TITLE: Existence of a weight of a representation in the fundamental Weyl chamber
QUESTION [6 upvotes]: Let $\mathfrak g$ be a complex simple Lie algebra. 
Fix a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$, let $\Delta$ denote the corresponding root system. 
Pick a partial order on $\mathfrak h$, which induces a positive root system $\Delta^+$, fundamental Weyl chamber, etc. 
I wish a reference for the following result, which must be well known. 
My proof is case-by-case and a bit long. 

For all but finitely many finite dimensional irreducible representations $(\pi, V)$ of $\mathfrak g$, there is a weight $\mu$ of $\pi$ lying in the fundamental Weyl chamber. 

Recall that $\mu\in\mathfrak h^*$ is called a weight of $\pi$ if $V(\mu):=\{v\in V: \pi(X)\cdot v=\mu(X)\, v\quad\forall \, X\in\mathfrak h\}$ is non-zero. 
Furthermore, the fundamental Weyl chamber is given by elements $\mu\in\mathfrak h^*$ satisfying $\langle \mu,\alpha\rangle >0$ for all $\alpha\in\Delta^+$. 
The proof is immediately reduced to consider irreducible representations of $\mathfrak g$ with highest weight lying in a face of the fundamental Weyl chamber. 
Furthermore, it reduces to check irreducible representations with highest weight a multiple of a fundamental weight. 
It is easy to see that the result is no longer valid when $\mathfrak g$ is assumed semisimple in place of simple.

REPLY [7 votes]: Here's maybe another (more conceptual?) way to think about it.
First of all, if $\mu_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_1}$, and $\mu_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda_2}$, then $\mu_1+\mu_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda_1+\lambda_2}$. (This is just saying that if $\lambda_1-\mu_1$ is a nonnegative sum of simple roots, and ditto for $\lambda_2-\mu_2$, then definitely $(\lambda_1+\lambda_2)-(\mu_1-\mu_2)$ is as well.)
The previous paragraph reduces the problem to showing that if $\lambda=\omega_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega_i}$ contains a weight strictly in the fundamental chamber. Let's show this. 
Recall the Weyl vector $\rho:=\frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum_{j=1}^{n}\omega_j$. Let $\Phi_i$ denote the maximal parabolic root system generated by the simple roots $\alpha_j$ for $j\neq i$, and let's use the notation $\rho_i :=\frac{1}{2}\sum_{\alpha\in\Phi_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho_i,\alpha^\vee_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho_i, \alpha^\vee_i \rangle < 0$ since no factor of $\alpha_i$ appears at all in $\rho_i$ but some $\alpha_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.
Therefore $2\rho - 2\rho_i =  \sum_{\alpha\in \Phi^+\setminus \Phi_i^+}\alpha= c\omega_i$ for some integer $c > 0$. But one of the $\alpha \in \Phi^+\setminus \Phi_i^+$ is the highest root $\theta$ of $\Phi$, which when written $\theta=\sum_{j=1}^{n}a_j\alpha_j$ satisfies $a_j \geq 1$. So there is some multiple $k$ of $2\rho - 2\rho_i$ for which $k(2\rho - 2\rho_i) - 2\rho$ is a nonnegative sum of simple roots. This is to say the strictly dominant weight $2\rho$ appears in $V^{kc\omega_i}$. And then for any $m\geq kc$, the strictly dominant weight $2\rho + (m-kc)\omega_i$ appears in $V^{m\omega_i}$, completing the proof.

REPLY [4 votes]: I don't know about a reference, but here is a short uniform proof. I assume that all of your representations are finite dimensional.
Let $\alpha_1$, ..., $\alpha_n$ be the simple roots and let $\omega_1$, ..., $\omega_n$ be the dual weights, so $\langle \alpha_i, \omega_j \rangle = \delta_{ij}$. So, if $\mu$ is an integral positive weight, then $\mu = \sum c_j \omega_j$ for $c_j \in \mathbb{Z}_{\geq 0}$.
Let $A_{ij}$ be the Cartan matrix. Let $\Gamma$ be the Dynkin diagram, so this is a graph with vertices $1$, $2$, ..., $n$ and an edge $(i,j)$ if $A_{ij} \neq 0$.
For a positive root $\mu$, let $a_r(\mu) = \{ j : c_j \leq r \}$. We partially order the integral positive weights as follows: We put $\mu \leq \nu$ if there is some index $q$ for which $a_q(\mu) > a_q(\nu)$ and $a_r(\mu) = a_r(\nu)$ for $0 \leq r < q$.
Fix your representation $V$ and let $\mu = \sum c_j \omega_j$ be minimal among the integral positive weights of $V$, with respect to the above order. 
Key lemma: For every edge $(i,j)$ of $\Gamma$, we have $c_i - c_j \leq 2$.
Proof: Suppose for the sake of contradiction that $c_i \geq c_j + 3$. Since $c_i = \langle \alpha_i, \mu \rangle > 0$, the weight $\mu' := \mu - \alpha_i$ is also a weight of $V$. We will verify that $\mu'>\mu$. We have $\mu' = \sum c'_k \omega_k$ where $c'_i = c_i-2$ and $c'_k = c_k + (-A_{ik})$ for $k \neq i$. 
Put $q = c_j$. I claim that $a_q(\mu') > a_q(\mu)$ and $a_r(\mu') \geq a_r(\mu)$ for $r < q$. Since $c'_i = c_i-2 \geq c_j+1=r+1$, the change between $c'_i$ and $c_i$ won't change the value of $a_q$ for $q \leq r$. Each coefficient other than the $i$-th gets larger going from $\mu$ to $\mu'$, so all the $a_q$ get weakly larger. Moreover, $c_j=r$ and $c'_j = r + (-A_{ij}) > r$, so $a_r$ gets strictly larger. $\square$
Let $\delta$ be the diameter of the graph $\Gamma$. (Here is where we use that the Lie algebra is simple, so $\Gamma$ is connected.) So, if $\mu = \sum c_j \omega_j$ is the above minimal weight, and one of the $c_j$ are $0$, then all of the $c_j$ are bounded by $2 \delta$. In particular, there are only finitely many choices for $\mu$.
Remark Experimentally, every representation seems to have a weight with $c_i - c_j \leq 1$, but it isn't always the minimum in the order I've defined, and I don't want to work hard enough to get this better bound.