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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.

Brethlosze
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2 Answers2

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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.

Mark Viola
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    @sangchullee I've generalized your approach a bit and identified a key issue, which is the non-uniqueness of, $H(0)$, the Heaviside function evaluated at $0$. This nonuniqueness is analogous to the regularization specification of $\delta_n(x)$. – Mark Viola May 10 '18 at 20:21
  • This is noteworthy. So $\delta(g(x))$ for $x_0$ a root of $g(x)$ over a test function $\phi(x)$ and with $\phi(0)=0$, and defining a proper value for $H(0)$ (as one normally do) could indeed be used as a "right vs left limit derivative differencial" operator. This is weird but makes a lot of sense now. This interpretation is suitable for interpreting other objects such as $\delta(1-\sin(x)/x)$. Though i am still greed about odd functions centered at 0 like $\delta(x^3)$ ("right & left limit derivative averager"?) and others... – Brethlosze May 10 '18 at 20:43
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    Thank you. It was a pleasure. – Mark Viola May 10 '18 at 21:03
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    For $\delta(x^3)$, we need $\phi(0)=\phi'(0)=0$. Then, $$\int_{-\infty}^\infty \delta(x^3)\phi(x),dx=\frac12\phi''(0)$$ – Mark Viola May 10 '18 at 21:11
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    @MarkViola: I believe it should be $\frac16\phi''(0)$ – robjohn May 11 '18 at 00:28
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    $$ \int_{-\infty}^\infty f(x),\delta!\left(x^3\right)\mathrm{d}x =\frac13\int_{-\infty}^\infty f!\left(x^{1/3}\right)x^{-2/3}\delta(x),\mathrm{d}x $$ – robjohn May 11 '18 at 00:44
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    @robjohn Indeed. And if $f(0)=f'(0)=0$ and $f''(0)\ne 0$, then $$\int_{-\infty}^\infty \delta(x^3)f(x),dx=\frac16 f''(0)$$ – Mark Viola May 11 '18 at 02:10
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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.

Sangchul Lee
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  • The last example got me lost. Why it converged to a value in that case? Is $\delta(x^2)$ capturing discontinuities? It is because the root of $x^2$ is double? I dont understand these possible relations... – Brethlosze May 10 '18 at 14:00
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    Using a regularized Dirac Delta is considered rigorous, is it not? But the result should be independent of the choice of regularization. – Mark Viola May 10 '18 at 14:21
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    The approximations of $\delta$ do not need to be symmetric around $0$. This leads to inconsistent values of $\delta!\left(x^2\right)$ against $|x|$. This can be seen by writing the integral as $$\int_0^\infty\frac{f!\left(\sqrt{x}\right)+f!\left(-\sqrt{x}\right)}{2\sqrt{x}},\delta(x),\mathrm{d}x$$ which for $f(x)=|x|$ gives $$\int_0^\infty1,\delta(x),\mathrm{d}x$$ and this is dependent on the choice of regularization. – robjohn May 10 '18 at 15:02
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    @hyprfrcb, The last formula is simply a result of computing the integral. Then I tried to interpret this result as 'capturing the sudden change of slopes of the test function', solely based on the fact that $\int x\delta_n(x^2),dx \to 0$ while $\int |x|\delta_n(x^2),dx\to\frac{1}{2}$. Of course this is merely a heuristic idea, and I am not making any formal claim about the nature of $\delta(x^2)$. Indeed, other users pointed out that a naive notion of $\delta(x^2)$ is not robust under general regularizations. – Sangchul Lee May 10 '18 at 15:38
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    @MarkViola, I agree that regularizing Dirac delta is a very rigorous way of understanding $\delta$ against nice test functions. What I am worried is whether we can provide a satisfactory theory on $\delta(x^2)$ or its generalizations which is still robust under various regulaziations. Indeed, robjohn's comment tells that the situation is not as simple as in the that of $\delta(x)$. – Sangchul Lee May 10 '18 at 15:43
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    @robjohn The notation $\int_0^\infty f(x)\delta(x),dx$ is interpreted to mean $\langle f,H, \delta\rangle$, where $H$ is the Heaviside function. Since the definition of $H(0)$ can be defined however one chooses, then the value of $\int_0^\infty f(x)\delta(x),dx$ is ambiguous up to that choice. Resolution of this ambiguity by choosing a value of $H(0)$ is analogous to a choice of regularization. – Mark Viola May 10 '18 at 16:08
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    @MarkViola: that is what I meant when I said that it was dependent on the choice of regularization. – robjohn May 10 '18 at 17:42
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    @robjohn Yes Rob, I know. I was only supplementing your comment. One other issue ... Does distribution theory extend to test functions that are not smooth and of compact support? This would exempt $|x|$ as a suitable test function. – Mark Viola May 10 '18 at 17:51
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    @MarkViola: standard distributions have $C_c^\infty$ test functions, tempered distributions have test functions in the Schwartz Class. $|x|$ is in neither of these classes. – robjohn May 10 '18 at 18:09
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    Yes and $|x|$ is not a Schwartz function. – Mark Viola May 10 '18 at 18:10
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    @MarkViola: yes, I just noticed that and was working on a new comment. – robjohn May 10 '18 at 19:54