4
$\begingroup$

In my course notes the support of a distribution (continous lineair functional) is defined as follows:

Definitions

First it defines something like open annihilation sets:

An open annihilation set $\omega$ of a distribution $T$ is an open set where $\langle T, \phi\rangle = 0$ if the compact support of $\phi$ is a subset of $\omega$.

Then

The support of a distribution $T$ is the complement of the open union of all open annihilation sets of $T$.

There are some examples provided: ($\mathcal{D}$ is the function space of $\mathscr{C}^\infty$ functions with compact support)

  • Choose a $\phi \in \mathcal{D}$ such that $0\not \in [\phi]$. Then $\langle \delta , \phi \rangle = \phi(0) = 0$. Which implies $[\delta]= \{0\}$.
  • Let $Y$ be the Heaviside distribution. Choose $\phi\in \mathcal{D}$ such that $[\phi]\subseteq ]-\infty, 0[$, then $$\langle Y, \phi\rangle = \int_{-\infty}^{+\infty}Y(x)\phi(x)\operatorname d x = 0$$ Which implies $[Y] = [0,+\infty[$

What does it all mean?

I find it hard to understand what support of a distribution really means. For example What does it mean for a distribution to have compact support?

If an ordinary function has compact support I can visualize this as some sort of bump function. But how should I look at the support of a distribution?

$\endgroup$
  • 1
    $\begingroup$ For a distribution given by integration against a function, the support of the distribution and the support of the corresponding function are the same. This definition is how you make sense of extending this to distributions not given by integration against functions (e.g. the Dirac delta). $\endgroup$ – Ian Dec 24 '15 at 14:53
12
$\begingroup$

I will try to start from the notion of support of a function and obtain the definitions above in a natural way.

If $f : \mathbb{R}^n \to R$ then its support is defined as $S = \overline{\{x \in \mathbb{R}^n : f(x) \neq 0\}} $ For the purpose of discussion it's easier to talk about $S^c$ instead of $S$, namely $S^c$ is the largest open set where $f = 0$.

So far, so good, but distributions are not functions, so it doesn't make sense to say that the value of a distribution at a point is $0, -1, \pi$, etc. However, distributions are linear functionals, so it's not unreasonable to define that a distribution $T$ is zero on an open set $\omega$ if it "doesn't do anything there". In other words, for an arbitrary $\phi$ smooth, compactly supported in $\omega$ then $\langle T, \phi \rangle = 0$. Thus, we have arrived at the definition of open annihilation set that you mentioned.

Now, to define the support of $T$ we take the complement of the largest open set where $T$ vanishes: just like in the case for support of a function $f$: look at the disussion about $S$ and $S^c$ above.

I hope this helps.

Note: it's worth checking that if $T$ is induced by a (locally) integrable function $f$ in the standard way, then the support of $T$ will be the support of $f$, in other words the two definitions are consistent.

$\endgroup$
  • $\begingroup$ Nice..............+1 $\endgroup$ – Bhaskara-III Dec 24 '15 at 15:26
  • $\begingroup$ @Milen Ivanov so what do you mean here support of the distribution $\lambda$ say denoted by $\text S_{\lambda}$ will be $\text S_{\lambda}=\{\text {W:}~ W^c \text {is the largest open set on which}~\lambda ~\text {vanishes} \}$ or $\text S_{\lambda}=\{\text {D(W):}~ W^c \text {is the largest open set on which}~ \lambda ~\text {vanishes}\}$ $\endgroup$ – bunny Aug 30 '17 at 4:41
2
$\begingroup$

Basically, it means that there exist a compact set $K$ such that if a function has support only outside $K$, than the pairing of the function and the distribution is zero. It is more or less the dual notion of "compactly supported function".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.