Are multiplicity-free representations weight multiplicity free? A rational representation $(G,V)$ of a complex reductive linear algebraic group is called multiplicity-free if the decomposition of $\mathbb C[V]$ into irreducible $G$-modules contains each irreducible $G$-module with multiplicity at most one. Part of the importance of such representations lies in the fact that they serve to explain several phenomena in invariant theory and harmonic analysis (e.g. Peter-Weyl theorem). 
There are several different characterizations of multiplicity free representations, among them: any Borel subgroup has an open (dense) orbit; the principal orbits of a compact real form are coisotropic submanifolds with respect to an invariant (real) symplectic structure. 
If $(G,V)$ is multiplicity free then $G$ has finitely many orbits in $V$. 
In particular, it is visible in the sense that $V$ has finitely many nilpotent orbits, and hence the multiplicity of any nonzero weight is at most one, see Kac, V. G. Some remarks on nilpotent orbits. J. Algebra 64 (1980), no. 1, 190–213.
It is apparent from the classification of (indecomposable) multiplicity free representations that they are weight multiplicity free, namely, the multiplicity of any weight is at most one. Is there a direct proof of this fact? 
 A: It is a theorem of Brion and, independently, of Vinberg that varieties with an open $B$-orbit (a.k.a. spherical varieties) have in fact only finitely many orbits. A shorter argument is due to Matsuki (see his ICM talk) and independently (using the same idea) by me (On the set of orbits $\ldots$). Thus multiplicity free spaces are visible. Kac implies that all non-zero weights have multiplicity one. On the other hand, the map from the zero weight space to the quotient $V/\!\!/ G$ is always finite (see e.g. Luna "Adherence d'orbites"). The latter is a point. So the zero-weight space of a multiplicity free space is in fact zero.
Edit 1: I realize that my reference to Adherance is over-kill in this situation but Dadok-Kac is so, as well, since it uses Hilbert-Mumford. There is an easier classical argument, probably due to either Rosenlicht or Borel.
Claim: Let $X$ be an affine $G$-variety and $x\in X^T$. Then $Gx$ is closed in $X$.
Proof: The key observation is due to Rosenlicht, saying that any orbit of a unipotent group on an affine variety is closed. Let $B=TU\subseteq G$ be a Borel. Then $Ux$ is closed. But then also $Bx=TUx=UTx=Ux$ is closed. The morphism $\phi:G\times^BBx\to X$ is proper since it factors through the closed immersion $G\times^BBx\hookrightarrow G/B\times X:[g,bx]\mapsto (gB,gbx)$ and the proper projection $G/B\times X\to X$. Thus $Gx={\rm image}(\phi)$ is closed.
Edit 2: I have erased a previous alternative proof since I couldn't recall all details. I do remember now. So it is possible to prove weight multiplicity freeness from the Local Structure Theorem (Brion-Luna-Vust) using the line of reasoning in my survey Some remarks on multiplicity free spaces.
