Consider the functions $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Use the fact that $f_n(x)$ $\rightarrow e^x $ for each $x \in \Bbb R $ to prove that $f_n(z) \rightarrow e^z$ normally in $\Bbb C$.
My work: Suppose $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Since $1+x \le e^x$ for $x \ge 0$, it follows that $|f_n(z)| \le e^{|z|}$, for all $z$. So, ${f_n}$ is a normal family. And $f_n(x)$ $\rightarrow e^x $ for each $x \ge 0$.
I am interested to perform following steps:
Since Uniformly Bounded Family is Locally Bounded, $\{f_n(z)\}$ is locally bounded on U. From Montel's Theorem, $\{f_n(z)\}$ is a normal family, From Vitali's Convergence Theorem, $\{f_n(z)\}$ converges uniformly on any compact subset of U. In particular, $\{f_n(z)\}$ converges uniformly on K. My intuition is to prove $f_n(z) \rightarrow e^z$ uniformly on compact sets. Your kind help will be appreciated. Thank you so much!
Vitali's Theorem does not tell you that $\{f_n\}$ converges on compact sets. It tells you that there is a subsequence which converges uniformly on compact sets. To show that $f_n(z) \to e^{z}$ 'normally' we have to show that $f_n(z) \to e^{z}$ uniformly on any compact set. Fix a compact set $K$. Consider any subsequence ${f_{n_k}}$. By Vitali's theorem there is a subsequence of this subsequence which converges uniformly on compact sets to some analytic function $g$. Now observe that $g(x)=e^{x}$ for $x \geq 0$. But then $g$ and $e^{z}$ are analytic functions which coincide on set with limit points, namely $[0,\infty)$. hence $g(z)=e^{z}$ for all $z$. We have proved that any subsequence of $\{f_n\}$ has a subsequence conveging uniformly on $K$ to $e^{z}$. This implies that the whole sequenec $\{F_n\}$ converges uniformly on $K$ to $e^{z}$.