Conjugacy classes of unipotent algebraic groups

Question

I want to prove the following statement:
Let $G$ be a unipotent linear algebraic group acting on an affine variety $X$. Then all $G$−orbits are closed in $X$.
The following is what I have done :
Let $Y$ be a non-closed orbit in $X$. Then $\overline{Y}-Y$ is a non-empty closed subset of $X$, which is transitive under the $G-$action. Hence $k[\overline{Y}-Y]\ne 0$ and $G$ acts on it by left translations. We can find a finite-dimensional vector subspace $V$ of $k[\overline{Y}-Y]$ stable under $G$. Since $G$ is unipotent, we can find a non-zero $f\in V$ fixed by all elements in $G$. In other words, $$f(g^{-1}x)=f(x)$$ for any $x\in\overline{Y}-Y$ and $g\in G$. Since $f\ne0$, there exists some $y\in\overline{Y}-Y$ such that $f(y)\ne 0$. Since $\overline{Y}-Y$ is $G-$transitive, we know $f(x)\ne0$ for any $x\in\overline{Y}-Y$. Then what should I do? I don't know how it leads to a contradiction.
If you have some ideas, please tell me. Thanks!

Answer 1

This is often called the Kostant--Rosenlicht theorem. It might be easiest to just read the original result. See Theorem 2 on Pg. 221 of this.