A version of the Hilbert-Mumford criterion states the following: Let $G$ be a linearly reductive group and $V$ a representation of $G$ over a field $k$ (alg. closed, char. zero). Suppose that $y \in \overline{Gx} - Gx$. Then, there is a one-parameter subgroup $\lambda : k^\times \to G$ such that
$$
\lim\_{t\to 0} \lambda(t)x \in \overline{Gy}.
$$


My question is: Is there an example where every one parameter subgroup misses the orbit of $y$? I.e., is there an example where, for every $\lambda: k^\times \to G$
$$
\lim\_{t\to 0} \lambda(t)x \in \overline{Gy} \implies
\lim\_{t\to 0} \lambda(t)x \in \overline{Gy}-Gy?
$$
If $G$ is a torus the answer is "no". What if $V$ is replaced by a more general scheme $X$ that is not itself a representation?



Hilbert-Mumford criterion and closedness
I have a counterexample now, thanks to some notes of Zinovy Reichstein I found. I think the counterexample is paraphrased as follows: Let $V$ be an irreducible representation of $G$ and suppose $x$ does not have a highest weight vector $y$ in its orbit, but $y \in \overline{Gx}$. There will be no way to get to $y$ from $x$ using only semi-simple elements. 


To be precise (and to take the example from Reichstein) let $G = {\rm SL}\_2(\mathbb{C})$ and $V = {\rm{Sym}} ^n \mathbb{C}^2$, $n \geq 2$. If $a$ and $b$ are the standard basis vectors of $\mathbb{C}^2$ then $a^{n-1} b \in V$ has the highest weight vector $a^n$ in its orbit closure, but not its orbit. Now check that there is no one-parameter subgroup $\lambda(t)$ such that $\lambda(t) a^{n-1} b \in G a^n$.


