Suppose $F,G \in \mathcal{D}'(\mathbb{R})$ with $\mathrm{supp}(F) \cap \mathrm{supp}(G) = \emptyset$, and $(F+G, \varphi) = 0$ for every $\varphi \in \mathcal{D}(\mathbb{R})$.
I am wondering if it is true that $(F, \varphi) = (G, \varphi) = 0 $ for all test functions $\varphi$?
I've been thinking about this and I'm not totally sure. If you have the equality $(f+g)(x) = 0$ pointwise for continuous functions $f,g$ on $\mathbb{R}$ and their supports are disjoint then the conclusion is clear, we can say $f(x) = g(x) = 0$. But when the equality is in the distributional sense, it is not immediately obvious. Intuitively it seems like the claimed result should be true...]
I tried to study the case where $F,G \in L^{1}_{loc}$ but this doesn't seem any easier. I also tried to come up with counterexamples by taking $F$ and/or $G$ to be deltas but couldn't find one.
I am sorry that yesterday I give you a wrong comment. Now, let me give a different idea(different from the above comment; I am not sure whether it is right but I am sure that it is close to the right answer). The different is that we should realize that the distribution is defined on $\mathscr{D}(\mathbb{R})$ NOT only $\mathscr{D}(suppF)$. If we restrict $F$ to $suppF$, we will loss some information(cutoff function can not help us to get $F=0$, see the comment above). We should enlarge the restriction domain of $F$(similar for $G$) according to $F+G=0$ on $\mathscr{D}(\mathbb{R})$(this means that $F$ is defined on $\mathscr{D}(\mathbb{R})$, similar for $G$, which is NON-trivial). So, how can we enlarge it? The following is my attemption.
Because both $suppF$ and $suppG$ are closed sets and $suppF\cap suppG=\emptyset$, we take $U_{F}$ to be a open neighborhood of $suppF$ and $U_{G}$ of $suppG$, which satisfy $U_{F}\cap suppG=\emptyset$ and $U_{G}\cap suppF=\emptyset$. We restrict $F+G$ to $U_{F}$, so we have $(F+G,\varphi)=0,\forall\varphi\in\mathscr{D}(U_{F})$, which means that $F|_{U_{F}}=0$. Let $\chi$ be a smooth cutoff function satisfying $\chi(x)=1,x\in suppF$ and $\chi(x)=0,x\notin U_{F}$. So taking any function $\psi\in\mathscr{D}(\mathbb{R})$, $(F,\psi)=(F,\chi\psi+(1-\chi)\psi)=(F,\chi\psi)+(F,(1-\chi)\psi)=0+0=0$. The first one $(F,\chi\psi)$ equals $0$ because $F|_{U_{F}}=0$ and $supp\chi\subset U_{F}$. The second term $(F,(1-\chi)\psi)=0$ because $supp(1-\chi)\cap suppF=\emptyset$. Similar proof for $G$.
Maybe the vague point is: Can we find neighborhood $U_{F}$(or $U_{G}$) of $suppF$(or $suppG$) such that $suppF\cap U_{G}=\emptyset$(or $suppG\cap U_{F}=\emptyset$)? Here is my attemption: for any $x\in suppF$, then $dist(x,suppG)>0$. We take $\delta_{x}=\frac{1}{8}dist(x,suppG)$, and take $U_{F}=\cup_{x\in suppF}B(x,\delta_{x})$. We can see that $U_{F}\cap suppG=\emptyset$ and similar for $suppG$.