What is the product of a Dirac delta function with itself?

Question

What is the product of a Dirac delta function with itself? What is the dot product with itself?

Answer 1

A distribution is actually a linear functional on the space of compactly supported infinitely differentiable functions (the so called "test functions"). A function $f$ is compactly supported if $\overline{\{x : f(x) \neq 0\}}$ is compact (the overline denotes the closure).

The $\delta$-distribution is a linear functional such that for all $\phi \in C_c^\infty(\mathbb{R}^n)$ we have that $\langle \delta, \phi \rangle = \phi(0)$.

When you want to compute the product of distributions the problem is that you don't have a property which you would really like to have, that is associativity. So for distributions $\alpha$, $\beta$ and $\gamma$ we usually have that $(\alpha \cdot \beta) \cdot \gamma \neq \alpha \cdot (\beta \cdot \gamma)$. Wikipedia gives an example. However this does not really turn out to be a problem in applications. What we do have is convolution.

When we want to do convolution we prefer a smaller class of distributions (for example because on the smaller class the Fourier transform of a distribution in this class is again a distribution in this class). This actually has a rougher class of test-functions, as test functions here we take the Schwartz functions, that are the smooth functions of which the function itself and all its derivatives are rapidly decreasing. $f$ is said to be rapidly decreasing if there are constants $M_n$ such that $|f(x)| \leq M_N |x|^{-N}$ as $x \to \infty$ for $N = 1,2,3,\ldots$.

To begin defining the convolution we first define what the convolution of Schwartz-function is with a tempered distribution. Let $f$ be our tempered distribution, then we can show that the following definition actually makes sense: $$\langle \phi * f, \psi \rangle := \langle f, \tilde{\phi} * \psi \rangle$$ where $\tilde{\phi}(x) = \phi(-x)$. Note that the RHS is well-defined. Convolution is a nice thing, we can see that if we start with a tempered distribution and convolute it with a test function, the result will be smooth. Now, $L_1 * L_2$ is the unique distribution $L$ with the property that $L * \phi = L_1 * (L_2 * \phi)$. We can show that this is commutative.

Fine, now note that $\delta * \phi(x) = \phi(x - y)|_{y = 0} = \phi(x)$. So we see that $\delta * \delta = \delta$.

If you want me to comment on the dot product of distributions, you first would have to explain what you mean with that.

So far for this short digression on distributions.

EDIT: Okay, you want to compute $\delta^2$. Let $\phi_n$ be an approximation to the identity and let it converge to $\delta$ in the sense of distributions, but $\phi_n^2$ does not converge at all since the integral against a test function that does not vanish at the origin blows up as $n \to \infty$.

Answer 2

Here is a heuristic to suggest that it will be difficult to define the square of the delta function.

The Fourier transform has the property that it takes the convolution of two functions to the product of their Fourier transforms, and vice versa, ie, it takes the product of two functions to their convolution.

Remember that the Fourier transform of the delta function is the constant function($=1$). Now suppose that $\delta^2$ exists. Then its Fourier transform would be the convolution of two constant functions. Such a convolution would shoot to infinity at every point. Even the theory of distributions can't handle such kind of stuff. So how would you imagine the inverse Fourier transform of this? How would it make sense?