About normal family of analytic functions Suppose $G$ is a connected domain in $\mathbb{C}$ and $0,1\in G$.Let $\mathbb{F}$ be the family of analytic functions $f$ defined on $G$,$f(0)=0$ and $|f|<2$ on G.Prove that there is $0<c<2$ such that $|f(1)|<c$ for all $f\in\mathbb{F}$.
Obviously $\mathbb{F}$ is locally bounded thus normal.If we use contradiction argument to assume that there is a sequence $f_n\in\mathbb{F}$ such that $f_n(1)>2-\frac{1}{2^n}$,then the limit $f$ will have value $\ge2$ at point 1.But the limit function is not necessarily in $\mathbb{F}$ and I can not draw the contradiction.Plus I don't know how to use the condition $f(0)=0$.
 A: 
Plus I don't know how to use the condition $f(0)=0$.

That ensures that the limit function does actually belong to $\mathbb{F}$.
Let $s = \sup\: \{ \lvert f(1)\rvert : f \in \mathbb{F}\}$, and $(f_n)_{n\in \mathbb{N}}$ a sequence in $\mathbb{F}$ with $\lim\limits_{n\to \infty} \lvert f_n(1)\rvert = s$. Since $\mathbb{F}$ is normal, we can extract a subsequence that converges locally uniformly on $G$. Without loss of generality we can assume that the full original sequence converges locally uniformly. Let $f_{\sup}$ be the limit function. By a theorem of Weierstraß, $f_{\sup}$ is analytic on $G$.
Since $\lvert f_n(z)\rvert < 2$ for all $z\in G$ and $n \in \mathbb{N}$, the pointwise convergence implies $\lvert f_{\sup}(z)\rvert \leqslant 2$ for all $z\in G$. Further, we have $f_{\sup}(0) = 0$. Now either $f_{\sup}$ is constant - then $f_{\sup}(z) = 0$ for all $z \in G$, and thus $f_{\sup} \in \mathbb{F}$ (this is the case for example if $\mathbb{C} \setminus G$ is finite) - or $f_{\sup}$ is not constant. In the latter case, the open mapping theorem says that $f_{\sup}(G)$ is an open subset of $\mathbb{C}$. Thus $f_{\sup}(G) \subset \{ w : \lvert w\rvert < 2\}$ (since that is the largest open subset of $\mathbb{C}$ contained in the closed disk $\{ w : \lvert w\rvert \leqslant 2\}$), hence $f_{\sup} \in \mathbb{F}$ also in this case.
Finally, we have $s = \lvert f_{\sup}(1)\rvert < 2$, so e.g. $c = 1 + \frac{s}{2}$ - any $c \in (s,2)$ would do - satisfies $\lvert f(1)\rvert < c$ for every $f \in \mathbb{F}$.
