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.

  • $\begingroup$ why $supp(F)\cap supp(G)=\emptyset\Rightarrow (F+G,\phi)=0,\forall\phi$? Is it an assumption? $\endgroup$ Nov 12, 2023 at 13:47
  • $\begingroup$ @monotoneoperator yes its an assumption sorry $\endgroup$
    – duelspace
    Nov 12, 2023 at 13:48
  • $\begingroup$ If it is an assumption, maybe you can take the restriction $\endgroup$ Nov 12, 2023 at 13:48
  • $\begingroup$ 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 $G$ $\endgroup$ Nov 12, 2023 at 13:52
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    $\begingroup$ Now, I write an idea in the following "Answer" and I wish it is useful. $\endgroup$ Nov 13, 2023 at 2:44

1 Answer 1


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 $suppG$.

  • $\begingroup$ Thank you for your answer! There is no need to apologise, even what you previously wrote was helpful because it gave me some ideas for tackling a similar problem :-) I want to think about this argument a bit more before accepting but I can't find any mistake in it yet $\endgroup$
    – duelspace
    Nov 13, 2023 at 22:55
  • $\begingroup$ Happy to help~Best wishes! $\endgroup$ Nov 14, 2023 at 1:11
  • $\begingroup$ 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). $\endgroup$ Nov 14, 2023 at 2:14
  • $\begingroup$ 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 $G$). $\endgroup$ Nov 14, 2023 at 2:29
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    $\begingroup$ @LorenzoPompili, Thank you, I learn a lot from your comment! $\endgroup$ Nov 14, 2023 at 9:51

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