Distribution annihilated by a function with simple zeros

If a smooth function $a$ on $\mathbb{R}$ has finitely many zeros $r_j$, all simple, then $aw=0$ exactly when $w={\sum }_j{c}_j{\delta }_{r_j}$. Away from the zeros, divide a test function by the nonvanishing $a$ to show $w$ vanishes. Near $r_j$, write $a(x)=(x-r_j)b_j(x)$ with smooth nonzero $b_j$. The kernel of multiplication by a coordinate, after translation and multiplication by $b_j$, says that the local distribution is a multiple of ${\delta }_{r_j}$. A partition of unity gives the global sum. Simplicity of the zeros excludes derivatives of deltas.

In one dimension, $xw=0$ if and only if $w=c{\delta }_0$. Choose a cutoff function $\eta =1$ near zero and write every test function as $\phi =\phi (0)\eta +x\psi$. The pairing with $x\psi$ vanishes, leaving $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$.