# Delta squared

1. Nov 10, 2015

### jk22

I consider the Dirac delta.

In physics the delta squared has an infinite norm : $$\int\delta (x)^2=\infty$$

But if i look at delta being a functional i could write : $$\delta [f]=f (0)$$ hence $$\delta^2 [f]=\delta [\delta [f]]=\delta [\underbrace {f (0)}_{constant function}]=f (0)$$

Thus in this view $$\delta^2=\delta$$ ?

2. Nov 10, 2015

### Khashishi

Dirac delta is defined by the integral
$\int \delta(c) g(x) dx = g(c)$
If you plug in $g(x) = \delta(x)$ then you get
$\int \delta(c) \delta(x) dx = \delta(c)$ which isn't finite. The integral fails.

You can think of the integral as a functional, but it doesn't make sense to think of delta as a functional.

3. Nov 11, 2015

### pwsnafu

Dirac delta is indeed a linear functional. Specifically it is the linear functional such that $f \mapsto f(0)$. Schwartz distributions are our most developed theory of generalized functions, and there all Schwartz distributions are linear functionals.

Now to OP. If you want $\delta^2 = \delta \cdot \delta$ to exist as a Schwartz distribution as well, then it turns out that $\delta^2 = c\delta$ for some constant $c$, but not for the reason you posted. The notation $\delta^2$ is reserved for the product of Dirac with itself, not the composition of Dirac with itself. As to what $c$ is, there's a lot of disagreement. If the product defined is a "normal product" then $c=0$. Some mathematicians have argued that non-zero c has physical meaning. Some argue $c = \infty$. Others define it to be $c = \delta(0)$ and not define what that means (as long as it cancels in the end they are happy with it). So yeah a lot of disagreement.

That being said, if you aren't working with Schwartz distributions, such as working in Colombeu algebra, then it is the operator $f(x) \mapsto f(-x)^2$ (technically the equivalence class of such operators), which in turn corresponds to the non-linear functional $f \mapsto f(0)^2$

Edit: One last thing. There are people who write $\delta^2$ for $\delta(x)\delta(y)$, i.e. the two dimensional Dirac delta. But then you wouldn't have $\int \delta^2 = \infty$ at the top of your post, so I'm ignoring that situation.

Last edited: Nov 11, 2015