I didn't understand step b) of this proof and would be happy if someone could help me with this.
Let dimV be finite. Let L be a semisimple Lie algebra, $\ L_\alpha $ a weight space. Let $\ \Delta $= {$\ \alpha_1 ,...,\alpha_l $} be a base, H the maximal toral subalgebra of L.
a) Then the Borel subalgebra $\ B(\Delta)=H\bigoplus\ L_\alpha$ has a common eigenvector (killed by all $\ L_\alpha, \alpha \succ 0$, thanks to Lie's Theorem.
b) This eigenvector has to lie in any $L_\lambda$ and it has to satisfy $L_\alpha v^+ \subseteq V_\lambda \bigcap V_{\alpha+\lambda}=0$ for all $\alpha \succ0$, hence it has to be a maximal vector. On the other hand it is obvious that every maximal vector is common eigenvector of B.