Let $u\in D'(\mathbb{R})$ and $f\in C^{\infty}$. I'm trying to figure which of the following statements is true:
I. If $f\restriction_{supp(u)}=1$ then $f\cdot u=u$.
II. If $f\restriction_{supp(u)}=0$ then $f\cdot u=0$.
Now, if I could prove that $\phi\restriction_{supp(u)}=0$ implies $u(\phi)=0$, then both I and II will follow. The problem is that I'm not sure it's true (I know that $supp(u)\cap supp(\phi)=\emptyset$ implies that $u(\phi)=0$, but this is not necessarily the case here). So, I have the following questions:
A. Does $\phi \restriction_{supp(u)}=0$ implies $u(\phi)=0$?
B. If the answer to (A) is "no", then which of the statements I and II is true and which is not?
It's worth stating the definition somewhere: the support of a distribution $u$ is defined as the complement of the maximal open set $U$ with the property that $u(\varphi)=0$ for every test function $\varphi$ with $\operatorname{supp}\varphi\subseteq U$.
Since $\operatorname{supp} (\varphi')\subseteq \operatorname{supp}\varphi$ for every test function, it follows that $\operatorname{supp} (u')\subseteq \operatorname{supp}u$. In particular, $\delta_0'$ (the derivative of the Dirac mass at $0$) has support $\{0\}$. It acts on test functions by evaluating their derivatives at $0$, with opposite sign: $$\delta_0'(\varphi) = \delta_0 (-\varphi')=-\varphi'(0)$$ Let $\varphi(x)=x\exp(-1/(1-x^2))$ for $|x|<1$, and $\varphi(x)=0$ for $|x|\ge 1$. Then $\varphi$ vanishes on the support of $\delta_0'$, but $\delta_0'(\varphi)=-1$. Thus, the answer to (A) is no.
Both I and II can also be disproved by taking $u=\delta_0'$ and letting $f$ be a smooth function with nonzero derivative at $0$. Indeed, for any test function $\varphi$ we have $$f\cdot \delta_0'(\varphi)=\delta_0'(f\varphi)=-f'(0)\varphi(0)-f(0)\varphi'(0)$$ The term $-f'(0)\varphi(0)$ is the problem here.