Complex analysis 2: $f \in \mathcal{H}(U,F)$ I have a problem:
Suppose $U$ is an open set in $E$ and $f \in \mathcal{H}(U,F)$. 
Prove that: 
$1/.$ If $U=E$ then $r_bf(x)=\infty, \forall x \in U$; 
$2/.$ If $U \ne E$ then $r_bf(x)< \infty, \forall x \in U$.
And case "$2$"; We have $$|r_bf(x)-r_bf(y)| \le \left \|x-y  \right \|, \forall x,y \in U$$.
Where $r_bf(x)=\sup \{r>0:f(x)\ \text{is bounded on}\ \overline{B}(x,r) \subset U\}$
 A: We assume $r_bf(x)=+\infty$ and $r_bf(y)<+\infty$.
Let $\forall r>0$ we have:
$$\overline{B}(y,r) \subset \overline{B}(x,r+\left \| x-y \right \|)$$
$\implies \sup \left \{\left \|f(t)  \right \|:t \in \overline{B}(y,r)  \right \} $
$\le \sup \left \{\left \|f(t)  \right \|: t \in \overline{B}(x,r+\left \| x-y \right \|) \right \} < + \infty$
$\implies r_bf(y)=\infty$ (Conflict).
2.1/ If $r_bf(x)< \infty,\ r_bf(y)< \infty,\ \forall x,y \in U$
How we can prove that: 
$$|r_bf(x)-r_bf(y)| \le \left \|x-y  \right \|, \forall x,y \in U$$ ?
A: Part 2: by symmetry it suffices to prove that 
$$r_bf(x)-r_bf(y)\le \|x-y\| \tag1$$
(exchanging $x$ and $y$ gives the other part of the desired inequality). If $r_bf(x)\le \|x-y\|$, we already have (1). Otherwise, take any $r$ such that $0<r<r_bf(x)- \|x-y\|$. Note that 
$f$ is bounded on $\overline B(x,r+\|x-y\|)$. It was  already observed that     $\overline{B}(y,r) \subset \overline{B}(x,r+\left \| x-y \right \|)$. Hence, $f$ is bounded on $\overline{B}(y,r)$. The conclusion is that $r_bf(y)\ge r - \|x-y\|$, and since $r$ can be arbitrarily close to $r_bf(x)$, (1) follows.
By the way, it's impossible to have infinite $r_bf(x)$ when $U$ is not the entire space. The definition of $r_bf(x)$ implies $r_bf(x)\le \operatorname{dist}(x,E\setminus U)$, and the quantity on the right is finite unless $E=U$.
