z) = \sum_{k=0}^{\infty} c_k z^k $$
Now, let $ C $ be a circle of radius $ r < R $ centered at $ 0 $. Observe the following estimates:
$$ |c_n| \leq \frac{1}{2\pi} \oint_{C} \frac{|f(z)|}{r^{n+1}}\, dz \leq \frac{1}{r^n} \max_{|z| = r} |f(z)| $$
which completes the proof. (We use the ML-lemma and the fact that the absolute value of the integral is less than or equal to the integral of the absolute value in the estimates.)
