Let $A$ be a commutative, regular, finitely generated $k$-algebra, where $k$ is a field of characteristic 0.


Then Grothendieck proved that in this case the algebra of differential operators on $A$ is simply the subalgebra of $End\_k(A)$ generated by multiplication by $A$ and $Der\_k(A)$, $k$-linear derivations. Denote this algebra by $D(A)$. 


I have read that $D(A)\subseteq End\_k(A)$ is 'strongly dense', i.e. given any finite-dimensional subspace $M\subseteq A$ and any linear operator $L\in End\_k(A)$, there exists a differential operator $D\in D(A)$ such that $L|\_M=D|\_M$. Does anyone know that proof of this fact?


My thought process has been to try and work locally first: we know that $Spec(A)$ has a cover by affines $Spec(A\_f)$ such that $D(A\_f)\cong A\_f[\frac{\partial}{\partial x\_1},\dots,\frac{\partial}{\partial x\_n}]$, where $x\_1,\dots,x\_n\in A\_f$ are a system of parameters at all closed points in $Spec(A\_f)$. Here the derivations $\frac{\partial}{\partial x\_i}$ are only required to satisfy $x\_j\mapsto \delta\_{ij}$ (if I understand things correctly). So it's easy to find a differential operator which does anything you like to a subspace generated by monomials in the $x\_i$, but I don't see what to do about other elements.



'Strong' density of differential operators in linear operators
I think it may be easier if you do not choose the open cover in advance, so that the cover may depend on $M$. Equivalently, let us work with local rings first. 


Here is a sketch of the argument. I am keeping your notation: $A$ is smooth over a characteristic zero field $k$, $D(A)$ is the algebra of differential operators, and $M\subset A$ is a finite-dimensional subspace. Consider the evaluation map $$ev:D(A)\to Hom\_k(M,A).$$ We want to show that $ev$ is surjective.


Note that $ev$ is a morphism of $A$-modules, so it is enough to check its surjectivity locally. (Explicitly, this step involves looking at partitions of unity.) That is to say, for every point $x\in Spec(A)$, we consider the local ring $A\_x$, an we need to prove surjectivity of the map 
$$ev\_{A\_x}:D(A\_x)=A\_x\otimes\_A D(A)\to Hom\_k(M,A\_x).$$
Moreover, $Hom\_k(M,A)$ is a finitely generated $A$-module (in fact, it is free of finite rank), so it suffices to consider closed points $x\in Spec(A)$. 


Let now $x\in Spec(A)$ be a closed point, and we want to prove surjectivity of $ev\_{A\_x}$. By Nakayama's Lemma, it suffices to check that the composition
$$D(A\_x)\to Hom\_k(M,A\_x)\to Hom\_k(M,k\_x)$$
is surjective, here $k\_x$ is the residue field of $x$. However, for this statement you can replace $A\_x$ by its completion, which turns it into a claim about Taylor series. 


P.S. It may be psychologically easier to extend scalars from $k$ to its algebraic closure at the very beginning... But if you resist the temptation, you get to think about things like $k$-linear differential operators on the ring of Taylor series with coefficients in a finite extension $k\_x\supset k$... 


