The answer is yes. We know G is reductive, take B=TU a Borel.

Decompose V into a direct sum of weight spaces for T: $V=\oplus V_\lambda$. I claim that C contains a non-zero wieght vector. Proof: First lets consider T=G<sub>m</sub>. If v=&Sigma;<sub>i</sub>v<sub>i</sub> in C with t in G<sub>m</sub> acting on v<sub>i</sub> by multiplication by t<sup>i</sup>, consider $\lim_{t\to \infty}t^{-N}(t\cdot v)$ (t<sup>-N</sup> is scalar multiplication, dot is action). Since C is closed  under action, scalar multiplication and taking limits, this is a weight vector in C for appropriate N. (this argument produces a weight vector of maximal possible weight in span(G<sub>m</sub>v).) A general T is a product of G<sub>m</sub>'s. We pick a vector in C, run the argument for each factor G<sub>m</sub> in succession, and end up with a weight vector for T.

If v in V<sub>&lambda;</sub> and u in U, then uv-v is in $\oplus V_\mu$ where only wieghts &mu; greater than &lambda; appear in the sum. Let v be a weight vector in C of maximal possible weight . Applying u to v and running the torus argument again, we see that for this maximalilty assumption to hold, v must be annihilated by all of U. Since stabaliser of v is reductive, containing U, it must contain derived group of G.

Let $\alpha^\vee$ be a simple coroot and &lambda; be a weight appearing in V. Suppose $\langle\alpha^\vee,\lambda\rangle$>0. Since C spans V, there exists c in C with non-zero projection onto V<sub>&lambda;</sub>. Run the G<sub>m</sub> argument with the vector c and the subgroup $\alpha^\vee(\mathbb{G}_m)$. Now run the torus argument on this output, we get a weight vector in C of weight &mu; where $\langle\alpha^\vee,\mu\rangle>0$.

Now under this assumption, the paragraph with the unipotent element shows that there is a weight vector v in V, of weight &nu; greater than &mu; with derived group in stabaliser. Also $\langle\alpha^\vee,\nu\rangle>0$. This is a contradiction. Simlilaraly $\langle\alpha^\vee,\lambda\rangle$<0 is a contradiction. So for every weight &lambda; and every simple coroot $\alpha^\vee$, we have $\langle\alpha^\vee,\lambda\rangle$=0, which forces the representation to be a direct sum of characters.