# Dirac delta and non-test functions

Normalization of the delta function (distribution) is often informally written as an integral $$\int_{-\infty}^{+\infty} \delta(x) \, dx = 1$$ An attempt to write this formally would be expression like $\delta[1] = 1$. However, the constant function $f(x) = 1$ has no compact support, therefore it is not a test function. Is there a precise way to translate such integral expression into the language of distributions?

Also, a related problem occurs with expressions like $\delta[\chi_{[a,b]}]$, where $\chi_{[a,b]}$ is a characteristic function over $[a,b] \subset \mathbb{R}$ (defined such that $\chi_{[a,b]}(x) = 1$ for $x \in [a,b]$ and $\chi_{[a,b]}(x) = 0$ for $x \notin [a,b]$). Informally, this would correspond to the integral $$\int_a^b \delta(x) \, dx$$ and, intuitively, one would like to get result $1$ for $0 \in [a,b]$ and $0$ otherwise. However, $\chi_{[a,b]}$ is not a test function (since it has discontinuities at $a$ and $b$), so we again have a formal problem as the one mentioned above.

• Well Dirac delta is not a function so you cannot use it in these ways. $\delta[\phi] = \phi(0)$, stop! Obviously if $\phi(0) = 1$ you get $1$. That "normalization property" isn't correct, and does not enter in distribution theory. Distributions eat test functions, not meat or vegetables, test functions. If you feed your delta with weird things you break up the theory. For the second question you could use mollifiers to smoothly approximate the function – EmarJ Apr 29 '15 at 12:57
• @Emar I absolutely agree with you, distributions are strictly test functions vegans! The potential issue raises because such integral expressions are widely used in physics, so I would just like to make more sense out of them. – Ivica Smolić Apr 29 '15 at 13:14

Suppose $d$ is a distribution with compact support $S$ and $f$ is a smooth function, not necessarily with compact support. Define a smooth, compactly supported cutoff function $k$ such that $k=1$ on a neighborhood of $S$. Then $\langle d,kf \rangle$ makes proper sense. One can prove that the value of this expression does not depend on $k$, under the preceding assumptions. So this defines a meaningful extension of $d$ to all of $C^\infty$. Now that we know the key idea is that of a compactly supported distribution, a quick google search finds a proof: http://www.math.chalmers.se/~hasse/distributioner_eng.pdf page 31.

Now the Dirac delta has support $\{ 0 \}$, so this applies to it.

The principle is the same if you want to apply $d$ to a function like $\chi_{[a,b]}$ which is smooth on a neighborhood of the support of $d$ but not globally smooth: use a smooth cutoff which keeps the function the same on a neighborhood of the support of $d$ but such that the result is globally smooth. So this procedure works for $\delta[\chi_{[a,b]}]$ provided $a,b \neq 0$.

• Since $\delta$ is of order $0$, you can even let it act on functions that are merely continuous, not necessarily smooth. – mrf Apr 29 '15 at 13:04
• Thank you Ian, your comments are very useful! – Ivica Smolić Apr 29 '15 at 16:02

You can see the Dirac not only as a distribution, but also as an element of the dual of continuous functions : a Radon measure.

This point of view gives sense to $\delta[1]$.

And if a measure $\mu$ is absolutely continuous in respect to the Lebesgue measure,

$$\mu( \phi ) = \int_{\mathbb{R}} \phi(x) d\mu(x) = \int_{\mathbb{R}} \phi(x) \mu(x) dx$$

(the first equality is Riesz representation theorem)

This gives, by abuse of notation (as $\delta$ is not absolutely continuous in respect to the Lebesgue measure)

$$\delta( 1 ) = \int_{\mathbb{R}} 1 d\delta(x) = \int_{\mathbb{R}} \delta(x) dx$$

For $\chi_{[a,b]}$, I'm not sure what could be the correct point of view.

• I see what you were going for now. I honestly didn't understand where you were going with your previous version. – Ian Apr 29 '15 at 13:04
• It was not clear, yes, but as $\int_{\mathbb{R}} \delta(x) dx$ is an abuse of notation, it was natural for me to consider the abuse of notation in the case where $\mu$ is not absolutely continuous in respect to the Lebesgue measure. But it's better when said explicitely ;) – Tryss Apr 29 '15 at 13:09