# Distribution with support equal to a prescribed set

Question: Let $$A \subset \mathbb{R}^d$$ be a closed set. Is there a distribution $$u \in D'(\mathbb{R}^d)$$ such that $$\mathrm{supp}(u) = A$$?

Thoughts: If the set $$A$$ is countable, then one can just take $$u$$ to be integration over $$A$$ with respect to the counting measure. EDIT: Actually this will only be true if the intersection of $$A$$ with any compact set contains only finitely many points, since $$u$$ is not well-defined otherwise. EDIT 2: As suggested in the comments, one could actually take $$\sum_{k=1}^\infty 2^{-k}\delta_{a_k}$$, if $$A = \{a_k\}_{k=1}^\infty$$ is (at most) countable.

If $$A$$ is connected and has interior, then one could take the $$u$$ to be the indicator function of $$A$$ (I was not sure what the best way to state the condition on $$A$$ is in this case).

Context: I was studying a theorem which says that if $$\psi$$ is a smooth function and $$u$$ is a distribution such that $$\psi u = 0$$, then $$\mathrm{supp}(u) \subset \psi^{-1}(0)$$.

I was wondering if this result was sharp, in the sense that for each $$\psi$$, there is a distribution $$u$$ with $$\psi u = 0$$ and $$\mathrm{supp}(u) = \psi^{-1}(0)$$.

• If $A = \{ a_k \}$ is countable, couldn't we then take $\langle u, \varphi \rangle = \sum_k 2^{-k} \varphi(a_k)$ to have $u$ well-defined? Aug 3, 2019 at 21:22
• @md2perpe That would indeed work as far as I can see, thanks for the comment
– MSDG
Aug 3, 2019 at 21:26
• Can the intersection of a closed set with a compact set have a countable infinity of points? Will it not then be dense, so that its closure (i.e. it self, since it's closed) contains some interval of points and thus be uncountable? Aug 3, 2019 at 21:28
• Yes, there would be no countably infinite compact sets.But I think that I was wrong; take $A = \{ 1/k \mid k = 1, 2, 3, \ldots \} \cup \{0\}.$ That's countably infinite and compact. Aug 3, 2019 at 21:50
• You could take a countable dense set inside $A$, which exists by the separability of $\mathbb{R}^d$, and use the idea above
– Del
Aug 6, 2019 at 3:31

Take any closed set $$A$$, for all $$N\in\mathbb N$$: $$A\cap[-N,N]\subset\bigcup_{x\in A} B(x,1/N)$$ so by compacity there exists a finite subset of $$A$$, say $$A_N$$, such that: $$A\cap[-N,N]\subset\bigcup_{x\in A_N} B(x,1/N)$$

$$A_\infty=\bigcup A_N$$ is a countable subset of $$A$$ which is dense in $$A$$, indeed for all $$x\in A$$ and $$\varepsilon > 0$$ by taking $$N>\max(|x|,1/\varepsilon)$$ there exists $$x_0 \in A_N\subset A$$ such that $$x\in B(x_0, 1/N) \subset B(x_0, \varepsilon)$$.

Now write $$A_\infty = \{ a_k, k\in \mathbb N\}$$ and let: $$T=\sum_{k\in \mathbb N} 2^{-k}\delta_{a_k}$$

By definition, the support of $$T$$ is the complement of the largest open set where $$T$$ vanishes.

If $$f$$ is a test function which support is included in $$A^c$$, clearly $$f(a_k)=0$$ for all $$k$$ and this $$T(f)=0$$. Hence $$T$$ vanishes on $$A^c$$

Conversely $$V$$ is an open set containing a point $$x \in A$$, then it contains a point $$a_k \in A_\infty$$ and one can build a positive test function $$f$$ such that $$f(a_k)>0$$ and $$\text{supp}(f)\subset V$$. Thus $$T(f)\geq 2^{-k}f(a_k) >0$$ and $$T$$ does not vanish on $$V$$.

So we can deduce that the support of a distribution can be any closed set $$A$$.

• I don't understand why you don't just say "let $\{a_k\}$ be a countable subset of $A$ whose closure is $A$ and $T = \sum_{k\ge 1} 2^{-k} \delta_{a_k}$ then its support is $A$" Aug 11, 2019 at 8:46
• I wanted to prove the separability of $A$, as it is not necessarily obvious that a subset of a separable space is separable (actually this is true for any metric separable space but not in general)
– FXV
Aug 11, 2019 at 9:03
• Let $a_0 \in A$, find a sequence of balls such that $\Bbb{R}^d = \bigcup_{k\ge 1} B(x_k,r_k)$ and $r_k \to 0$, for each $k$ pick $a_k \in A \cap B(x_k,r_k)$ if it is non-empty otherwise set $a_k = a_0$. For each $b \in A$ there is some $a_k$ arbitrary close to it. Aug 11, 2019 at 9:11