# Distributions with disjoint support

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.

• why $supp(F)\cap supp(G)=\emptyset\Rightarrow (F+G,\phi)=0,\forall\phi$? Is it an assumption? Nov 12, 2023 at 13:47
• @monotoneoperator yes its an assumption sorry Nov 12, 2023 at 13:48
• If it is an assumption, maybe you can take the restriction Nov 12, 2023 at 13:48
• Restriction is to say $(F+G)|_{supp F}=F$ and $(F+G)|_{supp G}=G$, so you take $\tilde{\phi}$ that has $supp\tilde{\phi}\subset suppF$ and $0=(F+G,\tilde{\phi})=(F,\tilde{\phi}),\forall \tilde{\phi}\in\mathscr{D}(supp F)$, and similar for $G$ Nov 12, 2023 at 13:52
• Now, I write an idea in the following "Answer" and I wish it is useful. Nov 13, 2023 at 2:44

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 $$suppG$$.
• In your answer you can simply choose $U_F$ as the complement of $\operatorname{supp} G$ and viceversa. Supports by definition are closed sets, so $U_F,U_G$ are open. The existence of $\chi$ follows from classical results about partitions of unity: it is always possible to find $\chi$ with support in $U_F$ such that $(1-\chi)$ has support in $U_G$, as $U_F,U_G$ are a covering of $\mathbb R$ (I don’t have a reference at the moment unfortunately). Nov 14, 2023 at 2:14
• Alternative proof: we said that $F|_{U_F}=0$, and trivially $G|_{U_F}=0$ (since the support of $G$ and $U_F$ are disjoint). Analogously, $F|_{U_G}=G|_{U_G}=0$. Take $F$: the first sentence says that the support of $F$ and $U_F$ are disjoint; the second sentence says that the support of $F$ and $U_G$ are disjoint. But $U_F\cup U_G=\mathbb R$, so the support of $F$ must be empty (analogously for $G$). Nov 14, 2023 at 2:29