# On support of a distribution vs support of a function

Given the following definitions, taken from Wikipedia:

Support of a function. When $$X$$ is a topological space and $$f : X \to \mathbb C$$ is a continuous function, the support of $$f$$ is defined topologically as the closure of the subset of $$X$$ where $$f$$ is non-zero. $${\displaystyle \operatorname {supp} (f):={\overline {\{x\in X\,|\,f(x)\neq 0\}}}={\overline {f^{-1}\left(\left\{0\right\}^{c}\right)}}.}$$

Support of a distribution. Suppose that $$u$$ is a distribution and that $$U$$ is an open set in Euclidean space such that, for all test functions $$f$$ such that $${\rm supp}\ f \subseteq U$$, $${\displaystyle u(f)=0}$$. Then $$u$$ is said to vanish on $$U$$. Now, if $$u$$ vanishes on an arbitrary family $$U_{\alpha}$$ of open sets, then for any test function $$f$$ supported in $$\bigcup \limits_\alpha U_{\alpha }$$, (...) $$u(f)=0$$ as well. Hence we can define the support of $$u$$ as the complement of the largest open set on which $$u$$ vanishes.

Question. For a given function $$g(x)$$ and a distribution $$u(f)$$, assuming $$g, f \in {\cal D}(\mathbb R^n)$$, how can we define the support of the product $$g(x)\ u(f)$$? $${\rm supp}\ g \ \cap {\rm supp} \ u$$?

• We're getting to the point where we have to be careful about things: if $g$ is as you say then $g(x)$ is not a function; also $u(f)$ is not a distribution; they are both scalars. What you probably wanted is the support of $gu$; yes, that's the intersection. (No, we don't "define" it to be that intersection! It's already defined; one can show that it's that intersection...) Jun 29, 2021 at 18:41
• Yes, that's what I wanted. How can this be shown? Is there a theorem underneath? Jun 29, 2021 at 18:46
• Well, since $gu$ is a distribution probably you start with the definition of the support of a distribution... Jun 29, 2021 at 18:48
• Right, I'm sorry. But anyway $u(gf) \neq gu(f)$, is it? Jun 29, 2021 at 18:50
• ?????????? What's the definition of $gu$? Jun 29, 2021 at 18:58

Say $$U$$ is the largest open set in which $$u$$ vanishes; let $$O$$ be the complement of the support of $$g$$, so $$O=(g^{-1}(0))^o$$.

The main step is

Proposition. $$gu$$ vanishes in $$U\cup O$$.

Proof: Say $$\phi\in C^\infty_c(U\cup O)$$. Say $$\psi_1,\psi_2$$ is a partition of unity: $$0\le\psi_j\le 1$$, $$supp(\psi_1)\subset U$$, $$supp(\psi_2)\subset O$$, and $$\psi_1+\psi_2=1$$. Let $$\phi_j=g\psi_j$$. Then $$\phi_1$$ is supported in $$U$$, so $$\phi_1g=0$$, hence by definition $$gu(\phi)=u(\phi_1g)=0$$. Similarly $$gu(\phi_2)=0$$, since $$g\phi_2$$ is supported in $$O$$ and $$u$$ vanishes in $$O$$. So $$gu(\phi)=0+0=0$$.

So now you just have to figure out why that's the largest open set in which $$gu$$ vanishes, and then see what that is qed. having just determined that we didn't learn the definition of $$gu$$ before asking the question I'll be stopping here.

The main idea is that "support" cannot be generally (meaning including both classical (=pointwise-defined) functions and "generalized functions" (=which need not have pointwise values...)) defined by reference to pointwise values. This is a hurdle, indeed. Of course, for classes of functions that do have meaningful and stable senses of pointwise values (like continuous functions), there should be a very good match of senses. But for $$L^2$$ functions? Almost-everywhere stuff? Sometimes we see "essential support" when talking about measurable functions...
By the way, the operation of smooth functions $$g$$ on distributions $$u$$ is exactly defined by duality: for test functions $$f$$, $$(g\cdot u)(f)=u(g\cdot f)$$, where the latter action of $$g$$ on $$f$$ is by pointwise multiplication. This multiplication can also be defined by taking distributional ("weak") limits of smooth functions multiplying test functions pointwise. :)