Let $V$ be a $\Bbbk$-variety such that $\Bbbk^\times$ (as an algebraic group) acts algebraically on $V$. Given any $f\in\Bbbk[V]$, let us call $f$ **homogeneous of degree $d$** if for all $v\in V$ and all $\lambda\in\Bbbk^\times$, we have $f(\lambda.v)=\lambda^d f(v)$.

My question is: Does this define a grading on $\Bbbk[V]$?

I was convinced that it is true, but I am running into difficulties. Let us first assume $\Bbbk=\mathbb{C}$, the ground field should not be an obstruction. The linear span of $\Bbbk^\times f$ decomposes since $\Bbbk^\times$ is reductive, but I don't see how to turn this into a grading on all of $\Bbbk[V]$.

If it is true, I would really like to see a proof - it should use as little machinery as possible.