I need to show that:
Let $R>0$ and let $f:\{ z\in \mathbb{C} \mid \lvert z \rvert < R \} \rightarrow \mathbb{C}$ be given by a convergent power series $\sum_0^\infty a_n z^n$. Show that $f \in C^\infty$.
I know the following facts:
Let the power series $\sum_0^\infty a_n z^n$ have convergence radius $R>0$. Then the function $f:\{ z\in \mathbb{C} \mid \lvert z \rvert < R \} \rightarrow \mathbb{C}$ given by $\sum_0^\infty a_n z^n$ is holomorphic and its derivative is $\sum_{n=1}^\infty na_nz^{n-1}$.
The convergence radius of $\sum_{n=1}^\infty na_nz^{n-1}$ is again $R$.
The following hint was given:
From real multivariate analysis we know that a continuous function, whose continuous first-order partial derivatives exist and are $C^k$, is also in $C^{k+1}$.
If I am not mistaken we can use the first two points to say that $f^\prime$ is holomorphic. If I am not mistaken, we can repeatedly use this argumentation to prove that $f \in C^\infty$; so I do not understand why this hint was given. Could you explain?
Best Answer
The radius of convergence $R$ of $\sum a_n z^n$ is given by
$$\frac{1}{R}= \lim \sup |a_n^{1/n}|.$$
So, the radius of convergence $R^\prime$ of $\sum n a_n z_n$ is
$$\frac{1}{R^\prime}= \lim \sup | (n a_n)^{1/n}|= \lim \sup | n^{1/n} a_n^{1/n}|=\frac{1}{R}$$
because $$\lim_{n\rightarrow \infty} n^{1/n} \rightarrow 1.$$
Let $q=n^{1/n}$. Then $\lim_{n\rightarrow \infty } \log q =\frac{\log n}{n} =0 $ by L'Hopital's rule and thus $q \rightarrow 1.$
For the last part ($f \in C^\infty$):
We have just shown that $f$ analytic in a disk of radius $R \Rightarrow f^\prime$ is analytic in the same disk of radius $R$. So $f^{k}$ analytic $\Rightarrow f^{k+1}$ analytic. By induction $f^{n}$ is analytic for all $n$. Thus $f\in C^\infty$.