How to simplify $\delta(x^2)$

Question

How to simplify the dirac delta of squared x?

How should i solve $\int f(x)\delta(x^2)dx$ for an enough smooth function $f$?

Is this related to $\int f'(x) \delta(x) dx$ ?

I can understand symbols like $\delta(x)$ and $\delta'(x)$ and $\delta(g(x))$ with $g$ having a root $x_0$ with $g'(x_0)\neq 0$ but i can't understand it when $g'(x_0)$ is zero: $\delta(x^2)$,$\delta(|x|)$, etc.

Answer 1

This is a very good question. Let's generalize the regularization approach that Sangchul presented.

Let $\delta_n(x)$ be a sequence of (at least piece-wise smooth) positive-valued functions such that

$$\lim_{n\to \infty}\delta_n(x)=\begin{cases}0&,x\ne 0\\\\\infty&,x=0\tag1\end{cases}$$

and for each $n$

$$\int_{-\infty}^\infty \delta_n(x)\,dx=1\tag2$$

and for all suitable test functions, $\phi(x)$, we have

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\phi(0)\tag2$$

We say that in distribution $\delta(x) \sim \lim_{n\to \infty}\delta_n(x)$. This is the Dirac Delta distribution and $\delta_n(x)$ is a regularization thereof.

Now, let's examine whether $\lim_{n\to \infty}\delta_n(x^2)$ has meaning in the distributional sense. We can write

$$\begin{align} \int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx&=\int_0^\infty \delta_n(x)\left(\frac{\phi(\sqrt{x}\,)+\phi(-\sqrt{x}\,)}{2\sqrt{x}}\right)\,dx\\\\ &=\int_{-\infty}^\infty \delta_n(x)\left(\frac{\phi(\sqrt{|x|}\,)+\phi(-\sqrt{|x|}\,)}{2\sqrt{|x|}}\,H(x)\right)\,dx \end{align}$$

where $H(x)$ is the Heaviside function. Letting $n\to \infty$, we find for suitable test functions (smooth and of compact support)

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\begin{cases}0&,\phi(0)=0\\\\\infty&,\phi(0)>0\\\\-\infty &,\phi(0)<0\end{cases}$$

So, $\delta(x^2)$ has meaning in distribution on the space of $C^\infty_C$ functions that are $0$ at the origin and the distribution assigns the value of $0$ to each of these functions. As Sangchul wrote, this is quite a "boring" distribution.

---

Now, if $\phi(x)$ is a test function such that $\phi(0)=0$, smooth a.e. except at $0$ where $\lim_{x\to 0^{\pm}}\phi'(x)=C^{\pm}$, then we see that

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x^2)\phi(x)\,dx=\left(\frac{C^+-C^-}{2}\right)\,H(0)$$

Inasmuch as $H(0)$ is not uniquely defined, then we cannot uniquely define $\delta(x^2)$ as a distribution acting on such functions. For example, if $\phi(x)=|x|$, then $C^\pm=\pm1$ and $\delta(x^2)\sim H(0)$, which is not uniquely defined.

Answer 2

Let $\delta_n (x) = \frac{n}{2}\mathbf{1}_{[-1/n,1/n]}(x)$. Then $\delta_n$ converges to $\delta$ in distribution sense. Let $f$ be smooth. Then by the Taylor's theorem, we have $f(x) = f(0) + f'(0)x + \mathcal{O}(x^2)$ near $x = 0$. So it follows that

\begin{align*} \int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx &= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} f(x) \, dx \\ &= \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} \left( f(0) + f'(0)x + \mathcal{O}\left(\frac{1}{n}\right) \right) \, dx \\ &= f(0)\sqrt{n} + \mathcal{O}\left(\frac{1}{\sqrt{n}}\right). \end{align*}

So if $f(0) \neq 0$, then this integral diverges as $n\to\infty$, and if $f(0) = 0$, then this integral converges to $0$ as $n\to\infty$. This tells that $\delta(x^2)$ is either meaningless or boring when paired with nice functions.

On the other hand, if $f(x) = |x|$ then

\begin{align*} \int_{\mathbb{R}} f(x)\delta_n(x^2) \, dx = \frac{n}{2} \int_{-\frac{1}{\sqrt{n}}}^{\frac{1}{\sqrt{n}}} |x| \, dx = \frac{1}{2} \end{align*}

This suggests that $\delta(x^2)$ will capture the jump discontinuity of $f'$. Of course, making this heuristics into a meaningful statement would necessitate an appropriate theory of distributions on 'not-so-regular functions', which I do not know much.