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 $?

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    $\begingroup$ 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...) $\endgroup$ Jun 29, 2021 at 18:41
  • $\begingroup$ Yes, that's what I wanted. How can this be shown? Is there a theorem underneath? $\endgroup$
    – ric.san
    Jun 29, 2021 at 18:46
  • $\begingroup$ Well, since $gu$ is a distribution probably you start with the definition of the support of a distribution... $\endgroup$ Jun 29, 2021 at 18:48
  • $\begingroup$ Right, I'm sorry. But anyway $u(gf) \neq gu(f)$, is it? $\endgroup$
    – ric.san
    Jun 29, 2021 at 18:50
  • $\begingroup$ ?????????? What's the definition of $gu$? $\endgroup$ Jun 29, 2021 at 18:58

2 Answers 2


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.


Following up on @DavidC.Ullrich's answer:

First, to understand these definitions, rather than trying to parse the formalities, it is better for many people to think about intentions and examples.

So, seriously, the support of a distribution should be an extension (in a colloquial sense) of the notion of support of a "function" (with some ambiguity about the latter! like "almost everywhere" stuff, for example).

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...

With this adaptation, the seemingly double-negative way to describe (let's not say "define"...) the support of a distribution is inevitable.

It is also correct for (integrate-against-) test functions (and more general classes).

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. :)


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