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On the smooth-function space, the product rule gives $ABf=(xf{)}^{\prime }=f+x{f}^{\prime }$ while $BAf=x{f}^{\prime }$. Hence $\boxed{[A,B]=I}$, and both maps preserve the stated smooth space.

For general operators with this commutator relation, use $[A,BC]=[A,B]C+B[A,C]$. Induction gives $[A,B^i]=i{B}^{i-1}$ for $i\ge 1$. Since $Ay=0$,Every displayed vector lies in $W$, so $W$ is invariant under $A$, as well as under $B$. Iterating the lowering formula gives $A^k{B}^{i}y=i!B^{i-k}y/(i-k)!$ for $k\le i$, and zero for $k>i$.

Suppose a nontrivial finite linear dependence has highest nonzero term $c_n{B}^{n}y$. Applying $A^n$ kills all lower terms and yields $c_{n}n!y=0$. The field is real, $n!\ne 0$, and $y\ne 0$, so this contradicts $c_n\ne 0$. Thus the entire sequence is linearly independent. This identity commutator cyclic ladder also proves that such a representation with a nonzero vector in the kernel of $A$ cannot be finite dimensional.