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Support of a distribution, what does it mean?

Writer Andrew Henderson
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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?

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2 Answers

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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.

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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".

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