Show that $G$ ( subgroup of $\mathrm{GL}(E)$) is finite. I came across with, I think, a difficult problem :

Let E a Hermitian space with a Hermitian norm $||\ ||$. We provide $\mathcal{L}(E)$ with the norm $|||\ \  |||$ subordinated to $||\ ||$.
Let $r \in ]0,2[$ and $G$ a subgroup of $\mathrm{GL}(E)$ such that $G \subset B_o (\mathrm{Id}_ E , r)$ (open ball) . Show that $G$ is finite.

I was able to show the following fact :

*

*if $\lambda \in \mathbb C$ is an eigenvalue of an element of $G$, then
$|\lambda -1| \leq r$
Proof.
For $X$ a eigenvector associated with the eigenvalue $\lambda$ of $f\in G$,
then $||f(x)-X||\leq  N(f_{Id}) ||X||$.

*

*The eigenvalues of elements of $G$ are the roots of unity, and there are finitely many eigenvalues.

Proof.
First, The eigenvalues of $f$ are  $\vert\  \vert = 1$
Otherwise, $f^n$ element of $G$ (group) and we have $N(f^n)>\vert a\vert ^n$
Thus with $N (f^n) - N(Id) \le N( f^n{_Id}$)
we get $\vert \lambda \vert^n \le r + 1$ winch is impossible when $n$ is big enough.
Furthermore $f^-1$ is an element of $G$(group) and its eigenvalues ​​are the inverse of those of $f$.
Therefore we deduce that all the eigenvalues ​​of $f$ are of modulus $1$.
EDIT : I forgot the rest of the proof,(I am not sure it's correct)
Let  $\lambda=\exp(i\theta)$ an eigenvalue of $f$, fixed.
If $\theta$ isn't of the form $\frac{2k \pi}n$ then all $\lambda^p$ with $p$ an integer will be dense on the points of the circle of radius 1 and we will find the eigenvalues of $f^p$ near $-1$ wich will contracdict the first inegality.

How can I continue ?
Thank you in advance
NB: I made this bounty because lie theory it's beyon my reach.
 A: Note that finite dimensionality is necessary for your claim to be true. Indeed, a counterexample
can be constructed very simply in infinite dimension, as follows : take $E$ with a
countable orthonormal basis $(e_k)_{k\geq 1}$, and take $G$ to be the subgroup of
all endomorphisms $f$ satisfying $f(e_k)=\lambda_ke_k (k\geq 1)$, where
$(\lambda_k)_{k\geq 1}$ is a sequence of complex numbers such that 
$\lambda_k \in \lbrace 1,e^{\frac{2\pi i}{3}},e^{\frac{4\pi i}{3}}\rbrace$ for every
$k$, and $\lambda_k=1$ for all but finitely many $k$.
In the finite dimensional case, your claim follows from 
Lemma. If $G$ is a subgroup of $GL(E)$ such that the set of traces 
${\sf Tr}(G)=\lbrace {\sf tr}(g) | g\in G\rbrace$ is finite. Then $G$ is itself finite.
Proof of lemma. Let $g_1,g_2,\ldots ,g_r$ be a maximal linearly independent
family in $G$ (thus $r\leq {\sf dim}(E)^2$). Consider the map
$\phi : G \to {\sf Tr}(G)^r, g\mapsto ({\sf tr}(g^{*}g_1),{\sf tr}(g^{*}g_2),\ldots,{\sf tr}(g^{*}g_r))$.
Then $\phi$ is injective (otherwise the family would not be maximal ; remember that ${\sf tr}(g^*g)=||g||^2$ is zero only when $g=0$), so
$G$ must be finite.
In the sequel I show the finiteness property, about which you seemed unsure. 
Let $U$ denote the set of the complex numbers with modulus one ; then $U$ is a multiplicative
subgroup of ${\mathbb C}^*$. Denote by $\Lambda$ the set of all eigenvalues of 
elements of $G$, and let 
$L=\lbrace \theta \in {\mathbb R} \ | \ e^{2\pi i\theta} \in \Lambda \rbrace$. The OP 
already shows that $\Lambda \subseteq U$, and that
$$
|\lambda-1| \leq r \ (\forall\lambda \in \Lambda)\tag{1}
$$
If $\lambda$ is an eigenvalue of some $g\in G$, then for any $j\in{\mathbb Z}$
$\lambda^j$ is an eigenvalue of $g^j$, so that
$$
j\theta \in L \ (\forall\theta \in L,\forall j\in{\mathbb Z})\tag{2}
$$
By definition of $L$, we have
$$
k+\theta \in L \ (\forall\theta \in L,\forall k\in{\mathbb Z})\tag{3}
$$
Let $\theta\in L$ and $\lambda=e^{2\pi i\theta} \in \Lambda$. Then $\theta$ cannot
be too near to $\frac{1}{2}$, else we would contradict (1). So there is an absolute 
integer constant $N$ such that
$$
\bigg|\theta-\frac{1}{2}\bigg| >  \frac{1}{2^N} (\forall\theta \in L) \tag{4}
$$
If $L$ contains a $\theta\in ]0,\frac{2^{N-1}-1}{2^{2(N-1)}}]$, then denoting by
$p$ the integer part of $\frac{2^{N-1}-1}{2^N\theta}$, we have $p\geq 2^{N-2}$ and also
$\frac{2^{N-1}-1}{2^Np} \leq \theta \leq \frac{2^{N-1}-1}{2^N(p+1)}$, whence
$\frac{2^{N-1}-1}{2^N} \leq (p+1)\theta \leq \frac{(2^{N-1}-1)p}{2^N(p+1)} \leq \frac{2^{N-1}+1}{2^N}$
so that $|(p+1)\theta-\frac{1}{2}| \leq \frac{1}{2^N}$, contradicting (4). So
$L\cap ]0,\frac{2^{N-1}-1}{2^{2(N-1)}}]$ must be empty. As (2) implies 
$\theta \in L \Leftrightarrow -\theta \in L$, we also see that
$L\cap [-\frac{2^{N-1}-1}{2^{2(N-1)}},0[$ must also be empty. To summarize,
$$
|\theta| >  \frac{2^{N-1}-1}{2^{2(N-1)}} (\forall\theta \in L\setminus \lbrace 0 \rbrace) \tag{5}
$$
Now, for $\theta\in{\mathbb Z}$, let $L_{\theta}={\mathbb Z}+\theta{\mathbb Z}$. Then
$L_{\theta} \subseteq L$ by (2) and (3), $L_{\theta}$ is an additive subgroup of $\mathbb R$, and a very famous result
(see for example here)
states that if $\theta$ is irrational then $L_{\theta}$ is dense in $\mathbb R$. This would
clearly contradict (5), so
$$
L \subseteq {\mathbb Q} \tag{6}
$$
Let $\theta\in L\cap(0,1)$. We can write $\theta=\frac{a}{b}$ where $a,b$ are positive
coprime integers. There are positive integers $u,v$ satisfying the Bezout relation
$au-bv=1$. Then $\frac{1}{b}=u\frac{a}{b}-v \in L$ by (2) and (3). By (5) we then
have $a \leq b \leq 2^{2(N-1)}$. So there are at most $2^{2(N-1)}$ possible values for
$a$ and $b$, and hence
$$
L\cap (0,1) \ \text{is finite} \tag{7}
$$
Combining (6) with (7), we have that $\Lambda$ is finite as wished.
A: There is probably also an elementary proof using Jordan decomposition (I would have to think about it), but I will use Lie theory instead. In particular, this proof will work only in the finite dimensional case. 
First of all, since norm is a continuous function, it suffices to consider the case when $G$ is closed (otherwise, replace $G$ with its closure, it will have the same "norm" property as in your question). Note that Closure of a subgroup is again a subgroup. Thus, $G$ is a Lie subgroup of $GL(E)$ (every closed subgroup of a Lie group is a Lie subgroup). Since $G$ is closed and bounded, it is compact. Thus, $G$ has only finite number of connected components. Therefore, it suffices to show that the identity  component of $G$ is trivial. (The $G$ would be finite.) I will assume, therefore, that $G$ is closed and connected; I will prove that it is trivial. Arguing by contradiction, assume that $G$ has positive dimension. Then, a maximal torus $T<G$ (a maximal connected abelian subgroup; it is unique up to conjugation) also has positive dimension. Therefore, it suffices to get a contradiction with the case when $G$ is abelian. The maximal torus $T$ splits as the product of circles $S^1\times ... \times S^1$. (In fact, if you like, you can assume that $G\cong U(1)$.) The action of $G$ on $E$ is completely reducible: $E$ splits as a direct sum of $G$-invariant 1-dimensional complex subspaces $L_k$. Since $G$ is nontrivial, its action on at least one of the lines $L=L_k$ is via the full unitary group $U(1)$ of rotations inside of $L$ (unitary with respect to the hermitian norm you are using). By our assumption, the norm of each $g\in U(1)$ (acting on $L$) is $\le r<2$. But this is absurd since $U(1)$ contains $-1$. QED
