I have a question about the following proof in Analysis on Symmetric Cones by Faraut and Koranyi (p.5). Let $G(\Omega)$ be the automorphism group of an open convex cone $\Omega$. For any point $a\in \Omega$ we define the stabiliser of $a$ in $G(\Omega)$ by $$G(\Omega)_a = \left\{g \in G(\Omega): ga = a \right\}.$$
Now my question concerns the proof of the following proposition.
If $\Omega$ is a proper open convex cone, then for every $a$ in $\Omega,$ $G(\Omega)_a$ is compact.
Proof. The set $\Omega \cap (a - \Omega)$ is bounded, open and non-empty. $G(\Omega)_a$ maps $\Omega \cap (a - \Omega)$ to itself. It follows easily (e.g. by choosing a basis of $V$ contained in $\Omega \cap (a- \Omega)$) that there exists a constant $C$ such that $$\frac{1}{C} ||x|| \leq ||gx|| \leq C ||x||$$ for all $g \in G(\Omega)_a$ and $x \in V.$ Hence $G(\Omega)_a$ is relatively compact in GL$(V)$, and is also clearly closed. (End of proof).
My question concerns the 'It follows easily' statement, which I just can't see. I will focus on the second inequality ($||gx|| \leq C||x||$). Let $x = c_1e_1+ \ldots c_ne_n$ where $\beta = \left\{e_1,\ldots,e_n \right\}$ is the basis we have chosen, which is contained in $\Omega \cap (a- \Omega).$ The fact that $g$ maps $\Omega \cap (a- \Omega)$ to itself suggests we could begin with something like $$||ge_i|| \leq C||e_i|| \text{ for all } 1 \leq i \leq n,$$ then say
\begin{equation} \begin{split} ||gx|| &= ||c_1ge_1+\cdots+c_nge_n|| \\ & \leq |c_1|\cdot||ge_1||+\cdots+|c_n|\cdot ||ge_n|| \\ & \leq C \big(|c_1|\cdot ||e_1||+\cdots |c_n| \cdot ||e_n|| \big) \end{split} \end{equation}
which is kinda close but obviously not good enough. Also, this suggests that bounding each basis vector isn't enough, but I can't seem to think of how to get the upper bound in terms of $||x||.$ Any help would be greatly appreciated.
$\|c_1e_1+...+c_ne_n\|_1=|c_1|\|e_1\|+...+|c_n|\|e_n\|$ is a norm defined on a finite dimensional space, and on a finite dimensional space norms are equivalent. There exists $D>0$ such that ${1\over D}\|x\|\leq \|x\|_1\leq D\|x\|$.