Let $(f_n)_{n\geq 1}$ be a sequence of holomorphic functions in $\mathbb{D}$. I want to prove that if:
- For some $M>0$, $|f_n(z)|\leq M$ for every $z\in \mathbb{D}$ and every $n\geq1$.
- $\lim_{n\rightarrow\infty}f^{(k)}_n(z)=0$ for all $k\geq 0$
then $f_n$ converges uniformly on compact sets to $0$. What I have to show is that for every $R\in (0,1)$ $$\lim_{n\rightarrow\infty}\sup_{z\in \overline{\mathbb{D}(0,R)}}|f_n(z)|=0$$ I don't know how to use the info that I have on the derivatives in this expression. Can someone help me?
In this answer, the convergence of functions is always supposed to be uniform on compact sets.
By Montel's theore, $\{f_n\}$ is a normal family so there's a subsequence such that $f_{n_k}\to f$. Cauchy's formula now implies that $f_{n_k}^{(m)}\to f^{(m)}$, so $f^{(k)}(z_0)=0$ for every $k\ge 0$, which implies $f\equiv 0$. The same reasoning proves that every subsequence of $f_n$ has a subsequence converging to $0$ and this proves by a standard argument in metric topology that $f_n\to 0$.