Assume that $G$ is a linear algebraic group, and let $B \subset G$ a Borel subgroup of it. Let $(V,\rho)$ a rational $G$-module. Define $$V^G := \{ v \in V \mid g \cdot v = v \quad \forall g \in G \}$$ prove that $V^B = V^G$.
My attempt Obviously $V^G \subset V^B$ because $B \subset G$. My strategy is to find a way to prove (if it is true!) that the action $G \to GL_n(V)$ can be restarted to an action $G\to GL_n(V^B)$. Assuming that true, then by hypothesis the morphism factor through the quotient $G/B$ which is a projective variety and whose image is still a closed subset of $GL_n(V^B)$, hence affine and hence the image is a finite set of points (because is proper) and hence by connectedness it's the identity. Hence $V^B$ is $G$-stable and hence $V^B \subset V^G$. The point is that I'm not able to prove that I can restrict the action to $V^B$. I hope the rest of the reasoning is clear and correct.
Thanks in advance!
As far as I can see, your reasoning is fine, and you don't even need an action of $G$ on $V^B$: It suffices to pick some $v\in V^B$ and to consider the map $G\to V$ given by $g\mapsto g.v$. Since $v\in V^B$, this map factors through $G/B\to V$, and your arguments show that it must be constant.