Suppose $G$ is a connected open set of $E$ and $f \in \mathcal{H}(G,F)= \{f: G \to F$, $f$ $\text{is holomorphic mapping} \}$.
Suppose there is a points $a \in G$, such that $\|f(x)\| \le \|f(a)\|$, $\forall x \in G$.
Prove that: $\|f\|$ is constant in $G$.
I have some ideas:
Let $A=\{x \in G: \|f(x)\|=\|f(a)\|\}$, we'll show that $A=G$.
(1). We have $A \ne \emptyset$, since $a \in A$ .
(2). We will show that $A$ is closed.
Which means $\{x_n\} \subset A, x_n \to x$ then $x\in A$ .
Because $f$ is holomorphic in $G$, hence $\|f(x)\|=\lim_{n \to \infty}\|f(x_n)\|=\|f(a)\| \implies x \in A$.
(3). We will show that $A$ be open set.
In this situation I'm having trouble. Any help (or hint or another solution) would be greatly appreciated. Thanks.
(3). We will show that A be open set.
$\forall x_0 \in A$, We choose $r>0$ such that $\overline{B}(x_0,r) \subset G$.
With $t \in E,\left \|t \right \|<r$ then: $x_0 +\xi t \in \overline{B}(x_0,r)\subset G,~~ \forall \left | \xi \right |=1$.
We now apply Cauchy's integral formula:
$f(x_0)=\frac{1}{2 \pi i}\int_{\left | \xi \right |=1}\frac{f(x_0+\xi t)}{\xi}\mathrm{d}\xi \implies \left \| f(x_0) \right \|=\frac{1}{2 \pi }\int_{\left | \xi \right |=1} \left \| \frac{f(x_0+\xi t)}{\xi}\mathrm{d}\xi \right \| \le \left \| f(a) \right \|.$
We have $\left \| f(x_0+\xi t) \right \|=\left \| f(a) \right \|,~~ \forall \left | \xi \right |=1$.
In particular,$\left \| f(x_0+t) \right \|=\left \| f(a) \right \|$
$\implies \left \| f(x) \right \|=\left \| f(x_0+t) \right \|=\left \| f(a) \right \|, \forall x=x_0+t \in B(x_0,r)$.
$\implies B(x_0,r) \subset A$.
Hence, $\emptyset \ne A$, $A$ is closed and open in a connected open set $G$.
Therefore, $A=G$. :p