How to prove that,
for any $w$ $\in$ $W$ (Weyl group), $ \delta - w \delta $ is in positive part (non negative part) of the root lattice $\mathbb{Z}[\Delta]$ ? where
$\Delta$ is a simple system in the root system $\Phi$
$\delta$ is the half sum of all positive roots (Weyl vector).
Thanks in Advance.
Prove that $$w\delta = \frac12\sum_{r \in \Phi^+}c_rr$$ where $c_r \in \{\pm 1\}$ for all $r$. You'll do this by induction on the length of $w$, so you just need to show that multiplying by a simple root will preserve the set of elements in that form (and you should know that a simple reflection sends one root to it's negative and permutes the remaining positive roots).
One you've done that, the difference $\delta - w\delta$ is just a sum of some subset of the positive roots, because the coefficients are either $\frac12 - \frac12 = 0$ or $\frac12 - (-\frac12) = 1$.
The statement is actually true for any vector $v$ in the identity chamber $C$. If you take a reduced expression $w=s_{i_1}\dots s_{i_r}$ and consider $C$, $s_{i_1}C$, $s_{i_1}s_{i_2}C$, $\dots$, at each step, you move to a chamber that is across one more hyperplane from the identity chamber. This moves each point in the chamber by some negative multiple of the positive root perpendicular to that hyperplane. So $v-wv$ will be the sum of a bunch of positive multiples of positive roots.
(It's easy to see that $\delta$ lies in the identity chamber using the fact that Jim mentioned: since $s_i$ permutes the positive roots except $\alpha_i$ and sends it to its negative, $s_i\delta=\delta-\alpha_i$, so $\langle\delta,\alpha_i\rangle >0$.)
