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Let $\mathcal{S}$ be a Schwartz space and $\delta_{a}$ the following distribution: $$\delta_{a}: \phi \rightarrow \phi(a) \ \ \ \ \text{ for each } \phi\in\mathcal{S}$$

Now, we routinely see something like: $$\int_{-\infty}^{\infty} f(x)\delta_a dx = f(a)$$ where $f$ is not necessarily $\in \mathcal{S}$ (e.g. $f \in L^2$). I'm having a hard time interpreting this integral using the language of distributions.
For example, if $f$ was in $\mathcal{S}$, I could say the integral is just giving me the value of the functional $\delta_a$ at a point in $\mathcal{S}$. But what about when $f\notin \mathcal{S}$?

Ali
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3 Answers3

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It's an abuse of notation. Essentially it is defined by $$ \int_{-\infty}^\infty \delta(x)\varphi(x) \,dx := \langle \delta, \varphi\rangle $$ for $\varphi\in C^\infty_0(\Bbb R)$, where $\langle \delta, \varphi\rangle$ denotes the dual paring. Note that this actually makes sense for any $\varphi\in C^\infty(\Bbb R)$ since $\delta$ is a distribution of compact support (and even for $\varphi\in C(\Bbb R)$ because $\delta$ can be identified with a Radon measure).

Note that $\int_{-\infty}^\infty \delta(x)f(x) \,dx$ does not make sense for $f \in L^2(\Bbb R)$ since elements of $L^2$ are only defined an equivalent classes and $f(0)$ is not well-defined for $f\in L^2$.

I gave a more thorough answer to a very similar question here.

BigbearZzz
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  • So let's say $\sqcap(x)$ which is $1$ on $I=[-1/2,1/2]$ and $0$ on $\bar{I}$ (so discontinuous at -1/2 and 1/2 so $\notin C^\infty$). Does that mean $\langle\delta,\sqcap\rangle$ is meaningles? It would be a bit weird cause intuitively we easily expect $\int_{-\infty}^{\infty} \sqcap(x)\delta dx = 1$. – Ali May 03 '19 at 21:02
  • @Ali When I said it could be extend to all $\varphi\in C(\Bbb R)$ I meant as a continuous linear functional on the space $X=C(\Bbb R)$ (with a suitable locally convex topology on $X$). While your function $\sqcap$ doesn't belong to this particular space, there's no particular reason why you couldn't just define it to be $\sqcap(0)$. (to be continued) – BigbearZzz May 03 '19 at 21:08
  • You can very well define $\int_{-\infty}^\infty \delta(x)f(x) ,dx$ for any function $f$ such that $f(0)$ is well-defined but you might need to justify as to why you may want to do that. Extending it to $C^\infty(\Bbb R)$ or $C(\Bbb R)$, on the other hand, are natural steps to take once you identify $\delta$ with the Dirac measure. – BigbearZzz May 03 '19 at 21:12
  • Then I would say a distribution $T$ acts on a test function $\phi$ not only with pairing but also with convolution $\phi \mapsto T \ast \phi$. If $T$ is a compactly supported distribution, $S$ is a tempered distribution and $\phi$ is Schwartz then $S \ast T$ is the distribution defined by $\phi \mapsto S \ast (T \ast \phi)$. In this context $f \ast \delta = f$ makes sense as well as its Fourier transform $\hat{f} 1 = \hat{f}$. @Ali – reuns May 04 '19 at 01:57
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Actually, the notation $\int T$ for a distribution $T$ was introduced by L. Schwartz in his book Theorie des distributions (on page 203 in the 1966 edition). By definition, it is given by $\int T = \langle T, 1 \rangle$, which is evaluation at the constant test function 1. This is of course not possible for all distributions but only for so-called integrable distributions $T \in \mathcal{D}'_{L^1} = (\dot{\mathcal{B}})'$, where $\dot{\mathcal{B}}$ is the space of smooth functions whose derivatives of any order vanish at infinity. Although the latter space does not contain $1$, one can show that $\mathcal{D}'_{L_1}$ also equals the dual space of $\mathcal{B}$, the space of all smooth functions which are bounded together with their derivatives of any order (much more details are available in Schwartz' book), hence this evaluation is well-defined.

With this background, integrals $$\int T(x) \varphi(x) dx$$ make sense whenever the product $T\varphi$ is an element of $\mathcal{D}'_{L^1}$.

Eduard Nigsch
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$\delta_a$ is a distribution with compact support {$a$}. (See this question for a definition and the first answer for an easy to understand explanation of the support of a distribution.)

Since such a distribution "doesn't do anything outside its support", it is irrelevant what the test functions do outside that support: you can evaluate the distribution on any $\varphi \in C^\infty$ by replacing $\varphi$ with a compactly supported $\psi \in D$ which is equal to $\varphi$ just on the support of the distribution, and vanishes further away.

Max
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