Corollary 1

Let $\| \|$ be a norm on $X(T)\otimes \mathbf{R}$. For all $\lambda \in X_{++}$, let $d(\lambda )$ be the dimension of the space of an irreducible representation of G of highest weight $\lambda$.

a) sup$_{\lambda\in X_{++}}d(\lambda )/\|\lambda +\rho \|^N<\infty$, where $N = 1/$2(dim $G-$ dim T).

b) If G is semi-simple, inf$_{\lambda\in X_{++}}d(\lambda )/\|\lambda +\rho \|>0$.

a) For all $\alpha \in R_+$, there exists $A_{\alpha}>0$ with $|\langle \lambda +\rho , K_{\alpha}\rangle | \leq A_{\alpha}\|\lambda +\rho \|$, hence $d(\lambda )/\|\lambda +\rho \|^N\leq \prod_{\alpha >0}A_{\alpha}/\langle \rho , K_{\alpha}\rangle$.

b) Assume that G is semi-simple, denote by $\beta_1, . . . , \beta_r$ the simple roots and put $N_i=K_{\beta_i}$. Then

since $\langle \lambda +\rho , N_i\rangle \geq \langle \rho , N_i\rangle = 1$, this implies that

$d(\lambda )\geq$ sup$_i|\langle \lambda +\rho , N_i\rangle |$.

If G is semi-simple, $x \rightarrow$ sup$|\langle x, N_i\rangle |$ is a norm on $X(T)\otimes \mathbf{R}$, necessarily equivalent to the given norm, hence b).