Uniform convergence of coefficients of power series

I am trying to solve some of my past exam papers and got stuck on this one.

Attempt: Since $f$ is analytic in $B(0,2)$, and for $c<1$, $\overline{B(0,c)} \subset B(0,2)$, for any $z$ in $\overline{B(0,c)}$, $f$ can be written as a power series expansion around $\zeta=z$, to get $f(\zeta)= \sum_{n=0}^{\infty} \frac{f^{(n)}(z)}{n!} (\zeta-z)^n$ for $|\zeta-z|<1$.

Let $\gamma$ denote the circle with center $z$ and radius $1$. Then

$ \frac{f^{(n)}(z)}{n!}= \frac{1}{2\pi \iota} \int_{\gamma} \frac{f(\zeta)}{(\zeta-z)^{n+1}}$. By estimates, $ \big|\frac{f^{(n)}(z)}{n!}\big| \leq M$ where $M$ is maximum of $f$ on trace of $\gamma$

How to proceed from here?

How should I do part (b)?

I know part (a) gives me that the function defined by the series in analytic in $\overline{B(0,c)}$ for every $c<1$ as it is represented by a convergent power series. How can I extend it to $B(0,1)$?

Solution 1:

Fix $0<c<1$ and let $C(0,r)$ be a closed, positively oriented circle of radius $ r := 2 - \frac{1 - c}{2} < 2$ centered at the origin. Assume that $|z|\leq c$. Then, since $r > \frac{3}{2} > c$, $$ \left| {\frac{{f^{(n)} (z)}}{{n!}}} \right| = \left| {\frac{1}{2\pi i}\oint_{C(0,r)} {\frac{{f(t)}}{{(t - z)^{n + 1} }}dt} } \right| \le \frac{1}{2\pi}\oint_{C(0,r)} {\frac{{\left| {f(t)} \right|}}{{\left| {t - z} \right|^{n + 1} }}dt} . $$ Now $$ \left| {t - z} \right| \ge \left| t \right| - \left| z \right| \ge r - c = 2 - \frac{{1 + c}}{2} =:q > 1. $$ Thus, if $M := \max _{t \in C(0,r)} \left| {f(t)} \right|$ then $$ \left| {\frac{{f^{(n)} (z)}}{{n!}}} \right| \le \frac{{Mr}}{{q^{n + 1} }}. $$ Therefore, $$ \left| {\sum\limits_{n = 0}^\infty {\frac{{f^{(n)} (z)}}{{n!}}} } \right| \le \sum\limits_{n = 0}^\infty {\frac{{\left| {f^{(n)} (z)} \right|}}{{n!}}} \le \sum\limits_{n = 0}^\infty {\frac{{Mr}}{{q^{n + 1} }}} = \frac{{Mr}}{{q - 1}} < + \infty . $$ For part $(b)$, use Morera's theorem (see, e.g., here).